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IB Maths Applications & Interpretation · SL · Topic 5

Calculus

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1IBMI-034-02-r1-01Integration and the trapezoidal rule Paper 1Easy5 marks

An environmental engineer measures the depth of a straight river channel at equal intervals across its width of 12 m. The depths, d metres, at horizontal distance x metres from one bank are shown below.

x (m): 0, 2, 4, 6, 8, 10, 12
d (m): 0, 1.4, 2.3, 2.8, 2.5, 1.6, 0

(a)

Use the trapezoidal rule with all the given data to estimate the cross-sectional area of the river channel.

(3)
(b)

The water flows through this cross-section at an average speed of 0.8 m s−1. Estimate the volume of water passing this point each second.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
h = 2✓ 1
Area ≈ \(\frac{2}{2}[0 + 0 + 2(1.4 + 2.3 + 2.8 + 2.5 + 1.6)]\)✓ 1
= 21.2 m2✓ 1
Part (b)
Volume per second = 21.2 × 0.8✓ 1
= 17.0 m3 per second (16.96)✓ 1

Exam tip: In the trapezoidal rule only the first and last readings appear once; every interior reading is doubled. Check that h is the gap between x-values, not the number of intervals.

2IBMI-034-02-r1-02Integration and the trapezoidal rule Paper 1Easy6 marks

A small bakery sells x loaves of speciality bread per day. The marginal profit, in dollars per loaf, is modelled by
\(P'(x) = 48 - 0.6x\).
When no loaves are sold the bakery makes a loss of $200 per day due to fixed costs, so $P(0) = -200$.

(a)

Find an expression for the daily profit P(x).

(4)
(b)

Find the daily profit when 50 loaves are sold.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Attempt to integrate P'(x)✓ 1
\(P(x) = 48x - 0.3x^2 + c\)✓ 1
Substituting P(0) = −200✓ 1
\(P(x) = 48x - 0.3x^2 - 200\)✓ 1
Part (b)
P(50) = 48(50) − 0.3(50)2 − 200✓ 1
= $1450.00✓ 1

Common error: Forgetting the constant of integration, or setting it to zero, loses the fixed-cost information. Always use the given condition to find c.

3IBMI-034-00-r1-01Gradients, tangents and rates of change Paper 1Easy6 marks

A car pulls away from a set of traffic lights. Its distance, \(s\) metres, from the lights \(t\) seconds after it starts moving is modelled by
\(s(t) = 0.4t^2 + 3t\), for \(0 \le t \le 10\).

(a)

Find an expression for the velocity of the car, \(\frac{ds}{dt}\), in terms of \(t\).

(2)
(b)

Find the velocity of the car when \(t = 5\).

(2)
(c)

Find the time at which the velocity of the car is 10 m s−1.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1: \(0.8t\)✓ 1
A1: \(+\,3\), giving \(\frac{ds}{dt} = 0.8t + 3\)✓ 1
Part (b)
M1: substituting \(t = 5\) into their \(\frac{ds}{dt}\), e.g. \(0.8(5) + 3\)✓ 1
A1: 7 m s−1✓ 1
Part (c)
M1: setting \(0.8t + 3 = 10\)✓ 1
A1: \(t = 8.75\) s✓ 1

Exam tip: The derivative of distance with respect to time is velocity. Always include units with a rate of change, here m s⁻¹.

4IBMI-034-07-r2-02Integration and the trapezoidal rule Paper 1Easy6 marks

A tree seedling is planted in a nursery. Its height, h cm, t weeks after planting is modelled by

\(\frac{dh}{dt} = 2.4 - 0.15t, \quad 0 \le t \le 16.\)

When it is planted, the seedling is 5 cm tall.

(a)

Find an expression for h in terms of t.

(4)
(b)

Find the height of the seedling 10 weeks after planting.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: attempt to integrate, at least one term correct✓ 1
A1: 2.4t✓ 1
A1: −0.075t2✓ 1
A1: uses h = 5 when t = 0 to obtain c = 5, so h = 2.4t − 0.075t2 + 5✓ 1
Part (b)
M1: substitutes t = 10, 2.4(10) − 0.075(100) + 5✓ 1
A1: 21.5 cm✓ 1

Exam tip: Always include the constant of integration and use the initial condition to find it; an answer without '+ c' resolved usually loses the final accuracy mark.

5IBMI-034-00-r2-02Gradients, tangents and rates of change Paper 1Easy7 marks

A section of a cycle path seen from above is modelled by the curve
\(y = 0.03x^2 - 1.2x + 15\), for \(0 \le x \le 40\),
where x and y are measured in metres from a fixed point.

(a)

Find \(\frac{dy}{dx}\).

(2)
(b)

Find the gradient of the path at the point where \(x = 30\).

(1)
(c)

A cyclist leaves the path at the point where \(x = 30\) and continues along the tangent to the path. Find the equation of this tangent, giving your answer in the form \(y = mx + c\).

(3)
(d)

Find the value of x at which the gradient of the path is zero.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Attempt to differentiate, with at least one term correct✓ 1
\(\frac{dy}{dx} = 0.06x - 1.2\)✓ 1
Part (b)
\(0.06 \times 30 - 1.2 = 0.6\)✓ 1
Part (c)
\(y = 0.03(900) - 1.2(30) + 15 = 6\)✓ 1
Uses point and gradient: \(y - 6 = 0.6(x - 30)\)✓ 1
\(y = 0.6x - 12\)✓ 1
Part (d)
\(0.06x - 1.2 = 0 \Rightarrow x = 20\)✓ 1

Common error: The tangent must pass through the point on the curve, so the y-coordinate comes from the original function, not from the derivative.

6IBMI-034-01-r2-01Optimisation in context Paper 1Easy5 marks

A small bakery sells x loaves of sourdough per day. Its daily profit, P dollars, is modelled by
\(P(x) = -0.4x^2 + 36x - 250\), for \(0 \le x \le 90\).

(a)

Find \(P'(x)\).

(2)
(b)

Hence find the number of loaves that maximises daily profit, and the maximum daily profit.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
−0.8x A1✓ 1
+36, with the constant −250 differentiating to 0 A1✓ 1
Part (b)
Sets \(P'(x)=0\): −0.8x + 36 = 0 M1✓ 1
x = 45 loaves A1✓ 1
Maximum profit = −0.4(45)2 + 36(45) − 250 = $560.00 per day A1✓ 1

Exam tip: Answer both parts of an optimisation question: the value of x that gives the maximum, and the maximum value itself. Give money to 2 decimal places.

7IBMI-034-01-r2-02Optimisation in context Paper 1Easy6 marks

A farmer uses 120 m of fencing to make a rectangular pen for goats. One side of the pen is formed by a long straight stone wall, so it needs no fencing. The two sides perpendicular to the wall each have length x metres.

(a)

Show that the area, A m2, of the pen is given by \(A = 120x - 2x^2\).

(2)
(b)

Find \(\frac{dA}{dx}\).

(1)
(c)

Find the maximum possible area of the pen, and state the dimensions that give it.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Side parallel to the wall = 120 − 2x M1✓ 1
A = x(120 − 2x) = 120x − 2x2 A1✓ 1
Part (b)
\(\frac{dA}{dx} = 120 - 4x\) A1✓ 1
Part (c)
Sets 120 − 4x = 0, so x = 30 M1✓ 1
Dimensions: 30 m by 60 m (the 60 m side parallel to the wall) A1✓ 1
Maximum area = 1800 m2 A1✓ 1

Common error: Using a perimeter of 2x + 2y = 120 ignores the wall. Only three sides are fenced, so the length of the fence is 2x + y.

8IBMI-034-06-r1-01Optimisation in context Paper 1Easy5 marks

A small bakery sells x loaves of speciality bread per day. The daily profit, P dollars, is modelled by
\(P(x) = -0.4x^2 + 36x - 300\), for \(0 \le x \le 90\).

(a)

Find \(P'(x)\).

(2)
(b)

Find the number of loaves the bakery should sell each day to maximise its profit.

(2)
(c)

Find the maximum daily profit.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(P'(x) = -0.8x\)✓ 1
\(+ 36\)✓ 1
Part (b)
Setting \(P'(x) = 0\), i.e. \(-0.8x + 36 = 0\) (M1)✓ 1
\(x = 45\) loaves (A1)✓ 1
Part (c)
$P(45) = -0.4(45)^2 + 36(45) - 300 = $ $510.00 (A1)✓ 1

Exam tip: Profit is money, so give the final answer to 2 decimal places with the currency, even when it is a whole number of dollars.

9IBMI-034-07-r1-01Integration and the trapezoidal rule Paper 1Easy5 marks

An engineer measures the depth of a river at equal intervals across its width. The river is 12 m wide. The depth, d metres, is recorded at a horizontal distance x metres from one bank.

x (m): 0, 2, 4, 6, 8, 10, 12
d (m): 0, 1.4, 2.1, 2.6, 2.3, 1.5, 0

(a)

Use the trapezoidal rule with all the data to estimate the cross-sectional area of the river, in m2.

(3)
(b)

The water flows through this cross-section with an average speed of 0.35 m s−1. Estimate the volume of water passing this point each second.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Uses interval width h = 2 (A1)✓ 1
Correct substitution into the trapezoidal rule: \(\frac{2}{2}[0+0+2(1.4+2.1+2.6+2.3+1.5)]\) (M1)✓ 1
Area ≈ 19.8 m2 (A1)✓ 1
Part (b)
Multiplies their area by 0.35: 19.8 × 0.35 (M1)✓ 1
6.93 m3 of water per second (A1)✓ 1

Common error: The first and last depths are counted once and every interior depth twice. Doubling the end values, or forgetting to double the middle ones, gives a wrong area even when h is correct.

10IBMI-034-07-r1-02Integration and the trapezoidal rule Paper 1Easy6 marks

A small company makes handmade chairs. The marginal cost of producing the xth chair is modelled by
\(C'(x) = 0.03x^2 - 1.2x + 25\) US dollars per chair.
The company has fixed costs of 400 US dollars, so that \(C(0) = 400\).

(a)

Find an expression for the total cost, \(C(x)\), of producing x chairs.

(4)
(b)

Find the total cost of producing 30 chairs.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(0.01x^3\) (A1)✓ 1
\(-0.6x^2\) (A1)✓ 1
\(+25x + c\) (A1)✓ 1
Uses \(C(0) = 400\) to obtain \(C(x) = 0.01x^3 - 0.6x^2 + 25x + 400\) (M1)✓ 1
Part (b)
Substitutes x = 30 into their C(x): \(0.01(30)^3 - 0.6(30)^2 + 25(30) + 400\) (M1)✓ 1
US$880.00 (A1)✓ 1

Exam tip: After integrating, always use the given condition to find the constant. Here the fixed cost supplies it directly. A cost answer must be given to 2 decimal places.

11IBMI-034-02-r2-02Integration and the trapezoidal rule Paper 1Easy5 marks

Water flows into a storage tank. The rate of flow, in litres per minute, is modelled by
\(r(t) = 40 + 6t - 0.5t^2\), for \(0 \le t \le 12\),
where t is the time in minutes after the inflow begins.

(a)

Find the rate of flow when t = 4.

(1)
(b)

Write down a definite integral for the volume of water that flows into the tank during the first 12 minutes, and hence find this volume.

(3)
(c)

The tank contained 150 litres of water at t = 0. Find the volume of water in the tank when t = 12.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
r(4) = 40 + 24 − 8 = 56 litres per minute✓ 1
Part (b)
Integral of r(t) with respect to t✓ 1
Correct limits: \(\int_0^{12} (40 + 6t - 0.5t^2)\,dt\)✓ 1
624 litres✓ 1
Part (c)
150 + 624 = 774 litres✓ 1

Common error: The integral gives only the volume added during the 12 minutes. The initial 150 litres must be added on to get the total in the tank.

12IBMI-034-05-r1-02Gradients, tangents and rates of change Paper 1Easy7 marks

An engineer models the vertical profile of a section of a mountain road by
\(y = 0.001x^3 - 0.06x^2 + 1.5x\), for \(0 \le x \le 50\),
where \(x\) is the horizontal distance in metres from the start of the section and \(y\) is the height in metres above the start.

(a)

Find \(\frac{dy}{dx}\).

(2)
(b)

Find the gradient of the road when \(x = 10\).

(2)
(c)

Find the equation of the tangent to the curve at \(x = 10\). Give your answer in the form \(y = mx + c\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Attempt to differentiate using the power rule M1✓ 1
\(\frac{dy}{dx} = 0.003x^2 - 0.12x + 1.5\) A1✓ 1
Part (b)
Substituting \(x = 10\): \(0.003(100) - 0.12(10) + 1.5\) M1✓ 1
Gradient \(= 0.6\) A1✓ 1
Part (c)
\(y = 0.001(1000) - 0.06(100) + 1.5(10) = 10\) A1✓ 1
Using a line with their gradient through \((10, 10)\): \(y - 10 = 0.6(x - 10)\) M1✓ 1
\(y = 0.6x + 4\) A1✓ 1

Common error: Using the gradient as the \(y\)-coordinate of the point of contact. The tangent passes through \((10, y(10))\), so substitute \(x = 10\) into the original function, not the derivative.

13IBMI-034-06-r2-02Optimisation in context Paper 1Easy7 marks

A farmer builds a rectangular chicken run against the long straight wall of a barn. The wall forms one side of the run. The other three sides use 80 m of fencing. The two sides perpendicular to the wall each have length \(x\) metres.

(a)

Show that the side of the run parallel to the wall has length \((80 - 2x)\) metres.

(1)
(b)

Write down an expression for the area, \(A\) m2, of the run in terms of \(x\).

(1)
(c)

Find \(\frac{dA}{dx}\).

(2)
(d)

Find the value of \(x\) that maximizes the area, and find this maximum area.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
The two perpendicular sides use \(2x\) m of fencing, so the remaining side is \(80 - 2x\) m R1✓ 1
Part (b)
\(A = x(80 - 2x) = 80x - 2x^2\) A1✓ 1
Part (c)
\(80\) A1✓ 1
\(-4x\) A1✓ 1
\(\frac{dA}{dx} = 80 - 4x\)—
Part (d)
Sets \(80 - 4x = 0\) M1✓ 1
\(x = 20\) m A1✓ 1
\(A = 20 \times 40 = 800\) m2 A1✓ 1

Common error: Using a perimeter of \(2x + 2y = 80\) ignores the barn wall. Only three sides are fenced, so the constraint is \(2x + y = 80\).

14IBMI-034-05-r2-02Gradients, tangents and rates of change Paper 1Easy7 marks

The profile of a small hill is modelled by
\(y = 12 + 0.3x^2 - 0.01x^3\), for \(0 \le x \le 31\),
where x is the horizontal distance in metres from a marker post and y is the height of the hill in metres. A straight support cable is fixed to the hill at the point where x = 30. The cable lies along the tangent to the hill at this point and is extended down the slope until it reaches ground level, y = 0.

(a)

Find \(\frac{dy}{dx}\).

(2)
(b)

Find the gradient of the hill at x = 30.

(1)
(c)

Find the equation of the tangent to the hill at x = 30.

(3)
(d)

Find the horizontal distance from the marker post at which the cable reaches the ground.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 for \(0.6x\)✓ 1
A1 for \(-0.03x^2\), giving \(\frac{dy}{dx} = 0.6x - 0.03x^2\)✓ 1
Part (b)
A1 gradient = 0.6(30) − 0.03(900) = 18 − 27 = −9✓ 1
Part (c)
A1 height at x = 30 is 12 + 270 − 270 = 12✓ 1
M1 using point (30, 12) with gradient −9: \(y - 12 = -9(x - 30)\)✓ 1
A1 \(y = -9x + 282\)✓ 1
Part (d)
A1 setting \(0 = -9x + 282\) gives x = 31.3 m (3 s.f.)✓ 1

Common error: The tangent needs a point on the curve as well as the gradient. Substituting x = 30 into the derivative gives only the gradient. The y-coordinate comes from the original function.

15IBMI-034-00-r1-02Gradients, tangents and rates of change Paper 1Easy8 marks

The profile of a hill is modelled by
\(y = -0.002x^3 + 0.12x^2\), for \(0 \le x \le 50\),
where \(x\) is the horizontal distance in metres from the foot of the hill and \(y\) is the height in metres.

(a)

Find \(\frac{dy}{dx}\).

(2)
(b)

Find the equation of the tangent to the profile at the point where \(x = 10\). Give your answer in the form \(y = mx + c\).

(3)
(c)

Hence find the maximum height of the hill, justifying that your answer is the maximum.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1: \(-0.006x^2\)✓ 1
A1: \(+\,0.24x\)✓ 1
Part (b)
A1: gradient at \(x = 10\): \(-0.006(100) + 0.24(10) = 1.8\)✓ 1
A1: height at \(x = 10\): \(y = -2 + 12 = 10\)✓ 1
A1: \(y - 10 = 1.8(x - 10)\), so \(y = 1.8x - 8\)✓ 1
Part (c)
M1: setting \(-0.006x^2 + 0.24x = 0\), giving \(x = 40\) (rejecting \(x = 0\))✓ 1
R1: valid justification, e.g. \(\frac{d^2y}{dx^2} = -0.012(40) + 0.24 = -0.24 < 0\), or the gradient changes from positive to negative at \(x = 40\), or comparing with the endpoints \(y(0) = 0\) and \(y(50) = 50\)✓ 1
A1: maximum height \(= -0.002(40)^3 + 0.12(40)^2 = 64\) m✓ 1

Common error: A tangent needs a point as well as a gradient. Calculate the \(y\)-coordinate from the original function, not from the derivative.

16IBMI-034-02-r2-01Integration and the trapezoidal rule Paper 1Easy6 marks

A farmer owns a field with one straight boundary, of length 80 m, along a road. The other boundary is curved. The perpendicular distance, d metres, from the road to the curved boundary is measured at points x metres along the road, as shown below.

x (m): 0, 20, 40, 60, 80
d (m): 32, 41, 45, 38, 27

The farmer uses the trapezoidal rule with the five measurements to estimate the area of the field.

(a)

Write down the width, h, of each interval.

(1)
(b)

Use the trapezoidal rule to estimate the area of the field.

(3)
(c)

The field is valued at USD 4.75 per m2. Using your answer to the previous part, estimate the value of the field. Give your answer to two decimal places.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
h = 20 m✓ 1
Part (b)
Correct substitution of h = 20 into the trapezoidal rule: \(\frac{20}{2}[\ldots]\)✓ 1
Correct bracket: 32 + 27 + 2(41 + 45 + 38) = 307✓ 1
Area ≈ 3070 m2✓ 1
Part (c)
Multiplies their area by 4.75, e.g. 3070 × 4.75✓ 1
USD 14582.50 (follow through from their area)✓ 1

Exam tip: In the trapezoidal rule the first and last values are counted once and every interior value is doubled. Writing the bracket out in full before calculating makes this mark easy to earn.

17IBMI-034-00-r2-01Gradients, tangents and rates of change Paper 1Easy5 marks

A ball is thrown upwards from a balcony. Its height above the ground, h metres, t seconds after it is thrown is modelled by
\(h(t) = 1.5 + 14t - 4.9t^2\), for \(0 \le t \le 2.9\).

(a)

Find \(h'(t)\).

(2)
(b)

Find the rate of change of the height of the ball when \(t = 2\).

(2)
(c)

Interpret your answer to part (b) in the context of the question.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Attempt to differentiate, with at least one term correct✓ 1
\(h'(t) = 14 - 9.8t\)✓ 1
Part (b)
Substitutes \(t = 2\) into their \(h'(t)\): \(14 - 9.8 \times 2\)✓ 1
\(h'(2) = -5.6\) m s−1✓ 1
Part (c)
2 seconds after being thrown, the height of the ball is decreasing (the ball is falling) at 5.6 metres per second✓ 1

Exam tip: When asked to interpret a rate of change, state what is changing, whether it is increasing or decreasing, the size of the rate and its units, all in the words of the context.

18IBMI-034-06-r2-01Optimisation in context Paper 1Easy5 marks

A small bakery sells handmade sourdough loaves at a weekend market. The weekly profit, \(P\) dollars, from selling \(x\) loaves is modelled by
\(P(x) = -0.4x^2 + 36x - 250\), for \(0 \le x \le 90\).

(a)

Find \(P'(x)\).

(2)
(b)

Find the number of loaves the bakery should sell each week to maximize its profit.

(2)
(c)

Find the maximum weekly profit.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(-0.8x\) A1✓ 1
\(+36\) (from \(36x\); the constant \(-250\) differentiates to 0) A1✓ 1
\(P'(x) = -0.8x + 36\)—
Part (b)
Sets \(P'(x) = 0\): \(-0.8x + 36 = 0\) M1✓ 1
\(x = 45\) loaves A1✓ 1
(This is a maximum because the coefficient of \(x^2\) is negative.)—
Part (c)
\(P(45) = -0.4(45)^2 + 36(45) - 250 = 560\) dollars A1✓ 1

Common error: To find the maximum profit, substitute \(x = 45\) into \(P(x)\), not into \(P'(x)\). Substituting into \(P'(x)\) always gives 0, which is the gradient, not the profit.

19IBMI-034-07-r2-01Integration and the trapezoidal rule Paper 1Easy5 marks

A hydrologist measures the depth of water across a straight river channel that is 12 m wide. Depths are recorded every 2 m from one bank to the other.

Distance from left bank (m): 0, 2, 4, 6, 8, 10, 12
Depth (m): 0, 1.2, 1.9, 2.4, 2.1, 1.5, 0

The cross-section of the river is modelled as the region between the water surface and the river bed.

(a)

Use the trapezoidal rule with all the data given to estimate the cross-sectional area of the river, in m2.

(3)
(b)

The water flows through this cross-section with a mean speed of 0.8 m s−1. Estimate the volume of water passing this point each second.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: strip width h = 2 used✓ 1
M1: correct substitution, \(\frac{2}{2}\left[0+0+2(1.2+1.9+2.4+2.1+1.5)\right]\)✓ 1
A1: area = 18.2 m2✓ 1
Part (b)
M1: multiplies their area by 0.8, 18.2 × 0.8✓ 1
A1: 14.6 m3 (accept 14.56)✓ 1

Common error: In the trapezoidal rule only the interior depths are doubled; the first and last depths appear once. Doubling every depth, or using h instead of h/2 as the multiplier, overestimates the area. Here either slip gives 36.4 m² instead of 18.2 m².

20IBMI-034-01-r1-01Optimisation in context Paper 1Easy5 marks

A small bakery sells \(x\) celebration cakes per week. The weekly profit, in dollars, is modelled by
\(P(x) = -0.04x^2 + 6x - 80\), for \(0 \le x \le 140\).

(a)

Find \(P'(x)\).

(2)
(b)

Find the number of cakes the bakery should sell each week to maximise its profit.

(2)
(c)

Find the maximum weekly profit.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate, with at least one term correct✓ 1
A1 \(P'(x) = -0.08x + 6\)✓ 1
Part (b)
M1 setting \(P'(x) = 0\), i.e. \(-0.08x + 6 = 0\)✓ 1
A1 \(x = 75\) cakes✓ 1
Part (c)
A1 \(P(75) = -0.04(75)^2 + 6(75) - 80 = $145\)✓ 1

Exam tip: 'Find the number of cakes' wants the value of x, and 'find the maximum profit' wants the value of P. Substitute your x-value back into P(x) to get the profit, rather than repeating x = 75.

21IBMI-034-01-r1-02Optimisation in context Paper 1Easy6 marks

A farmer builds a rectangular sheep pen against a long straight stone wall. The wall forms one side of the pen, and 84 m of fencing is used for the other three sides. The two sides perpendicular to the wall each have length \(x\) metres.

(a)

Show that the length of the side parallel to the wall is \((84 - 2x)\) metres.

(1)
(b)

Write down an expression for the area \(A\) m2 of the pen in terms of \(x\).

(1)
(c)

Find \(\frac{dA}{dx}\).

(2)
(d)

Hence find the maximum possible area of the pen.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
R1 the two perpendicular sides use \(2x\) m of fencing, so the remaining side is \(84 - 2x\) m✓ 1
Part (b)
A1 \(A = x(84 - 2x)\) or \(A = 84x - 2x^2\)✓ 1
Part (c)
M1 valid attempt to differentiate their expression, expanded or by the product rule, with at least one term correct✓ 1
A1 \(\frac{dA}{dx} = 84 - 4x\)✓ 1
Part (d)
M1 setting \(84 - 4x = 0\) and solving to get \(x = 21\)✓ 1
A1 \(A = 21 \times 42 = 882\) m2✓ 1
A justification such as \(\frac{d^2A}{dx^2} = -4 < 0\) is good practice but is not required for these marks.—

Common error: Using a perimeter of \(2x + 2y = 84\) ignores the wall. Read the context carefully: only three sides need fencing.

22IBMI-034-05-r1-01Gradients, tangents and rates of change Paper 1Easy5 marks

During the dry season the volume of water, \(V\) m3, in a small reservoir is modelled by
\(V(t) = 5000 - 12t^2 + 0.1t^3\), for \(0 \le t \le 20\),
where \(t\) is the number of days after the start of the dry season.

(a)

Find \(\frac{dV}{dt}\).

(2)
(b)

Find the value of \(\frac{dV}{dt}\) when \(t = 10\) and interpret your answer in context.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Attempt to differentiate, with at least one term correct M1✓ 1
\(\frac{dV}{dt} = -24t + 0.3t^2\) A1✓ 1
Part (b)
Substituting \(t = 10\) into their derivative: \(-24(10) + 0.3(10)^2\) M1✓ 1
\(-210\) (m3 per day) A1✓ 1
10 days after the start of the dry season, the volume of water in the reservoir is decreasing at a rate of 210 m3 per day. Accept equivalent wording such as 'after 10 days' or 'when \(t = 10\)'. R1✓ 1

Exam tip: An interpretation mark needs three things: what is changing (the volume of water), the direction (decreasing, shown by the negative sign) and the rate with units (210 m³ per day).

23IBMI3S-5-09-01Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Easy5 marks

The depth of water, d metres, in a reservoir t hours after a gate is opened is modelled by
d(t) = 0.02t3 − 0.3t2 + 1.5t + 2,   0 ≤ t ≤ 10.

(a)

Find d′(t).

(2)
(b)

Find the rate at which the depth is changing when t = 8. Interpret your answer in context.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 d′(t) = 0.06t2 − 0.6t + 1.5✓ 1
Part (b)
M1 substitute t = 8 into d′(t)✓ 1
A1 0.54✓ 1
R1 the depth is increasing at 0.54 m per hour at that moment✓ 1

Common error: A derivative gives a rate, so quote units (m h⁻¹) and say whether the quantity is rising or falling.

24IBMI3S-5-09-02Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Easy7 marks

The height, h metres above sea level, of a scenic road at a horizontal distance x km from its start is modelled by
h(x) = −x3 + 6x2 − 5x + 40,   0 ≤ x ≤ 5.

(a)

Find h′(x).

(2)
(b)

Find the equation of the tangent to the road profile at x = 2. Give your answer in the form y = mx + c.

(4)
(c)

Use the tangent to estimate the height of the road when x = 2.1.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 h′(x) = −3x2 + 12x − 5✓ 1
Part (b)
A1 h(2) = 46✓ 1
A1 h′(2) = 7✓ 1
M1 use of y − y1 = m(x − x1) with their values✓ 1
A1 y = 7x + 32✓ 1
Part (c)
A1 7(2.1) + 32 = 46.7 m (ECF from (b))✓ 1

Exam tip: A tangent is only a good approximation close to the point of contact, so use it for values of x near 2.

25IBMI3S-5-09-03Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Easy4 marks

A farmer wants to estimate the area of an irregular field. A straight road runs along one side. The width of the field, measured at right angles to the road, is recorded at 10 m intervals along the road.
Distance along road (m): 0, 10, 20, 30, 40
Width of field (m): 0, 14, 20, 17, 0

(a)

Use the trapezoidal rule with four intervals to estimate the area of the field.

(3)
(b)

Seed costs €0.35 per square metre. Find the cost of seeding the whole field, using your answer to (a).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct strip width h = 10 used in the trapezoidal rule✓ 1
A1 ½ × 10 × [0 + 0 + 2(14 + 20 + 17)]✓ 1
A1 510 m2✓ 1
Part (b)
A1 €178.50 (ECF from (a))✓ 1

Common error: In the trapezoidal rule only the first and last ordinates are counted once; every interior ordinate is doubled.

26IBMI3S-5-09-04Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Easy7 marks

A robot moves along a straight rail. Its displacement from a fixed point O is s metres at time t seconds, where
s(t) = t3 − 9t2 + 24t,   0 ≤ t ≤ 6.

(a)

Find an expression for the velocity of the robot.

(2)
(b)

Find the times at which the robot is momentarily at rest.

(3)
(c)

Find the acceleration of the robot when t = 4.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate s✓ 1
A1 v = 3t2 − 18t + 24✓ 1
Part (b)
M1 set v = 0✓ 1
A1 t = 2✓ 1
A1 t = 4✓ 1
Part (c)
M1 attempt to differentiate v (a = 6t − 18)✓ 1
A1 6 m s−2✓ 1

Common error: Being at rest means velocity is zero, not acceleration. Differentiate once for velocity and twice for acceleration.

27IBMI3S-5-09b-01Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Easy5 marks

In a video game, a player earns a bonus score B points, t seconds after activating a power-up, modelled by B(t) = 24t − 1.5t2, for 0 ≤ t ≤ 16.

(a)

Find dB/dt.

(2)
(b)

Find the time at which the bonus score is greatest, and the greatest bonus score.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 24 − 3t✓ 1
Part (b)
M1 set dB/dt = 0✓ 1
A1 t = 8 s✓ 1
A1 B = 96 points✓ 1

Common error: After finding the value of t that gives the maximum, many answers stop; the question also asks for the maximum bonus, so substitute back into B(t).

28IBMI3S-5-09b-02Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Easy6 marks

The main cable of a footbridge hangs in a curve modelled by y = 0.02x2 − 0.6x + 6, where x is the horizontal distance in metres from the left tower and y is the height of the cable in metres above the water.

(a)

Find dy/dx and hence the gradient of the cable at x = 5.

(3)
(b)

Find the equation of the tangent to the cable at x = 5. Give your answer in the form y = mx + c.

(2)
(c)

Write down the height at which this tangent meets the vertical axis (x = 0).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 dy/dx = 0.04x − 0.6✓ 1
A1 gradient = −0.4✓ 1
Part (b)
M1 height at x = 5 is 3.5 and use of y − 3.5 = −0.4(x − 5)✓ 1
A1 y = −0.4x + 5.5✓ 1
Part (c)
A1 5.5 m✓ 1

Exam tip: A tangent needs both a gradient and a point on the curve; find the y-value from the original function, not from the derivative.

29IBMI3S-5-09b-03Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Easy5 marks

Water flows into a treatment plant at a rate Q thousand litres per hour. The rate is recorded every two hours, as shown.

t (hours): 0, 2, 4, 6, 8
Q (thousand litres per hour): 12, 16, 20, 19, 14

(a)

Use the trapezoidal rule with four intervals to estimate the total volume of water that enters the plant in the 8 hours.

(3)
(b)

Hence find the mean rate of flow over the 8 hours.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of the trapezoidal rule with h = 2✓ 1
A1 (2/2)[12 + 14 + 2(16 + 20 + 19)]✓ 1
A1 136 thousand litres✓ 1
Part (b)
M1 divide the volume by 8✓ 1
A1 17 thousand litres per hour✓ 1

Common error: In the trapezoidal rule only the first and last readings are counted once; every interior reading is doubled.

30IBMI3S-5-09b-04Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Easy5 marks

A nurse programs an infusion pump so that the drug is delivered at a rate R(t) = 0.6t2 + 2 mL per minute, for 0 ≤ t ≤ 4, where t is in minutes.

(a)

Find the total volume of drug delivered during the first 4 minutes.

(3)
(b)

Find the rate at which the delivery rate is changing at t = 3.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to integrate R(t)✓ 1
A1 0.2t3 + 2t✓ 1
A1 20.8 mL✓ 1
Part (b)
M1 differentiate: R′(t) = 1.2t✓ 1
A1 3.6 mL min−2✓ 1

Exam tip: Integrating a rate gives a total amount; differentiating a rate gives how quickly the rate itself is changing, with units of the rate per unit time.

31IBMI3S-5-02r5-01Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy5 marks

A light aircraft climbs after take-off. Its height above the ground is modelled by h(t) = 900 + 240t − 6t2 metres, where t is the time in minutes and 0 ≤ t ≤ 30.

(a)

Find the rate at which the height is changing when t = 5.

(2)
(b)

Find the maximum height reached by the aircraft during this time.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 h′(t) = 240 − 12t✓ 1
A1 180 m per minute✓ 1
Part (b)
M1 set h′(t) = 0✓ 1
M1 t = 20 and substitution into h(t)✓ 1
A1 3300 m✓ 1

Exam tip: At a maximum the gradient is zero: solve h′(t) = 0 first, then substitute the t-value back into h(t) to get the height.

32IBMI3S-5-02r5-02Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy4 marks

A farmer models the profile of a sloping terrace by y = 0.01x3 − 0.15x + 3, where x is the horizontal distance in metres from the lower edge and y is the height in metres.

(a)

Find the gradient of the terrace at the point where x = 5.

(2)
(b)

Find the y-intercept of the tangent to the curve at the point where x = 5.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 dy/dx = 0.03x2 − 0.15✓ 1
A1 gradient = 0.6✓ 1
Part (b)
M1 y at x = 5 is 3.5 and tangent y − 3.5 = 0.6(x − 5)✓ 1
A1 y-intercept = 0.5✓ 1

Common error: Substitute the x-value into the derivative to get the gradient, but into the original function to get the y-coordinate.

33IBMI3S-5-02r5-03Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy5 marks

The amplitude A of a note played on a flute, in arbitrary units, is measured every 0.1 s after the note starts.

Time t (s)00.10.20.30.40.5
Amplitude A00.820.640.470.310.18

(a)

Use the trapezoidal rule with all six values to estimate the area under the amplitude–time graph from t = 0 to t = 0.5.

(3)
(b)

Hence find the mean amplitude over the first 0.5 s. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule with width 0.1: (0.1/2)(first + last + 2 × sum of others)✓ 1
A1 correct substitution (0.05)(0 + 0.18 + 2 × 2.24)✓ 1
A1 0.233✓ 1
Part (b)
M1 mean = area ÷ 0.5 (ECF from (a))✓ 1
A1 0.466✓ 1

Exam tip: The first and last ordinates are counted once; every middle ordinate is doubled.

34IBMI3S-5-02r5-04Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy4 marks

A farmer irrigates a field using a sprinkler. The rate at which water is delivered is modelled by r(t) = 15 − 0.03t2 litres per minute, for 0 ≤ t ≤ 20, where t is the time in minutes after the sprinkler is switched on.

(a)

State what the definite integral ∫020 r(t) dt represents.

(1)
(b)

Calculate the value of this integral.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 the total volume of water (in litres) delivered in the first 20 minutes✓ 1
Part (b)
M1 ∫(15 − 0.03t2) dt with an attempt to integrate✓ 1
A1 15t − 0.01t3✓ 1
A1 300 − 80 = 220 litres✓ 1

Common error: Integrate each term separately: 0.03t² becomes 0.01t³, not 0.06t.

35IBMI3S-5-02r6-01Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy4 marks

During an outbreak at a school, the total number of pupils who have caught an infection is modelled by \(N(t) = 240 + 36t - 1.5t^2\), where \(t\) is the time in days since monitoring began, for \(0 \le t \le 10\).

(a)

Find \(\frac{dN}{dt}\) and hence the rate at which the number of infected pupils is increasing when \(t = 4\).

(2)
(b)

Find the value of \(t\) at which the number of infected pupils is increasing at 9 pupils per day.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\frac{dN}{dt} = 36 - 3t\)✓ 1
A1 24 pupils per day✓ 1
Part (b)
M1 set \(36 - 3t = 9\)✓ 1
A1 \(t = 9\) days✓ 1

Common error: The rate of change is the derivative, not the value of N itself; substitute into \(\frac{dN}{dt}\) and give the units as pupils per day.

36IBMI3S-5-02r6-02Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy7 marks

The power output, \(P\) kW, of a small wind turbine is modelled by \(P = 0.05v^3\), where \(v\) m s−1 is the wind speed, for \(0 \le v \le 12\).

(a)

Find \(\frac{dP}{dv}\) and its value when \(v = 8\).

(2)
(b)

Find the equation of the tangent to the curve at the point where \(v = 8\).

(2)
(c)

Use the tangent to estimate the power output when \(v = 8.5\).

(1)
(d)

Find the power given by the model at \(v = 8.5\), to 3 significant figures, and explain whether the tangent estimate is too high or too low.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate, \(0.15v^2\)✓ 1
A1 9.6 kW per m s−1✓ 1
Part (b)
M1 \(P(8) = 25.6\) used in \(P - 25.6 = 9.6(v - 8)\)✓ 1
A1 \(P = 9.6v - 51.2\)✓ 1
Part (c)
A1 30.4 kW✓ 1
Part (d)
A1 \(0.05 \times 8.5^3 = 30.7\) kW✓ 1
R1 the gradient of the curve is increasing, so the curve lies above its tangent and the estimate is too low✓ 1

Exam tip: A tangent estimate is only reliable close to the point of tangency; comment on whether the curve is bending upwards or downwards to decide if the estimate is high or low.

37IBMI3S-5-02r6-03Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy4 marks

A solar panel array records its power output, \(P\) kW, every two hours after 08:00, as shown.

Time after 08:00 (hours): 0, 2, 4, 6, 8
Power \(P\) (kW): 0.4, 2.9, 4.6, 5.1, 3.2

(a)

Use the trapezoidal rule with four intervals to estimate the energy produced, in kWh, between 08:00 and 16:00.

(3)
(b)

Find the mean power output over the 8 hours.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct formula with width 2: \(\frac{2}{2}\left[\text{first} + \text{last} + 2(\text{rest})\right]\)✓ 1
M1 correct substitution, \(0.4 + 3.2 + 2(2.9 + 4.6 + 5.1)\)✓ 1
A1 28.8 kWh✓ 1
Part (b)
A1 \(\frac{28.8}{8} = 3.6\) kW✓ 1

Common error: Energy is the area under the power–time graph, so the width of each strip must be the time gap in hours, here 2, not 1.

38IBMI3S-5-02r6-08Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy6 marks

A hydroelectric intake channel has a cross-section whose depth is measured at 2 m intervals across its 10 m width.

Distance from left bank (m): 0, 2, 4, 6, 8, 10
Depth (m): 0, 0.9, 1.6, 1.8, 1.1, 0

(a)

Use the trapezoidal rule with five intervals to estimate the area of the cross-section.

(3)
(b)

Suggest one way to improve the accuracy of the estimate.

(1)
(c)

Water flows through the channel at a mean speed of 1.8 m s−1. Estimate the volume of water passing a point each second, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct formula with strip width 2✓ 1
M1 correct substitution, \(\frac22[0 + 0 + 2(0.9 + 1.6 + 1.8 + 1.1)]\)✓ 1
A1 10.8 m2✓ 1
Part (b)
R1 take depth measurements at smaller intervals (more strips)✓ 1
Part (c)
M1 multiply the area by 1.8✓ 1
A1 19.4 m3 per second✓ 1

Common error: With six data points there are five intervals, not six; check the number of strips before applying the rule.

39IBMI3S-5-01r8-01Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy7 marks

A wind farm's cumulative energy output is modelled by E(t) = 0.5t3 − 6t2 + 30t, where E is in MWh and t is the time in hours after midnight, 0 ≤ t ≤ 10.

(a)

Find E′(t).

(2)
(b)

Find the rate at which energy is being produced when t = 4.

(2)
(c)

Find the time at which the rate of energy production is least. Hence explain why the farm's cumulative energy never decreases during this period.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 E′(t) = 1.5t2 − 12t + 30✓ 1
Part (b)
M1 substitute t = 4 into E′✓ 1
A1 6 MWh per hour✓ 1
Part (c)
M1 E″(t) = 3t − 12 = 0✓ 1
A1 t = 4 hours✓ 1
R1 least rate is 6 MWh per hour, which is positive, so E is always increasing✓ 1

Common error: The rate of change is E′, not E; to find when the rate is least, set the derivative of the rate (E″) to zero.

40IBMI3S-5-01r8-03Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy6 marks

A wind farm maintenance compound is to be a rectangle fenced on three sides, the fourth side being the straight wall of a storage building. There is 48 m of fencing. The sides perpendicular to the wall each have length x metres.

(a)

Show that the area of the compound is A = 48x − 2x2 m2.

(2)
(b)

Find the value of x that gives the maximum area.

(3)
(c)

Find the maximum area.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 side parallel to the wall = 48 − 2x✓ 1
A1 A = x(48 − 2x) = 48x − 2x2 (AG)✓ 1
Part (b)
M1 differentiate: A′ = 48 − 4x✓ 1
M1 set A′ = 0✓ 1
A1 x = 12✓ 1
Part (c)
A1 288 m2✓ 1

Exam tip: In a 'show that' part, every step must appear; the final line should match the expression given.

41IBMI3S-5-01r8-04Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy7 marks

The power output P (kW) of a wind turbine was recorded every two hours during a 12-hour period.

Time t (hours)024681012
P (kW)016042061058034090

(a)

Use the trapezoidal rule with all the data to estimate the energy produced, in kWh, during the 12 hours.

(3)
(b)

Hence find the mean power output over the 12 hours. Give your answer to the nearest kW.

(2)
(c)

The electricity is sold at €0.085 per kWh. Find the income from the energy in (a), giving your answer in euros to 2 decimal places.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct use of trapezoidal rule with h = 2✓ 1
M1 ½ × 2 × (0 + 90 + 2(160 + 420 + 610 + 580 + 340))✓ 1
A1 4310 kWh✓ 1
Part (b)
M1 4310 ÷ 12✓ 1
A1 359 kW✓ 1
Part (c)
M1 4310 × 0.085✓ 1
A1 €366.35 (ECF from (a))✓ 1

Common error: The interior values are doubled, but the first and last values are used once only.

42IBMI3S-5-01r8-08Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy7 marks

A wind farm lies on a ridge whose height is modelled by h(x) = 4x − 0.25x2, 0 ≤ x ≤ 10, where x is the horizontal distance and h is the height, both in units of 100 m. A turbine stands at the point where x = 3.

(a)

Find h′(x).

(2)
(b)

Find the equation of the tangent to the ridge at the turbine, in the form y = mx + c.

(3)
(c)

Find the equation of the normal to the ridge at the turbine, in the form y = mx + c.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 h′(x) = 4 − 0.5x✓ 1
Part (b)
A1 gradient h′(3) = 2.5✓ 1
A1 h(3) = 9.75✓ 1
A1 y = 2.5x + 2.25✓ 1
Part (c)
M1 gradient of normal = −1/2.5 = −0.4✓ 1
A1 y = −0.4x + 10.95✓ 1

Common error: The normal's gradient is the negative reciprocal of the tangent's gradient, not just its negative.

43IBMI3S-5-01r13-01Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy4 marks

The total number of books issued by a city library t hours after it opens is modelled by N(t) = 4t2 − 0.1t3, for 0 ≤ t ≤ 10.

(a)

Find an expression for dN/dt.

(2)
(b)

Find the rate at which books are being issued 5 hours after opening. Include units in your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate, at least one term correct✓ 1
A1 dN/dt = 8t − 0.3t2✓ 1
Part (b)
M1 substitute t = 5 into their derivative (ECF from (a))✓ 1
A1 32.5 books per hour✓ 1

Exam tip: The derivative of a model gives its instantaneous rate of change, so the units are the units of the output divided by the units of the input.

44IBMI3S-5-01r13-02Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy4 marks

A wind farm records its power output P (in MW) once an hour for four hours:
Time t (hours): 0, 1, 2, 3, 4
Power P (MW): 18.0, 24.5, 31.0, 27.5, 21.0

(a)

Use the trapezoidal rule with four intervals to estimate the energy generated, in MWh, over the four hours.

(3)
(b)

The operator now takes readings every 30 minutes instead. State, with a reason, how the new estimate is likely to compare in accuracy.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct use of the trapezoidal rule with strip width 1✓ 1
M1 correct substitution: ½ × 1 × (18.0 + 21.0 + 2 × (24.5 + 31.0 + 27.5))✓ 1
A1 102.5 MWh✓ 1
Part (b)
R1 more accurate (closer to the true energy), because narrower strips follow the curve more closely✓ 1

Common error: The width of each strip is the gap between readings, not the number of readings.

45IBMI3S-5-01r13-03Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy6 marks

A swimming club ropes off a rectangular practice zone in a lake. The shore forms one side of the rectangle, so rope is needed for the other three sides only. The club has 120 m of rope. The sides perpendicular to the shore each have length x metres.

(a)

Show that the area of the zone is A = 120x − 2x2 square metres.

(2)
(b)

Find the value of x that maximises the area.

(3)
(c)

Find the maximum area.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 side parallel to the shore = 120 − 2x✓ 1
A1 A = x(120 − 2x) = 120x − 2x2 (AG)✓ 1
Part (b)
M1 dA/dx = 120 − 4x✓ 1
M1 set derivative equal to zero✓ 1
A1 x = 30✓ 1
Part (c)
A1 1800 m2 (ECF from (b))✓ 1

Exam tip: In an optimisation question, always finish by answering the quantity actually asked for, not just the value of the variable.

46IBMI3S-5-01r13-04Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy7 marks

The profile of a water slide is modelled by y = 0.04x2 − 0.8x + 5, for 0 ≤ x ≤ 20, where x is the horizontal distance in metres from the top and y is the height in metres above the pool surface.

(a)

Find the gradient of the slide when x = 3.

(2)
(b)

Find the equation of the tangent to the slide at x = 3. Give your answer in the form y = mx + c.

(3)
(c)

Find the value of x at which the slide is horizontal.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 dy/dx = 0.08x − 0.8✓ 1
A1 gradient = −0.56✓ 1
Part (b)
M1 y = 2.96 when x = 3✓ 1
M1 use of y − 2.96 = −0.56(x − 3)✓ 1
A1 y = −0.56x + 4.64✓ 1
Part (c)
M1 set 0.08x − 0.8 = 0✓ 1
A1 x = 10✓ 1

Common error: A horizontal tangent means the gradient is zero, not that y is zero.

47IBMI3S-5-03r21-01Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Easy7 marks

A grey seal makes a dive from a research boat. Its depth below the surface is modelled by D(t) = 0.04t3 − 1.2t2 + 9t, where D is in metres and t is the time in seconds after the dive begins, for 0 ≤ t ≤ 15.

(a)

Find D′(t).

(2)
(b)

Find the rate at which the depth is changing when t = 4. Include units in your answer.

(2)
(c)

Show that the seal is momentarily at rest, vertically, when t = 5, and find its depth at that time.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 D′(t) = 0.12t2 − 2.4t + 9✓ 1
Part (b)
M1 substitute t = 4 into D′✓ 1
A1 1.32 m s-1✓ 1
Part (c)
M1 substitute t = 5 into D′✓ 1
A1 0.12(25) − 12 + 9 = 0, so at rest✓ 1
A1 depth = 20 m✓ 1

Exam tip: A rate of change is a derivative. Substitute into the derivative, not the original function, and quote units such as m s⁻¹.

48IBMI3S-5-03r21-02Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Easy6 marks

The roof of a market canopy has a cross-section modelled by h(x) = −0.04x2 + 0.9x + 3 for 0 ≤ x ≤ 20, where h is the height above the ground in metres and x is the horizontal distance in metres from the left-hand wall.

(a)

Find h′(x).

(2)
(b)

A support cable lies along the tangent to the roof profile at the point where x = 5. Find the equation of this tangent, giving your answer in the form y = mx + c.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 h′(x) = −0.08x + 0.9✓ 1
Part (b)
A1 gradient h′(5) = 0.5✓ 1
A1 h(5) = 6.5✓ 1
M1 use of y − 6.5 = 0.5(x − 5) or substitution to find c✓ 1
A1 y = 0.5x + 4✓ 1

Common error: Using h(5) as the gradient, or h′(5) as the y-coordinate. The gradient comes from the derivative; the point comes from the original function.

49IBMI3S-5-03r21-03Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Easy5 marks

A tidal channel in a marine reserve is 8 m wide. The depth of water is measured at equal intervals across the channel.
Distance from bank (m): 0, 2, 4, 6, 8
Depth (m): 0, 1.2, 1.9, 1.5, 0

(a)

Use the trapezoidal rule with four intervals to estimate the cross-sectional area of the water in the channel.

(3)
(b)

Water flows through the channel at a constant speed of 0.8 m s-1. Hence estimate the volume of water passing through the channel each second. Use the unrounded value from part (a).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of trapezoidal rule with interval width 2✓ 1
A1 ½ × 2 × [0 + 0 + 2(1.2 + 1.9 + 1.5)]✓ 1
A1 9.2 m2✓ 1
Part (b)
M1 area × speed, 9.2 × 0.8✓ 1
A1 7.36 m3 per second (ECF from (a))✓ 1

Exam tip: In the trapezoidal rule the two end values are counted once and every inner value is doubled. Check that the number of intervals matches the width you use.

50IBMI3S-5-03r21-04Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Easy4 marks

The temperature of seawater at a monitoring station varies with depth. It is modelled by T(d) = 20 − 0.05d + 0.00005d2, where T is in °C and d is the depth in metres, 0 ≤ d ≤ 300.

(a)

Find T′(d).

(2)
(b)

Find the value of T′(200) and interpret this value in context.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 T′(d) = −0.05 + 0.0001d✓ 1
Part (b)
A1 T′(200) = -0.03✓ 1
R1 at a depth of 200 m the temperature is decreasing by 0.03 °C for each additional metre of depth✓ 1

Exam tip: When interpreting a gradient, state the quantity that is changing, the direction of change, the size, and the units, all in the words of the context.

51IBMI3S-5-02r24-01Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy6 marks

A landing module descends towards a moon. Its height above the surface is modelled by h(t) = 0.05t2 − 14t + 1000, where h is in metres and t is the time in seconds after the parachute opens, for 0 ≤ t ≤ 120.

(a)

Find h′(t).

(2)
(b)

Find the rate of change of height when t = 60, and state what the sign of your answer tells you about the module.

(2)
(c)

Find the value of t when the module is descending at 4 m s−1.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiating a term correctly✓ 1
A1 h′(t) = 0.1t − 14✓ 1
Part (b)
A1 −8 m s−1✓ 1
R1 negative gradient: the height is decreasing, so the module is descending at 8 m s−1✓ 1
Part (c)
M1 0.1t − 14 = −4✓ 1
A1 t = 100 s✓ 1

Exam tip: A negative rate of change of height means the height is falling; the speed is the size of that rate, so −4 m s⁻¹ is a speed of 4 m s⁻¹.

52IBMI3S-5-02r24-02Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy7 marks

The main cable of a footbridge is modelled by the curve y = 0.01x2 − 0.8x + 20 for 0 ≤ x ≤ 80, where x is the horizontal distance in metres from the left tower and y is the height of the cable above the deck in metres.

(a)

Find the gradient of the cable at the point where x = 10.

(2)
(b)

Find the equation of the tangent to the cable at the point where x = 10. Give your answer in the form y = mx + c.

(3)
(c)

Find the value of x at which the gradient of the cable is 0.4.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 dy/dx = 0.02x − 0.8✓ 1
A1 −0.6✓ 1
Part (b)
A1 y(10) = 13✓ 1
M1 y − 13 = −0.6(x − 10) (ECF from (a))✓ 1
A1 y = −0.6x + 19 (accept 19 − 0.6x)✓ 1
Part (c)
M1 0.02x − 0.8 = 0.4✓ 1
A1 x = 60✓ 1

Common error: Substitute the x-value into the derivative to get the gradient, but into the original function to get the y-coordinate; mixing these up gives the wrong tangent.

53IBMI3S-5-02r24-03Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy7 marks

During a coffee roast the electrical power P kW supplied to the roaster is measured every 2 minutes.
t (minutes): 0, 2, 4, 6, 8, 10
P (kW): 4.0, 5.2, 6.1, 6.6, 6.8, 6.9
The graph of P against t is increasing, and its gradient decreases throughout the 10 minutes.

(a)

Use the trapezoidal rule with 5 intervals to estimate the area under the graph of P against t between t = 0 and t = 10.

(3)
(b)

The energy supplied equals the area in part (a), converted using 1 kW = 1 kJ s−1. Find the energy supplied in kJ.

(2)
(c)

State whether your estimate is an underestimate or an overestimate of the true energy, giving a reason.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of trapezoidal rule with width 2✓ 1
A1 (2 ÷ 2)[4.0 + 6.9 + 2(5.2 + 6.1 + 6.6 + 6.8)]✓ 1
A1 60.3 kW min✓ 1
Part (b)
M1 60.3 × 60 (minutes converted to seconds)✓ 1
A1 3618 kJ✓ 1
Part (c)
A1 underestimate✓ 1
R1 the graph is concave down (gradient decreasing), so each trapezium lies below the curve✓ 1

Exam tip: The area under a power–time graph is energy; check the units of time before converting, as kW × minutes is not yet kJ.

54IBMI3S-5-01r27-01Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy7 marks

The number of people, N, in a town who are infected with a virus is modelled by N(t) = 40 + 18t − 0.6t2, for 0 ≤ t ≤ 25, where t is the time in days since the first case was recorded.

(a)

Find dN/dt.

(2)
(b)

Find the rate at which the number of infected people is increasing when t = 5.

(2)
(c)

Find the value of t at which the number of infected people stops increasing.

(2)
(d)

Find the greatest number of infected people predicted by the model.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiating at least one term correctly✓ 1
A1 dN/dt = 18 − 1.2t✓ 1
Part (b)
M1 substituting t = 5 into dN/dt✓ 1
A1 12 people per day✓ 1
Part (c)
M1 setting dN/dt = 0✓ 1
A1 t = 15 days✓ 1
Part (d)
A1 N(15) = 175 people (ECF from (c))✓ 1

Exam tip: A turning point of a model is found where the derivative is zero; always say what the value means in the context of the question.

55IBMI3S-5-01r27-02Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy6 marks

A shop sells a desk lamp at a price of €x. The daily profit, €P, is modelled by P(x) = −2x2 + 56x − 150, for 5 ≤ x ≤ 25.

(a)

Find dP/dx.

(2)
(b)

Find the price that gives the maximum daily profit.

(2)
(c)

Find the maximum daily profit.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiating at least one term correctly✓ 1
A1 dP/dx = 56 − 4x✓ 1
Part (b)
M1 setting dP/dx = 0✓ 1
A1 x = 14, so the price is €14✓ 1
Part (c)
M1 substituting x = 14 into P(x) (ECF from (b))✓ 1
A1 €242✓ 1

Common error: Students often stop after finding the optimal price. Read the question again to see whether the maximum value itself is also required.

56IBMI3S-5-01r27-03Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy7 marks

The height, h metres, of a mountain road above sea level is modelled by h(x) = x3 − 6x2 + 9x + 20, for 0 ≤ x ≤ 4, where x is the horizontal distance in kilometres from the start of the road.

(a)

Find h′(x).

(2)
(b)

Find the equation of the tangent to the road profile at the point where x = 2. Give your answer in the form y = mx + c.

(3)
(c)

This tangent meets the horizontal line y = 20 at the point Q. Find the x-coordinate of Q, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiating at least two terms correctly✓ 1
A1 h′(x) = 3x2 − 12x + 9✓ 1
Part (b)
M1 gradient = h′(2) = −3✓ 1
A1 h(2) = 22✓ 1
A1 y = −3x + 28✓ 1
Part (c)
M1 setting −3x + 28 = 20 (ECF from (b))✓ 1
A1 x = 2.67✓ 1

Exam tip: To find a tangent, you need both a gradient (from the derivative) and a point on the curve (from the original function).

57IBMI3S-5-01r27-09Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy5 marks

The cumulative energy, E MWh, generated by a solar farm t hours after a test began satisfies dE/dt = 3t2 − 18t + 30. After 2 hours the farm had generated 60 MWh in total.

(a)

Find an expression for E in terms of t, including a constant of integration.

(3)
(b)

Use the condition E(2) = 60 to find the value of c.

(1)
(c)

Find the total energy generated after 4 hours.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 integrating at least one term correctly✓ 1
A1 t3 − 9t2 + 30t✓ 1
A1 + c✓ 1
Part (b)
A1 8 − 36 + 60 + c = 60, so c = 28✓ 1
Part (c)
A1 E(4) = 64 − 144 + 120 + 28 = 68 MWh (ECF from (b))✓ 1

Common error: Forgetting the constant of integration loses marks. Use the given condition straight away to find it before substituting a new value of t.

58IBMI3S-5-02r27-01Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy7 marks

The height, H km, of an ash plume above a volcano vent is modelled by H(t) = −0.05t2 + 1.2t, for 0 ≤ t ≤ 24, where t is the time in minutes after the start of the eruption.

(a)

Find dH/dt.

(2)
(b)

Find the rate at which the height of the plume is increasing when t = 5.

(2)
(c)

Find the time at which the plume reaches its greatest height.

(2)
(d)

Find the greatest height of the plume.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiating at least one term correctly✓ 1
A1 dH/dt = −0.1t + 1.2✓ 1
Part (b)
M1 substituting t = 5 into dH/dt✓ 1
A1 0.7 km per minute✓ 1
Part (c)
M1 setting dH/dt = 0✓ 1
A1 t = 12 minutes✓ 1
Part (d)
A1 H(12) = 7.2 km (ECF from (c))✓ 1

Exam tip: State the units of a rate of change: here the derivative of height with respect to time is in km per minute.

59IBMI3S-5-02r27-02Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy4 marks

A theatre sells tickets at €p each. The number of tickets sold for a performance is modelled by N = 600 − 20p, for 0 < p < 30.

(a)

Show that the revenue, €R, from ticket sales is R = 600p − 20p2.

(1)
(b)

Find the ticket price that gives the maximum revenue.

(2)
(c)

Find the maximum revenue.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 R = p × N = p(600 − 20p) = 600p − 20p2✓ 1
Part (b)
M1 dR/dp = 600 − 40p = 0✓ 1
A1 €15✓ 1
Part (c)
A1 €4500 (ECF from (b))✓ 1

Exam tip: Revenue is price multiplied by quantity sold. Write this product first, then expand it before differentiating.

60IBMI3S-5-02r27-03Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy6 marks

The fuel cost, €f per nautical mile, for a cargo ship travelling at a steady speed of v knots is modelled by f(v) = 2744/v + 0.5v2, for 5 ≤ v ≤ 25.

(a)

Find f′(v).

(2)
(b)

Show that the cost per nautical mile is a minimum when v = 14.

(3)
(c)

Find the minimum cost per nautical mile.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiating at least one term correctly✓ 1
A1 −2744/v2 + v✓ 1
Part (b)
M1 setting f′(v) = 0✓ 1
A1 v3 = 2744✓ 1
A1 v = 14✓ 1
Part (c)
A1 €294 (using v = 14)✓ 1

Common error: Writing 2744/v as 2744v⁻¹ before differentiating avoids errors with the power of v; the derivative has a negative sign.

61IBMI3S-5-02r27-11Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy5 marks

The depth, h cm, of ash on a village roof t hours after an eruption begins satisfies dh/dt = 0.12t2 − 1.2t + 3.6. After 5 hours the depth of ash is 14 cm.

(a)

Find an expression for h in terms of t, including a constant of integration.

(3)
(b)

Find the value of c.

(1)
(c)

Find the depth of ash after 10 hours.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 integrating at least one term correctly✓ 1
A1 h = 0.04t3 − 0.6t2 + 3.6t✓ 1
A1 + c✓ 1
Part (b)
A1 5 − 15 + 18 + c = 14, so c = 6✓ 1
Part (c)
A1 h(10) = 40 − 60 + 36 + 6 = 22 cm (ECF from (b))✓ 1

Common error: Substitute the known depth into your expression for h before finding any other value; the constant of integration is not zero unless the question says so.

62IBMI3S-5-03r27-01Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Easy6 marks

The concentration of a drug in a patient's blood is modelled by C(t) = −0.8t2 + 6.4t, where C is in mg/L and t is the time in hours after an injection, for 0 ≤ t ≤ 8.

(a)

Find an expression for dC/dt.

(2)
(b)

Find the rate of change of concentration when t = 1.5, and state what this value tells you about the concentration at that moment.

(2)
(c)

Find the maximum concentration of the drug.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 dC/dt = −1.6t + 6.4✓ 1
Part (b)
M1 substitute t = 1.5 into the derivative✓ 1
A1 4.00 mg/L per hour; the concentration is increasing at that moment✓ 1
Part (c)
M1 set dC/dt = 0, giving t = 4✓ 1
A1 C(4) = 12.8 mg/L✓ 1

Exam tip: A rate of change is a derivative. Give its units as 'units of the output per unit of the input', for example mg/L per hour.

63IBMI3S-5-03r27-02Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Easy5 marks

Water flows into a storage tank at a rate of r(t) = 0.6t2 + 2 m3 per hour, where t is the time in hours, for 0 ≤ t ≤ 6.

(a)

Find the volume of water that flows into the tank during the first 6 hours.

(3)
(b)

The tank holds 20 m3 of water at t = 0 and has a capacity of 90 m3. Find the volume in the tank at t = 6 and state whether the tank overflows.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to integrate r(t) from 0 to 6✓ 1
A1 0.2t3 + 2t✓ 1
A1 55.2 m3✓ 1
Part (b)
M1 20 + 55.2 (ECF from (a))✓ 1
A1 75.2 m3, which is less than 90 m3, so the tank does not overflow✓ 1

Common error: Forgetting the starting volume: the integral gives only the water that flows in, not the total in the tank.

64IBMI3S-5-03r27-06Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Easy6 marks

A gardener uses 40 m of edging to mark three sides of a rectangular irrigation bed. The fourth side is a straight wall that needs no edging. The sides perpendicular to the wall each have length x m.

(a)

Write down an expression, in terms of x, for the length of the side parallel to the wall.

(1)
(b)

Show that the area of the bed, A m2, is A = 40x − 2x2.

(1)
(c)

Find the value of x that gives the largest area.

(3)
(d)

Find the largest possible area of the bed.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 40 − 2x✓ 1
Part (b)
M1 A = x(40 − 2x) expanded to the given form✓ 1
Part (c)
M1 attempt to differentiate✓ 1
A1 dA/dx = 40 − 4x✓ 1
A1 set equal to zero, x = 10✓ 1
Part (d)
A1 200 m2✓ 1

Exam tip: In optimisation, express the quantity in a single variable first, then differentiate.

65IBMI3S-5-03r27-12Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Easy6 marks

In an online game, the total experience points (XP) needed to reach level n is modelled by X(n) = 2n2 + 30n, for 1 ≤ n ≤ 30. Treat n as a continuous variable where needed.

(a)

Find the average rate of change of X between level 5 and level 10.

(2)
(b)

Find X′(n) and hence the instantaneous rate of change of X at n = 8.

(2)
(c)

Find the level at which the instantaneous rate of change is 90 XP per level.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 (X(10) − X(5))/(10 − 5) = (500 − 200)/5✓ 1
A1 60 XP per level✓ 1
Part (b)
M1 X′(n) = 4n + 30✓ 1
A1 62 XP per level✓ 1
Part (c)
M1 4n + 30 = 90✓ 1
A1 n = 15✓ 1

Exam tip: Average rate of change uses two points on the function. Instantaneous rate of change uses the derivative at one point.

66IBMI3S-5-03r28-01Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy6 marks

The density D of krill, in individuals per m3, at a monitoring station is modelled by
D(t) = 0.5t3 − 6t2 + 24t + 10,   0 ≤ t ≤ 10,
where t is the time in hours after sunset.

(a)

Find D′(t).

(2)
(b)

Find the rate of change of density when t = 3.

(1)
(c)

Find the value of t for which the rate of change of density is zero.

(2)
(d)

Find the density at this time.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 D′(t) = 1.5t2 − 12t + 24✓ 1
Part (b)
A1 1.5 individuals per m3 per hour (ECF from (a))✓ 1
Part (c)
M1 sets 1.5t2 − 12t + 24 = 0 (or equivalent, ECF from (a))✓ 1
A1 t = 4 (repeated root)✓ 1
Part (d)
A1 D(4) = 42 individuals per m3 (ECF from (c))✓ 1

Exam tip: "Rate of change at a point" means evaluate the derivative there, not the original function; include the units of the derivative.

67IBMI3S-5-03r28-02Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy7 marks

The roof of a small pavilion has a cross-section modelled by the curve
y = 6 − 0.04x2,   −12 ≤ x ≤ 12,
where x and y are measured in metres, with the origin at floor level directly below the highest point.

(a)

Find dy/dx.

(2)
(b)

Find the gradient of the roof at the point where x = 5.

(1)
(c)

Find the equation of the tangent to the roof at the point where x = 5. Give your answer in the form y = mx + c.

(2)
(d)

The tangent in part (c) is extended until it reaches floor level (y = 0). Find the x-coordinate of this point.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 dy/dx = −0.08x✓ 1
Part (b)
A1 gradient = −0.4 (ECF from (a))✓ 1
Part (c)
M1 finds y = 5 when x = 5 and uses y − 5 = (−0.4)(x − 5) (ECF from (b))✓ 1
A1 y = −0.4x + 7✓ 1
Part (d)
M1 sets −0.4x + 7 = 0 (ECF from (c))✓ 1
A1 x = 17.5 m✓ 1

Common error: The gradient formula gives the slope at a point; the tangent equation also needs the y-value at that point, found from the original curve and not from the derivative.

68IBMI3S-5-03r28-03Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy4 marks

Delivery vans arrive at a distribution depot at a rate that is recorded every two hours during an eight-hour shift.

Time since shift start, t (hours)02468
Arrival rate (vans per hour)1218252115

(a)

Use the trapezoidal rule with four intervals to estimate the total number of vans that arrive during the shift.

(3)
(b)

Each van delivers on average 1.2 tonnes of goods. Use your answer to part (a) to estimate the total mass delivered during the shift.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of the trapezoidal rule with width 2✓ 1
A1 (2/2)[12 + 15 + 2(18 + 25 + 21)]✓ 1
A1 155 vans✓ 1
Part (b)
A1 155 × 1.2 = 186 tonnes (ECF from (a))✓ 1

Exam tip: For the trapezoidal rule, write the formula with the interval width, the first and last values, and the doubled sum of the interior values so each method mark is visible.

69IBMI3S-5-03r28-04Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy5 marks

Seawater flows into a tidal pool at a rate of r(t) = 6 + 0.9t2 litres per minute, where t is the time in minutes after the tide starts to rise, 0 ≤ t ≤ 5.

(a)

Find ∫ r(t) dt.

(2)
(b)

Hence find the volume of water that flows into the pool during the first 5 minutes.

(2)
(c)

The pool contains 40 litres of water when t = 0. Find the volume of water in the pool when t = 5.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to integrate (at least one term correct)✓ 1
A1 6t + 0.3t3 (+ c)✓ 1
Part (b)
M1 substitutes limits 5 and 0 into the integral from (a)✓ 1
A1 67.5 litres✓ 1
Part (c)
A1 40 + 67.5 = 107.5 litres (ECF from (b))✓ 1

Common error: Integrating a rate gives the change in quantity, not the quantity itself; any starting amount must be added separately.

70IBMI3S-5-02r30-01Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Easy6 marks

The depth d metres below ground level of a drainage channel at an archaeological dig is modelled by \(d = 0.25x^2 - 2x + 6\), where x is the horizontal distance in metres from the start of the channel, for 0 ≤ x ≤ 12.

(a)

Find \(\frac{dd}{dx}\).

(2)
(b)

Find the value of x at which the floor of the channel is level.

(2)
(c)

Find the gradient of the channel floor when x = 7, and interpret this value.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 0.5x − 2✓ 1
Part (b)
M1 0.5x − 2 = 0✓ 1
A1 x = 4 m✓ 1
ECF from (a).—
Part (c)
A1 0.5 × 7 − 2 = 1.5✓ 1
R1 the channel gets 1.5 m deeper for each extra metre of horizontal distance✓ 1
ECF from (a).—

Common error: Writing the derivative of 0.25x² as 0.25x. Multiply by the power and reduce the power by one: 2 × 0.25 = 0.5.

71IBMI3S-5-02r30-02Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Easy7 marks

The side of an excavation trench is modelled by the curve \(y = 0.25x^2 - 3x + 10\), where x and y are in metres. The point P(2, 5) lies on the curve.

(a)

Find the gradient of the curve at P.

(2)
(b)

Find the equation of the tangent to the curve at P. Give your answer in the form y = mx + c.

(2)
(c)

The normal to the curve at P meets the x-axis at the point Q. Find the x-coordinate of Q.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\frac{dy}{dx} = 0.5x - 3\)✓ 1
A1 −2✓ 1
Part (b)
M1 y − 5 = −2(x − 2)✓ 1
A1 y = −2x + 9✓ 1
ECF from (a).—
Part (c)
M1 gradient of normal = 1/2✓ 1
M1 normal: y − 5 = 0.5(x − 2), so y = 0.5x + 4, then set y = 0✓ 1
A1 x = −8✓ 1
ECF from (a).—

Exam tip: The gradient of a normal is the negative reciprocal of the gradient of the tangent. Check that the product of the two gradients is −1.

72IBMI3S-5-02r30-06Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Easy7 marks

A cooling plate in a data centre has temperature T °C at time t minutes after it is switched off, given by \(T = 25 + \frac{144}{t}\) for t ≥ 2.

(a)

Find the temperature of the plate when t = 4.

(1)
(b)

Find \(\frac{dT}{dt}\).

(2)
(c)

Find the rate of change of the temperature when t = 4 and interpret your answer.

(2)
(d)

Find the value of t when the plate is cooling at exactly 1 °C per minute.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 25 + 144/4 = 61 °C✓ 1
Part (b)
M1 attempt to differentiate 144t−1✓ 1
A1 \(-144t^{-2}\)✓ 1
Part (c)
A1 −144/16 = −9✓ 1
R1 the plate is cooling at 9 °C per minute at that moment✓ 1
ECF from (b).—
Part (d)
M1 −144t−2 = −1✓ 1
A1 t2 = 144, so t = 12 minutes (reject −12)✓ 1
ECF from (b).—

Common error: Writing the derivative of 144/t as 144/t² without the minus sign. A cooling object must have a negative rate of change of temperature.

73IBMI3S-5-02r30-10Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Easy6 marks

The cross-section of an ancient river channel is measured at 2 m intervals across its 12 m width. The table shows the depth d of water at each position.

Distance from bank (m): 0, 2, 4, 6, 8, 10, 12
Depth (m): 0, 1.1, 2.3, 3.0, 2.2, 1.4, 0

(a)

Use the trapezoidal rule with 6 intervals to estimate the area of the cross-section.

(3)
(b)

The water flows at a mean speed of 0.8 m s−1. Find the volume of water passing the cross-section each second.

(1)
(c)

Find the volume of water passing the cross-section in one hour. Give your answer in the form a × 10k where 1 ≤ a < 10 and k is an integer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of the trapezoidal rule with width 2✓ 1
A1 \(\frac{2}{2}\left[0 + 0 + 2(1.1 + 2.3 + 3.0 + 2.2 + 1.4)\right]\)✓ 1
A1 20.0 m2✓ 1
Part (b)
A1 20.0 × 0.8 = 16 m3 s−1✓ 1
ECF from (a).—
Part (c)
M1 16 × 3600✓ 1
A1 57600 = 5.76 × 104 m3✓ 1
ECF from (b).—

Exam tip: Keep units consistent: convert the hour to 3600 seconds before multiplying by a rate in cubic metres per second.

74IBMI3S-5-01r31-01Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy4 marks

Calculator

A robotic arm moves its gripper along a rail. The distance, x centimetres, of the gripper from the start of the rail at time t seconds is modelled by x(t) = 0.5t3 − 3t2 + 8t, for 0 ≤ t ≤ 6.

(a)

Find x′(t).

(2)
(b)

Find the speed of the gripper when t = 4.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 0.5 × 3t2 = 1.5t2✓ 1
A1 −6t + 8 (accept x′(t) = 1.5t2 − 6t + 8)✓ 1
Part (b)
M1 substituting t = 4 into x′ (ECF from (a)): 1.5(16) − 6(4) + 8✓ 1
A1 8 cm s−1✓ 1

Common error: When differentiating, multiply the coefficient by the power before reducing the power; 0.5t³ becomes 1.5t², not 0.5t².

75IBMI3S-5-01r31-02Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy6 marks

Calculator

The number of fish, N thousand, in a coastal fishery t months after a new quota is introduced is modelled by N(t) = −2t2 + 24t + 150, for 0 ≤ t ≤ 10.

(a)

Find N′(t).

(2)
(b)

Find the rate of change of the fish population when t = 3, and state what this value means in context.

(2)
(c)

Find the time at which the fish population is greatest.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 −4t✓ 1
A1 + 24 (accept N′(t) = −4t + 24)✓ 1
Part (b)
A1 N′(3) = 12✓ 1
R1 the population is increasing by 12 thousand fish per month (at 3 months after the quota)✓ 1
Part (c)
M1 setting N′(t) = 0 (ECF from (a))✓ 1
A1 t = 6 months✓ 1

Exam tip: A maximum or minimum of a model occurs where the gradient is zero. Set the derivative equal to zero and state the answer with its unit.

76IBMI3S-5-01r31-03Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy6 marks

Calculator

A designer models the curve of a neckline by y = 0.5x2 − 3x + 6, where x and y are measured in centimetres from a fixed origin. The point A on the curve has x-coordinate 4.

(a)

Find the gradient of the curve at A.

(2)
(b)

Find the equation of the tangent to the curve at A.

(2)
(c)

The normal to the curve at A crosses the y-axis at the point B. Find the y-coordinate of B.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 dy/dx = x − 3✓ 1
A1 gradient = 1✓ 1
Part (b)
A1 y-coordinate of A is 2✓ 1
A1 y = x − 2 (accept y − 2 = 1(x − 4))✓ 1
Part (c)
M1 normal gradient = −1 (ECF), so y − 2 = −1(x − 4)✓ 1
A1 y = 6✓ 1

Common error: The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent gradient. Do not simply change the sign when the gradient is not ±1.

77IBMI3S-5-01r31-07Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Easy4 marks

Calculator

The angle, θ degrees, turned by a robot's elbow joint after t seconds is modelled by θ(t) = 2t3 − 9t2 + 12t, for 0 ≤ t ≤ 3. The joint momentarily stops turning when dθ/dt = 0.

(a)

Find dθ/dt.

(2)
(b)

Find the times at which the joint momentarily stops turning.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 6t2 − 18t✓ 1
A1 + 12✓ 1
Part (b)
M1 setting 6t2 − 18t + 12 = 0 (ECF from (a))✓ 1
A1 t = 1 s and t = 2 s✓ 1

Exam tip: A quadratic derivative can be solved on your calculator or by factorising; give every solution that lies inside the stated time interval.

78IBMI3S-5-01r32-01Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy4 marks

A climber leaves a mountain hut and ascends a snow slope. Her height, h metres, above the valley floor is modelled by
h = 150 + 72t − 4.5t2, for 0 ≤ t ≤ 6,
where t is the time in hours since she left the hut.

(a)

Find an expression for dh/dt.

(2)
(b)

Find the rate at which the climber is gaining height when t = 4.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 dh/dt = 72 − 9t✓ 1
Part (b)
M1 substitute t = 4 into their dh/dt (ECF from (a))✓ 1
A1 36 m h−1✓ 1

Exam tip: Write the units of a rate of change as 'unit of output per unit of input', here metres per hour.

79IBMI3S-5-01r32-02Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy5 marks

A mobile network records the data rate of a video stream. During the first 10 seconds the rate is modelled by
r(t) = 12 + 3t − 0.2t2, for 0 ≤ t ≤ 10,
where r is in megabits per second (Mbit s−1) and t is in seconds.

(a)

Find the total amount of data, in megabits, sent during the first 10 seconds. Give your answer to 3 significant figures.

(3)
(b)

There are 8 megabits in 1 megabyte (MB). Convert your unrounded answer to (a) into megabytes, giving your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct definite integral of r(t) from 0 to 10✓ 1
M1 attempt to evaluate it (by integration or using technology)✓ 1
A1 203 Mbit (exact value 203.33…)✓ 1
Part (b)
M1 divide by 8: 203.33… ÷ 8✓ 1
A1 25.4 MB✓ 1

Exam tip: The area under a rate-against-time graph gives the total amount transferred, so a definite integral of the rate over the time interval gives the total.

80IBMI3S-5-01r32-03Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy6 marks

A food scientist studies a fermentation process. The mass of product obtained from one batch is modelled by
y(T) = −0.8T2 + 52T − 700, for 20 ≤ T ≤ 45,
where y is in grams and T is the temperature of the batch in °C.

(a)

Find y′(T).

(2)
(b)

Find the temperature that gives the greatest mass of product.

(2)
(c)

Find the greatest mass of product obtained from one batch.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 y′(T) = −1.6T + 52✓ 1
Part (b)
M1 set y′(T) = 0✓ 1
A1 T = 32.5 °C✓ 1
Part (c)
M1 substitute T = 32.5 into y(T) (ECF from (b))✓ 1
A1 145 g✓ 1

Exam tip: At a maximum or minimum of a smooth function the gradient is zero, so solve the derivative equal to 0 and then substitute back into the original function for the value.

81IBMI3S-5-01r32-08Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy7 marks

A food manufacturer designs a closed cylindrical tin of volume 500 cm3. The tin has base radius r cm. The total outer surface area of the tin is given by
A(r) = 2πr2 + 1000r−1, for r > 0,
where A is in cm2. The manufacturer wants to use as little metal as possible.

(a)

Find dA/dr.

(2)
(b)

Find the value of r that makes the surface area a minimum. Give your answer to 3 significant figures.

(3)
(c)

Find the minimum surface area, giving your answer to 3 significant figures. Use the unrounded value of r.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 dA/dr = 4πr − 1000r−2✓ 1
Part (b)
M1 set dA/dr = 0✓ 1
M1 rearrange or solve (e.g. r3 = 250/π = 79.57…)✓ 1
A1 r = 4.30 cm✓ 1
Part (c)
M1 substitute r = 4.3012… into A(r)✓ 1
A1 349 cm2✓ 1

Exam tip: Always substitute the unrounded optimal value back into the original function; rounding r first can change the third significant figure.

82IBMI3S-5-01r33-01Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Easy7 marks

Ash falls steadily on a village near a volcano. The depth of ash on a flat roof, d mm, is modelled by d(t) = 0.05t3 − 0.9t2 + 5.4t, where t is the time in hours after the ash fall begins, for 0 ≤ t ≤ 10.

(a)

Find d′(t).

(2)
(b)

Find the rate at which the depth is increasing when t = 4. Include units in your answer.

(2)
(c)

Find the value of t at which the rate of increase of the depth is least.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 0.15t2 − 1.8t + 5.4✓ 1
Part (b)
M1 substitute t = 4 into d′(t) (ECF from (a))✓ 1
A1 0.6 mm per hour (mm h−1)✓ 1
Part (c)
M1 attempt to find the minimum of d′ (e.g. set d″(t) = 0.3t − 1.8 = 0, or use GDC minimum, or complete the square)✓ 1
A1 t = 6 (hours)✓ 1
A1 least rate is 0 mm h−1, since d′(6) = 0.15 × 36 − 10.8 + 5.4 = 0✓ 1

Common error: Do not substitute into d(t) when asked for a rate; the rate of change is the value of the derivative d′(t).

83IBMI3S-5-01r33-08Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Easy4 marks

The pressure, P MPa, of gas in a magma chamber is modelled by P = 2400/V, where V km3 is the volume of the chamber, for V ≥ 1.

(a)

Find dP/dV.

(2)
(b)

Find the value of dP/dV when V = 4, and interpret this value in context.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 write as 2400V−1 and attempt to differentiate✓ 1
A1 −2400V−2 (or −2400/V2)✓ 1
Part (b)
A1 −150✓ 1
R1 the pressure decreases by approximately 150 MPa for each 1 km3 increase in volume at this point✓ 1

Common error: A rate of change should be interpreted with its units and direction; a negative gradient means the quantity is decreasing.

84IBMI3S-5-01r33-13Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Easy5 marks

A scenery panel has a curved top edge. Its height, h m, at a horizontal distance x m from the left edge is modelled by h(x) = 4 + 0.5x − 0.05x2, for 0 ≤ x ≤ 6.

(a)

Find the area of the front face of the panel.

(2)
(b)

One litre of paint covers 8 m2. Paint is sold only in whole 1-litre cans. Find the number of cans needed to give the panel one coat.

(2)
(c)

Each can costs £12.80. Find the cost of the paint.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 ∫06 (4 + 0.5x − 0.05x2) dx✓ 1
A1 29.4 m2✓ 1
Part (b)
M1 29.4 ÷ 8 = 3.675 (ECF from (a))✓ 1
A1 4 cans (rounded up)✓ 1
Part (c)
A1 4 × 12.80 = £51.20✓ 1

Common error: When a quantity must be bought in whole units, round up even if the decimal part is less than 0.5.

85IBMI3S-5-02r33-01Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy7 marks

A road climbs a hill. The height of the road, h metres above sea level, at a distance x km along the climb is modelled by h(x) = 60 + 15x2 − x3, for 0 ≤ x ≤ 8. The gradient of the road is the rate of change of height with distance.

(a)

Find h′(x).

(2)
(b)

Find the gradient of the road, in metres per km, when x = 4. Express this as a percentage slope, where 10 m per km is 1%.

(2)
(c)

A cyclist finds the road is steepest at one point. Find the distance x at which this occurs and the percentage slope there.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 30x − 3x2✓ 1
Part (b)
A1 120 − 48 = 72 m per km✓ 1
A1 7.2%✓ 1
Part (c)
M1 set h″(x) = 30 − 6x = 0 (or use GDC maximum of h′)✓ 1
A1 x = 5 km✓ 1
A1 slope 150 − 75 = 75 m per km = 7.5%✓ 1

Common error: Gradient and height are different quantities: use h′(x) for steepness and h(x) for how high the road is.

86IBMI3S-5-02r33-04Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy5 marks

After a cooling failure the temperature of a server room, T °C, is modelled by T(t) = 22 + 0.8t + 0.05t2, where t is the time in minutes since the failure, 0 ≤ t ≤ 20.

(a)

Find dT/dt.

(2)
(b)

Find the rate at which the temperature is rising when t = 6.

(1)
(c)

Find the time at which the temperature is rising at 2.0 °C per minute.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 0.8 + 0.1t✓ 1
Part (b)
A1 1.4 °C per minute✓ 1
Part (c)
M1 set 0.8 + 0.1t = 2.0✓ 1
A1 t = 12 minutes✓ 1

Common error: Do not confuse the temperature at a time with the rate at which it is changing; a rate question needs the derivative.

87IBMI3S-5-02r33-09Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy4 marks

Coolant is pumped through a data centre rack at a rate of r(t) litres per minute, where r(t) = 0.6t2 + 2 and t is the time in minutes, 0 ≤ t ≤ 5.

(a)

Find the volume of coolant pumped in the first 5 minutes.

(3)
(b)

Find the mean rate of flow over the 5 minutes.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 ∫05 (0.6t2 + 2) dt✓ 1
A1 antiderivative 0.2t3 + 2t (or correct GDC evaluation)✓ 1
A1 35 litres✓ 1
Part (b)
A1 35 ÷ 5 = 7 litres per minute✓ 1

Common error: The integral of a rate gives the total amount, not another rate; divide by the time to get a mean rate.

88IBMI3S-5-02r33-11Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy6 marks

The concentration of dust in a burial chamber, c mg m−3, at a distance x metres from the chamber entrance is modelled by c(x) = 12 + 90/x, for x ≥ 1.

(a)

Find dc/dx.

(2)
(b)

Find the value of dc/dx at x = 3 and interpret it in context.

(2)
(c)

Find the value of x at which the concentration is decreasing at a rate of 2.5 mg m−3 per metre.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 write as 12 + 90x−1 and attempt to differentiate✓ 1
A1 −90x−2 (or −90/x2)✓ 1
Part (b)
A1 −10✓ 1
R1 the concentration is decreasing by about 10 mg m−3 for each extra metre from the entrance✓ 1
Part (c)
M1 set −90/x2 = −2.5✓ 1
A1 x = 6 m✓ 1

Common error: Rewrite 90/x as 90x⁻¹ before differentiating; differentiating the numerator and denominator separately is incorrect.

89IBMI3S-5-01r36-01Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; 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integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy6 marks

A space probe is moving away from Earth. Its distance from Earth is modelled by d(t) = 2t3 − 15t2 + 60t, where d is measured in thousands of kilometres and t is the time in days after a course correction, 0 ≤ t ≤ 6.

(a)

Find the rate at which the distance is changing when t = 4.

(3)
(b)

Find the values of t for which the distance is increasing at 24 thousand km per day.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate d(t)✓ 1
A1 d′(t) = 6t2 − 30t + 60✓ 1
A1 36 thousand km per day✓ 1
Part (b)
M1 set 6t2 − 30t + 60 = 24✓ 1
A1 t = 2✓ 1
A1 t = 3✓ 1

Exam tip: The derivative gives a rate of change, so quote units as (units of d) per (units of t).

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A coffee roaster burns gas at a rate of r(t) = 0.3t2 + 2 litres per minute, where t is the time in minutes after the start of a roast, 0 ≤ t ≤ 6.

(a)

Find the total volume of gas used during the first 6 minutes of the roast.

(3)
(b)

The gas costs £0.85 per litre. Find the cost of the gas used in these 6 minutes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to integrate r(t) from 0 to 6✓ 1
A1 [0.1t3 + 2t] evaluated from 0 to 6✓ 1
A1 33.6 litres✓ 1
Part (b)
M1 33.6 × 0.85 (or ECF from (a))✓ 1
A1 £28.56✓ 1

Common error: Integrating a rate over a time interval gives the total amount, not the rate; do not just substitute the end time into r(t).

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The depth of a river is measured at points across its width. The table shows the depth y metres at distance x metres from the left bank.

x (m): 0, 4, 8, 12, 16, 20
y (m): 0, 1.2, 2.6, 3.1, 2.4, 0

(a)

Use the trapezoidal rule with all six data points to estimate the area of the cross-section of the river.

(3)
(b)

The water flows at an average speed of 0.8 m s−1. Estimate the volume of water passing a point each second.

(2)
(c)

State one way to improve the accuracy of the estimate in (a).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to apply trapezoidal rule with h = 4✓ 1
A1 (4/2)[0 + 0 + 2(1.2 + 2.6 + 3.1 + 2.4)]✓ 1
A1 37.2 m2✓ 1
Part (b)
M1 37.2 × 0.8 (or ECF from (a))✓ 1
A1 29.76 m3 s−1 (accept 29.8)✓ 1
Part (c)
A1 take depth measurements at more points (smaller intervals) across the river✓ 1

Exam tip: In the trapezoidal rule the first and last values are counted once and all the middle values are doubled.

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Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; 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integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy6 marks

The arch of a footbridge has height y = 0.6x − 0.02x2 metres, where x is the horizontal distance in metres from one end, 0 ≤ x ≤ 30.

(a)

Find the equation of the normal to the arch at the point where x = 10. Give your answer in the form y = mx + c.

(4)
(b)

Find the x-coordinate of the point where this normal meets the horizontal ground (y = 0).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiate: dy/dx = 0.6 − 0.04x✓ 1
A1 gradient of tangent = 0.2 and y = 4 at x = 10✓ 1
M1 gradient of normal = −1/0.2 = −5✓ 1
A1 y = −5x + 54✓ 1
Part (b)
M1 −5x + 54 = 0 (or ECF from (a))✓ 1
A1 x = 10.8✓ 1

Common error: The normal is perpendicular to the tangent: its gradient is the negative reciprocal of the tangent gradient, not simply the negative.

93IBMI3S-5-04r36-01Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Easy5 marks

A cyclist accelerates gently along a straight road. Her distance from the start is modelled by s = 0.3t2 + 2t, where s is in metres and t is the time in seconds, for 0 ≤ t ≤ 30.

(a)

Find an expression for her speed v = ds/dt.

(2)
(b)

Find her speed when t = 10.

(1)
(c)

Find the time at which her speed is 11 m s−1.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiate a term correctly✓ 1
A1 v = 0.6t + 2✓ 1
Part (b)
A1 8 m s−1 (ECF from (a))✓ 1
Part (c)
M1 set 0.6t + 2 = 11✓ 1
A1 t = 15 s✓ 1

Exam tip: Speed is the gradient of the distance–time model, so differentiate first and only then substitute values of t.

94IBMI3S-5-04r36-02Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Easy8 marks

The wall of an excavation trench has a profile modelled by y = 0.1x2 − 0.8x + 3, where x is the horizontal distance in metres along the trench and y is the height in metres above a fixed datum.

(a)

Find dy/dx.

(2)
(b)

Find the gradient of the profile at the point where x = 6.

(1)
(c)

Find the equation of the tangent to the profile at x = 6.

(3)
(d)

Find the value of x at which this tangent crosses the line y = 0.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiate term by term✓ 1
A1 dy/dx = 0.2x − 0.8✓ 1
Part (b)
A1 0.4✓ 1
Part (c)
A1 y = 1.8 when x = 6✓ 1
M1 use y − 1.8 = 0.4(x − 6)✓ 1
A1 y = 0.4x − 0.6✓ 1
Part (d)
M1 set 0.4x − 0.6 = 0 (ECF from (c))✓ 1
A1 x = 1.5 m✓ 1

Common error: Substituting x into the derivative gives the gradient only; the y-coordinate must come from the original function.

95IBMI3S-5-04r36-06Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Easy7 marks

The rate at which data passes through a data centre is modelled by r(t) = −0.5t2 + 6t + 20 terabytes per hour, where t is the time in hours after 08:00, for 0 ≤ t ≤ 12.

(a)

Find the rate of data flow at 11:00.

(2)
(b)

Find the total amount of data passing through the centre between 08:00 and 20:00.

(3)
(c)

Find the mean rate of data flow over the 12 hours.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitute t = 3✓ 1
A1 33.5 terabytes per hour✓ 1
Part (b)
M1 attempt to find the integral of r(t) from 0 to 12✓ 1
A1 −t3/6 + 3t2 + 20t evaluated at 12 (or GDC)✓ 1
A1 384 terabytes✓ 1
Part (c)
M1 384 ÷ 12 (ECF from (b))✓ 1
A1 32 terabytes per hour✓ 1

Exam tip: The area under a rate graph is a total amount; divide that total by the time to get the mean rate.

96IBMI3S-5-04r36-08Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Easy7 marks

A cyclist's speed is recorded every 10 seconds while accelerating along a flat road.
Time t (s): 0, 10, 20, 30, 40, 50, 60
Speed v (m s−1): 0, 3.1, 5.4, 7.0, 7.9, 8.3, 8.4

(a)

Use the trapezoidal rule with six intervals to estimate the distance travelled in the 60 seconds.

(3)
(b)

Find the mean speed over the 60 seconds, correct to 3 significant figures.

(2)
(c)

The speed increases by smaller amounts in each successive interval. State whether your estimate in (a) is likely to be an underestimate or an overestimate, giving a reason.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule with width 10✓ 1
M1 5 × (0 + 8.4 + 2(3.1 + 5.4 + 7.0 + 7.9 + 8.3))✓ 1
A1 359 m✓ 1
Part (b)
M1 359 ÷ 60 (ECF from (a))✓ 1
A1 5.98 m s−1✓ 1
Part (c)
A1 underestimate✓ 1
R1 the speed–time graph is concave (curving downwards), so the trapeziums lie below the curve✓ 1

Exam tip: For data in a table, the trapezoidal rule uses the first and last values once and every other value twice.

97IBMI3S-5-01r39-01Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Easy6 marks

The surface area A cm2 of a coral colony on a reef is modelled by A(t) = 0.5t3 − 4t2 + 15t + 20, for 0 ≤ t ≤ 8, where t is the time in months since monitoring began.

(a)

Find A′(t).

(2)
(b)

Find the rate at which the area is growing when t = 4.

(2)
(c)

The colony grows most slowly at one moment during the 8 months. Find the minimum value of the rate of growth, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 A′(t) = 1.5t2 − 8t + 15✓ 1
Part (b)
M1 substitute t = 4 into A′: 1.5(16) − 8(4) + 15✓ 1
A1 7 cm2 per month✓ 1
Part (c)
M1 set A″(t) = 3t − 8 = 0 (or use GDC minimum of A′)✓ 1
A1 t = 8/3 and minimum rate = 4.33 cm2 per month✓ 1

Exam tip: "Rate of change" means differentiate; substitute the time only after you have the derivative, and include the unit of the rate.

98IBMI3S-5-01r39-02Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Easy5 marks

A harbour channel is 100 m long. Its width w was measured at 20 m intervals along the channel, as shown in the table.

Distance along channel (m): 0, 20, 40, 60, 80, 100
Width w (m): 0, 14, 22, 25, 18, 0

The surface of the channel is to be covered by a floating barrier, and its area must be estimated.

(a)

Use the trapezoidal rule with all six ordinates to estimate the surface area of the channel.

(3)
(b)

The average depth of the channel is 3.5 m. Using your answer to (a), estimate the volume of water in the channel.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct use of trapezoidal rule with h = 20 (first + last, and twice the sum of the others)✓ 1
A1 correct substitution: ½ × 20 × [0 + 0 + 2(14 + 22 + 25 + 18)]✓ 1
A1 1580 m2✓ 1
Part (b)
M1 multiply area by 3.5✓ 1
A1 5530 m3 (ECF from (a))✓ 1

Common error: Writing the factor as h instead of h/2 in front of the bracket doubles the estimate; the end ordinates are counted once and the inner ordinates twice.

99IBMI3S-5-01r39-03Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Easy5 marks

A filter-feeding manta ray takes in plankton at a rate of R(t) = −0.5t2 + 6t + 10 grams per hour, for 0 ≤ t ≤ 10, where t is the time in hours after the ray starts feeding.

(a)

Find the total mass of plankton taken in during the 10 hours.

(3)
(b)

Hence find the mean rate of feeding over the 10 hours, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct integral ∫010 (−0.5t2 + 6t + 10) dt✓ 1
A1 antiderivative −t3/6 + 3t2 + 10t (or GDC evaluation)✓ 1
A1 233 g (233.33…)✓ 1
Part (b)
M1 divide the total by 10 (use the unrounded total)✓ 1
A1 23.3 g per hour✓ 1

Exam tip: A mean rate is the total change divided by the time taken; keep the unrounded total in your calculator for the next step.

100IBMI3S-5-01r39-04Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Easy7 marks

The curved profile of a stone archway is modelled by h(x) = 6 − 0.04x2, where h is the height in metres above the ground and x is the horizontal distance in metres from the centre line of the arch.

(a)

Find the gradient of the arch at the point where x = 3.

(2)
(b)

Find the equation of the tangent to the arch at this point, giving your answer in the form y = mx + c.

(3)
(c)

Calculate the angle that the tangent makes with the horizontal, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate: h′(x) = −0.08x✓ 1
A1 gradient = −0.24✓ 1
Part (b)
A1 h(3) = 5.64✓ 1
M1 use of y − 5.64 = m(x − 3) with their gradient✓ 1
A1 y = −0.24x + 6.36✓ 1
Part (c)
M1 angle = tan−1(0.24) (ignore sign, ECF from (a))✓ 1
A1 13.5°✓ 1

Common error: Substituting x = 3 into the original function gives the height, not the gradient; the gradient comes from the derivative.

101IBMI3S-5-02r41-01Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; 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Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Easy6 marks

A space probe descends towards a moon. Its height above the surface is \(h = 500 - 4t - 0.05t^2\) metres, where t is the time in seconds after the descent engine starts.

(a)

Find an expression for the velocity of the probe, and hence its velocity when \(t=20\).

(2)
(b)

Find the value of t when the probe is descending at 9 m s−1.

(2)
(c)

Find the time at which the probe reaches the surface. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\frac{dh}{dt}=-4-0.1t\)✓ 1
A1 \(-6\) m s−1 (downwards at 6 m s−1)✓ 1
Part (b)
M1 \(-4-0.1t=-9\)✓ 1
A1 \(t=50\) s✓ 1
Part (c)
M1 \(500-4t-0.05t^2=0\) solved with a GDC or the quadratic formula✓ 1
A1 \(t=67.7\) s (reject the negative root)✓ 1

Common error: A negative velocity here means downward motion; the speed is the positive value.

102IBMI3S-5-02r41-02Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; 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Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Easy4 marks

During a roast, the temperature of coffee beans is modelled by \(T = 20 + 30t - 1.5t^2\) °C for \(0\le t\le10\), where t is the time in minutes after the beans enter the roaster.

(a)

Find the rate of change of temperature when \(t=4\).

(2)
(b)

The roast ends when the temperature stops rising. Find the temperature of the beans at this moment.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\frac{dT}{dt}=30-3t\)✓ 1
A1 18 °C per minute✓ 1
Part (b)
M1 \(30-3t=0\) giving \(t=10\) (and substitute into T)✓ 1
A1 170 °C✓ 1

Exam tip: 'Stops rising' means the rate of change is zero: set the derivative equal to 0.

103IBMI3S-5-02r41-03Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; 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Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Easy5 marks

The main cable of a footbridge hangs in the shape of the curve \(y = 0.004x^2 - 0.4x + 25\), where x is the horizontal distance in metres from the left tower and y is the height in metres above the deck.

(a)

Find the gradient of the cable at the point where \(x=20\).

(2)
(b)

Find the equation of the tangent to the cable at \(x=20\), giving your answer in the form \(y=mx+c\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\frac{dy}{dx}=0.008x-0.4\)✓ 1
A1 \(-0.24\)✓ 1
Part (b)
M1 \(y(20)=0.004\times400-8+25=18.6\)✓ 1
M1 \(y-18.6=-0.24(x-20)\) (ECF from (a))✓ 1
A1 \(y=-0.24x+23.4\)✓ 1

Exam tip: Find the y-coordinate of the point first; you need a point and a gradient to write a tangent.

104IBMI3S-5-02r41-04Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; 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Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Easy5 marks

A space probe fires a braking engine. Its speed is measured every 10 seconds:

Time (s): 0, 10, 20, 30, 40
Speed (m s−1): 0, 12.4, 21.1, 26.8, 29.2

(a)

Use the trapezoidal rule with 4 intervals to estimate the distance travelled in the 40 seconds.

(3)
(b)

Hence find the average speed over the 40 seconds. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 strip width \(h=10\)✓ 1
M1 \(\frac{10}{2}\left(0+29.2+2(12.4+21.1+26.8)\right)\)✓ 1
A1 749 m✓ 1
Part (b)
M1 \(\frac{749}{40}\) (ECF from (a))✓ 1
A1 18.7 m s−1✓ 1

Exam tip: In the trapezoidal rule the first and last values are used once; all the values between them are doubled.

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The power output P kW of a wind turbine for wind speeds between 3 m s−1 and 12 m s−1 is modelled by P = 0.4v3, where v is the wind speed in m s−1.

(a)

Find the power output when the wind speed is 8 m s−1.

(1)
(b)

Find dP/dv when v = 8.

(2)
(c)

Interpret your answer to part (b) in the context of the question.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 204.8 kW✓ 1
Part (b)
M1 dP/dv = 1.2v2✓ 1
A1 76.8✓ 1
Part (c)
A1 when the wind speed is 8 m s−1, the power output is increasing at 76.8 kW for each additional 1 m s−1 of wind speed✓ 1

Common error: A rate of change needs units that combine both variables, here kW per m s⁻¹, and the interpretation should name the context rather than just ‘the gradient is 76.8’.

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optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Easy4 marks

A library records the rate at which books are borrowed. The rate, R books per hour, t hours after the library opens is modelled by R = 60 + 12t − t2, for 0 ≤ t ≤ 6.

(a)

Find the number of books borrowed during the first 6 hours after opening.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 definite integral ∫06(60 + 12t − t2) dt seen✓ 1
M1 attempt to integrate (at least two terms correct) or use of a calculator✓ 1
A1 correct antiderivative 60t + 6t2 − t3/3 (or correct value from calculator)✓ 1
A1 504 books✓ 1

Exam tip: The area under a rate-of-change graph is the total change in the quantity, so integrate the rate over the given time interval.

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The power output of a wind farm is recorded every hour for 6 hours after a storm front passes. The table shows the power output P in megawatts (MW) at time t hours.
t (h): 0, 1, 2, 3, 4, 5, 6
P (MW): 0, 2.1, 3.4, 4.0, 3.8, 2.9, 1.4

(a)

Use the trapezoidal rule with 6 intervals to estimate the energy, in MWh, produced in the 6 hours.

(3)
(b)

Use your answer to part (a) to find the mean power output over the 6 hours. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule with h = 1, e.g. (1/2)(0 + 1.4 + 2(2.1 + 3.4 + 4.0 + 3.8 + 2.9))✓ 1
A1 (1/2)(1.4 + 32.4) or equivalent✓ 1
A1 16.9 MWh✓ 1
Part (b)
M1 energy ÷ 6 (ECF from (a))✓ 1
A1 2.82 MW✓ 1

Common error: In the trapezoidal rule the first and last ordinates are counted once and all the middle ordinates are counted twice.

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Optimisation in context; Optimisation in context Paper 1Easy6 marks

A track cyclist accelerates from rest in a training session. Her distance from the start, s metres, after t seconds is modelled by \(s=0.9t^2-0.02t^3\) for \(0\le t\le 20\).

(a)

Find an expression for the cyclist's velocity, v m s−1, at time t.

(2)
(b)

Find her velocity when \(t=5\).

(1)
(c)

Find the greatest velocity she reaches during the 20 seconds, and the time at which it occurs.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one power reduced correctly)✓ 1
A1 \(v=1.8t-0.06t^2\)✓ 1
Part (b)
A1 \(v=1.8(5)-0.06(25)=7.5\) m s−1✓ 1
Part (c)
M1 set the acceleration to zero: \(\frac{dv}{dt}=1.8-0.12t=0\)✓ 1
A1 \(t=15\) s✓ 1
A1 \(v=13.5\) m s−1✓ 1

Common error: The greatest velocity occurs where the derivative of velocity is zero; setting the velocity itself to zero finds when she is stationary.

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Optimisation in context; Optimisation in context Paper 1Easy6 marks

A railway line runs over a hill. The height of the track above sea level is modelled by \(y=0.0002x^3-0.015x^2+0.3x+60\), where x metres is the horizontal distance from a signal box and \(0\le x\le 100\). The gradient of the track is the value of \(\frac{dy}{dx}\).

(a)

Find \(\frac{dy}{dx}\).

(2)
(b)

Find the gradient of the track at \(x=40\), giving your answer as a percentage.

(1)
(c)

Find the equation of the tangent to the track profile at \(x=40\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least two terms correct)✓ 1
A1 \(\frac{dy}{dx}=0.0006x^2-0.03x+0.3\)✓ 1
Part (b)
A1 \(0.0006(1600)-0.03(40)+0.3=0.06\), which is a 6% gradient✓ 1
Part (c)
A1 \(y=60.8\) when \(x=40\)✓ 1
M1 \(y-60.8=0.06(x-40)\) (ECF from (b))✓ 1
A1 \(y=0.06x+58.4\)✓ 1

Exam tip: To find a tangent, you need both the gradient and the y-value at the point, so substitute into the original function as well as the derivative.

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Optimisation in context; Optimisation in context Paper 1Easy6 marks

A telescope records the flux (power per square metre) from a variable star every 2 hours. The readings, in μW m−2, are shown.

Time (h)0246810
Flux (μW m−2)12.415.118.917.213.811.5

(a)

Use the trapezoidal rule with all the data to estimate the area under the flux–time graph, in μW m−2 h.

(3)
(b)

Hence estimate the total energy received per square metre during the 10 hours, in joules.

(2)
(c)

State one way to improve the accuracy of this estimate.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule with 5 intervals of width 2✓ 1
A1 \(\frac{2}{2}\left[12.4+11.5+2(15.1+18.9+17.2+13.8)\right]\)✓ 1
A1 153.9✓ 1
Part (b)
M1 multiply by 3600 to convert hours to seconds (and μ to 10−6)✓ 1
A1 \(153.9\times 3600\div 10^6=0.554\) J m−2✓ 1
Part (c)
R1 take readings at shorter time intervals (more, narrower trapezia)✓ 1

Common error: Remember that the width h in the trapezoidal rule is the gap between readings, and the area is in flux × hours until you convert units.

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Optimisation in context; Optimisation in context Paper 1Easy6 marks

The intensity I of starlight reaching a detector at distance d astronomical units (AU) from a star is modelled by \(I=2400d^{-2}\) for \(0.5\le d\le 10\), where I is in arbitrary units.

(a)

Find \(\frac{dI}{dd}\) when \(d=1.5\), giving your answer to 3 significant figures.

(3)
(b)

Interpret your answer to part (a) in context.

(1)
(c)

Find the distance at which \(\frac{dI}{dd}=-150\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (power reduced, multiplied by −2)✓ 1
A1 \(\frac{dI}{dd}=-4800d^{-3}\)✓ 1
A1 \(-4800\div 1.5^3=-1420\)✓ 1
Part (b)
R1 at 1.5 AU the intensity is decreasing at about 1420 units for each extra AU of distance✓ 1
Part (c)
M1 \(-4800d^{-3}=-150\) so \(d^3=32\)✓ 1
A1 \(d=3.17\) AU✓ 1

Common error: A negative gradient means the quantity is falling; when interpreting it, say 'decreasing' rather than quoting a negative rate of increase.

112IBMI3S-5-01r45-01Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Easy6 marks

A beekeeper models the number of bees in a hive during spring. The number of bees, N thousand, t weeks after the start of spring is N(t) = −0.4t2 + 5t + 20, for 0 ≤ t ≤ 10.

(a)

Find N′(t).

(2)
(b)

Find the rate at which the number of bees is changing when t = 4.

(2)
(c)

Find the value of t at which the number of bees is greatest.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 N′(t) = −0.8t + 5✓ 1
Part (b)
M1 substitute t = 4 into N′(t)✓ 1
A1 1.8 thousand bees per week✓ 1
Part (c)
M1 set N′(t) = 0✓ 1
A1 t = 6.25 weeks✓ 1

Exam tip: The greatest value of a quadratic model occurs where the gradient is zero; always give the units of a rate.

113IBMI3S-5-01r45-02Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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After an injection, the concentration of a drug in a patient's blood is modelled by C(t) = 12t − 1.5t2 mg per litre, where t is the time in hours after the injection, for 0 ≤ t ≤ 8.

(a)

Find dC/dt.

(2)
(b)

Find the equation of the tangent to the graph of C at the point where t = 2. Give your answer in the form y = mt + c.

(3)
(c)

Find the time after the injection at which the concentration is greatest.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 12 − 3t✓ 1
Part (b)
M1 C(2) = 18 and gradient C′(2) = 6✓ 1
M1 substitute into y − 18 = 6(t − 2)✓ 1
A1 y = 6t + 6✓ 1
Part (c)
M1 set 12 − 3t = 0✓ 1
A1 t = 4 hours✓ 1

Common error: Substituting the t-value into C instead of C′ gives the height of the curve, not its gradient.

114IBMI3S-5-01r45-09Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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The flow rate of a hospital infusion pump is measured every 5 minutes. The flow rate, f mL per minute, is shown in the table.

t (min)05101520
f (mL/min)4.05.26.05.64.4

(a)

Use the trapezoidal rule with 4 intervals to estimate the volume of liquid delivered in the first 20 minutes.

(3)
(b)

Hence find the mean flow rate over the 20 minutes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use trapezoidal rule with h = 5✓ 1
A1 (5/2)[4.0 + 4.4 + 2(5.2 + 6.0 + 5.6)]✓ 1
A1 105 mL✓ 1
Part (b)
M1 105 ÷ 20 (ECF from (a))✓ 1
A1 5.25 mL per minute✓ 1

Exam tip: The area under a rate–time graph is the total quantity; dividing by the time gives the mean rate.

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integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Easy6 marks

A beekeeper sells jars of honey. The cost, in euros, of producing x jars in a week is C(x) = 0.02x2 + 1.5x + 40. The marginal cost is the rate of change of the cost with respect to x.

(a)

Find an expression for the marginal cost, C′(x).

(2)
(b)

Find the marginal cost when x = 50 and explain what this value means in context.

(2)
(c)

Find the number of jars for which the marginal cost is €3.90.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 0.04x + 1.5✓ 1
Part (b)
A1 €3.50✓ 1
R1 approximately the extra cost of producing one more jar when 50 jars are being produced✓ 1
Part (c)
M1 0.04x + 1.5 = 3.90✓ 1
A1 x = 60 jars✓ 1

Exam tip: When asked to interpret a derivative, state the quantity, its units and the point at which it applies.

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After a patient takes a tablet, the concentration of a drug in the blood is modelled by C(t) = 0.9t2 − 0.1t3 mg per litre, where t is the time in hours since the tablet was taken, for 0 ≤ t ≤ 9.

(a)

Find the concentration of the drug after 2 hours.

(1)
(b)

Find C′(t).

(2)
(c)

Find the rate at which the concentration is changing after 2 hours, including units.

(1)
(d)

Find the time at which the concentration is greatest, and the greatest concentration.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 C(2) = 0.9 × 4 − 0.1 × 8 = 2.8 mg per litre✓ 1
Part (b)
M1 differentiates a term correctly✓ 1
A1 C′(t) = 1.8t − 0.3t2✓ 1
Part (c)
A1 C′(2) = 3.6 − 1.2 = 2.4 mg per litre per hour✓ 1
Part (d)
M1 sets C′(t) = 0, so t(1.8 − 0.3t) = 0✓ 1
A1 t = 6 hours✓ 1
A1 greatest concentration C(6) = 10.8 mg per litre✓ 1

Exam tip: The maximum of a smooth model occurs where the derivative is zero, but state the value of the function as well as the time when the question asks for the greatest concentration.

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In a video game, a thrown grenade follows the path y = −0.04x2 + 1.2x + 1.5, where x is the horizontal distance in metres from the player and y is the height in metres above the ground.

(a)

Find dy/dx.

(2)
(b)

Find the equation of the tangent to the path at the point where x = 5. Give your answer in the form y = mx + c.

(3)
(c)

Find the angle that the grenade's path makes with the horizontal at x = 5, to the nearest 0.1°.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiates at least one term correctly✓ 1
A1 dy/dx = −0.08x + 1.2✓ 1
Part (b)
A1 y = −0.04 × 25 + 6 + 1.5 = 6.5 when x = 5, and gradient = −0.4 + 1.2 = 0.8✓ 1
M1 y − 6.5 = 0.8(x − 5)✓ 1
A1 y = 0.8x + 2.5✓ 1
Part (c)
A1 angle = tan−1(0.8) = 38.7°✓ 1

Common error: Substituting x into the original function and calling the result the gradient; the gradient comes from the derivative.

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A water-supply engineer records the flow rate of water through a pipe every 2 hours.
Time t (hours): 0, 2, 4, 6, 8
Flow rate (m3 per hour): 12.0, 15.5, 19.0, 16.5, 11.0

(a)

Use the trapezoidal rule with 4 intervals to estimate the total volume of water that flows through the pipe in the 8 hours.

(3)
(b)

Water is charged at €1.35 per m3. Find the cost of the water in part (a), giving your answer to the nearest cent.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule with strip width 2: ½ × 2 × [first + last + 2(sum of middle values)]✓ 1
A1 correct substitution: 1 × [12 + 11 + 2(15.5 + 19 + 16.5)]✓ 1
A1 volume = 125 m3✓ 1
Part (b)
A1 125 × 1.35 = €168.75✓ 1

Common error: Doubling the first and last values instead of the middle values when applying the trapezoidal rule.

119IBMI3S-5-02r46-04Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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A drug is given to a patient through an intravenous drip. The rate at which the drug enters the body is r(t) = 0.04t2 + 0.5 mg per minute, where t is the time in minutes since the drip started.

(a)

Write down a definite integral for the total mass of drug delivered in the first 10 minutes.

(1)
(b)

Evaluate this integral, giving your answer to 3 significant figures.

(2)
(c)

Find the mean rate of delivery over the first 10 minutes, using the unrounded answer to part (b).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 integral from 0 to 10 of (0.04t2 + 0.5) dt (correct limits and integrand)✓ 1
Part (b)
M1 integrates: 0.04t3/3 + 0.5t (or uses a calculator)✓ 1
A1 mass = 18.3 mg✓ 1
Part (c)
M1 mean rate = (integral) / 10✓ 1
A1 1.83 mg per minute✓ 1

Exam tip: A mean value of a rate over an interval is the total accumulated amount divided by the length of the interval.

120IBMI3S-5-03r47-01Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; 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The number of plankton, N, in a sample of seawater t hours after it is placed in a laboratory tank is modelled by N(t) = 500 + 40t − 0.5t2, for 0 ≤ t ≤ 60.

(a)

Find N′(t).

(2)
(b)

Find the rate at which the number of plankton is changing when t = 12.

(1)
(c)

Find the value of t at which the number of plankton stops increasing.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 N′(t) = 40 − t✓ 1
Part (b)
A1 N′(12) = 40 − 12 = 28 plankton per hour (ECF from (a))✓ 1
Part (c)
M1 set N′(t) = 0 (ECF from (a))✓ 1
A1 t = 40 hours✓ 1

Common error: A rate of change at a given time needs the derivative evaluated at that time, not the original function.

121IBMI3S-5-03r47-02Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; 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Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Easy7 marks

The entrance to a pedestrian tunnel is an arch. Taking the ground as the x-axis and the centre of the arch as x = 0, the height of the arch above the ground is modelled by h(x) = 6 − 0.04x2, where x and h are in metres.

(a)

Find the gradient of the arch at the point where x = 5.

(2)
(b)

Find the equation of the tangent to the arch at the point where x = 5. Give your answer in the form y = mx + c.

(2)
(c)

The arch meets the ground at two points. Find the width of the base of the arch, correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate: h′(x) = −0.08x✓ 1
A1 gradient = −0.4✓ 1
Part (b)
M1 h(5) = 5 and use of y − 5 = −0.4(x − 5) (ECF from (a))✓ 1
A1 y = −0.4x + 7✓ 1
Part (c)
M1 set 6 − 0.04x2 = 0✓ 1
A1 x = ±12.247… (x2 = 150)✓ 1
A1 width = 24.5 m✓ 1

Exam tip: A tangent needs both a gradient and a point on the curve; find the y-value first, then substitute into the equation of a line.

122IBMI3S-5-03r47-03Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; 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Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; 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Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Easy5 marks

Cartons are loaded onto a delivery lorry at a rate of r cartons per minute, where r(t) = 48 − 3t2 and t is the time in minutes after loading begins, for 0 ≤ t ≤ 4.

(a)

Find the total number of cartons loaded in the first 4 minutes.

(3)
(b)

Hence find the mean loading rate over the first 4 minutes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to integrate r(t) (or a correct definite integral from 0 to 4)✓ 1
A1 antiderivative 48t − t3✓ 1
A1 128 cartons✓ 1
Part (b)
M1 divide total by 4 (ECF from (a))✓ 1
A1 32 cartons per minute✓ 1

Common error: Integrating gives a total amount, not a rate. Divide by the time interval to turn a total into a mean rate.

123IBMI3S-5-03r47-13Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Easy4 marks

A warehouse's monthly cost, €C, of handling q pallets is modelled by C(q) = 4000 + 12q + 0.01q2.

(a)

Find C′(q).

(2)
(b)

Find the rate of change of cost with respect to the number of pallets when q = 300.

(1)
(c)

Find the number of pallets for which the rate of change of cost is €26 per pallet.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 C′(q) = 12 + 0.02q✓ 1
Part (b)
A1 €18 per pallet✓ 1
Part (c)
A1 q = 700✓ 1

Common error: The units of a rate of change are the units of the output per unit of the input, here euros per pallet.

124IBMI-034-00-r1-04Gradients, tangents and rates of change Paper 1Medium7 marks

The fuel consumption of a delivery van, \(F\) litres per 100 km, when travelling at a constant speed of \(v\) km h−1 is modelled by
\(F(v) = 0.0005v^2 + \dfrac{343}{v}\), for \(30 \le v \le 120\).

(a)

Find \(F'(v)\).

(2)
(b)

Find \(F'(50)\) and interpret your answer in context.

(3)
(c)

Find the speed at which \(F'(v) = 0\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1: \(0.001v\)✓ 1
A1: \(-\,343v^{-2}\) (accept \(-\frac{343}{v^2}\))✓ 1
Part (b)
M1: substituting \(v = 50\): \(0.05 - \frac{343}{2500}\)✓ 1
A1: \(F'(50) = -0.0872\)✓ 1
R1: at 50 km h−1, fuel consumption is decreasing by about 0.0872 litres per 100 km for each 1 km h−1 increase in speed✓ 1
Part (c)
M1: \(0.001v = \frac{343}{v^2}\), i.e. \(v^3 = 343\,000\), or using the GDC✓ 1
A1: \(v = 70\) km h−1✓ 1

Common error: Write \(\frac{343}{v}\) as \(343v^{-1}\) before differentiating. The derivative is \(-343v^{-2}\), which is negative, so do not lose the minus sign.

125IBMI-034-07-r2-04Integration and the trapezoidal rule Paper 1Medium5 marks

Water is pumped into a storage tank. The rate at which water enters the tank is modelled by

\(R(t) = 12 + 6t - 0.5t^2, \quad 0 \le t \le 12,\)

where R is in litres per minute and t is the time in minutes after the pump is switched on.

(a)

Find the total volume of water that enters the tank during the 12 minutes.

(3)
(b)

At t = 0 the tank already contains 150 litres. Throughout the 12 minutes, water also drains from the tank at a constant rate of 20 litres per minute. Find the volume of water in the tank at t = 12.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: \(\int_0^{12} (12 + 6t - 0.5t^2)\,dt\)✓ 1
A1: correct antiderivative 12t + 3t2 − t3/6, or correct GDC use✓ 1
A1: 288 litres✓ 1
Part (b)
M1: 150 + 288 − 20 × 12✓ 1
A1: 198 litres✓ 1

Exam tip: An integral of a rate gives a change in quantity, not the quantity itself. Add the starting amount and subtract any outflow to find the final amount.

126IBMI-034-01-r2-04Optimisation in context Paper 1Medium7 marks

A haulage firm plans a 300 km delivery. The truck travels at a constant speed of v km h−1. The fuel cost is modelled as 0.005v2 dollars per hour. The driver is paid $36 per hour.

(a)

Show that the total cost of the trip, C dollars, is given by \(C(v) = 10800v^{-1} + 1.5v\).

(2)
(b)

Find the speed that minimises the cost of the trip, and the minimum cost.

(3)
(c)

The truck has a legal speed limit of 80 km h−1. Find the cost of the trip at 80 km h−1, and explain why this is the cheapest legal speed.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Time for the trip = \(\frac{300}{v}\) hours M1✓ 1
\(C = \frac{300}{v}(36 + 0.005v^2) = \frac{10800}{v} + 1.5v\) A1✓ 1
Part (b)
\(C'(v) = -10800v^{-2} + 1.5\), set equal to 0 M1✓ 1
v2 = 7200, so v = 84.9 km h−1 A1✓ 1
Minimum cost = $254.56 A1✓ 1
Part (c)
C(80) = 135 + 120 = $255.00 A1✓ 1
C′(v) < 0 for v < 84.9, so the cost decreases as speed rises up to the limit. 80 km h−1 is therefore the cheapest legal speed, and it costs only $0.44 more than the unrestricted minimum R1✓ 1

Exam tip: When the stationary point lies outside the allowed domain, check the sign of the derivative. The optimum is then at the boundary of the domain.

127IBMI-034-07-r1-04Integration and the trapezoidal rule Paper 1Medium5 marks

During a storm, rainwater flows from a roof into a storage tank. The rate of flow is modelled by
\(r(t) = 40t\,e^{-0.2t}\) litres per minute,
where t is the time in minutes after the flow begins. The tank is empty at t = 0.

(a)

Find the maximum rate of flow into the tank.

(2)
(b)

Find the volume of water in the tank after the first 10 minutes.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Uses a GDC to find the maximum of r, or solves \(r'(t)=0\), giving t = 5 (M1)✓ 1
Maximum rate 73.6 litres per minute (A1)✓ 1
Part (b)
Recognises that volume is the integral of the rate (M1)✓ 1
\(\int_0^{10} 40t\,e^{-0.2t}\,dt\) (A1)✓ 1
594 litres (A1)✓ 1

Common error: Substituting t = 10 into r(t) gives a rate in litres per minute, not a volume. To find the total amount collected, integrate the rate over the time interval.

128IBMI-034-07-r1-06Integration and the trapezoidal rule Paper 1Medium8 marks

The front face of a concrete footbridge arch lies between two curves. Taking the ground as the x-axis, with measurements in metres, the curves are
outer edge: \(y = 6 - 0.15x^2\)
inner edge: \(y = 4 - 0.1x^2\).
The face of the arch is the region between the two curves and above the ground.

(a)

Show that both curves meet the ground at the same two points, and find the x-coordinates of these points.

(2)
(b)

Write down an integral for the area of the front face of the arch.

(2)
(c)

Find the area of the front face of the arch.

(2)
(d)

The arch is a solid of uniform thickness 0.8 m. The concrete has a density of 2400 kg m−3. Find the mass of the arch.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Solves \(6 - 0.15x^2 = 0\) and \(4 - 0.1x^2 = 0\); both give \(x^2 = 40\) (M1)✓ 1
\(x = \pm 6.32\) (±√40) (A1)✓ 1
Part (b)
Integrand is (outer − inner): \((6 - 0.15x^2) - (4 - 0.1x^2)\) (M1)✓ 1
\(\int_{-6.32}^{6.32}(2 - 0.05x^2)\,dx\), with the correct limits (A1)✓ 1
Part (c)
Evaluates their integral using a GDC or integration (M1)✓ 1
16.9 m2 (16.865…) (A1)✓ 1
Part (d)
Mass = area × thickness × density = 16.865… × 0.8 × 2400 (M1)✓ 1
32 400 kg (32 381…) (A1)✓ 1

Exam tip: For the area between two curves, subtract the lower curve from the upper curve before you integrate. Use the unrounded area in part (d) so that rounding errors do not change the final answer.

129IBMI-034-02-r2-03Integration and the trapezoidal rule Paper 1Medium7 marks

The side profile of a skateboard ramp is modelled by
\(y = 0.08x^2 + 0.5\), for \(0 \le x \le 5\),
where x is the horizontal distance and y is the height, both in metres. The side face of the ramp is the region bounded by the curve, the x-axis and the lines x = 0 and x = 5.

(a)

Use the trapezoidal rule with 5 strips of equal width to estimate the area of the side face.

(2)
(b)

Write down an integral for the exact area of the side face and find its value.

(2)
(c)

Find the percentage error in the estimate from part (a).

(2)
(d)

Explain why the trapezoidal rule gives an overestimate for this area.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Correct substitution with h = 1: \(\frac{1}{2}[0.5 + 2.5 + 2(0.58 + 0.82 + 1.22 + 1.78)]\)✓ 1
5.90 m2✓ 1
Part (b)
\(\int_0^5 (0.08x^2 + 0.5)\,dx\)✓ 1
5.83 m2 (5.8333…)✓ 1
Part (c)
\(\left|\frac{5.9 - 5.8333}{5.8333}\right| \times 100\)✓ 1
1.14 %✓ 1
Part (d)
The curve is concave up, so the top edge of each trapezoid lies above the curve, giving extra area✓ 1

Exam tip: Use the unrounded exact value (5.8333…) in the percentage error. Rounding to 5.83 first changes the third significant figure of the answer.

130IBMI-034-05-r1-04Gradients, tangents and rates of change Paper 1Medium7 marks

On a landscape plan, the edge of a decorative pond is modelled by the curve
\(y = \frac{12}{x} + 0.5x\), for \(1 \le x \le 10\),
where \(x\) and \(y\) are measured in metres from a fixed corner of the garden.

(a)

Find \(\frac{dy}{dx}\).

(2)
(b)

A straight stepping-stone path is to touch the edge of the pond at a point where the gradient of the curve is \(0.2\). Find the \(x\)-coordinate of this point.

(2)
(c)

A water pipe runs along the normal to the curve at the point where \(x = 2\). Find the equation of the normal, giving your answer in the form \(y = mx + c\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(-12x^{-2}\) (or \(-\frac{12}{x^2}\)) A1✓ 1
\(+0.5\) A1✓ 1
Part (b)
\(-\frac{12}{x^2} + 0.5 = 0.2\) M1✓ 1
\(x = \sqrt{40} = 6.32\) m A1✓ 1
Part (c)
Gradient of tangent at \(x = 2\) is \(-2.5\), so gradient of normal is \(0.4\) (M1 for negative reciprocal) M1✓ 1
Point is \((2, 7)\) A1✓ 1
\(y - 7 = 0.4(x - 2) \Rightarrow y = 0.4x + 6.2\) A1✓ 1

Common error: Using the tangent gradient for the normal, or taking only the reciprocal and forgetting the sign change. The normal gradient is \(-\frac{1}{m}\), so \(-2.5\) becomes \(+0.4\).

131IBMI-034-02-r1-03Integration and the trapezoidal rule Paper 1Medium7 marks

The opening of a tunnel in a garden wall is an arch. Taking the base of the opening as the x-axis, the arch is modelled by
\(y = 4 - 0.25x^2\), for \(-4 \le x \le 4\),
where x and y are in metres.

(a)

Use the trapezoidal rule with 4 intervals of equal width to estimate the area of the opening.

(2)
(b)

Write down an integral for the exact area of the opening and find its value.

(2)
(c)

Using the exact value of the integral from part (b), find the percentage error in the estimate from part (a).

(2)
(d)

Explain why the trapezoidal rule gives an underestimate in this context.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
y-values at x = −4, −2, 0, 2, 4 are 0, 3, 4, 3, 0 with h = 2✓ 1
Area ≈ \(\frac{2}{2}[0 + 0 + 2(3 + 4 + 3)] = 20\) m2✓ 1
Part (b)
\(\int_{-4}^{4} (4 - 0.25x^2)\,dx\)✓ 1
= \(\frac{64}{3}\) ≈ 21.3 m2✓ 1
Part (c)
\(\left|\frac{20 - \frac{64}{3}}{\frac{64}{3}}\right| \times 100\)✓ 1
= 6.25%✓ 1
(An answer of 6.10% comes from using the rounded value 21.3.)—
Part (d)
The arch is concave down, so the straight top edges of the trapezia lie below the curve and miss some of the area✓ 1

Exam tip: Percentage error is measured relative to the exact value, so divide by the unrounded integral, not by the estimate or a rounded value.

132IBMI-034-00-r2-04Gradients, tangents and rates of change Paper 1Medium7 marks

The cross-section of a curved sculpture is modelled by
\(f(x) = \frac{20}{x} + \frac{x}{5}\), for \(x > 0\),
where x and \(f(x)\) are measured in metres. The ground is horizontal and lies along the x-axis. A vertical wall lies along the y-axis. A light is installed at the point on the sculpture where \(x = 4\). Its beam is directed along the normal to the curve at that point, towards the wall.

(a)

Find \(f'(x)\).

(2)
(b)

Find the gradient of the curve at \(x = 4\).

(1)
(c)

Find the equation of the normal to the curve at \(x = 4\).

(3)
(d)

Find the height above the ground at which the beam hits the wall.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Writes \(20x^{-1}\) and differentiates to obtain \(-20x^{-2}\)✓ 1
\(f'(x) = -\frac{20}{x^2} + \frac{1}{5}\) (accept \(+0.2\))✓ 1
Part (b)
\(f'(4) = -1.25 + 0.2 = -1.05\)✓ 1
Part (c)
Gradient of normal \(= \frac{1}{1.05} = \frac{20}{21}\) \((= 0.952)\)✓ 1
\(f(4) = 5.8\)✓ 1
\(y - 5.8 = 0.952(x - 4)\), i.e. \(y = 0.952x + 1.99\) (accept equivalent forms, including exact)✓ 1
Part (d)
Substitutes \(x = 0\): height \(= 1.99\) m✓ 1

Common error: The normal gradient is the negative reciprocal of the tangent gradient. Changing only the sign, or only taking the reciprocal, gives the wrong line.

133IBMI3S-5-09-06Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium6 marks

The concentration of a drug in a patient's blood, C mg per litre, t hours after an injection is modelled by
C(t) = 0.02t3 − 0.48t2 + 2.88t,   0 ≤ t ≤ 12.

(a)

Find C′(t).

(2)
(b)

Find the time at which the concentration is greatest. Justify that this time gives a maximum.

(3)
(c)

Find the maximum concentration.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 0.06t2 − 0.96t + 2.88✓ 1
Part (b)
M1 set C′(t) = 0 and attempt to solve (roots 4 and 12)✓ 1
A1 t = 4✓ 1
R1 C″(4) = 0.12(4) − 0.96 = −0.48 < 0, so a maximum✓ 1
Part (c)
A1 5.12 mg per litre✓ 1

Common error: Setting the derivative to zero gives stationary points only. State why the point is a maximum, using the second derivative or a sign change.

134IBMI3S-5-09-07Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium7 marks

A section of a model roller-coaster track is described by
y = 0.1x3 − 0.9x2 + 2.4x + 1,   0 ≤ x ≤ 6,
where x and y are the horizontal and vertical distances measured in units of 10 m.

(a)

Find \(\frac{dy}{dx}\).

(2)
(b)

Find the values of x at which the track is horizontal.

(2)
(c)

The track descends most steeply where the gradient is least. Find the value of x at which this occurs, and state the gradient at that point.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 0.3x2 − 1.8x + 2.4✓ 1
Part (b)
M1 set \(\frac{dy}{dx}=0\)✓ 1
A1 x = 2 and x = 4✓ 1
Part (c)
M1 differentiate the gradient function and set equal to zero (0.6x − 1.8 = 0)✓ 1
A1 x = 3✓ 1
A1 gradient = −0.3✓ 1

Exam tip: To find where a gradient is greatest or least, differentiate the gradient function and set that to zero.

135IBMI3S-5-09-08Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium6 marks

During an 8-minute storm, rainwater runs from a roof into a tank at a rate r(t) = 8t − t2 litres per minute, where t is the time in minutes after the storm begins.

(a)

Find the total volume of water that enters the tank during the storm.

(4)
(b)

Hence find the mean rate of flow into the tank over the storm.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to integrate r(t)✓ 1
A1 4t2 − t3/3✓ 1
M1 substitute limits 8 and 0✓ 1
A1 85.3 litres✓ 1
Part (b)
M1 total volume ÷ 8 (ECF)✓ 1
A1 10.7 litres per minute✓ 1

Common error: A definite integral of a rate gives the total change. Do not divide by the time again unless a mean rate is asked for.

136IBMI3S-5-09-09Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium6 marks

The speed v of a sprinter is recorded at 2-second intervals after the start of a race.
Time, t (s): 0, 2, 4, 6, 8, 10
Speed, v (m s−1): 0, 5.1, 7.6, 9.0, 9.6, 9.8

(a)

Use the trapezoidal rule with five intervals to estimate the distance run in the first 10 seconds.

(3)
(b)

The speed–time graph is a smooth curve that bends downward (concave). State, with a reason, whether your answer to (a) is an underestimate or an overestimate.

(1)
(c)

Find the average acceleration of the sprinter between t = 0 and t = 4.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct strip width h = 2 used in the trapezoidal rule✓ 1
A1 ½ × 2 × [0 + 9.8 + 2(5.1 + 7.6 + 9.0 + 9.6)]✓ 1
A1 72.4 m✓ 1
Part (b)
R1 underestimate, because the straight-line tops of the trapezoids lie below a concave curve✓ 1
Part (c)
M1 change in speed ÷ change in time✓ 1
A1 1.9 m s−2✓ 1

Exam tip: For a concave curve the trapezoids sit under it (underestimate); for a convex curve they sit over it (overestimate).

137IBMI3S-5-09-10Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium7 marks

The cross-section of a skate bowl is modelled by the curve y = 0.25x2 − x + 3, 0 ≤ x ≤ 8, where x and y are in metres. A support rod is fixed to the surface at the point where x = 6. The rod is perpendicular to the surface and extends down to the ground, which is the line y = 0.

(a)

Find the gradient of the surface at the point where x = 6.

(2)
(b)

Find the equation of the line along which the rod lies.

(3)
(c)

Find the value of x at which the rod meets the ground.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (0.5x − 1)✓ 1
A1 2✓ 1
Part (b)
M1 perpendicular gradient −½ (ECF)✓ 1
M1 point (6, 6) used in the equation of a line✓ 1
A1 y = −0.5x + 9✓ 1
Part (c)
M1 set y = 0✓ 1
A1 x = 18 m✓ 1

Common error: A normal has gradient −1 divided by the tangent gradient. Using the tangent gradient itself is a frequent slip.

138IBMI3S-5-09-11Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium7 marks

A company produces x hundred phone cases per day. The average cost per case, C euros, is modelled by
C(x) = 0.5x + 200/x,   5 ≤ x ≤ 50.

(a)

Find C′(x).

(2)
(b)

Find the value of x that minimises the average cost.

(2)
(c)

Show that this value gives a minimum.

(2)
(d)

Find the minimum average cost per case.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate, using 200x−1✓ 1
A1 C′(x) = 0.5 − 200x−2✓ 1
Part (b)
M1 set C′(x) = 0 and rearrange to x2 = 400✓ 1
A1 x = 20 (negative root outside the domain)✓ 1
Part (c)
M1 C″(x) = 400x−3✓ 1
R1 C″(20) > 0, so a minimum✓ 1
Part (d)
A1 €20.00✓ 1

Exam tip: Rewrite 200/x as 200x⁻¹ before differentiating, and check the stationary value lies in the stated domain.

139IBMI3S-5-09b-05Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium7 marks

After a patient takes a tablet, the concentration of a drug in the blood is modelled by C(t) = −0.2t3 + 2.4t2 mg/L, where t is the time in hours and 0 ≤ t ≤ 12.

(a)

Find dC/dt.

(2)
(b)

Hence find the time at which the concentration is greatest.

(2)
(c)

The drug is effective while the concentration exceeds 40 mg/L. Find, to 3 significant figures, how long the drug is effective.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 −0.6t2 + 4.8t✓ 1
Part (b)
M1 set dC/dt = 0 and factorise or solve✓ 1
A1 t = 8 hours (rejecting t = 0, where C is a minimum)✓ 1
Part (c)
M1 solve C(t) = 40 using a graphing calculator✓ 1
A1 t = 5.58 and t = 10✓ 1
A1 4.42 hours✓ 1

Exam tip: When the concentration is above a threshold, the duration is the difference between the two crossing times, not the later time.

140IBMI3S-5-09b-06Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium6 marks

During a dry spell the volume of water in a reservoir changes at a rate of dV/dt = 3t2 − 24t + 36 thousand m3 per day, where t is the number of days since the dry spell began and 0 ≤ t ≤ 8. At t = 0 the reservoir holds 150 thousand m3.

(a)

Find an expression for V in terms of t.

(3)
(b)

Find the volume of water in the reservoir at the instant when the rate of change of volume is at its least.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to integrate✓ 1
A1 V = t3 − 12t2 + 36t + c✓ 1
A1 c = 150 from V(0) = 150✓ 1
Part (b)
M1 minimise dV/dt, e.g. set 6t − 24 = 0 or use the graph of dV/dt✓ 1
A1 t = 4✓ 1
A1 V = 166 thousand m3✓ 1

Common error: Do not forget the constant of integration; the initial volume is needed to fix it before substituting any later time.

141IBMI3S-5-09b-07Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium7 marks

The depth d metres of a stream at a horizontal distance x metres from one bank is modelled by d(x) = 0.03x(10 − x), for 0 ≤ x ≤ 10. The cross-sectional area of the stream is the area under this curve.

(a)

Use the trapezoidal rule with five equal intervals to estimate the cross-sectional area.

(3)
(b)

Use integration to find the exact cross-sectional area.

(2)
(c)

Find the percentage error in the estimate from part (a).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule with h = 2✓ 1
A1 depths 0.48, 0.72, 0.72, 0.48 at x = 2, 4, 6, 8✓ 1
A1 4.8 m2✓ 1
Part (b)
M1 ∫010 0.03x(10 − x) dx✓ 1
A1 5 m2✓ 1
Part (c)
M1 |5 − 4.8| / 5 × 100✓ 1
A1 4%✓ 1

Exam tip: Percentage error is measured relative to the exact value, not the estimate.

142IBMI3S-5-09b-09Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium7 marks

In a game, the cross-section of the terrain is given by y = 0.5x2 + 8/x, for x > 0, where both axes are in map units. A laser is fired from the point P on the terrain with x = 4, along the normal to the terrain at P.

(a)

Find dy/dx.

(2)
(b)

Find the y-coordinate of P and the gradient of the terrain at P.

(2)
(c)

Find the equation of the normal at P.

(2)
(d)

The ground level is the x-axis (y = 0). Find the x-coordinate where the laser line meets the x-axis.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate, with 8/x written as 8x−1✓ 1
A1 dy/dx = x − 8/x2✓ 1
Part (b)
A1 y = 10✓ 1
A1 gradient = 3.5✓ 1
Part (c)
M1 normal gradient = −1/3.5 = −2/7✓ 1
A1 y − 10 = −(2/7)(x − 4), or 2x + 7y = 78✓ 1
Part (d)
A1 x = 39✓ 1

Common error: The normal is perpendicular to the tangent: take the negative reciprocal of the gradient, not just the negative.

143IBMI3S-5-09b-11Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium6 marks

During an outbreak of a stomach virus at a school, the total number of students who have fallen ill after t days is modelled by N(t) = −0.04t3 + 1.8t2 + 5t, for 0 ≤ t ≤ 30.

(a)

Find N′(t).

(2)
(b)

Find the time at which new cases are appearing most quickly.

(2)
(c)

Find the greatest rate at which students are falling ill, stating the units.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 −0.12t2 + 3.6t + 5✓ 1
Part (b)
M1 set N″(t) = −0.24t + 3.6 = 0✓ 1
A1 t = 15 days✓ 1
Part (c)
M1 substitute t = 15 into N′(t)✓ 1
A1 32 students per day✓ 1

Exam tip: The fastest growth of a cumulative total occurs where its second derivative is zero, because that is where the rate N′ peaks.

144IBMI3S-5-02r5-05Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium6 marks

The fuel cost of a small aircraft depends on its cruising speed. The cost, C euros per 10 km flown, is modelled by C(v) = 0.0015v2 + 2400v−1, where v km per hour is the speed and 50 ≤ v ≤ 200.

(a)

Find C′(v).

(2)
(b)

Find the speed that minimises the cost per 10 km. Give your answer to 3 significant figures.

(2)
(c)

Show that this speed gives a minimum value of C.

(1)
(d)

Find the minimum cost per 10 km, to the nearest cent.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiation of a power term✓ 1
A1 C′(v) = 0.003v − 2400v−2✓ 1
Part (b)
M1 set C′(v) = 0✓ 1
A1 v3 = 800000, v = 92.8 km per hour✓ 1
Part (c)
R1 C″(v) = 0.003 + 4800v−3 > 0 at v = 92.8 (or sign change of C′ from negative to positive)✓ 1
Part (d)
A1 €38.78✓ 1

Exam tip: Use the unrounded speed when finding the minimum cost; rounding v first can change the cent value.

145IBMI3S-5-02r5-06Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium6 marks

A farmer has 330 m of fencing to make a rectangular pen. The pen is divided into two equal parts by an internal fence parallel to its width, so the 330 m of fencing is used for two lengths L and three widths W (all measured in metres).

(a)

Show that the area of the pen is A = 165W − 1.5W2.

(2)
(b)

Find the value of W that maximises the area.

(2)
(c)

Justify that this value of W gives a maximum.

(1)
(d)

Find the maximum area of the pen.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 2L + 3W = 330 so L = 165 − 1.5W✓ 1
A1 A = LW = 165W − 1.5W2✓ 1
Part (b)
M1 dA/dW = 165 − 3W = 0✓ 1
A1 W = 55 m✓ 1
Part (c)
R1 d2A/dW2 = −3 < 0✓ 1
Part (d)
A1 4537.5 m2✓ 1

Common error: Count every length of fence: an internal fence adds an extra length to the total, so the perimeter formula is not 2L + 2W.

146IBMI3S-5-02r5-07Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium6 marks

A speaker at an outdoor music festival produces a sound intensity of I(d) = 480d−2 units at a distance of d metres from the speaker, for d ≥ 2.

(a)

Find I′(d) and hence the rate of change of intensity with distance at d = 4.

(3)
(b)

Find the distance at which the intensity is decreasing at a rate of 0.5 units per metre. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiation of a power term✓ 1
A1 I′(d) = −960d−3✓ 1
A1 I′(4) = −15 units per metre✓ 1
Part (b)
M1 set −960d−3 = −0.5✓ 1
M1 d3 = 1920✓ 1
A1 d = 12.4 m✓ 1

Common error: A negative gradient shows the intensity is falling; when a question gives a rate of decrease, set the derivative equal to a negative value.

147IBMI3S-5-02r5-08Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium6 marks

The speed v of an aircraft along a runway is recorded every 20 s after the start of its take-off run.

Time t (s)020406080100
Speed v (m s−1)03562809094

(a)

Use the trapezoidal rule with all six values to estimate the distance travelled in the first 100 s.

(3)
(b)

The speed is increasing, but by smaller and smaller amounts in each 20 s interval. State, with a reason, whether the estimate in (a) is an overestimate or an underestimate of the true distance.

(2)
(c)

Find the mean speed over the first 100 s, using your answer to (a).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule with width 20: (20/2)(first + last + 2 × sum of others)✓ 1
A1 10(0 + 94 + 2 × 267)✓ 1
A1 6280 m✓ 1
Part (b)
A1 underestimate✓ 1
R1 the speed–time graph is concave down, so each trapezium lies below the curve✓ 1
Part (c)
A1 6280 ÷ 100 = 62.8 m s−1✓ 1

Exam tip: The area under a speed–time graph is a distance; divide it by the time taken to get a mean speed.

148IBMI3S-5-02r5-09Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

The height of a sunflower, H cm, is modelled by H(t) = −0.0004t3 + 0.03t2, where t is the number of days after the seedling emerges and 0 ≤ t ≤ 50.

(a)

Find H′(t).

(2)
(b)

Find the day on which the sunflower is growing fastest.

(2)
(c)

Find the greatest rate of growth, in cm per day.

(1)
(d)

The sunflower stops growing when H′(t) = 0 for t > 0. Find its final height.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiation of a polynomial✓ 1
A1 H′(t) = −0.0012t2 + 0.06t✓ 1
Part (b)
M1 set H″(t) = −0.0024t + 0.06 = 0 (or maximum of H′ using a GDC)✓ 1
A1 t = 25✓ 1
Part (c)
A1 0.75 cm per day✓ 1
Part (d)
M1 t = 50 and substitution into H(t)✓ 1
A1 25 cm✓ 1

Exam tip: Fastest growth is a maximum of the rate H′(t), so differentiate H′ again (or use the maximum function on the graph of H′).

149IBMI3S-5-02r5-10Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

The side profile of a harp soundboard is modelled by the region between the x-axis and the curve y = 9 − 0.01x2, where x and y are in centimetres.

(a)

Find the values of x where the curve meets the x-axis.

(2)
(b)

Find the area of the region.

(3)
(c)

The soundboard is a flat plate of thickness 0.8 cm cut to this shape from wood of density 0.45 g cm−3. Find its mass.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 9 − 0.01x2 = 0✓ 1
A1 x = −30 and x = 30✓ 1
Part (b)
M1 area = ∫−3030 (9 − 0.01x2) dx (or 2 × ∫030)✓ 1
A1 antiderivative 9x − (0.01/3)x3✓ 1
A1 360 cm2✓ 1
Part (c)
M1 volume = 360 × 0.8 = 288 cm3 and mass = volume × density (ECF from (b))✓ 1
A1 129.6 g✓ 1

Exam tip: Units follow through: area in cm², times thickness in cm gives a volume in cm³, which matches a density in g per cm³.

150IBMI3S-5-02r5-11Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium6 marks

During take-off an aircraft's acceleration is modelled by a(t) = 3 − 0.05t m s−2, where t is the time in seconds after it starts to move from rest, for 0 ≤ t ≤ 40.

(a)

Find an expression for the speed v(t) of the aircraft.

(3)
(b)

Find the speed at t = 20.

(1)
(c)

Find the distance travelled in the first 20 s. Give your answer to 1 decimal place.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 integration of a(t)✓ 1
A1 v = 3t − 0.025t2 + c✓ 1
A1 c = 0 since v(0) = 0✓ 1
Part (b)
A1 50 m s−1✓ 1
Part (c)
M1 ∫020 (3t − 0.025t2) dt✓ 1
A1 533.3 m✓ 1

Common error: Always determine the constant of integration from the initial condition before using the velocity formula.

151IBMI3S-5-02r6-04Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

A shop sells reusable water bottles. When the selling price is €\(p\), the number sold per week is \(480 - 12p\). Each bottle costs the shop €8 to buy.

(a)

Show that the weekly profit, €\(P\), is \(P = -12p^2 + 576p - 3840\).

(2)
(b)

Find \(\frac{dP}{dp}\) and hence the selling price that gives the greatest weekly profit.

(3)
(c)

Find the greatest weekly profit.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(P = (p - 8)(480 - 12p)\)✓ 1
A1 expand to reach the given expression✓ 1
Part (b)
A1 \(\frac{dP}{dp} = -24p + 576\)✓ 1
M1 set derivative equal to zero✓ 1
A1 \(p = 24\)✓ 1
Part (c)
M1 substitute \(p = 24\)✓ 1
A1 €3072✓ 1

Exam tip: In an optimisation problem, state the quantity you are maximising as a function of one variable before differentiating.

152IBMI3S-5-02r6-05Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

During an epidemic, the number of new cases reported per day is modelled by \(c(t) = 0.04t^3 - 1.8t^2 + 24t\), where \(t\) is the number of days since the first report, for \(0 \le t \le 30\).

(a)

Find \(c'(t)\).

(2)
(b)

Find the values of \(t\) for which \(c'(t) = 0\).

(2)
(c)

Show that \(t = 10\) gives a local maximum of \(c(t)\).

(1)
(d)

Find the greatest number of new cases reported in one day during the first 30 days.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate a polynomial✓ 1
A1 \(c'(t) = 0.12t^2 - 3.6t + 24\)✓ 1
Part (b)
M1 set \(0.12t^2 - 3.6t + 24 = 0\) and solve✓ 1
A1 \(t = 10\) and \(t = 20\)✓ 1
Part (c)
R1 \(c'(t)\) changes from positive to negative at \(t = 10\) (or \(c''(10) = -1.2 < 0\))✓ 1
Part (d)
M1 compare \(c(10) = 100\) with the end value \(c(30)\)✓ 1
A1 180 cases, on day 30✓ 1

Common error: A local maximum is not always the greatest value on an interval; also check the values at the ends of the domain.

153IBMI3S-5-02r6-06Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium6 marks

A company installs \(n\) solar panels on a farm. The average cost per panel, in euros, is modelled by \(A(n) = 0.4n + \frac{9000}{n}\) for \(n > 0\).

(a)

Find \(A'(n)\).

(2)
(b)

Find the number of panels that makes the average cost per panel a minimum.

(2)
(c)

Verify that this is a minimum.

(1)
(d)

Find the minimum average cost per panel.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate, including \(\frac{9000}{n} = 9000n^{-1}\)✓ 1
A1 \(0.4 - \frac{9000}{n^2}\)✓ 1
Part (b)
M1 set \(A'(n) = 0\)✓ 1
A1 \(n = 150\)✓ 1
Part (c)
R1 \(A''(n) = \frac{18000}{n^3} > 0\) for \(n > 0\)✓ 1
Part (d)
A1 €120✓ 1

Exam tip: Rewrite terms such as \(\frac{9000}{n}\) as \(9000n^{-1}\) before differentiating.

154IBMI3S-5-02r6-07Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium6 marks

The rate at which new infections are reported, in thousands per day, is modelled by \(r(t) = 2 + 0.6t - 0.03t^2\), where \(t\) is the number of days since the start of the study.

(a)

Find \(\int_2^9 r(t)\,dt\), giving your answer to 3 significant figures.

(3)
(b)

State what your answer to part (a) represents.

(1)
(c)

Find the mean rate at which infections were reported between day 2 and day 9, using the unrounded value from part (a).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct integral setup with limits 2 and 9✓ 1
M1 attempt to integrate, \(2t + 0.3t^2 - 0.01t^3\)✓ 1
A1 29.9✓ 1
Part (b)
R1 the total number of new infections, in thousands, reported between day 2 and day 9✓ 1
Part (c)
M1 divide the integral by the interval length 7✓ 1
A1 4.27 thousand per day✓ 1

Exam tip: An integral of a rate gives a total amount; the units are the rate's units multiplied by the units of time.

155IBMI3S-5-02r6-12Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium6 marks

After a treatment starts, the concentration of a virus in a patient's blood is modelled by \(V(t) = \frac{1200}{t^2} + 15\) units per mL, where \(t \ge 1\) is the time in hours since treatment began.

(a)

Find \(V'(t)\).

(2)
(b)

Find \(V'(2)\) and interpret its value in context.

(2)
(c)

Find the time at which the concentration is falling at exactly 9.6 units per mL per hour, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate \(1200t^{-2}\)✓ 1
A1 \(V'(t) = -\frac{2400}{t^3}\)✓ 1
Part (b)
A1 \(V'(2) = -300\)✓ 1
R1 at \(t = 2\) hours the concentration is falling at 300 units per mL per hour✓ 1
Part (c)
M1 set \(-\frac{2400}{t^3} = -9.6\)✓ 1
A1 \(t = 6.30\) hours✓ 1

Common error: A negative derivative means the quantity is decreasing; in a sentence of interpretation say 'falling at', not 'rising at a negative rate'.

156IBMI3S-5-02r6-13Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

A wind farm is planned on a hill whose height above sea level, \(y\) metres, at a horizontal distance of \(x\) metres from a reference point is modelled by \(y = -0.004x^2 + 0.5x + 20\), for \(0 \le x \le 100\).

(a)

Find \(\frac{dy}{dx}\).

(1)
(b)

Find the gradient of the hill at \(x = 40\) and the angle this slope makes with the horizontal, to 3 significant figures.

(2)
(c)

Find the equation of the normal to the hill at \(x = 40\), giving \(m\) and \(c\) to 3 significant figures.

(2)
(d)

Find the \(x\)-coordinate where the normal meets the line \(y = 0\), to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(-0.008x + 0.5\)✓ 1
Part (b)
A1 gradient 0.18✓ 1
A1 angle \(= \tan^{-1}(0.18) = 10.2°\)✓ 1
Part (c)
M1 gradient of normal \(= -\frac{1}{0.18}\) and point (40, 33.6)✓ 1
A1 \(y = -5.56x + 256\)✓ 1
Part (d)
M1 set \(y = 0\) in the unrounded equation of the normal✓ 1
A1 \(x = 46.0\)✓ 1

Common error: The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the gradient of the tangent, not just the negative.

157IBMI3S-5-02r6-14Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

A pop-up shop's profit rate, in £ thousand per month, is modelled by \(p(t) = -0.5t^2 + 4t - 6\), where \(t\) is the number of months since opening, for \(0 \le t \le 6\).

(a)

Find the values of \(t\) for which \(p(t) = 0\).

(2)
(b)

Find \(\int_0^6 p(t)\,dt\) and explain what this value means for the shop.

(3)
(c)

Find the total profit made during the period when the profit rate is positive, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 set \(-0.5t^2 + 4t - 6 = 0\) and solve✓ 1
A1 \(t = 2\) and \(t = 6\)✓ 1
Part (b)
M1 antiderivative \(-\frac{t^3}{6} + 2t^2 - 6t\)✓ 1
A1 the integral is 0✓ 1
R1 the loss made in the first 2 months exactly cancels the profit made in the last 4 months, so total profit over 6 months is zero✓ 1
Part (c)
M1 \(\int_2^6 p(t)\,dt\)✓ 1
A1 £5.33 thousand✓ 1

Common error: When a function changes sign, the integral over the whole interval combines positive and negative areas; split the integral at the roots when you need the total of one sign.

158IBMI3S-5-01r8-02Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

The centre line of a water slide at a swimming pool is modelled by h(x) = 0.02x3 − 0.3x2 + 12, 0 ≤ x ≤ 10, where x is the horizontal distance in metres from the top and h is the height in metres above the pool surface.

(a)

Find h′(x).

(2)
(b)

Find the equation of the tangent to the slide at the point where x = 4. Give your answer in the form y = mx + c.

(3)
(c)

The slide is steepest at one point between x = 0 and x = 10. Find the value of x at this point.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 h′(x) = 0.06x2 − 0.6x✓ 1
Part (b)
M1 gradient = h′(4) = −1.44✓ 1
A1 h(4) = 8.48✓ 1
A1 y = −1.44x + 14.24✓ 1
Part (c)
M1 h″(x) = 0.12x − 0.6 = 0 (minimum of h′)✓ 1
A1 x = 5✓ 1

Exam tip: A tangent needs both a gradient and a point on the curve: find h(4) as well as h′(4).

159IBMI3S-5-01r8-05Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

The rate at which people enter a library is modelled by r(t) = 60 + 18t − 2t2 people per hour, where t is the number of hours after the library opens, 0 ≤ t ≤ 8.

(a)

Find the time at which people enter at the greatest rate, and the greatest rate.

(3)
(b)

Find the total number of people who enter during the first 8 hours. Give your answer to the nearest whole number.

(2)
(c)

Use the unrounded value from (b) to find the mean rate of entry over the 8 hours. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 r′(t) = 18 − 4t = 0✓ 1
A1 t = 4.5 hours✓ 1
A1 100.5 people per hour✓ 1
Part (b)
M1 ∫08 (60 + 18t − 2t2) dt✓ 1
A1 715 people✓ 1
Part (c)
M1 714.666… ÷ 8✓ 1
A1 89.3 people per hour✓ 1

Exam tip: Total change comes from integrating a rate; a mean rate is the total divided by the time interval.

160IBMI3S-5-01r8-06Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

After a pool is shock-dosed, the chlorine concentration is modelled by C(t) = 0.8 + 9/t mg per litre, where t ≥ 1 is the time in hours since dosing.

(a)

Find the concentration when t = 3.

(1)
(b)

Find the rate of change of concentration when t = 3.

(2)
(c)

Find the time at which the concentration is decreasing at 0.25 mg per litre per hour.

(2)
(d)

The pool may be used when C ≤ 1.5. Find the earliest time at which it may be used, giving your answer in hours to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 C(3) = 0.8 + 3 = 3.8 mg per litre✓ 1
Part (b)
M1 C′(t) = −9/t2✓ 1
A1 −1 mg per litre per hour✓ 1
Part (c)
M1 9/t2 = 0.25✓ 1
A1 t = 6 hours✓ 1
Part (d)
M1 0.8 + 9/t = 1.5 (or equivalent graph intersection)✓ 1
A1 t = 12.9 hours✓ 1

Common error: Write 9/t as 9t⁻¹ before differentiating; the derivative is −9t⁻², not 9t⁰.

161IBMI3S-5-01r8-13Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

The power output of a wind farm is modelled by P(t) = −0.5t2 + 6t + 8 MW, where t is the time in hours since 06:00, 0 ≤ t ≤ 8.

(a)

Find the time at which the output is greatest, and the greatest output.

(3)
(b)

Find the total energy produced between t = 0 and t = 8, in MWh, to 3 significant figures.

(2)
(c)

Using the unrounded answer from (b), find the mean output over the 8 hours, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 P′(t) = −t + 6 = 0✓ 1
A1 t = 6 hours✓ 1
A1 26 MW✓ 1
Part (b)
M1 ∫08 (−0.5t2 + 6t + 8) dt✓ 1
A1 171 MWh✓ 1
Part (c)
M1 170.666… ÷ 8✓ 1
A1 21.3 MW✓ 1

Exam tip: Mean value of a function over an interval = (definite integral) ÷ (length of interval).

162IBMI3S-5-01r8-14Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

A library orders n copies of a new title at one time (n ≥ 1). The cost per copy, in pounds, is modelled by C(n) = 240/n + 0.3n + 5.

(a)

Find C′(n).

(2)
(b)

Find the value of n for which C′(n) = 0, to 3 significant figures.

(2)
(c)

The library can only order a whole number of copies. Determine the number of copies that gives the lowest cost per copy, and state that cost to 2 decimal places.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiate 240n−1✓ 1
A1 C′(n) = −240/n2 + 0.3✓ 1
Part (b)
M1 n2 = 800✓ 1
A1 n = 28.3✓ 1
Part (c)
M1 evaluate C at n = 28 and n = 29✓ 1
A1 C(28) = 21.97 and C(29) = 21.98✓ 1
A1 28 copies, with cost £21.97✓ 1

Common error: The stationary point may not be a whole number; compare the two nearest whole numbers before choosing.

163IBMI3S-5-01r13-05Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium6 marks

Water flows into a swimming pool at a rate of r(t) = 30 + 9t − 0.6t2 litres per minute, where t is the time in minutes after the tap is opened, for 0 ≤ t ≤ 10.

(a)

Find the volume of water that flows into the pool during the first 10 minutes.

(3)
(b)

Find the maximum rate of flow, in litres per minute, during the first 10 minutes.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to integrate r(t) from 0 to 10✓ 1
A1 antiderivative 30t + 4.5t2 − 0.2t3✓ 1
A1 550 litres✓ 1
Part (b)
M1 set dr/dt = 9 − 1.2t = 0✓ 1
A1 t = 7.5✓ 1
A1 63.75 litres per minute✓ 1

Exam tip: Integrating a rate over a time interval gives the total change in the quantity; differentiating the rate gives how fast the rate itself changes.

164IBMI3S-5-01r13-06Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

A swimming club needs an open-topped storage box with a square base of side x metres and vertical sides of height h metres. The box must have a volume of 32 m3. The material for the base and sides costs €25 per square metre.

(a)

Show that the total outer surface area of the box is S = x2 + 128/x square metres.

(2)
(b)

Find the value of x that minimises S.

(3)
(c)

Find the minimum cost of the material, in euros.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 h = 32/x2 from the volume condition✓ 1
A1 S = x2 + 4xh = x2 + 128/x (AG)✓ 1
Part (b)
M1 dS/dx = 2x − 128x−2✓ 1
M1 set derivative equal to zero✓ 1
A1 x = 4✓ 1
Part (c)
A1 minimum surface area 48 m2✓ 1
A1 €1200✓ 1

Common error: When a constraint links two variables, substitute it into the quantity being optimised before differentiating.

165IBMI3S-5-01r13-08Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium6 marks

The cross-section of a swimming lake is surveyed at 5 m intervals across its width. The depth d (in metres) is measured at distance x metres from the west bank:
x (m): 0, 5, 10, 15, 20, 25
d (m): 0, 1.8, 3.1, 3.6, 2.4, 0
The lake has a uniform cross-section along its length of 40 m.

(a)

Use the trapezoidal rule with five intervals to estimate the area of the cross-section.

(3)
(b)

The depth profile curves downwards (is concave) between x = 0 and x = 25. State, with a reason, whether the estimate in part (a) is an overestimate or an underestimate of the true area.

(2)
(c)

Use your answer to part (a) to estimate the volume of water in the lake.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct use of the trapezoidal rule with strip width 5✓ 1
M1 correct substitution: ½ × 5 × (0 + 0 + 2 × (1.8 + 3.1 + 3.6 + 2.4))✓ 1
A1 54.5 m2✓ 1
Part (b)
A1 underestimate✓ 1
R1 each trapezium's sloping top lies below the concave curve, so the trapeziums cover less area than the lake cross-section✓ 1
Part (c)
A1 54.5 × 40 = 2180 m3✓ 1

Common error: The trapezoidal rule gives an area under a curve; a volume needs that area multiplied by the length of the uniform section.

166IBMI3S-5-01r13-10Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

A swimmer in a 400 m race has distance from the start modelled by s(t) = 1.5t − 0.001t2 metres, where t is the time in seconds, valid until the swimmer finishes.

(a)

Find an expression for the swimmer's speed, v(t) = ds/dt, in m s−1.

(2)
(b)

Find the time taken to swim 400 m. Give your answer to the nearest second.

(3)
(c)

Find the swimmer's speed at the finish. Use the unrounded time from part (b), and give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 v(t) = 1.5 − 0.002t✓ 1
Part (b)
M1 set 1.5t − 0.001t2 = 400✓ 1
M1 solve the quadratic (using technology or the quadratic formula) and reject the root that is too large (t ≈ 1153)✓ 1
A1 t = 347 s (unrounded 346.887…)✓ 1
Part (c)
M1 substitute the time into v(t) (ECF from (a) and (b))✓ 1
A1 0.806 m s−1✓ 1

Common error: A quadratic can have two roots; check each against the domain of the model before choosing one.

167IBMI3S-5-01r13-11Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

A library courtyard lies between a curved path and a straight path. Using coordinates in metres, the curved path is modelled by y = 2x − 0.05x2 and the straight path by y = 0.6x. The two paths meet at the origin and at one other point.

(a)

Find the x-coordinate of the other point where the paths meet.

(2)
(b)

Find the area of the courtyard.

(3)
(c)

The courtyard is covered with turf costing €8 per square metre. Using the unrounded area from part (b), find the cost to the nearest cent.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 set 2x − 0.05x2 = 0.6x✓ 1
A1 x = 28✓ 1
Part (b)
M1 area = ∫028 (1.4x − 0.05x2) dx✓ 1
M1 antiderivative 0.7x2 − (0.05/3)x3 evaluated between 0 and 28✓ 1
A1 182.93… m2 (183 m2 to 3 s.f.)✓ 1
Part (c)
M1 multiply area by 8✓ 1
A1 €1463.47✓ 1

Exam tip: For the area between two curves, integrate the upper function minus the lower function between the intersection points.

168IBMI3S-5-03r21-05Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium5 marks

A coral colony spreads across a reef. The rate at which its surface area is increasing is modelled by r(t) = 4 + 0.6t − 0.03t2 cm2 per month, where t is the time in months after monitoring begins, for 0 ≤ t ≤ 12.

(a)

Find the increase in the surface area of the colony during the first 12 months.

(3)
(b)

At the start of monitoring the surface area was 150 cm2. Find the surface area after 12 months, giving your answer to 2 decimal places.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to integrate r(t) (at least one term with its power increased)✓ 1
A1 4t + 0.3t2 − 0.01t3✓ 1
A1 73.92 cm2✓ 1
Part (b)
M1 150 + 73.92✓ 1
A1 223.92 cm2 (ECF from (a))✓ 1

Common error: Forgetting that integrating a rate gives the change in the quantity, not its final value. The starting amount must be added separately.

169IBMI3S-5-03r21-07Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium6 marks

The curved side wall of a museum extension is to be covered in cladding. The height of the wall is measured at equal horizontal intervals along its 6 m length.
Distance along wall (m): 0, 1.5, 3, 4.5, 6
Height (m): 3.2, 4.1, 4.6, 4.2, 3.5

(a)

Use the trapezoidal rule with four intervals to estimate the area of the wall. Give your answer to 3 significant figures.

(3)
(b)

Cladding costs €48 per m2. Use the unrounded value from part (a) to find the estimated cost of cladding the wall.

(2)
(c)

State whether using eight intervals instead of four is likely to give a more accurate estimate, giving a reason.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of trapezoidal rule with interval width 1.5✓ 1
A1 ½ × 1.5 × [3.2 + 3.5 + 2(4.1 + 4.6 + 4.2)] = 24.375✓ 1
A1 24.4 m2✓ 1
Part (b)
M1 24.375 × 48✓ 1
A1 €1170 (ECF from (a))✓ 1
Part (c)
R1 yes, narrower trapezia follow the curved top edge more closely✓ 1

Common error: Rounding the area before using it in the next part. Keep the full calculator value for later working and round only the final answer.

170IBMI3S-5-03r21-08Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium6 marks

A dolphin leaps out of the sea. Its height above the water surface is modelled by h(x) = 2.4x − 0.3x2, where h is in metres and x is the horizontal distance in metres from the point of take-off.

(a)

Find the horizontal distance from take-off at which the dolphin re-enters the water.

(2)
(b)

Find the gradient of the dolphin's path at the point of re-entry.

(2)
(c)

Hence find the acute angle between the dolphin's path and the water surface as it re-enters. Give your answer to the nearest 0.1°.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 set h(x) = 0, x(2.4 − 0.3x) = 0✓ 1
A1 x = 8 m (reject x = 0)✓ 1
Part (b)
M1 attempt to differentiate, h′(x) = 2.4 − 0.6x✓ 1
A1 gradient = -2.4✓ 1
Part (c)
M1 tan θ = 2.4 or θ = tan−1(2.4)✓ 1
A1 67.4° (ECF from (b))✓ 1

Exam tip: The gradient of a path is the tangent of the angle it makes with the horizontal. Use the size of the gradient to find an acute angle.

171IBMI3S-5-03r21-10Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium7 marks

A port stores shipping containers in a holding yard. The number of containers in the yard, S, is modelled by S(t) = 1200 − 60t + 0.8t2, where t is the number of days after a restocking, for 0 ≤ t ≤ 60.

(a)

Find S′(t).

(2)
(b)

Find the value of t at which the number of containers is decreasing at a rate of 28 containers per day.

(2)
(c)

Find the smallest number of containers in the yard during this period.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 −60 + 1.6t✓ 1
Part (b)
M1 set −60 + 1.6t = −28✓ 1
A1 t = 20 days✓ 1
Part (c)
M1 set S′(t) = 0✓ 1
A1 t = 37.5✓ 1
A1 75 containers✓ 1

Common error: Setting S′(t) = 28 for a decreasing quantity. A decrease means the derivative is negative, so use −28.

172IBMI3S-5-03r21-11Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium7 marks

A loading ramp at a distribution centre has a profile modelled by h(x) = 0.004x3 − 0.06x2 + 0.4x for 0 ≤ x ≤ 20, where h is the height above the dock floor in metres and x is the horizontal distance in metres from the foot of the ramp.

(a)

Find h′(x).

(2)
(b)

Find the gradient of the ramp when x = 10.

(2)
(c)

The ramp is least steep at one point. Find the value of x at this point and the gradient of the ramp there.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 0.012x2 − 0.12x + 0.4✓ 1
Part (b)
M1 substitute x = 10 into h′✓ 1
A1 0.4✓ 1
Part (c)
M1 minimise h′, for example set h″(x) = 0.024x − 0.12 = 0, or find the minimum of h′ on a graph✓ 1
A1 x = 5✓ 1
A1 gradient = 0.1✓ 1

Exam tip: Least steep means the smallest value of the gradient, so you minimise the derivative function. That means finding where the second derivative is zero.

173IBMI3S-5-02r24-08Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

A space probe's acceleration during a manoeuvre is a(t) = 0.4t − 0.03t2 m s−2, where t is the time in seconds after the manoeuvre starts. At t = 0 the speed of the probe is 12 m s−1.

(a)

Show that the speed of the probe is given by v(t) = 12 + 0.2t2 − 0.01t3.

(3)
(b)

Find the speed of the probe when t = 8.

(1)
(c)

Find the distance travelled by the probe between t = 0 and t = 8. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 integrating: v = 0.2t2 − 0.01t3 + c✓ 1
M1 using v(0) = 12 to find c✓ 1
A1 c = 12, hence the given result✓ 1
Part (b)
A1 19.68 m s−1✓ 1
Part (c)
M1 ∫08(12 + 0.2t2 − 0.01t3) dt✓ 1
A1 [12t + (0.2 ÷ 3)t3 − 0.0025t4]08✓ 1
A1 120 m✓ 1

Exam tip: After integrating an acceleration, always use the initial condition to find the constant before substituting; leaving out + c loses the starting speed.

174IBMI3S-5-02r24-11Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium6 marks

The depth of a river was measured at points 4 m apart across a channel 28 m wide, where the banks are at depth 0.
Distance from bank (m): 0, 4, 8, 12, 16, 20, 24, 28
Depth (m): 0, 1.8, 3.1, 3.9, 4.2, 3.6, 2.2, 0

(a)

Use the trapezoidal rule with 7 intervals to estimate the area of the cross-section of the river.

(3)
(b)

The water flows at an average speed of 1.4 m s−1. Find the volume of water passing under a bridge per second. Give your answer to 3 significant figures.

(2)
(c)

Suggest one way to obtain a more accurate estimate of the cross-sectional area.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of trapezoidal rule with width 4✓ 1
A1 (4 ÷ 2)[0 + 0 + 2(1.8 + 3.1 + 3.9 + 4.2 + 3.6 + 2.2)]✓ 1
A1 75.2 m2✓ 1
Part (b)
M1 75.2 × 1.4 (ECF from (a))✓ 1
A1 105 m3 s−1✓ 1
Part (c)
R1 take depth measurements closer together (more intervals)✓ 1

Common error: With 8 data values there are 7 intervals, not 8; the number of intervals is always one fewer than the number of ordinates.

175IBMI3S-5-02r24-12Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

Two roast profiles use different heating power. For 0 ≤ t ≤ 4, where t is in minutes, Roast A has power PA(t) = 6t − t2 kW and Roast B has power PB(t) = 2t kW.

(a)

Find the values of t at which the two powers are equal.

(2)
(b)

Find the area of the region enclosed by the graphs of PA and PB between these values of t.

(3)
(c)

Hence find how much more energy, in kJ, Roast A uses than Roast B over these 4 minutes. Use 1 kW = 1 kJ s−1.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 6t − t2 = 2t✓ 1
A1 t = 0 and t = 4✓ 1
Part (b)
M1 ∫04(PA − PB) dt = ∫04(4t − t2) dt✓ 1
A1 [2t2 − t3 ÷ 3]04✓ 1
A1 32/3 = 10.7 kW min✓ 1
Part (c)
M1 multiply by 60 to convert minutes to seconds✓ 1
A1 640 kJ✓ 1

Exam tip: Between two curves, subtract the lower from the upper before integrating, and check which one is higher over the whole interval.

176IBMI3S-5-01r27-05Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium5 marks

During an outbreak, the rate at which new infections are reported, r people per day, is recorded every two days.

Time t (days): 0, 2, 4, 6, 8
Rate r (people per day): 12, 30, 48, 40, 22

(a)

Use the trapezoidal rule with four intervals to estimate the total number of new infections reported between t = 0 and t = 8.

(3)
(b)

The actual number of new infections reported in this period was 280. Find the percentage error in the estimate from part (a). Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 using the trapezoidal rule with width 2✓ 1
A1 (2/2)[12 + 22 + 2(30 + 48 + 40)]✓ 1
A1 270 people✓ 1
Part (b)
M1 |280 − 270| ÷ 280 × 100 (ECF from (a))✓ 1
A1 3.57%✓ 1

Common error: The trapezoidal rule uses the width of each interval (here 2), not the number of intervals. The first and last values appear once; every other value is doubled.

177IBMI3S-5-01r27-08Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium6 marks

A curved glass canopy over a shop entrance has a height, h metres, modelled by h(x) = x2 − 4x + 7, for 0 ≤ x ≤ 5, where x is the horizontal distance in metres from the shop wall. A support strut is fixed perpendicular to the canopy surface at the point where x = 3 and is extended in a straight line back to the wall at the point P.

(a)

Find dh/dx.

(2)
(b)

Find the equation of the line along which the strut lies.

(3)
(c)

Find the height of P above the ground.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiating at least one term correctly✓ 1
A1 2x − 4✓ 1
Part (b)
M1 gradient of canopy at x = 3 is 2 and h(3) = 4✓ 1
A1 gradient of the normal = −1/2✓ 1
A1 h = −0.5x + 5.5✓ 1
Part (c)
A1 5.5 m (ECF from (b))✓ 1

Exam tip: The gradient of a normal is the negative reciprocal of the gradient of the tangent. Check that your normal passes through the given point.

178IBMI3S-5-01r27-12Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

The weekly sales, S thousand units, of a seasonal retail product are modelled by S(t) = pt2 + qt, where t is the number of weeks after launch and p and q are constants. After 2 weeks, sales are 10 thousand units per week and are rising at a rate of 4 thousand units per week, per week.

(a)

Write down two equations connecting p and q.

(2)
(b)

Find the value of p and the value of q.

(2)
(c)

Find the time at which weekly sales are greatest, and the greatest weekly sales.

(2)
(d)

Find the number of weeks after launch at which weekly sales fall back to zero.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 S(2) = 10 gives 4p + 2q = 10✓ 1
M1 S′(t) = 2pt + q and S′(2) = 4 gives 4p + q = 4✓ 1
Part (b)
M1 solving the equations simultaneously✓ 1
A1 p = −0.5, q = 6✓ 1
Part (c)
M1 setting S′(t) = 0 (ECF from (b))✓ 1
A1 t = 6 weeks and S = 18 thousand units✓ 1
Part (d)
A1 t = 12 weeks (ECF from (b))✓ 1

Common error: The rate at which sales are rising is the gradient, S′(2), not the value S(2). Label which fact gives which equation.

179IBMI3S-5-01r27-13Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium6 marks

During an outbreak the rate of new infections, r people per day, is modelled by r(t) = 0.3t(20 − t), for 0 ≤ t ≤ 20, where t is the time in days since the outbreak began.

(a)

Find the time at which the rate of new infections is greatest.

(2)
(b)

Find the total number of new infections during the 20 days.

(3)
(c)

Find the mean rate of new infections over the 20 days.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 setting r′(t) = 6 − 0.6t = 0 (or using symmetry about the midpoint of the roots)✓ 1
A1 t = 10 days✓ 1
Part (b)
M1 integral of r between 0 and 20✓ 1
A1 antiderivative 3t2 − 0.1t3✓ 1
A1 400 people✓ 1
Part (c)
A1 400 ÷ 20 = 20 people per day (ECF from (b))✓ 1

Exam tip: The total change in a quantity is the definite integral of its rate of change. Expand the brackets before integrating by hand, or use the GDC.

180IBMI3S-5-02r27-04Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium6 marks

Lava leaves a vent at a rate of R thousand cubic metres per hour, where R(t) = 3t − 0.1t2, for 0 ≤ t ≤ 30, and t is the time in hours since the eruption began. The flow stops at the end of this period.

(a)

Find the value of t at which the flow rate is zero, other than t = 0.

(2)
(b)

Find the total volume of lava, in thousand cubic metres, that leaves the vent.

(3)
(c)

Find the mean flow rate over the 30 hours.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 setting 3t − 0.1t2 = 0 and factorising or using a GDC✓ 1
A1 t = 30✓ 1
Part (b)
M1 integral of R between 0 and 30✓ 1
A1 antiderivative 1.5t2 − t3/30✓ 1
A1 450 thousand m3✓ 1
Part (c)
A1 450 ÷ 30 = 15 thousand m3 per hour (ECF from (b))✓ 1

Exam tip: The area under a rate graph is the total quantity. Divide by the length of the time interval to find a mean rate.

181IBMI3S-5-02r27-05Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium5 marks

People enter a theatre before a show. The rate of entry, r people per minute, is recorded every 10 minutes.

Time t (minutes): 0, 10, 20, 30, 40, 50
Rate r (people per minute): 0, 18, 30, 22, 9, 2

(a)

Use the trapezoidal rule with five intervals to estimate the number of people who entered the theatre in the 50 minutes.

(3)
(b)

The ticket scanner shows that 812 people actually entered. Find the percentage error in the estimate from part (a), correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 using the trapezoidal rule with width 10✓ 1
A1 (10/2)[0 + 2 + 2(18 + 30 + 22 + 9)]✓ 1
A1 800 people✓ 1
Part (b)
M1 |812 − 800| ÷ 812 × 100 (ECF from (a))✓ 1
A1 1.48%✓ 1

Common error: Percentage error is measured relative to the exact (or true) value, not relative to the estimate.

182IBMI3S-5-02r27-06Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

The profile of the flank of a volcano is modelled by h(x) = 900 − 0.002x2, for 0 ≤ x ≤ 600, where h is the height in metres above sea level and x is the horizontal distance in metres from the summit axis.

(a)

Find dh/dx.

(2)
(b)

At the point P on the flank the gradient is −0.8. Find the x-coordinate of P.

(2)
(c)

Find the height of P above sea level.

(1)
(d)

Find the angle that the flank makes with the horizontal at P. Give your answer in degrees, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiating at least one term correctly✓ 1
A1 −0.004x✓ 1
Part (b)
M1 setting −0.004x = −0.8✓ 1
A1 x = 200 m✓ 1
Part (c)
A1 h(200) = 820 m (ECF from (b))✓ 1
Part (d)
M1 tan θ = 0.8✓ 1
A1 38.7°✓ 1

Exam tip: The gradient of a curve at a point equals the tangent of the angle between the tangent line and the horizontal; ignore the negative sign when asked for the size of the angle.

183IBMI3S-5-02r27-07Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

A ship's planned course on a map, where distances are in kilometres, is modelled by the curve y = 0.25x2 − 2x + 8. The coastline lies along the x-axis. At the point A where x = 6, the ship's engine fails and it drifts in a straight line in the direction of the tangent to the course, towards the coast.

(a)

Find dy/dx.

(2)
(b)

Find the equation of the tangent to the course at A.

(3)
(c)

Find the distance the ship drifts from A to the coastline, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiating at least one term correctly✓ 1
A1 0.5x − 2✓ 1
Part (b)
M1 gradient at x = 6 is 1✓ 1
A1 y(6) = 5✓ 1
A1 y = x − 1✓ 1
Part (c)
M1 the tangent meets y = 0 at (1, 0) and distance = √(52 + 52) (ECF from (b))✓ 1
A1 7.07 km✓ 1

Exam tip: After finding the point where a tangent meets an axis, use Pythagoras for the distance travelled along the line, not the difference in one coordinate.

184IBMI3S-5-02r27-13Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium6 marks

The cross-section of a shipping channel is modelled by the depth d(x) = 0.02x(30 − x), for 0 ≤ x ≤ 30, where d is the depth of water in metres and x is the distance in metres from one bank.

(a)

Find the greatest depth of the channel.

(2)
(b)

Find the area of the cross-section of the channel.

(3)
(c)

Water flows along the channel at a steady 0.8 m/s. Find the volume of water passing a fixed point each second.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 maximum at the midpoint x = 15 (or setting d′(x) = 0)✓ 1
A1 4.5 m✓ 1
Part (b)
M1 integral of d between 0 and 30✓ 1
A1 antiderivative 0.3x2 − x3/150✓ 1
A1 90 m2✓ 1
Part (c)
A1 90 × 0.8 = 72 m3 per second (ECF from (b))✓ 1

Exam tip: Flow rate (volume per second) is cross-sectional area multiplied by speed; check that the units combine to m³ per second.

185IBMI3S-5-03r27-03Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Medium7 marks

A survey of a river channel measures the depth of the water at points across its 10 m width, as shown.

Distance from left bank (m): 0, 2, 4, 6, 8, 10
Depth (m): 0, 0.8, 1.4, 1.7, 1.1, 0

The cross-sectional area of the water is estimated using the trapezoidal rule with five strips.

(a)

Calculate the estimated cross-sectional area.

(3)
(b)

The mean speed of the water across the channel is 0.45 m/s. Find the discharge, in m3/s, given by (area × mean speed).

(2)
(c)

Find the volume of water that passes a fixed point in 25 minutes, using the discharge from (b).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule structure with strip width 2✓ 1
A1 (2/2)(0 + 0 + 2(0.8 + 1.4 + 1.7 + 1.1))✓ 1
A1 10.0 m2✓ 1
Part (b)
M1 10.0 × 0.45✓ 1
A1 4.5 m3/s (ECF from (a))✓ 1
Part (c)
M1 4.5 × 25 × 60✓ 1
A1 6750 m3 (ECF from (b))✓ 1

Exam tip: Convert minutes to seconds before multiplying by a rate given per second.

186IBMI3S-5-03r27-08Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Medium8 marks

A garden designer plans a decorative pond. In a coordinate system with units in metres, the pond is the region between the curve y = 2.4x − 0.4x2 and the straight line y = 0.4x, where the curve is above the line.

(a)

Find the x-coordinates of the points where the curve meets the line.

(2)
(b)

Write down an integral that gives the area of the pond.

(2)
(c)

Find the area of the pond in m2, to 3 significant figures.

(2)
(d)

The pond liner costs €18 per m2. Using the unrounded area, find the cost of the liner.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 equate 2.4x − 0.4x2 = 0.4x✓ 1
A1 x = 0 and x = 5✓ 1
Part (b)
M1 integral of (upper curve − lower curve)✓ 1
A1 ∫05 (2x − 0.4x2) dx✓ 1
Part (c)
M1 attempt to evaluate the integral✓ 1
A1 8.33 m2✓ 1
Part (d)
M1 (25/3) × 18✓ 1
A1 €150✓ 1

Exam tip: For the area between two curves, subtract the lower function from the upper function before integrating.

187IBMI3S-5-03r27-09Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Medium7 marks

The flow rate of water through a main pipe is measured every 5 minutes. The flow rate, r m3 per minute, at time t minutes is shown.

t (min): 0, 5, 10, 15, 20
r (m3/min): 0.8, 1.6, 2.2, p, 2.4

(a)

Show that the trapezoidal rule with four strips gives an estimate of the total volume of water that passed through the pipe in 20 minutes of 27 + 5p.

(3)
(b)

The estimate of the volume is 41.5 m3. Find the value of p.

(2)
(c)

A tank has a capacity of 50 m3. Find the percentage of the tank that would be filled by the estimated volume.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule structure with strip width 5✓ 1
A1 (5/2)(0.8 + 2.4 + 2(1.6 + 2.2 + p))✓ 1
A1 (5/2)(10.8 + 2p) = 27 + 5p✓ 1
Part (b)
M1 27 + 5p = 41.5✓ 1
A1 p = 2.9✓ 1
Part (c)
M1 41.5/50 × 100✓ 1
A1 83.0 %✓ 1

Exam tip: In a 'show that' question, write every line of the substitution so the given result follows clearly.

188IBMI3S-5-03r27-13Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Medium8 marks

A pharmacy infusion pump delivers a drug at a rate of r(t) = kt(6 − t) mg per minute, where t is the time in minutes, for 0 ≤ t ≤ 6, and k is a constant. The pump delivers a total of 90 mg in the 6 minutes.

(a)

Find the value of k.

(3)
(b)

Find the maximum rate of delivery.

(2)
(c)

Find the percentage of the total dose delivered in the first 2 minutes. Give your answer to 3 significant figures, using unrounded intermediate values.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 ∫06 kt(6 − t) dt = 90✓ 1
A1 integral evaluates to 36k✓ 1
A1 k = 2.5✓ 1
Part (b)
M1 maximum at t = 3 (by symmetry or from r′(t) = 0)✓ 1
A1 22.5 mg per minute✓ 1
Part (c)
M1 ∫02 2.5t(6 − t) dt✓ 1
A1 23.3 mg✓ 1
A1 25.9 %✓ 1

Exam tip: Write the integral with its limits before evaluating it, so the method mark is secure even if the arithmetic slips.

189IBMI3S-5-03r28-05Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

A logistics company makes closed wooden crates in the shape of a cuboid with a square base of side x metres and height h metres. Each crate must have a volume of 1.5 m3.

(a)

Show that the total outer surface area of a crate is S = 2x2 + 6/x square metres.

(2)
(b)

Find the value of x that makes the surface area a minimum. Give your answer to 3 significant figures.

(3)
(c)

Find the minimum surface area. Use the unrounded value of x from part (b) and give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses volume to write h = 1.5/x2 and S = 2x2 + 4xh✓ 1
A1 substitutes to obtain 2x2 + 4x(1.5/x2) = 2x2 + 6/x (AG)✓ 1
Part (b)
M1 differentiates: dS/dx = 4x − 6x−2✓ 1
M1 sets dS/dx = 0 (or finds the minimum using a graph)✓ 1
A1 x = 1.14 m✓ 1
Part (c)
M1 substitutes x = 1.51/3 (= 1.1447…) into S (ECF from (b))✓ 1
A1 S = 7.86 m2✓ 1

Exam tip: In a "show that" question every step must appear; here the substitution for h must be shown before the final expression is stated.

190IBMI3S-5-03r28-06Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

A survey of an underwater sandbank gives its height h metres above the surrounding seabed as
h(x) = 12xe−0.25x,   0 ≤ x ≤ 20,
where x is the horizontal distance in metres along the survey line from the western edge.

(a)

Find h′(x).

(2)
(b)

Find the value of x at which the sandbank is highest.

(2)
(c)

Find the maximum height of the sandbank. Give your answer to 3 significant figures.

(1)
(d)

Find the gradient of the sandbank when x = 10. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use the product rule✓ 1
A1 h′(x) = 12e−0.25x − 3xe−0.25x (= 12e−0.25x(1 − 0.25x))✓ 1
Part (b)
M1 sets h′(x) = 0 (ECF from (a))✓ 1
A1 x = 4✓ 1
Part (c)
A1 h(4) = 48e−1 = 17.7 m✓ 1
Part (d)
M1 substitutes x = 10 into h′(x) (ECF from (a))✓ 1
A1 −1.48 (the height is decreasing)✓ 1

Exam tip: Set your calculator to radians or check exponentials carefully; a quick sketch of the graph helps to confirm that the stationary point is a maximum.

191IBMI3S-5-03r28-07Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium6 marks

The cross-section of a river channel is surveyed by measuring its depth at points 3 m apart across a width of 15 m.

Distance from bank, x (m)03691215
Depth (m)01.83.13.92.60

(a)

Use the trapezoidal rule with five intervals to estimate the area of the cross-section.

(3)
(b)

Assume that the graph of depth against distance is concave down for the whole width. State, with a reason, whether your estimate is an overestimate or an underestimate of the true area.

(1)
(c)

Water flows through the channel at a mean speed of 0.85 m s−1. Use your answer to part (a) to estimate the volume of water passing a point each second, in m3 s−1. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of the trapezoidal rule with width 3✓ 1
A1 (3/2)[0 + 0 + 2(1.8 + 3.1 + 3.9 + 2.6)]✓ 1
A1 34.2 m2✓ 1
Part (b)
R1 underestimate, because each chord (trapezium top) lies below a concave-down curve, so each trapezium has less area than the region under the curve✓ 1
Part (c)
M1 multiplies area by speed: 34.2 × 0.85 (ECF from (a))✓ 1
A1 29.1 m3 s−1✓ 1

Common error: Do not mix up the number of intervals and the number of data points: six measurements give five intervals, and the width must be the interval width (3 m).

192IBMI3S-5-03r28-08Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

A freight lorry makes a journey of 400 km at a constant speed of v km h−1. The running cost of the lorry is (18 + 0.004v2) euros per hour, for 40 ≤ v ≤ 100.

(a)

Show that the total cost of the journey is T(v) = 7200/v + 1.6v euros.

(2)
(b)

Find the speed that minimises the total cost, giving your answer to 3 significant figures. Justify that this speed gives a minimum.

(3)
(c)

Find the minimum total cost of the journey, using the unrounded speed from part (b). Give your answer to the nearest cent.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 journey time = 400/v hours and cost = time × hourly cost✓ 1
A1 (400/v)(18 + 0.004v2) = 7200/v + 1.6v (AG)✓ 1
Part (b)
M1 differentiates and sets T′(v) = −7200/v2 + 1.6 = 0✓ 1
A1 v = 67.1 km h−1✓ 1
R1 T″(v) = 14400/v3 > 0, so a minimum (or a valid sign-change or graph argument)✓ 1
Part (c)
M1 substitutes v = √4500 (= 67.08…) into T(v) (ECF from (b))✓ 1
A1 €214.66✓ 1

Exam tip: When justifying a minimum, show the sign of the second derivative at the stationary point rather than just stating that the graph is U-shaped.

193IBMI3S-5-03r28-13Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

The main cable of a footbridge hangs between two towers. Its height y metres above the deck is modelled by
y = 0.015x2 − 0.45x + 7,   0 ≤ x ≤ 40,
where x is the horizontal distance in metres from the left tower.

(a)

Find dy/dx.

(2)
(b)

Find the coordinates of the lowest point of the cable.

(3)
(c)

A support hanger is fixed at the point where the gradient of the cable is 0.3. Find the height of the cable at this point.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 dy/dx = 0.03x − 0.45✓ 1
Part (b)
M1 sets dy/dx = 0 (ECF from (a))✓ 1
A1 x = 15✓ 1
A1 y = 3.625, so the point is (15, 3.625)✓ 1
Part (c)
M1 solves 0.03x − 0.45 = 0.3 to obtain x = 25 (ECF from (a))✓ 1
A1 y = 5.125 m✓ 1

Exam tip: Read what each part asks for: "coordinates" needs both values, while "height" needs only the y-value found by substituting back into the original equation.

194IBMI3S-5-03r28-14Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium5 marks

An underwater robot descends vertically with speed v(t) = 0.8e−0.04t m s−1, where t is the time in seconds after it is released.

(a)

Find the acceleration of the robot when t = 10. Give your answer to 3 significant figures.

(2)
(b)

Find the distance descended during the first 50 seconds. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiates: v′(t) = −0.032e−0.04t✓ 1
A1 −0.0215 m s−2 (the robot is slowing down)✓ 1
Part (b)
M1 sets up ∫050 0.8e−0.04t dt✓ 1
A1 antiderivative −20e−0.04t✓ 1
A1 17.3 m✓ 1

Common error: Acceleration is the derivative of speed and distance is the integral of speed; swapping these two operations is a frequent source of lost marks.

195IBMI3S-5-02r30-04Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium6 marks

During a 50-second sprint, a cyclist's speed v m s−1 is recorded every 10 seconds.

Time (s): 0, 10, 20, 30, 40, 50
Speed (m s−1): 0, 3.2, 5.6, 7.4, 8.2, 8.6

(a)

Use the trapezoidal rule with 5 intervals to estimate the distance travelled in the 50 seconds.

(3)
(b)

Hence find the cyclist's average speed over the 50 seconds.

(1)
(c)

The speed increases at a decreasing rate throughout the sprint. State whether your answer to (a) is an overestimate or an underestimate of the true distance, and give a reason.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of the trapezoidal rule with width 10✓ 1
A1 \(\frac{10}{2}\left[0 + 8.6 + 2(3.2 + 5.6 + 7.4 + 8.2)\right]\)✓ 1
A1 287 m✓ 1
Part (b)
A1 287 ÷ 50 = 5.74 m s−1✓ 1
ECF from (a).—
Part (c)
A1 underestimate✓ 1
R1 the speed–time curve is concave, so each straight chord lies below the curve and the trapezia have less area than the region under the curve✓ 1

Exam tip: The trapezoidal rule adds the first and last values once and every middle value twice. Writing the bracket out in full makes it easy to check.

196IBMI3S-5-02r30-05Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium7 marks

The cross-section of a ditch at an archaeological site is the region between ground level, which is the x-axis, and the ditch floor with equation \(y = x^2 - 9\). The units are metres.

(a)

Show that the ditch is 6 m wide at ground level.

(1)
(b)

Find the area of the cross-section.

(3)
(c)

The ditch is 25 m long. Find the volume of soil that was removed to make it.

(1)
(d)

Soil takes up 20% more volume once it has been dug out. Find the volume of the loose soil.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 x2 − 9 = 0, so x = ±3, hence width = 6 m✓ 1
Part (b)
M1 \(\int_{-3}^{3}(9 - x^2)\,dx\)✓ 1
M1 antiderivative \(9x - \frac{x^3}{3}\)✓ 1
A1 36 m2✓ 1
Part (c)
A1 36 × 25 = 900 m3✓ 1
ECF from (b).—
Part (d)
M1 900 × 1.2✓ 1
A1 1080 m3✓ 1
ECF from (c).—

Common error: Integrating x² − 9 instead of 9 − x². Area under the x-axis gives a negative integral, so reverse the subtraction or take the modulus.

197IBMI3S-5-02r30-07Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium6 marks

The electrical power demand of a data centre is modelled by \(P(t) = 250 + 80\sin\left(\frac{\pi t}{12}\right)\) kW, where t is the time in hours after midnight, for 0 ≤ t ≤ 24, and the angle is in radians. The energy used between times a and b is \(\int_a^b P(t)\,dt\) kWh.

(a)

Find the energy used between midnight and noon. Give your answer to 3 significant figures.

(2)
(b)

Show that the energy used in the whole day is 6000 kWh.

(2)
(c)

Electricity costs €0.14 per kWh. Find the cost of the electricity used in the whole day.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\int_0^{12} P(t)\,dt\)✓ 1
A1 3610 kWh (3611.15… before rounding)✓ 1
Part (b)
M1 \(\int_0^{24} P(t)\,dt\)✓ 1
A1 6000 (accept an argument that the sine term contributes zero over one full period)✓ 1
Part (c)
M1 6000 × 0.14✓ 1
A1 €840✓ 1

Exam tip: Check that your calculator is in radian mode before evaluating an integral that contains sin(πt/12).

198IBMI3S-5-02r30-08Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium7 marks

A cycling route is modelled by the height \(H = x^3 - 6x^2 + 9x + 20\) metres, where x is the horizontal distance in units of 100 m from the start, for 0 ≤ x ≤ 4. The gradient H′(x) is the change in height, in metres, for each 100 m travelled horizontally.

(a)

Find H′(x).

(2)
(b)

By writing H′(x) in the form \(3(x - 2)^2 - 3\), or otherwise, find the steepest downhill gradient on the route and the value of x at which it occurs.

(2)
(c)

Find the equation of the tangent to the route at the point where the gradient is steepest downhill. Give your answer in the form y = mx + c.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least two terms correct)✓ 1
A1 \(3x^2 - 12x + 9\)✓ 1
Part (b)
M1 completing the square: \(3x^2 - 12x + 9 = 3(x-2)^2 - 3\)✓ 1
A1 steepest downhill gradient −3 (metres per 100 m) at x = 2✓ 1
ECF from (a).—
Part (c)
M1 H(2) = 22✓ 1
M1 y − 22 = −3(x − 2)✓ 1
A1 y = −3x + 28✓ 1
ECF from (b).—

Exam tip: The gradient function is itself a quadratic here, so its least value is found from the vertex, with no need for a second derivative.

199IBMI3S-5-02r30-11Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium8 marks

During a test ride, a cyclist starts from rest and the speed is modelled by \(v = -0.1t^2 + 2t\) m s−1, for 0 ≤ t ≤ 20, where t is the time in seconds. The cyclist is at rest again at the end of the ride.

(a)

Show that the cyclist comes to rest when t = 20.

(2)
(b)

Find the maximum speed of the cyclist.

(3)
(c)

Find the distance travelled during the ride. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 −0.1t2 + 2t = 0, so t(−0.1t + 2) = 0✓ 1
A1 t = 0 or t = 20, so rest again at t = 20✓ 1
Part (b)
M1 \(\frac{dv}{dt} = -0.2t + 2\)✓ 1
A1 −0.2t + 2 = 0, so t = 10✓ 1
A1 v = −10 + 20 = 10 m s−1✓ 1
Part (c)
M1 \(\int_0^{20}(-0.1t^2 + 2t)\,dt\)✓ 1
M1 antiderivative \(-\frac{t^3}{30} + t^2\)✓ 1
A1 133 m (133.33… before rounding)✓ 1

Exam tip: Maximum or minimum speed comes from setting the derivative of speed to zero, whereas distance comes from integrating speed.

200IBMI3S-5-01r31-04Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium5 marks

Calculator

The rate at which a trawler brings fish on board is modelled by R(t) = 12 − 0.5t2 tonnes per hour, where t is the time in hours since the nets were hauled, for 0 ≤ t ≤ 4.
Find the total mass of fish brought on board during the first 4 hours. Give your answer to 3 significant figures.

(a)

Find the total mass of fish, in tonnes, brought on board in the first 4 hours.

(5)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 setting up ∫04 (12 − 0.5t2) dt✓ 1
A1 12t✓ 1
A1 − t3/6✓ 1
M1 substituting limits 4 and 0 into the antiderivative: 48 − 64/6✓ 1
A1 37.3 tonnes✓ 1

Exam tip: If a question gives a rate and asks for a total, integrate the rate over the stated time interval. Include the units of the answer.

201IBMI3S-5-01r31-05Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium5 marks

Calculator

The depth of a river channel is measured at 5 m intervals across its width. The depth is zero at both banks.
x (m): 0, 5, 10, 15, 20, 25
d (m): 0, 1.2, 2.1, 2.6, 1.8, 0

(a)

Use the trapezoidal rule with all the data in the table to estimate the area of the cross-section of the channel.

(3)
(b)

Water flows through the channel at a constant speed of 0.8 m s−1. Using your answer to (a), find the volume of water that passes the cross-section in one minute.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule with h = 5 (5/2)[0 + 0 + 2(1.2 + 2.1 + 2.6 + 1.8)]✓ 1
A1 correct substitution of the interior values (sum 7.7)✓ 1
A1 38.5 m2✓ 1
Part (b)
M1 volume per second = 38.5 × 0.8 = 30.8 m3 (ECF from (a))✓ 1
A1 1848 m3 (accept 1850 to 3 s.f.)✓ 1

Exam tip: In the trapezoidal rule the first and last values are used once and every interior value is doubled. Count the strips from the number of intervals, not the number of data points.

202IBMI3S-5-01r31-06Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

Calculator

A fish farm owner has 60 m of fencing to build a rectangular holding pen. One side of the pen is the straight wall of a quay and needs no fencing. The two sides perpendicular to the quay each have length x metres.

(a)

Show that the area, A m2, of the pen is given by A = 60x − 2x2.

(2)
(b)

Find the value of x that gives the maximum area.

(3)
(c)

Find the maximum area of the pen and justify that it is a maximum.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 side parallel to the quay = 60 − 2x✓ 1
A1 A = x(60 − 2x) = 60x − 2x2 (AG)✓ 1
Part (b)
A1 dA/dx = 60 − 4x✓ 1
M1 setting dA/dx = 0✓ 1
A1 x = 15✓ 1
Part (c)
A1 A = 450 m2✓ 1
R1 d2A/dx2 = −4 < 0 (or gradient changes from positive to negative), so a maximum✓ 1

Exam tip: In optimisation questions, write the quantity to be optimised in terms of one variable before you differentiate, then justify that the stationary point is a maximum or minimum.

203IBMI3S-5-01r31-08Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium6 marks

Calculator

A textile printer cuts fabric panels whose upper edge follows the curve y = 6x − x2, with the x-axis as the lower edge. Both x and y are measured in decimetres.

(a)

Find the x-coordinates of the points where the curve meets the x-axis.

(1)
(b)

Find the area of one panel.

(3)
(c)

The fabric costs €0.85 per dm2. Find the cost of the fabric for 12 panels. Give your answer to 2 decimal places.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 x = 0 and x = 6✓ 1
Part (b)
M1 ∫06 (6x − x2) dx (ECF from (a))✓ 1
A1 antiderivative 3x2 − x3/3✓ 1
A1 36 dm2✓ 1
Part (c)
M1 12 × 36 × 0.85 (ECF from (b))✓ 1
A1 €367.20✓ 1

Common error: Check the limits come from the intersection points of the curve with the axis. Substituting the lower limit 0 gives zero here, but state it to show the method.

204IBMI3S-5-01r31-09Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

Calculator

A robotics factory finds that when it assembles x robots per hour, the cost per robot is C(x) = 0.02x + 80x−1 thousand euros, for x > 0.

(a)

Find the cost per robot when x = 40.

(1)
(b)

Find C′(x) and hence the gradient of the graph of C at x = 40.

(3)
(c)

Find the equation of the tangent to the graph of C at x = 40.

(2)
(d)

Interpret the value of C′(40) in context.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 C(40) = 0.8 + 2 = 2.8 thousand euros✓ 1
Part (b)
A1 0.02✓ 1
A1 − 80x−2✓ 1
A1 C′(40) = −0.03✓ 1
Part (c)
M1 y − 2.8 = −0.03(x − 40) (ECF from (a) and (b))✓ 1
A1 y = −0.03x + 4✓ 1
Part (d)
R1 cost per robot decreases by about 0.03 thousand euros (€30) for each extra robot assembled per hour, near x = 40✓ 1

Exam tip: Rewrite terms such as 80/x as 80x⁻¹ before differentiating. A negative gradient means the quantity is decreasing.

205IBMI3S-5-01r31-12Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Medium7 marks

Calculator

The hemline of a skirt is modelled by the curve y = px2 − 4x + q, where x and y are in centimetres and p and q are constants. At the point (3, 10) on the curve the gradient is 8.

(a)

Find the value of p.

(3)
(b)

Find the value of q.

(2)
(c)

Find the equation of the tangent to the curve at (3, 10).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 dy/dx = 2px − 4✓ 1
M1 2p(3) − 4 = 8✓ 1
A1 p = 2✓ 1
Part (b)
M1 substituting (3, 10) and p = 2: 10 = 2(9) − 12 + q (ECF from (a))✓ 1
A1 q = 4✓ 1
Part (c)
M1 y − 10 = 8(x − 3)✓ 1
A1 y = 8x − 14✓ 1

Exam tip: When a question gives the gradient at a point and a point on the curve, use the derivative for one equation and the coordinates for the other.

206IBMI3S-5-01r32-04Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

A telecoms cable is strung between two masts that are 18 m apart. The height of the cable above the ground is modelled by
f(x) = 0.05x2 − 0.9x + 9, for 0 ≤ x ≤ 18,
where x is the horizontal distance in metres from the first mast and f(x) is in metres.

(a)

Find the gradient of the cable at the point where x = 3.

(2)
(b)

Find the equation of the tangent to the cable at x = 3. Give your answer in the form y = mx + c.

(3)
(c)

The normal to the cable at x = 3 meets the y-axis at the point P. Find the y-coordinate of P.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 f′(x) = 0.1x − 0.9✓ 1
A1 gradient = −0.6✓ 1
Part (b)
A1 f(3) = 6.75✓ 1
M1 use of y − 6.75 = −0.6(x − 3) or equivalent (ECF from (a))✓ 1
A1 y = −0.6x + 8.55✓ 1
Part (c)
M1 gradient of normal = 1 ÷ 0.6 = 5/3 (accept 1.67 or better)✓ 1
A1 y = 1.75 (from y = (5/3)x + 1.75)✓ 1

Exam tip: The gradient of the normal is the negative reciprocal of the gradient of the tangent; both lines pass through the same point on the curve.

207IBMI3S-5-01r32-05Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

The strength of a radio signal received at a distance d kilometres from a transmitter is modelled by
S(d) = 900d−2, for d ≥ 1,
where S is measured in microwatts (μW).

(a)

Find dS/dd.

(2)
(b)

Find the value of dS/dd when d = 3, giving your answer to 3 significant figures, and interpret this value in context.

(3)
(c)

At a distance D km the rate of change of signal strength is −2.5 μW km−1. Find D, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate using the power rule (power reduced by 1)✓ 1
A1 dS/dd = −1800d−3✓ 1
Part (b)
M1 substitute d = 3 (ECF from (a))✓ 1
A1 −66.7✓ 1
R1 at 3 km the signal strength is decreasing at 66.7 μW per kilometre of extra distance✓ 1
Part (c)
M1 set −1800D−3 = −2.5 (and solve, e.g. D3 = 720)✓ 1
A1 D = 8.96 km✓ 1

Exam tip: Rewrite 1/d² as d⁻² before differentiating; a negative rate of change means the quantity is decreasing.

208IBMI3S-5-01r32-11Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

Glaciologists measure the depth of a glacier at seven points along a straight line across its width. The points are 20 m apart.

Distance from west edge (m)020406080100120
Depth of ice (m)04.27.89.58.15.20

(a)

Use the trapezoidal rule with all seven measurements to estimate the area of the cross-section of the glacier along this line.

(3)
(b)

The ice moves past this line at an average speed of 0.35 m per day. Find the volume of ice passing the line each day, in m3, to 3 significant figures. Use your answer to (a).

(2)
(c)

The depth profile is smooth and curves downward throughout (concave down). State whether the trapezoidal rule gives an overestimate or an underestimate of the true area, and justify your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of the trapezoidal rule with interval width 20✓ 1
A1 correct substitution: (20/2)(0 + 0 + 2(4.2 + 7.8 + 9.5 + 8.1 + 5.2))✓ 1
A1 696 m2✓ 1
Part (b)
M1 multiply area by speed: 696 × 0.35 (ECF from (a))✓ 1
A1 244 m3 per day (243.6)✓ 1
Part (c)
A1 underestimate✓ 1
R1 each straight-line chord lies below a concave-down curve, so each trapezium has less area than the region under the curve✓ 1

Exam tip: A trapezoidal estimate is an underestimate for a concave-down curve and an overestimate for a concave-up curve.

209IBMI3S-5-01r33-03Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Medium7 marks

A theatre company paints a curved backdrop. The height of the backdrop above the floor is 4 m, and its width, w m, is measured at heights of 0 m, 1 m, 2 m, 3 m and 4 m:

Height (m)01234
Width w (m)6.27.07.47.16.0

(a)

Use the trapezoidal rule with four intervals to estimate the area of the backdrop.

(3)
(b)

Paint costs €4.50 per m2. Calculate the cost of covering the estimated area with one coat of paint.

(2)
(c)

The exact area of the backdrop is 28.1 m2. Find the percentage error in the estimate from part (a). Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct strip width 1 and attempt at trapezoidal rule structure ½ × 1 × (first + last + 2 × others)✓ 1
A1 correct substitution ½(6.2 + 6.0 + 2(7.0 + 7.4 + 7.1))✓ 1
A1 27.6 m2✓ 1
Part (b)
M1 27.6 × 4.50 (ECF from (a))✓ 1
A1 €124.20✓ 1
Part (c)
M1 |28.1 − 27.6|/28.1 × 100 (ECF from (a), use the unrounded value of the estimate)✓ 1
A1 1.78%✓ 1

Common error: In the trapezoidal rule the first and last values are used once and every middle value is doubled; doubling every value overestimates the area.

210IBMI3S-5-01r33-05Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Medium6 marks

During an eruption, ash is released at a rate of E(t) tonnes per hour, where E(t) = 80 + 24t − 3t2 and t is the time in hours since the eruption began, 0 ≤ t ≤ 8.

(a)

Find the time at which the rate of release is greatest, and state the greatest rate.

(3)
(b)

Find the total mass of ash released during the first 8 hours.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to find maximum (e.g. 24 − 6t = 0 or GDC maximum)✓ 1
A1 t = 4 hours✓ 1
A1 greatest rate = 128 tonnes per hour✓ 1
Part (b)
M1 integral ∫08 (80 + 24t − 3t2) dt✓ 1
A1 correct antiderivative 80t + 12t2 − t3 (or correct GDC evaluation)✓ 1
A1 896 tonnes✓ 1

Exam tip: The total amount of a quantity released is the area under its rate graph, so integrate the rate function over the given time interval.

211IBMI3S-5-01r33-07Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Medium6 marks

A shipping channel is 100 m wide. Its depth, d m, is measured at 20 m intervals across the channel:

Distance from west bank (m)020406080100
Depth d (m)04.26.87.55.90

(a)

Use the trapezoidal rule with five intervals to estimate the area of the cross-section of the channel.

(3)
(b)

The cross-section is assumed to be the same along a 1.5 km length of the channel. Calculate the volume of water in this length. Give your answer in the form a × 10k where 1 ≤ a < 10 and k is an integer.

(2)
(c)

State one way in which the estimate in part (a) could be improved.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule structure with strip width 20✓ 1
A1 correct substitution ½ × 20 × (0 + 0 + 2(4.2 + 6.8 + 7.5 + 5.9))✓ 1
A1 488 m2✓ 1
Part (b)
M1 488 × 1500 (units consistent; ECF from (a))✓ 1
A1 7.32 × 105 m3✓ 1
Part (c)
R1 use more depth measurements (narrower intervals)✓ 1

Exam tip: When a length is given in kilometres and the area in square metres, convert to metres before multiplying to get a volume in cubic metres.

212IBMI3S-5-01r33-10Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Medium7 marks

A theatre sells tickets for a new show. The total number of tickets sold after t weeks is modelled by N(t) = −2t3 + 36t2 + 10t, for 0 ≤ t ≤ 12.

(a)

Find N′(t).

(2)
(b)

Find the rate at which tickets are being sold after 3 weeks.

(2)
(c)

Find the time at which tickets are being sold at the greatest rate, and calculate this greatest rate.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 −6t2 + 72t + 10✓ 1
Part (b)
M1 substitute t = 3 into N′ (ECF from (a))✓ 1
A1 172 tickets per week✓ 1
Part (c)
M1 set N″(t) = −12t + 72 = 0 or use GDC to find maximum of N′✓ 1
A1 t = 6 weeks✓ 1
A1 greatest rate = 226 tickets per week✓ 1

Common error: The rate of selling tickets is N′(t), not N(t); the greatest rate occurs where the derivative N′ is largest, found from N″ = 0.

213IBMI3S-5-02r33-02Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

An archaeologist records the cross-section of an ancient ditch. The ditch is 3 m wide. Its depth, d m, is measured at 0.5 m intervals across the ditch:

Distance across (m)00.51.01.52.02.53.0
Depth d (m)00.620.951.080.910.580

(a)

Use the trapezoidal rule with six intervals to estimate the area of the cross-section.

(3)
(b)

The ditch is 40 m long and has this cross-section throughout. Find the volume of soil that must be removed to empty it.

(2)
(c)

Removing soil costs £18.50 per m3. Find the cost, to the nearest pound.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule structure with strip width 0.5✓ 1
A1 ½ × 0.5 × (0 + 0 + 2(0.62 + 0.95 + 1.08 + 0.91 + 0.58))✓ 1
A1 2.07 m2✓ 1
Part (b)
M1 2.07 × 40 (ECF from (a))✓ 1
A1 82.8 m3✓ 1
Part (c)
M1 82.8 × 18.50 = 1531.8✓ 1
A1 £1532✓ 1

Exam tip: Keep the unrounded intermediate value (82.8) when it is used in a later part, and round only the final money answer as instructed.

214IBMI3S-5-02r33-06Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

A cyclist accelerates from a rolling start. Her speed, v m s−1, is modelled by v(t) = 4 + 1.5t − 0.06t2, where t is the time in seconds, 0 ≤ t ≤ 12.

(a)

Find her acceleration when t = 5.

(2)
(b)

Find the distance she travels in the first 12 seconds.

(3)
(c)

Find her average speed over the 12 seconds.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiate: v′(t) = 1.5 − 0.12t✓ 1
A1 0.9 m s−2✓ 1
Part (b)
M1 ∫012 v(t) dt✓ 1
A1 antiderivative 4t + 0.75t2 − 0.02t3 (or correct GDC evaluation)✓ 1
A1 121.44 m✓ 1
Part (c)
M1 121.44 ÷ 12 (ECF from (b))✓ 1
A1 10.12 m s−1✓ 1

Exam tip: Distance is the integral of speed; average speed is that distance divided by the time taken, not the mean of the first and last speeds.

215IBMI3S-5-02r33-07Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium6 marks

The power drawn by a data centre, P kW, is recorded every 4 hours over one day:

Time (h)04812162024
Power P (kW)410380350430520480420

(a)

Use the trapezoidal rule with six intervals to estimate the energy used in the 24 hours, in kWh.

(3)
(b)

Hence find the mean power drawn over the 24 hours. Give your answer to 3 significant figures.

(1)
(c)

Electricity costs €0.14 per kWh. Find the cost of the energy used in the 24 hours.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule structure with strip width 4✓ 1
A1 ½ × 4 × (410 + 420 + 2(380 + 350 + 430 + 520 + 480))✓ 1
A1 10 300 kWh✓ 1
Part (b)
A1 10 300 ÷ 24 = 429 kW (429.17)✓ 1
Part (c)
M1 10 300 × 0.14 (ECF from (a))✓ 1
A1 €1442.00✓ 1

Common error: Energy is power multiplied by time, so the area under a power–time graph is in kWh; do not divide by the interval width.

216IBMI3S-5-02r33-10Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

A cycling company runs guided tours. When the price of a ticket is €p, the number of tickets sold per week is 400 − 8p. Each ticket costs the company €12 to provide.

(a)

Show that the weekly profit, P euros, is P = −8p2 + 496p − 4800.

(2)
(b)

Find dP/dp.

(2)
(c)

Find the ticket price that maximises the weekly profit.

(1)
(d)

Justify that this is a maximum and find the maximum weekly profit.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 profit = (p − 12)(400 − 8p)✓ 1
A1 expansion to −8p2 + 496p − 4800 (AG)✓ 1
Part (b)
M1 attempt to differentiate✓ 1
A1 −16p + 496✓ 1
Part (c)
A1 −16p + 496 = 0 gives p = €31✓ 1
Part (d)
R1 d2P/dp2 = −16 < 0 (or sign change)✓ 1
A1 maximum profit = 19 × 152 = €2888✓ 1

Exam tip: Profit per ticket is price minus cost; forgetting the cost per ticket gives the revenue-maximising price instead.

217IBMI3S-5-02r33-13Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

A cycle path crosses a field. Using a coordinate grid in metres, the centre line of the path is modelled by y = 0.02x2 − 0.4x + 5, for 0 ≤ x ≤ 30. A drainage pipe is to be laid perpendicular to the path at the point where x = 15.

(a)

Find dy/dx.

(2)
(b)

Find the y-coordinate of the point on the path where x = 15, and the gradient of the path at this point.

(2)
(c)

Find the equation of the line of the pipe (the normal to the path at this point), in the form y = mx + c.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 0.04x − 0.4✓ 1
Part (b)
A1 y = 4.5 − 6 + 5 = 3.5✓ 1
A1 gradient = 0.6 − 0.4 = 0.2✓ 1
Part (c)
A1 normal gradient = −1/0.2 = −5 (ECF from (b))✓ 1
M1 y − 3.5 = −5(x − 15)✓ 1
A1 y = −5x + 78.5✓ 1

Common error: The gradient of the normal is the negative reciprocal of the gradient of the tangent, not just its negative.

218IBMI3S-5-01r36-07Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; 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integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; 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During a roast, the temperature of the beans in °C is modelled by T(t) = 20 + 17t − 0.5t2, where t is the time in minutes after the beans enter the drum, 0 ≤ t ≤ 12.

(a)

Find the temperature of the beans when t = 8.

(2)
(b)

Find the rate at which the temperature is increasing when t = 8.

(3)
(c)

The roaster stops the roast when the temperature first reaches 150 °C. Find the time at which this happens, giving your answer in minutes correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitute t = 8✓ 1
A1 124 °C✓ 1
Part (b)
M1 differentiate T(t)✓ 1
A1 T′(t) = 17 − t✓ 1
A1 9 °C per minute✓ 1
Part (c)
M1 set 20 + 17t − 0.5t2 = 150✓ 1
A1 t = 11.6148… (the other root, 22.4, is outside the domain)✓ 1
A1 11.6 minutes✓ 1

Common error: When a quadratic equation gives two solutions, check each against the stated domain before choosing the answer.

219IBMI3S-5-01r36-11Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; 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integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium6 marks

The solar panels of a space probe produce electrical power P kW. The power was recorded every 2 hours over a 10-hour period.

Time t (hours): 0, 2, 4, 6, 8, 10
Power P (kW): 0, 1.8, 3.1, 3.6, 2.9, 1.3

(a)

Use the trapezoidal rule to estimate the total energy, in kWh, produced during the 10 hours.

(3)
(b)

Hence find the mean power output over the 10 hours.

(2)
(c)

The instruments need a mean power of at least 2.5 kW. State whether the mean power in (b) meets this requirement, giving a reason.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use the trapezoidal rule with h = 2✓ 1
A1 (2/2)[0 + 1.3 + 2(1.8 + 3.1 + 3.6 + 2.9)]✓ 1
A1 24.1 kWh✓ 1
Part (b)
M1 24.1 ÷ 10 (or ECF from (a))✓ 1
A1 2.41 kW✓ 1
Part (c)
R1 No, because 2.41 kW is less than 2.5 kW (or ECF from (b))✓ 1

Exam tip: Energy is the area under a power–time graph, so the trapezoidal rule applied to power readings gives energy.

220IBMI3S-5-01r36-13Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; 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integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; 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After a heater is switched off, the temperature of a component on a space probe is modelled by T(t) = 15 + 120t−1 °C, where t is the time in hours after switch-off, t ≥ 1.

(a)

Find the temperature of the component when t = 2.

(1)
(b)

Find T′(t) and hence the rate of change of temperature when t = 2.

(2)
(c)

Find the time at which the temperature is falling at 7.5 °C per hour.

(3)
(d)

Find the temperature of the component at this time.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 75 °C✓ 1
Part (b)
M1 T′(t) = −120t−2✓ 1
A1 −30 °C per hour✓ 1
Part (c)
M1 set −120t−2 = −7.5✓ 1
A1 t2 = 16✓ 1
A1 t = 4 hours✓ 1
Part (d)
A1 45 °C (or ECF from (c))✓ 1

Exam tip: A negative derivative means the quantity is decreasing; when a question says 'falling at', set the derivative equal to a negative value.

221IBMI3S-5-04r36-11Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Medium8 marks

An open-topped storage crate for pottery has a square base of side x metres and height h metres. The total external surface area, including the base but no lid, is 24 m2.

(a)

Show that h = (24 − x2) ÷ 4x.

(2)
(b)

Hence show that the volume of the crate is V = 6x − 0.25x3.

(2)
(c)

Find the value of x that gives the greatest volume, correct to 3 significant figures.

(2)
(d)

Show that this value of x gives a maximum, and find the maximum volume correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 surface area = x2 + 4xh = 24✓ 1
A1 rearrange to h = (24 − x2) ÷ 4x (AG)✓ 1
Part (b)
M1 V = x2h✓ 1
A1 V = x(24 − x2) ÷ 4 = 6x − 0.25x3 (AG)✓ 1
Part (c)
M1 dV/dx = 6 − 0.75x2 = 0✓ 1
A1 x = √8 = 2.83 m✓ 1
Part (d)
R1 d2V/dx2 = −1.5x < 0 at x = √8 (or sign change of dV/dx)✓ 1
A1 V = 11.3 m3✓ 1

Common error: Setting the derivative to zero finds a stationary point only; show it is a maximum before stating the greatest volume.

222IBMI3S-5-04r36-12Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Medium8 marks

A cyclist travels at a constant speed v km h−1 on a flat road. The energy she uses per kilometre is modelled by E = 0.05v2 + 400v−1 kJ per km, for 5 ≤ v ≤ 40.

(a)

Find dE/dv.

(2)
(b)

Find the speed that minimises the energy used per kilometre. Give your answer correct to 3 significant figures.

(3)
(c)

Find the minimum energy used per kilometre. Use the unrounded speed and give your answer correct to 3 significant figures.

(2)
(d)

Justify that the value in (b) gives a minimum.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiate a term✓ 1
A1 dE/dv = 0.1v − 400v−2✓ 1
Part (b)
M1 set 0.1v − 400v−2 = 0✓ 1
M1 v3 = 4000✓ 1
A1 v = 15.9 km h−1✓ 1
Part (c)
M1 substitute the unrounded value of v into E✓ 1
A1 37.8 kJ per km✓ 1
Part (d)
R1 second derivative 0.1 + 800v−3 > 0 for v > 0 (or gradient changes from negative to positive)✓ 1

Common error: Writing 400/v as 400v rather than 400v⁻¹ before differentiating gives a derivative with the wrong sign and power.

223IBMI3S-5-04r36-14Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Medium8 marks

A sample tube for archaeological soil is a closed cylinder of radius r cm and height h cm with volume 500 cm3. The plastic used has a total surface area A cm2, including both circular ends.

(a)

Show that A = 2πr2 + 1000r−1.

(2)
(b)

Find dA/dr, and hence the value of r that minimises the surface area. Give r correct to 3 significant figures.

(3)
(c)

Find the height of the tube for this value of r, and the minimum surface area. Use the unrounded value of r and give each answer correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 h = 500 ÷ πr2 and A = 2πr2 + 2πrh✓ 1
A1 2πr × 500 ÷ πr2 = 1000r−1, so A = 2πr2 + 1000r−1 (AG)✓ 1
Part (b)
A1 dA/dr = 4πr − 1000r−2✓ 1
M1 set equal to zero: r3 = 1000 ÷ 4π✓ 1
A1 r = 4.30 cm✓ 1
Part (c)
A1 h = 8.60 cm✓ 1
M1 substitute the unrounded r into A✓ 1
A1 A = 349 cm2✓ 1

Exam tip: Express the quantity to be optimised in one variable first; the fixed volume is what lets you eliminate h.

224IBMI3S-5-01r39-05Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium7 marks

A logistics firm makes open-topped boxes from rectangular cardboard sheets measuring 60 cm by 40 cm. A square of side x cm is cut from each corner and the sides are folded up to form a box of height x cm, where 0 < x < 20.

(a)

Show that the volume of the box is V = 4x3 − 200x2 + 2400x cm3.

(2)
(b)

Find the value of x that maximises the volume. Give your answer to 3 significant figures.

(3)
(c)

Find the maximum volume, to 3 significant figures. Use the unrounded value of x.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 length = 60 − 2x and width = 40 − 2x (stated or used)✓ 1
A1 V = x(60 − 2x)(40 − 2x) expanded correctly to the given result (AG)✓ 1
Part (b)
M1 V′ = 12x2 − 400x + 2400 and set equal to zero✓ 1
A1 x = 7.85 (or 25.5)✓ 1
R1 reject 25.5 as it is outside 0 < x < 20✓ 1
Part (c)
M1 substitute unrounded x into V✓ 1
A1 8450 cm3✓ 1

Exam tip: When a derivative equals zero gives two roots, check each against the stated domain before using it.

225IBMI3S-5-01r39-06Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium6 marks

A research submarine accelerates from rest during a 20-second launch. Its speed v m/s was recorded every 4 seconds.

Time (s): 0, 4, 8, 12, 16, 20
Speed (m/s): 0, 2.1, 3.6, 4.5, 5.0, 5.2

The speed–time graph is concave down (curving downwards) throughout the 20 seconds.

(a)

Use the trapezoidal rule with all six values to estimate the distance travelled in the 20 seconds.

(3)
(b)

State, with a reason, whether your estimate is an overestimate or an underestimate of the true distance.

(1)
(c)

Hence calculate the mean speed over the 20 seconds, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule with h = 4✓ 1
A1 ½ × 4 × [0 + 5.2 + 2(2.1 + 3.6 + 4.5 + 5.0)]✓ 1
A1 71.2 m✓ 1
Part (b)
R1 underestimate, because the graph is concave down so each trapezium lies below the curve✓ 1
Part (c)
M1 distance ÷ 20 (ECF from (a))✓ 1
A1 3.56 m/s✓ 1

Common error: Mean speed is total distance divided by total time, not the average of the first and last speeds.

226IBMI3S-5-01r39-07Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium7 marks

The number of fish N in a protected marine reserve is modelled by N(t) = 2000 + at − 8t2, where t is the time in months since the reserve opened and a is a constant. The population is greatest when t = 10.

(a)

Find the value of a.

(3)
(b)

Find the greatest number of fish in the reserve.

(2)
(c)

Find the rate of change of the population when t = 15 and interpret your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 N′(t) = a − 16t✓ 1
M1 set N′(10) = 0✓ 1
A1 a = 160✓ 1
Part (b)
M1 substitute t = 10 into N✓ 1
A1 2800✓ 1
Part (c)
A1 N′(15) = 160 − 240 = −80✓ 1
R1 the population is decreasing by 80 fish per month✓ 1

Exam tip: A negative rate of change means the quantity is falling; say so in the context of the question.

227IBMI3S-5-01r39-09Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium7 marks

The vertical cross-section of a curved glass canopy over a plaza entrance lies between x = 0 and x = 8 (metres) and is bounded above by y = −0.25x2 + 2x + 1 and below by the line y = 0. An architect estimates its area using the trapezoidal rule with 4 equal intervals.

(a)

Use the trapezoidal rule with 4 intervals to estimate the area of the cross-section.

(3)
(b)

Use your GDC to find the exact value of the area, correct to 3 significant figures.

(1)
(c)

Calculate the percentage error in the estimate from part (a), to 3 significant figures. Use the unrounded exact area.

(2)
(d)

Suggest how the architect could obtain a more accurate estimate.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 ordinates 1, 4, 5, 4, 1✓ 1
M1 trapezoidal rule with h = 2✓ 1
A1 28 m2✓ 1
Part (b)
A1 29.3 m2 (29.333…)✓ 1
Part (c)
M1 |29.333… − 28| / 29.333… × 100✓ 1
A1 4.55%✓ 1
Part (d)
R1 use more (narrower) intervals✓ 1

Exam tip: Percentage error is measured against the exact value, not against the estimate.

228IBMI3S-5-01r39-10Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium7 marks

The main cable of a footbridge hangs between two towers and is modelled by y = 0.02x2 − 0.6x + 9, for 0 ≤ x ≤ 30, where y is the height in metres above the deck and x is the horizontal distance in metres from the left tower.

(a)

Find dy/dx.

(2)
(b)

A support rod is fixed at the point where the gradient of the cable is 0.3. Find the coordinates of this point.

(3)
(c)

Find the height above the deck of the lowest point of the cable.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 0.04x − 0.6✓ 1
Part (b)
M1 set 0.04x − 0.6 = 0.3✓ 1
A1 x = 22.5✓ 1
A1 y = 5.625✓ 1
Part (c)
M1 set gradient = 0 giving x = 15✓ 1
A1 4.5 m✓ 1

Exam tip: At the lowest or highest point of a smooth curve the gradient is zero; solve dy/dx = 0 and then find the height.

229IBMI3S-5-01r39-11Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 1Medium6 marks

At a distribution centre, pallets are loaded onto lorries at a rate of R(t) = 40 + 12t − t2 pallets per hour, for 0 ≤ t ≤ 8, where t is the number of hours after the shift starts.

(a)

Find the total number of pallets loaded during the first 8 hours, to the nearest whole pallet.

(3)
(b)

Find the number of pallets loaded during the last four hours of the shift (4 ≤ t ≤ 8), to the nearest whole pallet.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 ∫08 (40 + 12t − t2) dt✓ 1
A1 antiderivative 40t + 6t2 − t3/3 (or GDC)✓ 1
A1 533 pallets (533.33…)✓ 1
Part (b)
M1 ∫48 R(t) dt set up with the correct limits✓ 1
M1 evaluate, e.g. [40t + 6t2 − t3/3] from 4 to 8, or use GDC✓ 1
A1 299 pallets (298.67…)✓ 1

Exam tip: The limits of a definite integral give the time interval being measured; a total over part of the shift uses only those limits.

230IBMI3S-5-02r41-05Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; 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integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium8 marks

A coffee roaster designs a closed rectangular box with a square base of side x cm and height h cm. The box must hold a volume of 2000 cm3. The surface area of the box is A cm2.

(a)

Show that \(A = 2x^2 + \frac{8000}{x}\).

(2)
(b)

Find \(\frac{dA}{dx}\).

(2)
(c)

Find the value of x that minimises the surface area. Give your answer to 3 significant figures.

(2)
(d)

Find the minimum surface area. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(x^2h=2000\) so \(h=\frac{2000}{x^2}\)✓ 1
A1 \(A=2x^2+4xh = 2x^2+\frac{8000}{x}\) (AG)✓ 1
Part (b)
M1 differentiate \(2x^2\) to \(4x\) and \(8000x^{-1}\) to a term in \(x^{-2}\)✓ 1
A1 \(\frac{dA}{dx}=4x-8000x^{-2}\) (or \(4x-\frac{8000}{x^2}\))✓ 1
Part (c)
M1 \(4x-\frac{8000}{x^2}=0\), so \(x^3=2000\)✓ 1
A1 \(x=12.6\) cm✓ 1
Part (d)
M1 substitute the unrounded value of x into A✓ 1
A1 952 cm2✓ 1

Common error: Write \(\frac{8000}{x}\) as \(8000x^{-1}\) before differentiating; the derivative is \(-8000x^{-2}\), not \(-8000\).

231IBMI3S-5-02r41-06Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; 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Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium7 marks

The strength of the signal received from a space probe is modelled by \(S = \frac{4800}{d^2}\) units, where d is the distance from the probe in millions of kilometres and \(d\ge10\).

(a)

Find the signal strength when \(d=20\).

(1)
(b)

Find \(\frac{dS}{dd}\) and hence the rate of change of signal strength when \(d=20\).

(3)
(c)

Find the value of d at which the signal strength is decreasing at a rate of 0.3 units per million km. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(S=\frac{4800}{400}=12\) units✓ 1
Part (b)
M1 \(S=4800d^{-2}\)✓ 1
A1 \(\frac{dS}{dd}=-9600d^{-3}\)✓ 1
A1 \(-1.2\) units per million km✓ 1
Part (c)
M1 \(-9600d^{-3}=-0.3\)✓ 1
M1 \(d^3=32000\)✓ 1
A1 \(d=31.7\)✓ 1

Exam tip: 'Decreasing at a rate of 0.3' means the derivative equals \(-0.3\).

232IBMI3S-5-02r41-07Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; 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integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium7 marks

The opening under an arched bridge has height \(h = 12 - 0.03x^2\) metres at a horizontal position x metres from the centre line, for \(-20\le x\le20\). The cross-sectional area of the opening is the area under this curve.

(a)

Use your GDC or integration to find the exact area of the opening.

(2)
(b)

Use the trapezoidal rule with 4 equal intervals to estimate the area.

(3)
(c)

Find the percentage error in the trapezoidal estimate.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\int_{-20}^{20}(12-0.03x^2)\,dx\)✓ 1
A1 320 m2✓ 1
Part (b)
M1 values of h at \(x=-20,-10,0,10,20\): \(0,\,9,\,12,\,9,\,0\)✓ 1
M1 \(\frac{10}{2}\left(0+0+2(9+12+9)\right)\)✓ 1
A1 300 m2✓ 1
Part (c)
M1 \(\frac{320-300}{320}\times100\)✓ 1
A1 6.25 %✓ 1

Exam tip: The percentage error is the difference divided by the exact value, then multiplied by 100.

233IBMI3S-5-02r41-09Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; 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integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium6 marks

A coffee roaster sells x hundred kilograms of beans per week, for \(0\le x\le8\). The weekly profit is \(P = -x^3 + 9x^2 - 15x - 10\) thousand euros.

(a)

Find \(\frac{dP}{dx}\).

(2)
(b)

Find the values of x for which \(\frac{dP}{dx}=0\).

(2)
(c)

Find the greatest weekly profit that can be made, in euros, justifying that it is the greatest.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiate term by term✓ 1
A1 \(-3x^2+18x-15\)✓ 1
Part (b)
M1 \(-3x^2+18x-15=0\)✓ 1
A1 \(x=1\) and \(x=5\)✓ 1
Part (c)
M1 compare P at the stationary points and at the ends of the domain: values \(-10\) at \(x=0\), \(-17\) at \(x=1\), \(15\) at \(x=5\), \(-66\) at \(x=8\)✓ 1
A1 greatest profit is 15 thousand euros, i.e. €15 000 (at \(x=5\))✓ 1

Exam tip: When a domain is given, check the end points as well as the stationary points before naming a maximum.

234IBMI3S-5-02r41-13Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; 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Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Medium7 marks

The gas flow to a coffee roaster is measured every 2 minutes during a roast:

Time (min): 0, 2, 4, 6, 8, 10
Flow rate (litres per minute): 0, 3.2, 4.4, 4.8, 3.6, 0.9

(a)

Use the trapezoidal rule with 5 intervals to estimate the total volume of gas used.

(3)
(b)

The gas costs €0.90 per litre. Find the estimated cost of the gas used.

(2)
(c)

A meter shows that the roast actually used 30.5 litres. Taking this as the exact value, find the percentage error in the estimate from part (a). Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 strip width \(h=2\)✓ 1
M1 \(\frac{2}{2}\left(0+0.9+2(3.2+4.4+4.8+3.6)\right)\)✓ 1
A1 32.9 litres✓ 1
Part (b)
M1 \(32.9\times0.90\) (ECF from (a))✓ 1
A1 €29.61✓ 1
Part (c)
M1 \(\frac{32.9-30.5}{30.5}\times100\)✓ 1
A1 7.87 %✓ 1

Exam tip: State how many intervals you are using; with 6 data values there are 5 intervals.

235IBMI3S-5-02r42-05Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium6 marks

A library designs an open-topped storage box with a square base of side x metres and height h metres. The total outer surface area of the box (the base and the four sides) is 1.2 m2.

(a)

Show that the volume of the box is V = 0.3x − x3/4.

(2)
(b)

Find the value of x that maximises the volume. Give your answer to 3 significant figures.

(3)
(c)

Find the maximum volume. Give your answer to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 x2 + 4xh = 1.2, so h = (1.2 − x2)/(4x)✓ 1
A1 V = x2h = x(1.2 − x2)/4 = 0.3x − x3/4 (shown)✓ 1
Part (b)
M1 dV/dx = 0.3 − 0.75x2✓ 1
M1 setting dV/dx = 0✓ 1
A1 x = √0.4 = 0.632 m✓ 1
Part (c)
A1 0.126 m3✓ 1

Exam tip: In a ‘show that’ question, state the constraint equation first, rearrange it for one variable, and only then substitute into the quantity being optimised.

236IBMI3S-5-02r42-06Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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A heater warms the water in a training pool. The temperature, T °C, of the water t hours after the heater is switched on is modelled by T = 20 + 3.2t − 0.4t2, for 0 ≤ t ≤ 8.

(a)

Find the rate at which the temperature is changing when t = 1.5.

(2)
(b)

Find the maximum temperature of the water and the time at which it occurs.

(3)
(c)

Find the time at which the temperature is falling at a rate of 0.8 °C per hour.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 dT/dt = 3.2 − 0.8t✓ 1
A1 2.0 °C per hour✓ 1
Part (b)
M1 3.2 − 0.8t = 0✓ 1
A1 t = 4 hours✓ 1
A1 T(4) = 26.4 °C✓ 1
Part (c)
M1 3.2 − 0.8t = −0.8✓ 1
A1 t = 5 hours✓ 1

Common error: ‘Falling at 0.8’ means the derivative equals −0.8, not +0.8.

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A cable runs along a ridge whose height, y metres above sea level, at horizontal distance x metres from a wind farm substation is modelled by y = 0.05x3 − 0.6x2 + 2x + 4, for 0 ≤ x ≤ 10. Point A on the ridge has x = 4.

(a)

Find the equation of the tangent to the ridge at A. Give your answer in the form y = mx + c.

(4)
(b)

The normal to the ridge at A meets the horizontal line y = 0 at the point B. Find the x-coordinate of B.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 dy/dx = 0.15x2 − 1.2x + 2✓ 1
A1 gradient at A is 0.15 × 16 − 4.8 + 2 = −0.4✓ 1
A1 height at A is 5.6✓ 1
A1 y = −0.4x + 7.2 (equivalent form)✓ 1
Part (b)
M1 gradient of the normal = −1/(−0.4) = 2.5✓ 1
M1 equation of the normal y − 5.6 = 2.5(x − 4), then setting y = 0✓ 1
A1 x = 1.76✓ 1

Common error: The gradient of the normal is the negative reciprocal of the gradient of the tangent, so −0.4 becomes +2.5, not +0.4 or −2.5.

238IBMI3S-5-02r42-08Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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A pool is 10 m long and 25 m wide. Along its length, the depth d metres at distance x metres from one end is modelled by d = 0.04x2 − 0.4x + 2, for 0 ≤ x ≤ 10, and the depth does not vary across the width.

(a)

Find the area of the vertical cross-section of the pool along its length.

(3)
(b)

Find the volume of water in the full pool, in litres. Use the unrounded area from part (a), note that 1 m3 = 1 000 litres, and give your answer in the form a × 10k where 1 ≤ a < 10 and k ∈ ℤ, with a given to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 ∫010(0.04x2 − 0.4x + 2) dx✓ 1
A1 correct antiderivative 0.04x3/3 − 0.2x2 + 2x (or correct value from calculator)✓ 1
A1 13.3 m2✓ 1
Part (b)
M1 area × 25 × 1 000 (ECF from (a))✓ 1
A1 3.33 × 105 litres✓ 1

Exam tip: For a prism with a uniform width, volume = cross-sectional area × width. Convert units at the end: 1 m³ = 1 000 litres.

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The power output P MW of a wind farm is measured at irregular times t hours after midnight.
t (h): 0, 2, 5, 9
P (MW): 1.8, 3.4, 4.6, 3.2

(a)

Use the trapezoidal rule, applied to each of the three time intervals, to estimate the energy produced between t = 0 and t = 9. Give your answer in MWh.

(3)
(b)

Using your answer to part (a), find the mean power output between t = 0 and t = 9. Give your answer to 3 significant figures.

(2)
(c)

A further reading of p MW is taken at t = 12. Including this reading, the trapezoidal estimate of the energy produced between t = 0 and t = 12 is 40.6 MWh. Find the value of p.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 area of a trapezium (1/2) × width × (sum of parallel sides) applied to at least one interval✓ 1
A1 5.2, 12.0 and 15.6 for the three intervals✓ 1
A1 32.8 MWh✓ 1
Part (b)
M1 energy ÷ 9 (ECF from (a))✓ 1
A1 3.64 MW✓ 1
Part (c)
M1 (1/2) × 3 × (3.2 + p) = 40.6 − 32.8 (ECF from (a))✓ 1
A1 p = 2.0✓ 1

Common error: With unequal intervals the formula (h/2)(y₀ + 2y₁ + …) does not apply directly: find the area of each trapezium separately using its own width.

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Optimisation in context; Optimisation in context Paper 1Medium7 marks

During a 20 m shuttle drill, an athlete starts from rest and stops exactly at the far line. Her velocity v m s−1 at time t seconds is modelled by \(v=1.2t-0.12t^2\), valid from the start until she stops.

(a)

Find the time at which she stops.

(2)
(b)

Show, by integration, that she covers exactly 20 m.

(3)
(c)

Find her greatest speed.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 set \(1.2t-0.12t^2=0\) and solve (for example by factorising)✓ 1
A1 \(t=10\) s (rejecting \(t=0\))✓ 1
Part (b)
M1 \(\int_0^{10}(1.2t-0.12t^2)\,dt\) (ECF from (a))✓ 1
A1 \(0.6t^2-0.04t^3\)✓ 1
A1 \(60-40=20\) m (shown)✓ 1
Part (c)
M1 \(\frac{dv}{dt}=1.2-0.24t=0\) so \(t=5\)✓ 1
A1 \(v=6-3=3.0\) m s−1✓ 1

Common error: The distance travelled is the integral of velocity, not the integral of position or the derivative of velocity.

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Optimisation in context; Optimisation in context Paper 1Medium7 marks

A section of a cycling training bank is modelled by \(f(x)=0.5x^3-2x^2+3\), where x and \(f(x)\) are in metres.

(a)

Find \(f'(x)\).

(2)
(b)

Find the equation of the normal to the curve at the point where \(x=2\). Give your answer in the form \(y=mx+c\).

(4)
(c)

Find the x-coordinate of the point where this normal meets the x-axis.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 \(f'(x)=1.5x^2-4x\)✓ 1
Part (b)
A1 \(f(2)=-1\)✓ 1
A1 \(f'(2)=-2\)✓ 1
M1 normal gradient is the negative reciprocal, \(\frac{1}{2}\)✓ 1
A1 \(y=0.5x-2\)✓ 1
Part (c)
A1 \(x=4\)✓ 1

Common error: The normal is perpendicular to the tangent: use \(-\frac{1}{f'(x)}\), and do not forget the minus sign.

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Optimisation in context; Optimisation in context Paper 1Medium6 marks

The volume of fuel, V litres, in a diesel locomotive's tank t minutes after it starts a shunting task changes at the rate \(\frac{dV}{dt}=-(2+0.4t)\). Initially the tank holds 450 litres.

(a)

Find an expression for V in terms of t.

(3)
(b)

Find the volume of fuel left after 15 minutes.

(1)
(c)

Find the time at which the tank holds 100 litres.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to integrate (power increased)✓ 1
A1 \(V=-2t-0.2t^2+c\)✓ 1
A1 \(c=450\), so \(V=450-2t-0.2t^2\)✓ 1
Part (b)
A1 \(450-30-45=375\) litres✓ 1
Part (c)
M1 \(450-2t-0.2t^2=100\)✓ 1
A1 \(t=37.1\) minutes✓ 1

Common error: After integrating, always use the initial condition to find the constant; leaving out \(+c\) loses the final marks.

243IBMI3S-5-01r45-03Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Medium6 marks

Honey flows from an extractor into a storage tank at a rate of R(t) = 0.3t2 + 2 litres per minute, where t is the time in minutes after the extractor is switched on, for 0 ≤ t ≤ 10.

(a)

Find ∫ (0.3t2 + 2) dt.

(2)
(b)

Calculate the volume of honey that flows into the tank during the first 10 minutes.

(2)
(c)

Hence find the mean rate of flow over the first 10 minutes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to integrate (at least one term correct)✓ 1
A1 0.1t3 + 2t + c✓ 1
Part (b)
M1 substitute limits 10 and 0 into the integral✓ 1
A1 120 litres✓ 1
Part (c)
M1 120 ÷ 10 (ECF from (b))✓ 1
A1 12 litres per minute✓ 1

Exam tip: A definite integral of a rate of flow gives the total amount; divide by the time taken to get the mean rate.

244IBMI3S-5-01r45-04Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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During a test run, the speed of a Formula 1 car is recorded every 2 seconds. The speed, v m s−1, is shown in the table.

t (s)0246810
v (m s−1)03152667478

(a)

Use the trapezoidal rule with 5 intervals to estimate the distance travelled by the car in the first 10 seconds.

(3)
(b)

The speed increases at a decreasing rate throughout the 10 seconds. State, with a reason, whether your answer to (a) is an overestimate or an underestimate.

(2)
(c)

Using your answer to (a), find the mean speed of the car over the 10 seconds, giving your answer in km h−1 correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use trapezoidal rule with h = 2✓ 1
A1 (2/2)[0 + 78 + 2(31 + 52 + 66 + 74)]✓ 1
A1 524 m✓ 1
Part (b)
A1 underestimate✓ 1
R1 the graph is concave down (curved below the chords), so the trapeziums lie below the curve✓ 1
Part (c)
M1 (524 ÷ 10) × 3.6✓ 1
A1 189 km h−1✓ 1

Common error: Remember to use the interval width h in the trapezoidal rule, not the number of intervals, and to double only the interior values.

245IBMI3S-5-01r45-06Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Medium7 marks

A landscaped bank beside a hospital car park has a height profile modelled by h(x) = 0.002x3 − 0.06x2 + 0.5x + 2, where h is the height in metres above the car park and x is the horizontal distance in metres from the car park wall, for 0 ≤ x ≤ 30.

(a)

Find h′(x).

(2)
(b)

Find the gradient of the bank when x = 10, and state whether the bank is rising or falling at this point.

(2)
(c)

The tangent to the bank's profile at x = 10 meets the line x = 0 at the height H metres. Find H.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 0.006x2 − 0.12x + 0.5✓ 1
Part (b)
A1 gradient = −0.1✓ 1
R1 the gradient is negative, so the bank is falling✓ 1
Part (c)
M1 h(10) = 3✓ 1
M1 tangent: y − 3 = −0.1(x − 10)✓ 1
A1 H = 4 m✓ 1

Common error: A negative gradient means the curve is falling; do not confuse the sign of the gradient with the sign of the height.

246IBMI3S-5-01r45-08Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Medium7 marks

A raised flower bed is built on a hospital roof garden. Its cross-section is the region between the horizontal ground (the x-axis) and the curve y = 0.1x(6 − x), where x and y are in metres and y ≥ 0.

(a)

Find the width of the bed at ground level.

(1)
(b)

Find the area of the cross-section.

(3)
(c)

The bed is 12 m long. Soil costs €38 per m3. Calculate the cost of filling the bed with soil.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 6 m (from the zeros x = 0 and x = 6)✓ 1
Part (b)
M1 ∫06 (0.6x − 0.1x2) dx✓ 1
A1 [0.3x2 − x3/30]06✓ 1
A1 3.6 m2✓ 1
Part (c)
M1 volume = 3.6 × 12 = 43.2 m3 (ECF from (b))✓ 1
M1 43.2 × 38✓ 1
A1 €1641.60✓ 1

Common error: The width of the region comes from where the curve meets the axis, not from the maximum height of the curve.

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A water-supply company makes an open storage box by cutting a square of side x cm from each corner of a rectangular metal sheet measuring 80 cm by 50 cm and then folding up the sides.

(a)

State the range of possible values of x.

(1)
(b)

Show that the volume of the box is V = 4x3 − 260x2 + 4000x cm3.

(2)
(c)

Find dV/dx.

(1)
(d)

Find the value of x that gives the maximum volume, justifying that this value is a maximum.

(3)
(e)

Find the maximum volume, in litres. (1 litre = 1000 cm3)

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 0 < x < 25✓ 1
Part (b)
M1 base dimensions (80 − 2x) by (50 − 2x) and height x✓ 1
A1 V = x(80 − 2x)(50 − 2x) = 4x3 − 260x2 + 4000x (AG)✓ 1
Part (c)
A1 12x2 − 520x + 4000✓ 1
Part (d)
M1 sets dV/dx = 0✓ 1
A1 solutions x = 10 and x = 33.3… (33.3 rejected as outside the range)✓ 1
R1 justifies a maximum, for example d2V/dx2 = 24x − 520 = −280 < 0 at x = 10, or sign change of the gradient✓ 1
Part (e)
A1 V(10) = 10 × 60 × 30 = 18 000 cm3 = 18 litres✓ 1

Common error: Leaving both solutions of dV/dx = 0 in your answer; one of them lies outside the domain and must be rejected.

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Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Medium7 marks

A games studio sells a map pack online. When the price is €p, the number of downloads per week is 1200 − 80p. The platform takes a fee of €1.50 per download, and this is the studio's only cost.

(a)

Show that the weekly profit is P(p) = −80p2 + 1320p − 1800 euros.

(2)
(b)

Find the price that maximises the profit and the maximum weekly profit.

(3)
(c)

The studio decides to charge a whole number of euros. Find the price that gives the greatest weekly profit and state this profit.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 profit = (price − 1.50) × downloads = (p − 1.5)(1200 − 80p)✓ 1
A1 expands to −80p2 + 1320p − 1800 (AG)✓ 1
Part (b)
M1 P′(p) = −160p + 1320 = 0✓ 1
A1 p = €8.25✓ 1
A1 maximum profit = €3645✓ 1
Part (c)
M1 compares P(8) and P(9) (the whole-euro prices either side of 8.25)✓ 1
A1 €8 gives €3640, which is greater than €3600 at €9✓ 1

Exam tip: When a model's best value is not allowed (here, a whole number), check the nearest allowed values on both sides rather than just rounding.

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The volume of water in a reservoir is modelled by V(t) = 0.05t3 − 1.8t2 + 14t + 120 megalitres, where t is the time in months since the start of the year, for 0 ≤ t ≤ 24.

(a)

Find the rate of change of the volume when t = 10, and state whether the reservoir is filling or emptying at that time.

(2)
(b)

Find the two values of t at which the volume is neither increasing nor decreasing. Give your answers to 3 significant figures.

(2)
(c)

Find the minimum volume of water in the reservoir during the 24 months, justifying your answer.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 V′(t) = 0.15t2 − 3.6t + 14✓ 1
A1 V′(10) = −7 megalitres per month, so the reservoir is emptying (volume decreasing)✓ 1
Part (b)
M1 sets V′(t) = 0✓ 1
A1 t = 4.88 and t = 19.1✓ 1
Part (c)
M1 compares V at the stationary point t = 19.1… and at the end points: V(0) = 120, V(24) = 110.4✓ 1
A1 V(19.1…) = 79.1✓ 1
R1 this is the smallest of the candidate values, so the minimum volume is 79.1 megalitres✓ 1

Common error: Stopping at the stationary points. A greatest or least value on an interval must also be compared with the values at the end points.

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A section of a track in a racing game is modelled by the curve y = 0.5x3 − 3x2 + 4x + 5, for 0 ≤ x ≤ 5, where x and y are measured in units of 100 m.

(a)

Find dy/dx.

(2)
(b)

Find the equation of the tangent to the track at the point where x = 1, giving your answer in the form y = mx + c.

(3)
(c)

Find the values of x at which the gradient of the track is 4.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiates at least one term correctly✓ 1
A1 dy/dx = 1.5x2 − 6x + 4✓ 1
Part (b)
A1 y = 0.5 − 3 + 4 + 5 = 6.5 at x = 1✓ 1
A1 gradient = 1.5 − 6 + 4 = −0.5✓ 1
A1 y − 6.5 = −0.5(x − 1), so y = −0.5x + 7✓ 1
Part (c)
M1 sets 1.5x2 − 6x + 4 = 4, so 1.5x(x − 4) = 0✓ 1
A1 x = 0 and x = 4✓ 1

Exam tip: To find where a gradient takes a given value, set the derivative equal to that value and solve; do not substitute into the original function.

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The depth of a river is measured every 2 m across its 12 m width, starting at one bank.
Distance from bank (m): 0, 2, 4, 6, 8, 10, 12
Depth (m): 0, 0.8, 1.5, 2.1, 1.9, 1.0, 0

(a)

Use the trapezoidal rule to estimate the area of the river's cross-section.

(3)
(b)

The water flows at an average speed of 0.85 m s−1. Find the volume of water passing through the cross-section each second, in m3.

(2)
(c)

Estimate the volume of water passing through the cross-section in one day, in m3, giving your answer in the form a × 10k where 1 ≤ a < 10 and k is an integer, with a to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule with strip width 2 and six strips✓ 1
A1 correct substitution: ½ × 2 × [0 + 0 + 2(0.8 + 1.5 + 2.1 + 1.9 + 1.0)]✓ 1
A1 area = 14.6 m2✓ 1
Part (b)
M1 discharge = cross-sectional area × speed = 14.6 × 0.85✓ 1
A1 12.41 m3 per second✓ 1
Part (c)
M1 12.41 × 86 400 (24 × 60 × 60 seconds in a day)✓ 1
A1 1.07 × 106 m3✓ 1

Exam tip: Check that your units agree before multiplying: speed in m s⁻¹ and area in m² give a flow in m³ per second.

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The rate at which new players join an online game during the first 12 hours after a launch is modelled by r(t) = 120 + 90 sin(πt/12) players per hour, where t is the time in hours after the launch and the angle is in radians.

(a)

Write down an integral for the total number of players who join in the first 12 hours, and evaluate it to 3 significant figures.

(2)
(b)

Find the mean rate at which players joined over the 12 hours. Use the unrounded total.

(2)
(c)

Find the times, in the first 12 hours, at which the rate equals 200 players per hour. Give your answers to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 integral from 0 to 12 of (120 + 90 sin(πt/12)) dt✓ 1
A1 total = 2128, so 2130 players to 3 s.f. (accept 2128)✓ 1
Part (b)
M1 mean rate = total / 12✓ 1
A1 177.3 = 177 players per hour✓ 1
Part (c)
M1 sets 120 + 90 sin(πt/12) = 200 and solves graphically or algebraically✓ 1
A1 t = 4.18 and t = 7.82, so 4.18 and 7.82 hours✓ 1

Common error: Entering the angle in degrees mode on a calculator. The model uses radians, so check the calculator setting before evaluating.

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The concentration of a medicine in a patient's blood is modelled by C(t) = −0.15t2 + 1.8t + 0.2 mg per litre, where t is the time in hours after the dose, for 0 ≤ t ≤ 6.

(a)

Find the average rate of change of the concentration between t = 1 and t = 5, including units.

(2)
(b)

Find C′(t).

(1)
(c)

Find the time at which the instantaneous rate of change of the concentration equals the average rate found in part (a).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 C = 1.85 when t = 1 and C = 5.45 when t = 5✓ 1
A1 average rate = (5.45 − 1.85) / 4 = 0.9 mg per litre per hour✓ 1
Part (b)
A1 C′(t) = −0.3t + 1.8✓ 1
Part (c)
M1 sets −0.3t + 1.8 = 0.9✓ 1
A1 t = 3 hours✓ 1

Common error: Mixing up the two ideas: the average rate uses two points on the curve (change in C over change in t), but the instantaneous rate uses the derivative at one point.

254IBMI3S-5-03r47-04Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; 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Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Medium7 marks

A logistics company designs an open-topped storage box (no lid) with a square base of side x cm and height h cm. The box must have a volume of 4000 cm3. Let S cm2 be the total outside surface area of the box.

(a)

Show that S = x2 + 16000/x.

(2)
(b)

Find the value of x for which S is a minimum.

(3)
(c)

Find the minimum surface area, and justify that it is a minimum.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 h = 4000/x2 from x2h = 4000 and S = x2 + 4xh✓ 1
A1 substitution leading to S = x2 + 16000/x (AG)✓ 1
Part (b)
M1 S′ = 2x − 16000/x2✓ 1
M1 set S′ = 0, so x3 = 8000✓ 1
A1 x = 20✓ 1
Part (c)
A1 S = 400 + 800 = 1200 cm2✓ 1
R1 valid test, e.g. S″ = 2 + 32000/x3 > 0 at x = 20, or sign change of S′ from negative to positive✓ 1

Exam tip: In a "show that" question every step must appear; the printed result is the end of your working, not the start.

255IBMI3S-5-03r47-05Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; 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Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Medium6 marks

A marine biologist surveys a channel that is 60 m wide. The depth of the water is measured at 10 m intervals across the channel, starting at one bank.

Distance from bank (m)0102030405060
Depth (m)02.43.84.54.12.90

The bed of the channel rises smoothly from the deepest point to each bank, so that the depth curve is concave down throughout.

(a)

Use the trapezoidal rule with all the intervals to estimate the area of the cross-section of the channel.

(3)
(b)

The channel has the same cross-section along a length of 250 m. Estimate the volume of water in this length of channel.

(1)
(c)

State, with a reason, whether your estimate in (a) is an overestimate or an underestimate of the true area.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule with h = 10✓ 1
A1 ½ × 10 × [0 + 0 + 2(2.4 + 3.8 + 4.5 + 4.1 + 2.9)] correct substitution✓ 1
A1 177 m2✓ 1
Part (b)
A1 177 × 250 = 44250 m3 (ECF from (a))✓ 1
Part (c)
R1 underestimate✓ 1
R1 for a concave-down curve each trapezium lies below (inside) the curve✓ 1

Common error: In the trapezoidal rule the first and last values are used once and every other value is doubled.

256IBMI3S-5-03r47-10Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; 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optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Medium7 marks

The height, H cm, of a coral colony is monitored. Its growth rate is modelled by dH/dt = 0.6t + 0.9 cm per year, where t is the time in years since monitoring began. At t = 0 the height is 12 cm.

(a)

Find an expression for H in terms of t.

(3)
(b)

Find the height of the colony when t = 5.

(2)
(c)

Find how many years after monitoring began the colony first reaches a height of 30 cm. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to integrate (power rule)✓ 1
A1 H = 0.3t2 + 0.9t + c✓ 1
A1 c = 12, so H = 0.3t2 + 0.9t + 12✓ 1
Part (b)
M1 substitute t = 5 (ECF from (a))✓ 1
A1 24 cm✓ 1
Part (c)
M1 solve 0.3t2 + 0.9t + 12 = 30 (ECF from (a))✓ 1
A1 t = 6.39 years✓ 1

Common error: Do not omit the constant of integration; the starting height is what fixes it.

257IBMI3S-5-03r47-11Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; 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Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Medium8 marks

Vehicles arrive at a distribution depot. The number of vehicles arriving per hour was counted at two-hourly intervals during an 8-hour shift.

Time t (hours)02468
Arrival rate (vehicles per hour)1229373113

The depot manager models the arrival rate by r(t) = −1.5t2 + 12t + 12.

(a)

Use the trapezoidal rule with four intervals to estimate the total number of vehicles that arrived during the shift.

(3)
(b)

Use the model to find the total number of vehicles expected to arrive during the shift.

(3)
(c)

Find the percentage error in the trapezoidal estimate, taking the model value as exact. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule with h = 2✓ 1
A1 ½ × 2 × [12 + 13 + 2(29 + 37 + 31)] correct substitution✓ 1
A1 219 vehicles✓ 1
Part (b)
M1 ∫08 r(t) dt✓ 1
A1 antiderivative −0.5t3 + 6t2 + 12t✓ 1
A1 224 vehicles✓ 1
Part (c)
M1 (224 − 219)/224 × 100 (ECF from (a) and (b))✓ 1
A1 2.23%✓ 1

Exam tip: Percentage error divides the difference by the exact (or accepted) value, not by the estimate.

258IBMI3S-5-03r47-14Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; 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Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 1Medium7 marks

The area, A m2, of the sea surface covered by an algal bloom is modelled by A(t) = 4.5t2 − 0.5t3, where t is the number of days since the bloom was first detected, for 0 ≤ t ≤ 9.

(a)

Find A′(t).

(2)
(b)

Find the maximum area covered by the bloom, and the day on which it occurs.

(3)
(c)

Find the rate at which the area is changing on day 8, and state whether the bloom is growing or shrinking then.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 9t − 1.5t2✓ 1
Part (b)
M1 set A′(t) = 0✓ 1
A1 t = 6✓ 1
A1 54 m2✓ 1
Part (c)
M1 substitute t = 8 into A′✓ 1
A1 −24 m2 per day, so shrinking✓ 1

Exam tip: A negative value of the derivative means the quantity is decreasing; say so in words when the question is set in context.

259IBMI3S-5-09b-14Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 1Hard7 marks

The cross-section of a curved solar reflector is modelled by the curve y = 6/x, for x > 0, with both axes in metres. A support strut lies along the tangent to the curve at the point P(a, 6/a), where a > 0. The strut, the horizontal axis and the vertical axis enclose a triangle.

(a)

Find the gradient of the curve at P in terms of a.

(2)
(b)

Show that the equation of the tangent at P is y = −(6/a2)x + 12/a.

(2)
(c)

Show that the area of the triangle is 12 m2, whatever the value of a.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 write y = 6x−1 and differentiate✓ 1
A1 gradient = −6/a2✓ 1
Part (b)
M1 y − 6/a = −(6/a2)(x − a)✓ 1
A1 simplify: 6/a + 6/a = 12/a✓ 1
Part (c)
A1 vertical intercept 12/a✓ 1
A1 horizontal intercept 2a (set y = 0)✓ 1
R1 area = ½ × 2a × 12/a = 12, independent of a✓ 1

Exam tip: To show that a quantity does not depend on a parameter, simplify until the parameter cancels completely and state this conclusion.

260IBMI3S-5-01r27-14Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 1Hard7 marks

The power output, P MW, of an offshore wind farm during a 15-hour storm is modelled by P(t) = 0.06t(15 − t), for 0 ≤ t ≤ 15, where t is the time in hours since the storm began.

(a)

Find the total energy, in MWh, generated during the 15 hours.

(3)
(b)

The wind farm has generated exactly 20 MWh by time T hours. Find the value of T, correct to 3 significant figures.

(3)
(c)

Find the mean power output over the first T hours, correct to 3 significant figures. Use the unrounded value of T.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 integral of P between 0 and 15✓ 1
A1 antiderivative 0.45t2 − 0.02t3✓ 1
A1 33.75 MWh✓ 1
Part (b)
M1 setting up 0.45T2 − 0.02T3 = 20✓ 1
M1 solving using a GDC, with 0 < T < 15✓ 1
A1 T = 8.43 hours✓ 1
Part (c)
A1 20 ÷ 8.4307… = 2.37 MW (ECF from (b))✓ 1

Exam tip: An equation such as a cubic in T should be solved on the GDC. Write down the equation you entered so that method marks are secured even if the value is slightly out.

261IBMI3S-5-02r27-09Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Hard7 marks

A volcanology team's base A lies on a straight road. An observation post B is 6 km from the road, and the point C on the road nearest to B is 10 km from A. The team drives along the road from A to a point P, which is x km from C, and then travels straight across open ground from P to B. The vehicle travels at 60 km/h on the road and 20 km/h on open ground, for 0 ≤ x ≤ 10.

(a)

Show that the total journey time is T = (10 − x)/60 + √(x2 + 36)/20 hours.

(2)
(b)

Use your GDC to find the value of x that minimises T. Give your answer correct to 3 significant figures.

(2)
(c)

Find how many minutes shorter the minimum journey is than driving directly from A to B across open ground. Give your answer correct to 2 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 distance AP = 10 − x and distance PB = √(x2 + 62)✓ 1
A1 time = distance ÷ speed for each stage, added to give the result✓ 1
Part (b)
M1 graphing T or solving dT/dx = 0✓ 1
A1 x = 2.12 km✓ 1
Part (c)
M1 minimum T = 0.4495 h = 26.97 minutes (unrounded x)✓ 1
M1 direct time = √(102 + 62) ÷ 20 × 60 = 34.99 minutes✓ 1
A1 8.0 minutes✓ 1

Exam tip: For a 'minimum time' model, sketch the situation first and label each distance with a variable. Keep full accuracy from the GDC until the final answer.

262IBMI3S-5-02r27-12Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Hard7 marks

The total cost, €C, of staging a theatre performance for an audience of n people is modelled by C(n) = 5000 + 60n − 0.2n2 + 0.0004n3, for 50 ≤ n ≤ 500. The average cost per person is €A.

(a)

Write down an expression for A(n).

(1)
(b)

Find A′(n).

(2)
(c)

Use your GDC to solve A′(n) = 0 and hence find the audience size that minimises the average cost, to the nearest whole number.

(2)
(d)

Find the minimum average cost per person, correct to the nearest cent, using the audience size found in part (c).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 A(n) = 5000/n + 60 − 0.2n + 0.0004n2✓ 1
Part (b)
M1 differentiating at least two terms correctly✓ 1
A1 −5000/n2 − 0.2 + 0.0008n✓ 1
Part (c)
M1 solving −5000/n2 − 0.2 + 0.0008n = 0 on a GDC (n = 313.6…)✓ 1
A1 314 people✓ 1
Part (d)
M1 substituting n = 314 into A(n)✓ 1
A1 €52.56✓ 1

Exam tip: If the variable must be a whole number, check that the nearest whole number really gives the lowest value, and state clearly which value of n you substitute.

263IBMI3S-5-03r28-11Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 1Hard7 marks

An aquaculture company designs an open-topped cylindrical tank of internal radius r metres and height h metres to hold exactly 12 m3 of water. The base costs €40 per m2 to build and the curved wall costs €25 per m2.

(a)

Show that the total cost of the tank is C = 40πr2 + 600/r euros.

(2)
(b)

Find the value of r for which C is a minimum. Give your answer to 3 significant figures.

(2)
(c)

The height of the tank must not exceed 1.5 m. Show that the radius in part (b) is not allowed, and find the minimum possible cost of a tank that meets this condition. Give your answer to the nearest euro.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 h = 12/(πr2) and cost = 40(πr2) + 25(2πrh)✓ 1
A1 25 × 2πr × 12/(πr2) = 600/r, giving the result (AG)✓ 1
Part (b)
M1 dC/dr = 80πr − 600/r2 = 0✓ 1
A1 r = 1.34 m✓ 1
Part (c)
R1 at r = 1.34, h = 2.14 m > 1.5 m, so not allowed; and C increases for r larger than this, so the cheapest allowed tank has the smallest allowed radius✓ 1
M1 sets h = 1.5, so πr2 = 8, giving r2 = 8/π and r = 1.596…✓ 1
A1 C = €696✓ 1

Exam tip: When a design constraint is added, check whether the unconstrained optimum satisfies it; if not, the optimum lies on the boundary of the allowed values.

264IBMI3S-5-02r42-11Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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The frame of a library window consists of a rectangle of width 2r metres and height h metres, with a semicircle of radius r metres on top. The total length of the frame (the bottom edge, the two vertical sides and the semicircular arc) is 12 m. The area of the window is A m2.

(a)

Show that A = 12r − (2 + π/2)r2.

(3)
(b)

Find the value of r that gives the largest possible area. Give your answer to 3 significant figures.

(3)
(c)

Find the largest possible area of the window. Give your answer to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 2r + 2h + πr = 12, so 2h = 12 − 2r − πr✓ 1
M1 A = 2rh + πr2/2 with A expressed using r only✓ 1
A1 A = r(12 − 2r − πr) + πr2/2 = 12r − (2 + π/2)r2 (shown)✓ 1
Part (b)
M1 dA/dr = 12 − (4 + π)r✓ 1
M1 setting dA/dr = 0✓ 1
A1 r = 12/(4 + π) = 1.68 m✓ 1
Part (c)
A1 10.1 m2✓ 1

Exam tip: Express the area in terms of one variable using the perimeter condition before differentiating. Keep π in exact form until the final answer.

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Optimisation in context; Optimisation in context Paper 1Hard7 marks

A model rocket's height above the ground is \(h=at^2-bt^3\) metres, where t is the time in seconds after launch and a and b are positive constants. The rocket reaches its maximum height when \(t=8\), and its height is 108 m when \(t=6\).

(a)

Find an expression for \(\frac{dh}{dt}\).

(2)
(b)

Find the values of a and b.

(4)
(c)

Find the maximum height reached.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 \(2at-3bt^2\)✓ 1
Part (b)
M1 \(\frac{dh}{dt}=0\) at \(t=8\): \(16a-192b=0\)✓ 1
M1 \(h(6)=108\): \(36a-216b=108\)✓ 1
A1 \(a=6\)✓ 1
A1 \(b=0.5\)✓ 1
Part (c)
A1 \(6(64)-0.5(512)=128\) m✓ 1

Exam tip: Two pieces of information about an unknown model give two equations: use one for the gradient condition and one for a point on the curve.

266IBMI3S-5-01r45-14Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 1Hard7 marks

A beekeeper sells x jars of honey per week at a price of €(14 − 0.02x) per jar. The weekly cost of production is €(120 + 3x). The weekly profit is €P.

(a)

Show that P(x) = 11x − 0.02x2 − 120.

(2)
(b)

Find the number of jars that gives the greatest weekly profit if there is no restriction on production.

(2)
(c)

Find this greatest profit.

(1)
(d)

The beekeeper can produce at most 200 jars per week. Find the greatest weekly profit under this restriction, justifying your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 revenue = x(14 − 0.02x) and profit = revenue − cost✓ 1
A1 14x − 0.02x2 − 120 − 3x = 11x − 0.02x2 − 120 (AG)✓ 1
Part (b)
M1 P′(x) = 11 − 0.04x = 0✓ 1
A1 x = 275✓ 1
Part (c)
A1 €1392.50✓ 1
Part (d)
R1 P increases for x < 275, so the maximum for 0 ≤ x ≤ 200 occurs at x = 200✓ 1
A1 P(200) = €1280✓ 1

Common error: A stationary point only gives the answer if it lies inside the allowed domain; otherwise check the end point.

267IBMI-034-01-r1-03Optimisation in context Paper 2Medium13 marks

A company manufactures closed cylindrical tins, each with a volume of 750 cm3. A tin has radius \(r\) cm and height \(h\) cm.
Material for the top and bottom costs $0.005 per cm2. Material for the curved side costs $0.002 per cm2.
The cost of material for one tin is $\(C\).

(a)

Express \(h\) in terms of \(r\).

(2)
(b)

Show that \(C = 0.01\pi r^2 + \frac{3}{r}\).

(3)
(c)

Find \(\frac{dC}{dr}\).

(2)
(d)

Find the value of \(r\) that minimises the cost of one tin.

(3)
(e)

Justify that this value of \(r\) gives a minimum, and find the minimum cost of material for one tin.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 using \(\pi r^2 h = 750\)✓ 1
A1 \(h = \frac{750}{\pi r^2}\)✓ 1
Part (b)
M1 cost of top and bottom: \(0.005 \times 2\pi r^2\)✓ 1
M1 cost of curved side: \(0.002 \times 2\pi r h\)✓ 1
A1 substituting \(h\): \(0.004\pi r \times \frac{750}{\pi r^2} = \frac{3}{r}\), giving \(C = 0.01\pi r^2 + \frac{3}{r}\)✓ 1
Part (c)
A1 \(0.02\pi r\)✓ 1
A1 \(-\frac{3}{r^2}\)✓ 1
Part (d)
M1 setting \(0.02\pi r - \frac{3}{r^2} = 0\)✓ 1
A1 \(r^3 = \frac{150}{\pi}\) (= 47.7…)✓ 1
A1 \(r = 3.63\) cm✓ 1
Part (e)
R1 valid justification, e.g. \(\frac{d^2C}{dr^2} = 0.02\pi + \frac{6}{r^3} > 0\), or \(\frac{dC}{dr}\) changes from negative to positive at \(r = 3.63\), or a GDC graph showing a minimum✓ 1
M1 substituting \(r = 3.6278…\) into \(C\)✓ 1
A1 \(C = $1.24\)✓ 1

Exam tip: Keep the unrounded value of r in your calculator when finding the minimum cost. Rounding r to 3.6 first gives a slightly wrong cost.

268IBMI-034-01-r1-05Optimisation in context Paper 2Medium12 marks

An architect designs a window in the shape of a rectangle topped by a semicircle. The rectangle has width \(2r\) metres and height \(h\) metres, and the diameter of the semicircle is the top edge of the rectangle.
The total perimeter of the window (the two vertical sides, the bottom edge and the curved arc) is 6 m.

(a)

Show that \(h = \frac{6 - 2r - \pi r}{2}\).

(2)
(b)

Show that the area \(A\) m2 of the window is given by \(A = 6r - 2r^2 - \frac{\pi}{2}r^2\).

(3)
(c)

Find \(\frac{dA}{dr}\).

(2)
(d)

Find the value of \(r\) that maximises the area of the window.

(2)
(e)

Find the corresponding value of \(h\) and the maximum area of the window.

(2)
(f)

The glass for the window costs $85 per m2. Find the cost of the glass for the window of maximum area.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 perimeter \(2h + 2r + \pi r = 6\)✓ 1
A1 rearranging correctly to the given result✓ 1
Part (b)
M1 area of rectangle \(2rh\)✓ 1
M1 area of semicircle \(\frac{1}{2}\pi r^2\)✓ 1
A1 substituting for \(h\) and simplifying: \(r(6 - 2r - \pi r) + \frac{1}{2}\pi r^2 = 6r - 2r^2 - \frac{\pi}{2}r^2\)✓ 1
Part (c)
A1 \(6 - 4r\)✓ 1
A1 \(-\pi r\)✓ 1
Part (d)
M1 setting \(6 - 4r - \pi r = 0\)✓ 1
A1 \(r = \frac{6}{4 + \pi} = 0.840\) m✓ 1
Part (e)
A1 \(h = 0.840\) m✓ 1
A1 \(A = 2.52\) m2✓ 1
Part (f)
A1 \(2.5204… \times 85 = $214.24\)✓ 1

Exam tip: The diameter of the semicircle is not part of the perimeter, because it is inside the window. Only the arc of length πr is counted.

269IBMI-034-00-r1-03Gradients, tangents and rates of change Paper 2Medium14 marks

A small workshop makes handmade wooden chairs. The total cost, \(C\) dollars, of making \(x\) chairs in a month is modelled by
\(C(x) = 0.01x^3 - 0.6x^2 + 15x + 200\), for \(0 \le x \le 50\).
The derivative \(C'(x)\) is called the marginal cost.

(a)

Find \(C'(x)\).

(2)
(b)

Find \(C'(10)\) and interpret your answer in the context of the question.

(3)
(c)

Find the number of chairs for which the marginal cost is least, and state this least marginal cost.

(3)
(d)

Find the equation of the tangent to the graph of \(C\) at \(x = 30\).

(3)
(e)

The workshop uses this tangent to estimate the cost of making 32 chairs.
Find this estimate, and hence find the percentage error in the estimate compared with the value given by the model.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1: \(0.03x^2 - 1.2x\)✓ 1
A1: \(+\,15\)✓ 1
Part (b)
M1: substituting \(x = 10\): \(0.03(100) - 1.2(10) + 15\)✓ 1
A1: \(C'(10) = 6\)✓ 1
R1: when 10 chairs are being made, total cost is increasing at about $6.00 per additional chair✓ 1
Part (c)
M1: valid method for the minimum of the quadratic \(C'(x)\), e.g. \(x = \frac{1.2}{2(0.03)}\) or the GDC minimum of \(C'(x)\)✓ 1
A1: \(x = 20\) chairs✓ 1
A1: least marginal cost $C'(20) = 12 - 24 + 15 = $ $3.00 per chair✓ 1
Part (d)
A1: \(C(30) = 270 - 540 + 450 + 200 = 380\)✓ 1
A1: \(C'(30) = 27 - 36 + 15 = 6\)✓ 1
A1: \(y - 380 = 6(x - 30)\), i.e. \(y = 6x + 200\)✓ 1
Part (e)
A1: estimate $= 6(32) + 200 = $ $392.00✓ 1
M1: \(C(32) = 393.28\) used in \(\frac{|393.28 - 392|}{393.28} \times 100\)✓ 1
A1: percentage error \(= 0.325\%\)✓ 1

Exam tip: When interpreting a derivative, state what is changing, the rate, its units (here dollars per chair) and the value of \(x\) at which the rate applies.

270IBMI-034-05-r2-03Gradients, tangents and rates of change Paper 2Medium12 marks

The depth of water, D metres, in a reservoir t months after the start of a year is modelled by
\(D(t) = -0.01t^3 + 0.12t^2 + 0.3t + 12\), for \(0 \le t \le 12\).

(a)

Find \(D'(t)\).

(2)
(b)

Find the rate at which the depth is changing when t = 3.

(2)
(c)

Find the value of t at which the depth is a maximum. Justify that this gives a maximum.

(3)
(d)

Find the set of values of t for which the depth is increasing at a rate greater than 0.6 metres per month.

(3)
(e)

Find the maximum depth of water in the reservoir.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(D'(t) = -0.03t^2 + 0.24t + 0.3\)—
A1 for any two correct terms✓ 1
A1 for fully correct expression✓ 1
Part (b)
M1 substituting t = 3: −0.27 + 0.72 + 0.3✓ 1
A1 0.75 metres per month (increasing)✓ 1
Part (c)
M1 setting \(D'(t) = 0\)✓ 1
A1 t = 9.10 months✓ 1
R1 \(D'(t)\) changes from positive to negative at this value (e.g. D′(8) > 0 and D′(10) < 0), so the depth is a maximum✓ 1
Part (d)
M1 forming \(-0.03t^2 + 0.24t + 0.3 > 0.6\)✓ 1
A1 critical values 1.55 and 6.45✓ 1
A1 1.55 < t < 6.45✓ 1
Part (e)
M1 substituting t = 9.099… into D(t)✓ 1
A1 17.1 m✓ 1

Exam tip: Keep the unrounded value of t from your GDC when substituting back into D(t). Using t = 9.1 or 9.10 can be enough to shift the final 3 s.f. answer.

271IBMI-034-00-r2-03Gradients, tangents and rates of change Paper 2Medium12 marks

After heavy rainfall, the volume of water in a reservoir, V thousand m3, t days after the rain began is modelled by
\(V(t) = -0.02t^3 + 0.9t^2 + 5t + 200\), for \(0 \le t \le 40\).

(a)

Write down the volume of water in the reservoir when the rain began.

(1)
(b)

Find \(V'(t)\).

(2)
(c)

Find the rate at which the volume is changing when \(t = 10\), giving the units of your answer.

(2)
(d)

Find the average rate of change of the volume over the first 10 days.

(2)
(e)

Find the value of t at which the volume of water stops increasing.

(3)
(f)

Hence find the maximum volume of water in the reservoir.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
200 thousand m3✓ 1
Part (b)
Attempt to differentiate, with at least two terms correct✓ 1
\(V'(t) = -0.06t^2 + 1.8t + 5\)✓ 1
Part (c)
\(V'(10) = -6 + 18 + 5\)✓ 1
17 thousand m3 per day✓ 1
Part (d)
\(V(10) = 320\), so \(\frac{320 - 200}{10}\)✓ 1
12 thousand m3 per day✓ 1
Part (e)
Recognises the condition \(V'(t) = 0\)✓ 1
\(-0.06t^2 + 1.8t + 5 = 0\) solved using GDC✓ 1
\(t = 32.6\) days (rejecting the negative root)✓ 1
Part (f)
Substitutes their \(t = 32.559...\) into \(V(t)\)✓ 1
627 thousand m3✓ 1

Exam tip: Keep the full calculator value of t (32.559...) when substituting into V(t); using a rounded value can change the third significant figure.

272IBMI-034-06-r1-03Optimisation in context Paper 2Medium9 marks

A drinks company designs a closed cylindrical can with radius r cm and height h cm. The can must hold 330 cm3. The company wants to minimise the total surface area, A cm2, of metal used.

(a)

Show that \(A = 2\pi r^2 + \dfrac{660}{r}\).

(2)
(b)

Find \(\dfrac{\mathrm{d}A}{\mathrm{d}r}\).

(2)
(c)

Find the value of r that minimises the surface area.

(2)
(d)

Use the second derivative to justify that this value of r gives a minimum.

(2)
(e)

Find the minimum surface area.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(\pi r^2 h = 330 \Rightarrow h = \dfrac{330}{\pi r^2}\) (M1)✓ 1
\(A = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r\cdot\dfrac{330}{\pi r^2} = 2\pi r^2 + \dfrac{660}{r}\) (A1)✓ 1
Part (b)
\(4\pi r\) (A1)✓ 1
\(-\dfrac{660}{r^2}\) (A1)✓ 1
Part (c)
\(4\pi r - \dfrac{660}{r^2} = 0\) (or use of GDC on graph of derivative) (M1)✓ 1
\(r = 3.74\) cm (3.74494…) (A1)✓ 1
Part (d)
\(\dfrac{\mathrm{d}^2A}{\mathrm{d}r^2} = 4\pi + \dfrac{1320}{r^3}\) (A1)✓ 1
This is positive for \(r = 3.74\) (indeed for all \(r > 0\)), so the stationary point is a minimum (R1)✓ 1
Part (e)
\(A = 2\pi(3.74494)^2 + \dfrac{660}{3.74494} = 264\) cm2 (264.36…) (A1)✓ 1

Exam tip: Keep the unrounded value of r in your GDC memory when substituting; using r = 3.74 can shift the third significant figure of later answers.

273IBMI-034-06-r1-04Optimisation in context Paper 2Medium10 marks

A ferry makes a 120 km crossing at a constant speed of v km h−1. The running cost of the ferry is \((0.02v^3 + 160)\) dollars per hour, where the 160 dollars covers crew wages and the \(0.02v^3\) term covers fuel.

(a)

Show that the total cost, C dollars, of the crossing is \(C = 2.4v^2 + \dfrac{19200}{v}\).

(2)
(b)

Find \(\dfrac{\mathrm{d}C}{\mathrm{d}v}\).

(2)
(c)

Find the speed that minimises the cost of the crossing.

(2)
(d)

Find the minimum cost of the crossing.

(2)
(e)

A new regulation limits the ferry to a maximum speed of 12 km h−1. Explain why the ferry should now travel at exactly 12 km h−1, and find the cost of the crossing at this speed.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Time taken \(= \dfrac{120}{v}\) hours (M1)✓ 1
\(C = \dfrac{120}{v}(0.02v^3 + 160) = 2.4v^2 + \dfrac{19200}{v}\) (A1)✓ 1
Part (b)
\(4.8v\) (A1)✓ 1
\(-\dfrac{19200}{v^2}\) (A1)✓ 1
Part (c)
\(4.8v - \dfrac{19200}{v^2} = 0 \Rightarrow v^3 = 4000\) (M1)✓ 1
\(v = 15.9\) km h−1 (15.874…) (A1)✓ 1
Part (d)
Substituting \(v = 15.874…\) into C (M1)✓ 1
$C = $ $1814.29 (A1)✓ 1
Part (e)
\(\dfrac{\mathrm{d}C}{\mathrm{d}v} < 0\) for \(0 < v < 15.9\), so cost decreases as speed increases up to the limit; the cheapest permitted speed is the fastest, 12 km h−1 (R1)✓ 1
$C = 2.4(12)^2 + \dfrac{19200}{12} = $ $1945.60 (A1)✓ 1

Common error: When the optimum lies outside the permitted range, the answer is not the stationary point. Check whether the function is increasing or decreasing on the allowed interval and use the endpoint.

274IBMI-034-07-r1-03Integration and the trapezoidal rule Paper 2Medium11 marks

A gardener designs a flower bed against two perpendicular walls. One wall lies along the x-axis and the other lies along the y-axis. The curved edge of the bed is modelled by
\(f(x) = 4 + 1.5x - 0.25x^2\), for \(0 \le x \le 8\),
where x and y are measured in metres. The bed is the region enclosed by the curve, the x-axis and the y-axis.

A partially completed table of values is shown.
x: 0, 2, 4, 6, 8
f(x): 4, p, 6, q, 0

(a)

Find the value of p and the value of q.

(2)
(b)

Use the trapezoidal rule with the five values in the table to estimate the area of the flower bed.

(3)
(c)

Write down a definite integral for the area of the flower bed, and use it to find this area.

(2)
(d)

Find the percentage error in the estimate from part (b).

(2)
(e)

Explain, with reference to the shape of the curve, why the trapezoidal rule gives an underestimate in this context.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
p = 6 (A1)✓ 1
q = 4 (A1)✓ 1
Part (b)
Uses h = 2 (A1)✓ 1
Correct substitution: \(\frac{2}{2}[4 + 0 + 2(6 + 6 + 4)]\) (M1)✓ 1
36 m2 (A1)✓ 1
Part (c)
\(\int_0^8 (4 + 1.5x - 0.25x^2)\,dx\) (A1)✓ 1
37.3 m2 (accept 37.33…) (A1)✓ 1
Part (d)
Correct substitution: \(\left|\frac{36 - 37.33\ldots}{37.33\ldots}\right| \times 100\) (M1)✓ 1
3.57% (A1)✓ 1
Part (e)
The curve is concave down for \(0 \le x \le 8\) (it bends downwards, since \(f''(x) = -0.5 < 0\)) (R1)✓ 1
so the straight top of each trapezium lies below the curve, and each trapezium has less area than the strip under the curve (R1)✓ 1

Exam tip: For percentage error, divide by the exact value, not the estimate. Keep the unrounded exact area (37.333…) in your calculator until the final step.

275IBMI-034-06-r2-03Optimisation in context Paper 2Medium11 marks

A drinks company makes closed cylindrical aluminium cans that each hold 330 cm3. A can has base radius \(r\) cm and height \(h\) cm. The company wants to use the least possible area of aluminium, \(A\) cm2, for each can. Ignore the thickness of the metal.

(a)

Write down an expression for \(h\) in terms of \(r\).

(2)
(b)

Show that \(A = 2\pi r^2 + \frac{660}{r}\).

(2)
(c)

Find \(\frac{dA}{dr}\).

(2)
(d)

Find the value of \(r\) that minimizes \(A\). Justify that this value gives a minimum.

(3)
(e)

Find the minimum area of aluminium needed for one can.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(\pi r^2 h = 330\) M1✓ 1
\(h = \frac{330}{\pi r^2}\) A1✓ 1
Part (b)
Substitutes into \(A = 2\pi r^2 + 2\pi r h\): \(A = 2\pi r^2 + 2\pi r \times \frac{330}{\pi r^2}\) M1✓ 1
\(2\pi r \times \frac{330}{\pi r^2} = \frac{660}{r}\), so \(A = 2\pi r^2 + \frac{660}{r}\) A1✓ 1
Part (c)
\(4\pi r\) A1✓ 1
\(-660r^{-2}\) A1✓ 1
\(\frac{dA}{dr} = 4\pi r - \frac{660}{r^2}\)—
Part (d)
Sets \(4\pi r - \frac{660}{r^2} = 0\), giving \(r^3 = \frac{660}{4\pi}\) M1✓ 1
\(r = 3.74\) cm (3.7449…) A1✓ 1
\(\frac{d^2A}{dr^2} = 4\pi + \frac{1320}{r^3} > 0\) for \(r > 0\), so the value is a minimum (or a sign change of \(\frac{dA}{dr}\) from negative to positive) R1✓ 1
Part (e)
\(A = 2\pi(3.7449)^2 + \frac{660}{3.7449}\) M1✓ 1
\(A = 264\) cm2 (264.36…) A1✓ 1

Exam tip: Keep the unrounded value of \(r\) in your GDC memory when you substitute. Using \(r = 3.74\) still gives 264 here, but early rounding often costs the final accuracy mark.

276IBMI-034-06-r2-04Optimisation in context Paper 2Medium12 marks

A haulage company sends a lorry on a 360 km motorway delivery at a constant speed of \(v\) km h−1. While the lorry is being driven, it costs the company $54 per hour in driver's wages plus \(0.006v^2\) dollars per hour in fuel.

(a)

Write down an expression, in terms of \(v\), for the time in hours taken for the journey.

(1)
(b)

Show that the total cost of the journey, \(C\) dollars, is given by \(C = \frac{19440}{v} + 2.16v\).

(2)
(c)

Find \(\frac{dC}{dv}\).

(2)
(d)

Find the speed that minimizes the cost of the journey.

(2)
(e)

Find the minimum cost of the journey.

(2)
(f)

The legal speed limit for lorries on this motorway is 90 km h−1. Explain why the cheapest legal journey is made at a speed of 90 km h−1, and find the cost of the journey at this speed.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(\frac{360}{v}\) hours A1✓ 1
Part (b)
\(C = \frac{360}{v}(54 + 0.006v^2)\) M1✓ 1
\(= \frac{19440}{v} + 2.16v\) A1✓ 1
Part (c)
\(-19440v^{-2}\) A1✓ 1
\(+2.16\) A1✓ 1
Part (d)
Sets \(-\frac{19440}{v^2} + 2.16 = 0\), so \(v^2 = 9000\) M1✓ 1
\(v = 94.9\) km h−1 (94.868…) A1✓ 1
Part (e)
Substitutes \(v = 94.868…\) into \(C\) M1✓ 1
\(C = $409.83\) (409.831…) A1✓ 1
Part (f)
\(\frac{dC}{dv} < 0\) for \(v < 94.9\), so \(C\) is decreasing on \(0 < v \le 90\) R1✓ 1
The lowest legal cost is therefore at the highest legal speed, \(v = 90\) km h−1 A1✓ 1
\(C = \frac{19440}{90} + 2.16(90) = $410.40\) A1✓ 1

Exam tip: If the optimum lies outside the allowed domain, look at the sign of the derivative on the domain. The best allowed value is then at an endpoint.

277IBMI-034-02-r2-04Integration and the trapezoidal rule Paper 2Medium12 marks

A small company makes handmade lamps. The marginal cost of producing x lamps is modelled by
\(C'(x) = 0.03x^2 - 1.2x + 20\) dollars per lamp,
where \(C(x)\) is the total cost, in dollars, of producing x lamps. The company has fixed costs of 500 dollars, so \(C(0) = 500\).

(a)

Find an expression for \(C(x)\).

(4)
(b)

Find the total cost of producing 50 lamps.

(2)
(c)

Find the additional cost of increasing production from 50 lamps to 60 lamps.

(3)
(d)

Each lamp is sold for 35 dollars. Find the profit the company makes when it produces and sells 60 lamps.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Attempt to integrate \(C'(x)\)✓ 1
\(0.01x^3 - 0.6x^2 + 20x + c\)✓ 1
Uses \(C(0) = 500\) to find the constant✓ 1
\(C(x) = 0.01x^3 - 0.6x^2 + 20x + 500\)✓ 1
Part (b)
Substitutes x = 50: 1250 − 1500 + 1000 + 500✓ 1
1250 dollars✓ 1
Part (c)
Uses \(C(60) - C(50)\) or \(\int_{50}^{60} C'(x)\,dx\)✓ 1
Correct intermediate value: \(C(60) = 1700\) with \(C(50) = 1250\), or a correct evaluation of the integral, e.g. \([0.01x^3 - 0.6x^2 + 20x]_{50}^{60} = 1200 - 750\)✓ 1
450 dollars✓ 1
Part (d)
Revenue = 35 × 60 = 2100✓ 1
Profit = revenue − cost: 2100 − 1700✓ 1
400 dollars✓ 1

Common error: Forgetting the constant of integration loses the fixed cost. Every total cost found afterwards will then be 500 dollars too small, and the profit 500 dollars too large.

278IBMI-034-07-r2-03Integration and the trapezoidal rule Paper 2Medium11 marks

A landscape architect designs a lawn bounded by a straight path along the x-axis and a curved hedge. The hedge is modelled by

\(f(x) = 0.1x^2(9 - x), \quad 0 \le x \le 9,\)

where x and y are measured in metres. The architect records the following values.

x (m): 0, 1.5, 3, 4.5, 6, 7.5, 9
f(x) (m): 0, 1.6875, p, 9.1125, q, 8.4375, 0

(a)

Find the value of p and the value of q.

(2)
(b)

Use the trapezoidal rule with six strips to estimate the area of the lawn.

(3)
(c)

Write down a definite integral for the exact area of the lawn and find its value.

(3)
(d)

Explain, with reference to the shape of the hedge, why the estimate in part (b) is smaller than the value in part (c).

(1)
(e)

Find the percentage error in the trapezoidal estimate.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1: p = f(3) = 5.4✓ 1
A1: q = f(6) = 10.8✓ 1
Part (b)
M1: h = 1.5✓ 1
M1: correct substitution, \(\frac{1.5}{2}\left[0+0+2(1.6875+5.4+9.1125+10.8+8.4375)\right]\)✓ 1
A1: 53.2 m2 (53.15625)✓ 1
Part (c)
M1: \(\int_0^9 0.1x^2(9-x)\,dx\) with correct limits✓ 1
A1: correct integration or GDC use, e.g. \([0.3x^3 - 0.025x^4]_0^9\)✓ 1
A1: 54.7 m2 (54.675)✓ 1
Part (d)
R1: the curve is concave down for 3 < x ≤ 9 (most of the interval), so the chords forming the tops of the trapezia lie below the curve there. This underestimate outweighs the small overestimate on 0 ≤ x < 3, where the curve is concave up, so the net effect is an underestimate✓ 1
Part (e)
M1: \(\frac{54.675 - 53.15625}{54.675} \times 100\)✓ 1
A1: 2.78%✓ 1

Exam tip: Use the unrounded values from (b) and (c) when you find the percentage error. Rounding them first to 53.2 and 54.7 gives 2.74%, which is not accurate to 3 significant figures.

279IBMI-034-01-r2-03Optimisation in context Paper 2Medium11 marks

A drinks company designs a closed cylindrical can that must hold 500 cm3. The can has radius r cm and height h cm. To reduce costs, the company wants to use the least possible amount of metal, so it wants to minimise the total surface area, A cm2.

(a)

Write down an expression for h in terms of r.

(1)
(b)

Show that \(A = 2\pi r^2 + 1000r^{-1}\).

(2)
(c)

Find \(\frac{dA}{dr}\).

(2)
(d)

Find the value of r that minimises the surface area.

(2)
(e)

Find the minimum surface area.

(2)
(f)

Use the second derivative to justify that this value of r gives a minimum.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(h = \frac{500}{\pi r^2}\) A1✓ 1
Part (b)
States \(A = 2\pi r^2 + 2\pi r h\) M1✓ 1
Substitutes for h: \(2\pi r \times \frac{500}{\pi r^2} = \frac{1000}{r}\), giving the result A1✓ 1
Part (c)
4πr A1✓ 1
−1000r−2 A1✓ 1
Part (d)
Sets \(4\pi r - \frac{1000}{r^2} = 0\), so \(r^3 = \frac{250}{\pi}\), or solves dA/dr = 0 using a GDC M1✓ 1
r = 4.30 cm A1✓ 1
Part (e)
Substitutes r = 4.3013… into A M1✓ 1
A = 349 cm2 A1✓ 1
Part (f)
\(\frac{d^2A}{dr^2} = 4\pi + 2000r^{-3}\) A1✓ 1
This is positive for all r > 0 (it equals 12π ≈ 37.7 at r = 4.30), so the stationary point is a minimum R1✓ 1

Exam tip: Keep the unrounded value of r in your calculator when you find the minimum area. Substituting the rounded value 4.30 can give a small error in the final answer.

280IBMI-034-05-r1-03Gradients, tangents and rates of change Paper 2Medium12 marks

A company produces \(x\) hundred phone cases per week. The average cost per case, in US dollars, is modelled by
\(A(x) = 0.8x + 6 + \frac{45}{x}\), for \(x > 0\).

(a)

Find the average cost per case when 500 cases are produced in a week.

(2)
(b)

Find \(A'(x)\).

(2)
(c)

Find \(A'(5)\) and interpret your answer in context.

(3)
(d)

Solve \(A'(x) = 0\) and hence find the minimum average cost per case.

(3)
(e)

By considering the values of \(A'(7)\) and \(A'(8)\), justify that your answer to the previous part is a minimum.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Substituting \(x = 5\): \(0.8(5) + 6 + \frac{45}{5}\) M1✓ 1
$19.00 A1✓ 1
Part (b)
\(0.8\) A1✓ 1
\(-45x^{-2}\) (or \(-\frac{45}{x^2}\)) A1✓ 1
Part (c)
\(A'(5) = 0.8 - \frac{45}{25}\) M1✓ 1
\(A'(5) = -1\) A1✓ 1
When 500 cases are produced, the average cost per case is decreasing at about $1.00 per additional hundred cases produced R1✓ 1
Part (d)
\(0.8 - \frac{45}{x^2} = 0 \Rightarrow x^2 = 56.25\) M1✓ 1
\(x = 7.5\) (750 cases), since \(x > 0\) A1✓ 1
Minimum average cost $= A(7.5) = 6 + 6 + 6 = $ $18.00 A1✓ 1
Part (e)
\(A'(7) \approx -0.118\) and \(A'(8) \approx 0.0969\) M1✓ 1
Gradient changes from negative to positive at \(x = 7.5\), so the average cost has a minimum there R1✓ 1

Exam tip: Rewrite \(\frac{45}{x}\) as \(45x^{-1}\) before differentiating. With a GDC you can also check \(A'(x) = 0\) by graphing \(A(x)\) and finding its minimum, but show the derivative equation to earn the method mark.

281IBMI-034-02-r1-04Integration and the trapezoidal rule Paper 2Medium11 marks

Water flows into a reservoir during a storm. The rate of flow, r m3 per hour, is recorded every two hours, where t is the time in hours after the storm begins.

t (h): 0, 2, 4, 6, 8
r (m3 h−1): 120, 188, 222, 207, 157

A hydrologist proposes the model \(r(t) = 120 + 45t - 5t^2\), for \(0 \le t \le 8\).

(a)

Use the trapezoidal rule with the recorded data to estimate the volume of water that flowed into the reservoir in the first 8 hours.

(3)
(b)

Use the model to find the volume of water that flowed into the reservoir in the first 8 hours.

(3)
(c)

At the start of the storm the reservoir contains 5000 m3 of water. Use the model to find the time at which the reservoir contains 6000 m3. Assume no water leaves the reservoir.

(3)
(d)

Using your unrounded answers to parts 1 and 2, find the percentage difference between the trapezoidal estimate and the model's value, relative to the model's value.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
h = 2✓ 1
\(\frac{2}{2}[120 + 157 + 2(188 + 222 + 207)]\)✓ 1
= 1511 m3 (1510 to 3 s.f.)✓ 1
Part (b)
\(\int_0^8 (120 + 45t - 5t^2)\,dt\)✓ 1
\(= \left[120t + 22.5t^2 - \frac{5}{3}t^3\right]_0^8\) or GDC evaluation✓ 1
= 1546.67 m3 (1550 to 3 s.f.)✓ 1
Part (c)
\(5000 + \int_0^T r(t)\,dt = 6000\)✓ 1
\(120T + 22.5T^2 - \frac{5}{3}T^3 = 1000\), or the integral equation solved directly on a GDC✓ 1
T = 5.21 hours✓ 1
Part (d)
\(\frac{1546.67 - 1511}{1546.67} \times 100\)✓ 1
= 2.31%; the trapezoidal estimate is smaller than the model's value✓ 1

Exam tip: When the unknown is a limit of integration, set up the integral equation and solve it with the GDC rather than rearranging the cubic by hand. Carry unrounded values into later parts.

282IBMI3S-5-09-05Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 2Medium12 marks

An engineer studies the cross-section of a river 12 m wide. The depth of the river is measured every 2 m across the width.
Distance from bank, x (m): 0, 2, 4, 6, 8, 10, 12
Depth (m): 0, 1.1, 1.8, 2.4, 2.2, 1.3, 0

(a)

Use the trapezoidal rule with six intervals to estimate the area of the cross-section.

(3)
(b)

The water flows at a constant speed of 0.8 m s−1 through the cross-section. Calculate the volume of water passing the cross-section each second.

(2)
(c)

The engineer proposes the model d(x) = (12x − x2)/15 for the depth. Find the area of the cross-section predicted by this model, using integration.

(4)
(d)

Calculate the percentage difference between the trapezoidal estimate in (a) and the area in (c), as a percentage of the area in (c).

(2)
(e)

The engineer needs the trapezoidal estimate to be within 5% of the model area. State whether this requirement is met.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct strip width h = 2 in the trapezoidal rule✓ 1
A1 ½ × 2 × [0 + 0 + 2(1.1 + 1.8 + 2.4 + 2.2 + 1.3)]✓ 1
A1 17.6 m2✓ 1
Part (b)
M1 area × speed✓ 1
A1 14.1 m3 s−1 (14.08, ECF from (a))✓ 1
Part (c)
M1 attempt to integrate \(\int_0^{12} d(x)\,dx\)✓ 1
A1 antiderivative \(\frac{1}{15}\left(6x^2-\frac{x^3}{3}\right)\)✓ 1
M1 substitute limits 12 and 0✓ 1
A1 19.2 m2✓ 1
Part (d)
M1 \(\frac{19.2-17.6}{19.2}\times 100\) (ECF)✓ 1
A1 8.33%✓ 1
Part (e)
R1 not met, since 8.33% is greater than 5% (ECF)✓ 1

Exam tip: Check every number against the units asked for: area in m², flow in m³ s⁻¹, and a percentage difference relative to the stated reference value.

283IBMI3S-5-09b-08Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 2Medium12 marks

A cylindrical rainwater storage tank with a closed top has radius r metres and height h metres. Its volume must be 20 m3. The total outer surface area of the tank is A m2.

(a)

Show that h = 20 / (πr2).

(2)
(b)

Hence show that A = 2πr2 + 40/r.

(2)
(c)

Find dA/dr.

(2)
(d)

Find the value of r that gives the least surface area.

(2)
(e)

Use the second derivative to show that this value of r gives a minimum.

(2)
(f)

Find the least surface area of the tank.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 volume equation πr2h = 20✓ 1
A1 rearranging to h = 20 / (πr2)✓ 1
Part (b)
M1 A = 2πr2 + 2πrh and substitute for h✓ 1
A1 2πr × 20/(πr2) = 40/r, so A = 2πr2 + 40/r✓ 1
Part (c)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 dA/dr = 4πr − 40/r2✓ 1
Part (d)
M1 set dA/dr = 0, so r3 = 10/π✓ 1
A1 r = 1.47 m✓ 1
Part (e)
M1 d2A/dr2 = 4π + 80/r3✓ 1
R1 value 37.7 > 0 at r = 1.47, so a minimum✓ 1
Part (f)
M1 substitute r into A✓ 1
A1 40.8 m2✓ 1

Exam tip: In a 'show that' question, write the substitution step explicitly; the final line must match the printed result exactly.

284IBMI3S-5-01r8-09Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Medium13 marks

The depth of a swimming pool was measured every 4 m along its 24 m length, giving the table below. The pool is 12 m wide and its depth does not vary across the width.

Distance x (m)04812162024
Depth d (m)0.901.251.511.671.731.701.57

(a)

Use the trapezoidal rule to estimate the area of the longitudinal cross-section of the pool.

(3)
(b)

Hence estimate the volume of water in the full pool, giving your answer to the nearest m3.

(2)
(c)

Water enters at 25 m3 per hour. Using the unrounded volume 436.56 m3, find the time to fill the pool, in hours to 3 significant figures.

(2)
(d)

The depth is also modelled by d(x) = 0.9 + 0.1x − 0.003x2. Find the area under this model between x = 0 and x = 24, to 3 significant figures.

(2)
(e)

Using 36.576 for the model and 36.38 for the trapezoidal estimate, find the percentage by which the trapezoidal estimate is below the model's value. Give your answer to 3 significant figures.

(2)
(f)

Show that the trapezoidal rule must underestimate the area under this model, and explain why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule with h = 4✓ 1
M1 ½ × 4 × (0.90 + 1.57 + 2(1.25 + 1.51 + 1.67 + 1.73 + 1.70))✓ 1
A1 36.38 m2✓ 1
Part (b)
M1 36.38 × 12✓ 1
A1 437 m3✓ 1
Part (c)
M1 436.56 ÷ 25✓ 1
A1 17.5 hours✓ 1
Part (d)
M1 ∫024 (0.9 + 0.1x − 0.003x2) dx✓ 1
A1 36.6 m2 (exact 36.576)✓ 1
Part (e)
M1 (36.576 − 36.38) ÷ 36.576 × 100✓ 1
A1 0.536 %✓ 1
Part (f)
A1 d″(x) = −0.006 < 0, so the graph is concave (curves downwards)✓ 1
R1 each straight-line chord lies below the curve, so every trapezium has less area than the region under the curve✓ 1

Exam tip: Trapezoidal estimates fall below the true area for concave-down curves and above it for concave-up curves.

285IBMI3S-5-01r8-11Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Medium9 marks

A diver leaves a springboard at time t = 0. Her height above the water, in metres, is modelled by h(t) = 3 + vt − 4.9t2, where t is in seconds and v m s−1 is her initial upward speed. Her greatest height above the water is 3.9 m.

(a)

Write down an expression for h′(t).

(2)
(b)

Find the value of v.

(3)
(c)

Find the time at which she enters the water, giving your answer to 3 significant figures.

(2)
(d)

Using the unrounded time from (c), find her speed on entering the water, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 h′(t) = v − 9.8t✓ 1
Part (b)
M1 h′ = 0 gives t = v/9.8✓ 1
M1 3 + v2/19.6 = 3.9✓ 1
A1 v = 4.2✓ 1
Part (c)
M1 3 + 4.2t − 4.9t2 = 0 with t > 0✓ 1
A1 t = 1.32 s✓ 1
Part (d)
M1 |h′(t)| = |4.2 − 9.8t| at the unrounded t✓ 1
A1 8.74 m s−1✓ 1

Common error: Speed is the magnitude of the velocity: the derivative is negative as she falls, but the speed is positive.

286IBMI3S-5-01r13-07Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Medium15 marks

The number of people in a central library t hours after it opens is modelled by N(t) = −t3 + 9t2 + 48t, for 0 ≤ t ≤ 10.

(a)

Find the number of people in the library 5 hours after opening.

(1)
(b)

Find an expression for dN/dt.

(2)
(c)

Find the rate at which the number of people is changing when t = 2, and interpret your answer.

(2)
(d)

Find the equation of the tangent to the graph of N at t = 2, in the form N = mt + c.

(3)
(e)

Find the time at which the library is fullest, and the number of people then.

(4)
(f)

Find the time at which the number of people is increasing most rapidly, and state that greatest rate of increase.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 340✓ 1
Part (b)
M1 attempt to differentiate✓ 1
A1 dN/dt = −3t2 + 18t + 48✓ 1
Part (c)
M1 substitute t = 2 into dN/dt✓ 1
A1 72 people per hour (the number in the library is increasing at this rate)✓ 1
Part (d)
M1 N(2) = 124✓ 1
M1 use of N − 124 = 72(t − 2) (ECF from (c))✓ 1
A1 N = 72t − 20✓ 1
Part (e)
M1 set dN/dt = 0✓ 1
M1 solve −3t2 + 18t + 48 = 0 (the roots are t = 8 and t = −2)✓ 1
A1 t = 8✓ 1
A1 448 people✓ 1
Part (f)
M1 d2N/dt2 = −6t + 18 = 0 (or the vertex of the quadratic derivative)✓ 1
A1 t = 3✓ 1
A1 75 people per hour✓ 1

Exam tip: The greatest rate of change of a function occurs where its derivative is at a maximum, which is found by setting the second derivative to zero.

287IBMI3S-5-03r21-06Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 2Medium13 marks

An architect designs an open-topped steel planter for a roof garden. It is made from a square sheet of steel of side 1.2 m. A square of side x metres is cut from each corner and the sides are folded up to make a box with a square base and height x m.

(a)

Show that the volume of the planter is V = 4x3 − 4.8x2 + 1.44x m3.

(2)
(b)

Find dV/dx.

(2)
(c)

Find the value of x, with 0 < x < 0.6, that gives the greatest volume.

(3)
(d)

Find the greatest volume of the planter.

(2)
(e)

Use the second derivative to show that this volume is a maximum.

(2)
(f)

The planter is filled completely with soil costing €45 per m3. Find the cost of the soil.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 base side = 1.2 − 2x so V = x(1.2 − 2x)2✓ 1
A1 expansion (1.44 − 4.8x + 4x2)x to the given result✓ 1
Part (b)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 12x2 − 9.6x + 1.44✓ 1
Part (c)
M1 set dV/dx = 0✓ 1
A1 x = 0.2 or x = 0.6✓ 1
R1 reject x = 0.6 as the base side 1.2 − 2(0.6) = 0, or as outside the stated domain✓ 1
Part (d)
M1 substitute x = 0.2 into V✓ 1
A1 0.128 m3✓ 1
Part (e)
M1 d2V/dx2 = 24x − 9.6 evaluated at x = 0.2✓ 1
R1 value is −4.8 < 0, so a maximum✓ 1
Part (f)
M1 0.128 × 45✓ 1
A1 €5.76✓ 1

Exam tip: In optimisation questions always check the stated domain. A stationary point that gives a zero dimension must be rejected, and you should say why.

288IBMI3S-5-03r21-09Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 2Medium12 marks

A harbour authority plans an open-topped cylindrical storage tank of radius r metres and height h metres. The tank must hold exactly 8 m3 of water. The total outer surface area of the tank (curved surface and base) is A m2. The volume of a cylinder is πr2h and its curved surface area is 2πrh.

(a)

Show that A = πr2 + 16/r.

(3)
(b)

Find dA/dr.

(2)
(c)

Find the value of r that minimises the surface area. Give your answer to 3 significant figures.

(3)
(d)

Find the minimum surface area. Use the unrounded value of r and give your answer to 3 significant figures.

(2)
(e)

Show that, for this minimum, the height of the tank equals its radius.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 h = 8/(πr2) from the volume✓ 1
M1 A = πr2 + 2πrh✓ 1
A1 substitution and simplification to πr2 + 16/r✓ 1
Part (b)
M1 attempt to differentiate, with 16/r written as 16r−1✓ 1
A1 2πr − 16r−2✓ 1
Part (c)
M1 set dA/dr = 0✓ 1
A1 πr3 = 8, or r3 = 8/π✓ 1
A1 r = 1.37 m✓ 1
Part (d)
M1 substitute r = (8/π)1/3 = 1.3656… into A✓ 1
A1 17.6 m2✓ 1
Part (e)
M1 h = 8/(πr2) with πr3 = 8✓ 1
R1 h = πr3/(πr2) = r✓ 1

Exam tip: In a "show that" question, write every step. Marks are given for the substitution and the simplification, not only for the stated result.

289IBMI3S-5-02r24-05Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Medium11 marks

A coffee roaster designs an open-topped storage crate for green beans. The crate has a square base of side x metres and height h metres, and must have a volume of 0.75 m3. The base and four sides are made of sheet metal.

(a)

Show that the total area A m2 of sheet metal is given by A = x2 + 3 ÷ x.

(3)
(b)

Find dA/dx.

(2)
(c)

Find the value of x that minimises the area of sheet metal. Give your answer to 3 significant figures.

(2)
(d)

Show that this value of x gives a minimum.

(2)
(e)

The sheet metal costs €18 per m2. Find the minimum cost of the sheet metal. Use the unrounded value of x and give your answer to the nearest cent.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 x2h = 0.75, so h = 0.75 ÷ x2✓ 1
M1 A = x2 + 4xh✓ 1
A1 A = x2 + 3 ÷ x (shown)✓ 1
Part (b)
M1 rewriting 3 ÷ x as 3x−1✓ 1
A1 2x − 3x−2✓ 1
Part (c)
M1 2x − 3 ÷ x2 = 0✓ 1
A1 x = 1.14 m✓ 1
Part (d)
M1 d2A/dx2 = 2 + 6x−3 (or a sign change of dA/dx)✓ 1
R1 positive for x > 0 (gradient changes from negative to positive), so a minimum✓ 1
Part (e)
M1 18 × (1.1447…2 + 3 ÷ 1.1447…)✓ 1
A1 €70.76✓ 1

Exam tip: In an optimisation question, reduce the quantity to one variable using the constraint before differentiating, and check the end result against the context (here, a positive length).

290IBMI3S-5-02r24-06Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Medium9 marks

After a heater is switched on, the temperature T °C of an electronics box on a space probe is modelled by T(t) = −0.5t2 + 12t − 20, where t is the time in hours, for 0 ≤ t ≤ 20.

(a)

Find the rate of change of temperature when t = 3.

(2)
(b)

Find the maximum temperature of the box and the time at which it occurs. Justify that it is a maximum.

(4)
(c)

Find the first time at which the temperature of the box is 0 °C. Give your answer in hours to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 T′(t) = −t + 12✓ 1
A1 9 °C per hour✓ 1
Part (b)
M1 T′(t) = 0✓ 1
A1 t = 12 hours✓ 1
A1 T(12) = 52 °C✓ 1
R1 T″ = −1 < 0 (or the gradient changes from positive to negative)✓ 1
Part (c)
M1 −0.5t2 + 12t − 20 = 0✓ 1
M1 use of the quadratic formula or solver: t = 12 ± √104✓ 1
A1 t = 1.80 hours (the other root 22.2 is outside the domain)✓ 1

Exam tip: State the second-derivative or gradient-sign test explicitly when asked to justify a maximum; a value of the derivative equal to zero alone does not distinguish a maximum from a minimum.

291IBMI3S-5-02r24-07Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Medium12 marks

The speed of a space probe during a short engine burn is modelled by v(t) = 20 + 6t − 0.3t2, where v is in m s−1 and t is the time in seconds, for 0 ≤ t ≤ 10. The distance travelled during the burn is the area under the graph of v against t.

(a)

Find the values of v when t = 0, 2, 4, 6, 8 and 10.

(2)
(b)

Use the trapezoidal rule with 5 intervals to estimate the distance travelled.

(3)
(c)

Find the exact distance travelled by integration.

(3)
(d)

Find the percentage error of the estimate in part (b), taking the answer to part (c) as the exact value.

(2)
(e)

Explain why the trapezoidal rule gives an underestimate here.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substituting values of t into v(t)✓ 1
A1 20, 30.8, 39.2, 45.2, 48.8, 50✓ 1
Part (b)
M1 use of trapezoidal rule with width 2✓ 1
A1 (2 ÷ 2)[20 + 50 + 2(30.8 + 39.2 + 45.2 + 48.8)] (ECF from (a))✓ 1
A1 398 m✓ 1
Part (c)
M1 ∫010(20 + 6t − 0.3t2) dt✓ 1
A1 [20t + 3t2 − 0.1t3]010✓ 1
A1 400 m✓ 1
Part (d)
M1 (400 − 398) ÷ 400 × 100✓ 1
A1 0.5 % (ECF from (b) and (c))✓ 1
Part (e)
M1 v″(t) = −0.6 < 0, so the graph is concave down✓ 1
R1 the straight top of each trapezium lies below the curve, so the area is too small✓ 1

Common error: In the trapezoidal rule only the first and last ordinates are counted once; every middle ordinate is doubled.

292IBMI3S-5-02r24-09Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Medium10 marks

A coffee roaster's daily profit, P euros, from roasting x kg of beans is modelled by P(x) = −0.02x3 + 1.2x2 − 10x − 150, for 0 ≤ x ≤ 40.

(a)

Find P′(x).

(2)
(b)

Find the values of x for which P′(x) = 0. Give your answers to 3 significant figures.

(3)
(c)

Show that the profit is a local maximum when x = 35.3.

(2)
(d)

Find the maximum daily profit for 0 ≤ x ≤ 40, to the nearest euro. Justify that your answer is the greatest value in this domain.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiating at least two terms correctly✓ 1
A1 −0.06x2 + 2.4x − 10✓ 1
Part (b)
M1 −0.06x2 + 2.4x − 10 = 0✓ 1
A1 x = 4.72✓ 1
A1 x = 35.3✓ 1
Part (c)
M1 P″(x) = −0.12x + 2.4✓ 1
R1 P″(35.3) ≈ −1.84 < 0 (or the gradient changes from positive to negative)✓ 1
Part (d)
M1 P(35.275…) = 112.57…✓ 1
A1 €113✓ 1
R1 the endpoints give P(0) = −150 and P(40) = 90, both less than 112.57✓ 1

Exam tip: When the domain is restricted, compare the value at the stationary point with the values at the end points before claiming a greatest value.

293IBMI3S-5-01r27-06Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Medium12 marks

A company makes open-topped gift boxes with a square base of side x cm and height h cm. Each box is made from exactly 1200 cm2 of card, which forms the base and the four vertical sides.

(a)

Show that h = (1200 − x2) ÷ (4x).

(2)
(b)

Hence show that the volume of the box is V = 300x − x3/4.

(2)
(c)

Find dV/dx.

(2)
(d)

Find the value of x for which V is a maximum.

(2)
(e)

Show that this value of x gives a maximum volume.

(2)
(f)

Find the height of the box when x = 20.

(1)
(g)

Find the maximum volume of the box.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 x2 + 4xh = 1200✓ 1
A1 rearranging to the given result✓ 1
Part (b)
M1 V = x2h and substituting for h✓ 1
A1 simplifying to the given result✓ 1
Part (c)
M1 differentiating at least one term correctly✓ 1
A1 300 − 0.75x2✓ 1
Part (d)
M1 setting dV/dx = 0✓ 1
A1 x = 20 (reject x = −20)✓ 1
Part (e)
M1 d2V/dx2 = −1.5x✓ 1
R1 at x = 20 this is −30, which is negative, so a maximum✓ 1
Part (f)
A1 h = 800 ÷ 80 = 10 cm✓ 1
Part (g)
A1 4000 cm3 (ECF from (d) and (f))✓ 1

Exam tip: In a 'show that' question every step must appear, because the result is given. Use the given result in the following parts even if you could not prove it.

294IBMI3S-5-01r27-10Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Medium12 marks

The power output, S MW, of a solar farm is modelled by S(t) = −0.1t2 + 1.8t, for 0 ≤ t ≤ 18, where t is the time in hours after 04:00.

A table of values is shown.
Time t (hours): 0, 3, 6, 9, 12, 15, 18
Power S (MW): 0, a, b, 8.1, 7.2, 4.5, 0

(a)

Find the values of a and b.

(2)
(b)

Use the trapezoidal rule with six intervals to estimate the energy, in MWh, produced between t = 0 and t = 18.

(3)
(c)

Find the exact value of this energy by integration.

(3)
(d)

Calculate the percentage error in the estimate from part (b), using the exact value from part (c). Give your answer correct to 3 significant figures.

(2)
(e)

Explain why the trapezoidal rule gives an underestimate in this case.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 a = 4.5✓ 1
A1 b = 7.2✓ 1
Part (b)
M1 using the trapezoidal rule with width 3✓ 1
A1 (3/2)[0 + 0 + 2(4.5 + 7.2 + 8.1 + 7.2 + 4.5)]✓ 1
A1 94.5 MWh✓ 1
Part (c)
M1 integral of S between 0 and 18✓ 1
A1 antiderivative −t3/30 + 0.9t2✓ 1
A1 97.2 MWh✓ 1
Part (d)
M1 (97.2 − 94.5) ÷ 97.2 × 100 (ECF from (b) and (c))✓ 1
A1 2.78%✓ 1
Part (e)
R1 the graph of S is concave down (S″ = −0.2 < 0)✓ 1
R1 so each chord lies below the curve and each trapezium has less area than the region under the curve✓ 1

Exam tip: For a concave-down curve the trapezoidal rule underestimates, and for a concave-up curve it overestimates. Link your explanation to the shape of the graph.

295IBMI3S-5-02r27-08Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Medium13 marks

A theatre company builds a closed rectangular crate to transport scenery. The base of the crate measures x metres by 2x metres and the height is h metres. The volume of the crate must be 9 m3. The surface area of the crate is S m2.

(a)

Express h in terms of x.

(2)
(b)

Show that S = 4x2 + 27/x.

(3)
(c)

Find dS/dx.

(2)
(d)

Find the value of x that minimises S.

(2)
(e)

Show that this value of x gives a minimum surface area.

(2)
(f)

Paint for the crate costs €6.50 per square metre. Find the cost of painting the crate with the minimum surface area.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 2x2h = 9✓ 1
A1 h = 9 ÷ (2x2)✓ 1
Part (b)
M1 S = 2(2x2) + 2(xh) + 2(2xh)✓ 1
A1 S = 4x2 + 6xh✓ 1
A1 substituting for h to obtain the given result✓ 1
Part (c)
M1 differentiating at least one term correctly✓ 1
A1 8x − 27/x2✓ 1
Part (d)
M1 setting dS/dx = 0, giving x3 = 27/8✓ 1
A1 x = 1.5✓ 1
Part (e)
M1 d2S/dx2 = 8 + 54/x3✓ 1
R1 this is positive at x = 1.5, so a minimum✓ 1
Part (f)
M1 S(1.5) = 9 + 18 = 27 m2✓ 1
A1 27 × 6.50 = €175.50✓ 1

Common error: When listing the faces of a closed box, count every pair: the top and bottom (2 × 2x²) plus the four sides, which come in two different sizes.

296IBMI3S-5-02r27-14Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Medium12 marks

A volcano releases sulfur dioxide gas after an eruption. The emission rate, r kilotonnes per day, was measured during the first 12 days.

Time t (days): 0, 1.5, 3, 4.5, 6, 7.5, 9, 10.5, 12
Rate r (kt per day): 2.0, 3.8, 5.2, 7.1, 8.4, 7.9, 6.0, 3.7, 1.6

(a)

Use the trapezoidal rule with the values at t = 0, 3, 6, 9 and 12 only to estimate the total mass of gas released.

(3)
(b)

Use the trapezoidal rule with all nine values to estimate the total mass of gas released.

(3)
(c)

Later analysis showed that the true total mass released was 66.0 kt. Find the percentage error in each estimate, correct to 3 significant figures, and state which estimate is better.

(3)
(d)

Use the better estimate to find the mean emission rate over the 12 days, correct to 3 significant figures.

(2)
(e)

Explain why using narrower intervals usually improves a trapezoidal estimate.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 using the trapezoidal rule with width 3✓ 1
A1 (3/2)[2.0 + 1.6 + 2(5.2 + 8.4 + 6.0)]✓ 1
A1 64.2 kt✓ 1
Part (b)
M1 using the trapezoidal rule with width 1.5✓ 1
A1 (1.5/2)[2.0 + 1.6 + 2(3.8 + 5.2 + 7.1 + 8.4 + 7.9 + 6.0 + 3.7)]✓ 1
A1 65.85 kt✓ 1
Part (c)
M1 (66.0 − 64.2) ÷ 66.0 × 100 = 2.73%✓ 1
A1 (66.0 − 65.85) ÷ 66.0 × 100 = 0.227%✓ 1
A1 the nine-value estimate is better, because its error is smaller✓ 1
Part (d)
M1 65.85 ÷ 12✓ 1
A1 5.49 kt per day✓ 1
Part (e)
R1 narrower trapezia follow the curve more closely, so less area is missed or double-counted✓ 1

Exam tip: Doubling the number of intervals halves the width, so the interior values in the formula change. Check which ordinates are doubled each time.

297IBMI3S-5-03r27-04Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 2Medium13 marks

In a video game, the height of a hill above a flat reference level is modelled by h(x) = 0.05x3 − 0.6x2 + 1.8x + 2, for 0 ≤ x ≤ 8, where x is the horizontal distance in metres from the start of the hill and h is in metres.

(a)

Find h′(x).

(2)
(b)

Find the coordinates of the local maximum point and the local minimum point on the hill.

(3)
(c)

Find the gradient of the hill at x = 1.

(1)
(d)

Find the equation of the tangent to the hill at x = 1. Give your answer in the form y = mx + c.

(3)
(e)

Find the value of x at which the hill is steepest downhill, and state the gradient there.

(2)
(f)

There are two points on the hill where the gradient is 0.75. One is at x = 1. Find the x-coordinate of the other.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least two terms correct)✓ 1
A1 h′(x) = 0.15x2 − 1.2x + 1.8✓ 1
Part (b)
M1 set h′(x) = 0✓ 1
A1 x = 2 and x = 6✓ 1
A1 local maximum (2, 3.6) and local minimum (6, 2.0)✓ 1
Part (c)
A1 0.75✓ 1
Part (d)
M1 height at x = 1 is 3.25✓ 1
M1 use of y − 3.25 = 0.75(x − 1) or equivalent✓ 1
A1 y = 0.75x + 2.5 (ECF from (c))✓ 1
Part (e)
M1 attempt to find the minimum of h′(x)✓ 1
A1 x = 4, gradient −0.6✓ 1
Part (f)
M1 0.15x2 − 1.2x + 1.8 = 0.75✓ 1
A1 x = 7✓ 1

Exam tip: A tangent needs both the gradient and a point on the curve. Find the y-value from the original function, not from the derivative.

298IBMI3S-5-03r27-07Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 2Medium11 marks

The concentration of an antibiotic in a patient's bloodstream is modelled by C(t) = −0.05t3 + 0.6t2, where C is in mg/L and t is the time in hours after the dose, for 0 ≤ t ≤ 10.

(a)

Find an expression for C′(t).

(2)
(b)

Find the rate of change of concentration at t = 2, and interpret your answer.

(2)
(c)

Find the maximum concentration during this period.

(2)
(d)

The concentration equals 10 mg/L at two times during this period. Find both times, giving your answers to 3 significant figures.

(3)
(e)

Find the rate of change of concentration at the later of these two times.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 C′(t) = −0.15t2 + 1.2t✓ 1
Part (b)
M1 substitute t = 2✓ 1
A1 1.80 mg/L per hour; the concentration is rising at that time✓ 1
Part (c)
M1 C′(t) = 0 giving t = 8✓ 1
A1 12.8 mg/L✓ 1
Part (d)
M1 equation −0.05t3 + 0.6t2 = 10✓ 1
A1 t = 5.58✓ 1
A1 t = 10.0✓ 1
Part (e)
M1 substitute t = 10 into C′(t)✓ 1
A1 −3.00 mg/L per hour✓ 1

Common error: A negative rate of change means the quantity is decreasing. Do not drop the sign when interpreting.

299IBMI3S-5-03r27-11Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 2Medium9 marks

A manufacturer makes an open tray from a rectangular sheet of card measuring 24 cm by 18 cm. Squares of side x cm are cut from each corner and the sides are folded up.

(a)

Show that the volume of the tray, V cm3, is V = 4x3 − 84x2 + 432x.

(2)
(b)

Write down the range of possible values of x.

(1)
(c)

Find the value of x that maximises the volume. Give your answer to 3 significant figures.

(4)
(d)

Find the maximum volume, to 3 significant figures. Use the unrounded value of x.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 V = x(24 − 2x)(18 − 2x)✓ 1
A1 expansion leading to 4x3 − 84x2 + 432x✓ 1
Part (b)
A1 0 < x < 9✓ 1
Part (c)
M1 attempt to differentiate✓ 1
A1 dV/dx = 12x2 − 168x + 432✓ 1
M1 set dV/dx = 0✓ 1
A1 x = 3.39 (the other solution, 10.6, is outside the range)✓ 1
Part (d)
M1 substitute the unrounded value of x into V✓ 1
A1 655 cm3✓ 1

Common error: Reject stationary points that lie outside the physical domain, such as a cut larger than half the shortest side.

300IBMI3S-5-03r28-10Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Medium14 marks

The depth d metres of water in a tidal pool is modelled by
d(t) = 3.1 + 1.8 sin(0.5t),   0 ≤ t ≤ 12,
where t is the time in hours after midnight and the argument of sine is in radians.

(a)

Write down the maximum and minimum depths of water in the pool.

(2)
(b)

Find d′(t).

(2)
(c)

Find the rate at which the depth is changing at 02:00. Give your answer to 3 significant figures.

(2)
(d)

Find the first time at which the depth is decreasing at a rate of 0.5 m h−1. Give your answer in hours to 3 significant figures.

(3)
(e)

Find the depth of water at the time found in part (d). Use the unrounded value of t and give your answer to 3 significant figures.

(2)
(f)

Find the length of time, in hours, for which the depth exceeds 4 m during 0 ≤ t ≤ 12. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 maximum depth 4.9 m✓ 1
A1 minimum depth 1.3 m✓ 1
Part (b)
M1 attempt to differentiate, chain rule✓ 1
A1 d′(t) = 0.9 cos(0.5t)✓ 1
Part (c)
M1 substitutes t = 2 into d′(t) (ECF from (b))✓ 1
A1 0.486 m h−1 (increasing)✓ 1
Part (d)
M1 sets 0.9 cos(0.5t) = −0.5 (ECF from (b))✓ 1
A1 0.5t = 2.159… (the smallest positive solution)✓ 1
A1 t = 4.32 hours✓ 1
Part (e)
M1 substitutes the unrounded t into d(t) (ECF from (d))✓ 1
A1 4.60 m✓ 1
Part (f)
M1 solves 3.1 + 1.8 sin(0.5t) = 4✓ 1
A1 t = 1.05 and t = 5.24 (π/3 and 5π/3)✓ 1
A1 4.19 hours✓ 1

Exam tip: With trigonometric models, check that the calculator is in radian mode and remember that the first solution found may not be the only one in the given interval.

301IBMI3S-5-02r30-12Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 2Medium10 marks

A workshop makes n custom bicycle frames per week. The total weekly cost, in euros, is modelled by \(C = 200 + 0.5n^2\), for n > 0.

(a)

Show that the average cost per frame is \(A = \frac{200}{n} + 0.5n\).

(1)
(b)

Find \(\frac{dA}{dn}\).

(2)
(c)

Find the number of frames per week that gives the least average cost.

(2)
(d)

Find the least average cost per frame.

(1)
(e)

Show that this is a minimum.

(2)
(f)

Each frame sells for €32. Find the weekly profit when 20 frames are made.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 A = C/n = (200 + 0.5n2)/n, which simplifies to the given expression✓ 1
Part (b)
M1 attempt to differentiate 200n−1✓ 1
A1 \(-200n^{-2} + 0.5\)✓ 1
Part (c)
M1 set \(\frac{dA}{dn} = 0\)✓ 1
A1 n2 = 400, so n = 20✓ 1
Part (d)
A1 A(20) = 10 + 10 = €20✓ 1
Part (e)
M1 \(\frac{d^2A}{dn^2} = 400n^{-3}\)✓ 1
R1 positive when n = 20, so a minimum✓ 1
Part (f)
M1 20 × 32 − C(20) = 640 − 400✓ 1
A1 €240✓ 1

Exam tip: Average cost is total cost divided by the number made. Divide every term of the cost function, then differentiate.

302IBMI3S-5-01r31-13Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Medium14 marks

Calculator

A robotics company makes x hundred drones per month, for 0 ≤ x ≤ 7. Its monthly profit is modelled by P(x) = −x3 + 9x2 − 15x − 10, where P is measured in thousands of euros.

(a)

Find P(0) and explain what this value means in context.

(2)
(b)

Find P′(x).

(2)
(c)

Find the values of x for which P′(x) = 0.

(3)
(d)

Find P″(x) and use it to determine the nature of each stationary point.

(3)
(e)

Use your calculator to find the range of values of x for which the company makes a profit. Give your answers to 3 significant figures.

(2)
(f)

Find the maximum monthly profit, in thousands of euros.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 P(0) = −10✓ 1
R1 the company makes a loss of €10 000 (fixed costs) when it makes no drones✓ 1
Part (b)
A1 −3x2 + 18x✓ 1
A1 − 15✓ 1
Part (c)
M1 setting −3x2 + 18x − 15 = 0 (ECF from (b))✓ 1
A1 x = 1✓ 1
A1 x = 5✓ 1
Part (d)
A1 P″(x) = −6x + 18✓ 1
A1 at x = 1, P″ = 12 > 0, so a local minimum at x = 1✓ 1
A1 P″(5) = −12 < 0, so a local maximum at x = 5✓ 1
Part (e)
M1 solving P(x) = 0 or sketching the graph of P and finding where it is above the x-axis✓ 1
A1 3.08 < x < 6.42✓ 1
Part (f)
M1 evaluating P(5) (ECF from (c) and (d))✓ 1
A1 15 thousand euros (€15 000)✓ 1

Exam tip: Stationary points are found from the first derivative, but the second derivative (or a sign change of the gradient) is needed to justify whether each is a maximum or a minimum.

303IBMI3S-5-01r32-06Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Medium12 marks

A mountaineer follows a ridge route. Her altitude, h metres above sea level, t hours after setting out is modelled by
h(t) = 2.5t3 − 37.5t2 + 135t + 3200, for 0 ≤ t ≤ 10.

(a)

Find h′(t).

(2)
(b)

There are two times at which the mountaineer's altitude is momentarily neither increasing nor decreasing. Find these times, to 3 significant figures.

(3)
(c)

Find the altitude of the mountaineer at each of the times found in (b). Give each answer to the nearest metre.

(4)
(d)

Between these two times she is losing height. Find the greatest rate at which she loses height, and the time at which this occurs.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate (at least two terms correct)✓ 1
A1 h′(t) = 7.5t2 − 75t + 135✓ 1
Part (b)
M1 set h′(t) = 0✓ 1
A1 t = 2.35 hours✓ 1
A1 t = 7.65 hours✓ 1
Part (c)
M1 substitute a time from (b) into h(t) (use unrounded values; ECF from (b))✓ 1
A1 3343 m (at t = 2.35…)✓ 1
M1 substitute the second time into h(t)✓ 1
A1 3157 m (at t = 7.65…)✓ 1
Part (d)
M1 recognise the greatest rate of descent is the minimum of h′(t) (vertex of the quadratic or use of technology)✓ 1
A1 t = 5 hours✓ 1
A1 52.5 m h−1✓ 1

Exam tip: A turning point of h occurs where h′ = 0, but the quickest descent occurs at the lowest point of the graph of h′, not of h.

304IBMI3S-5-01r32-09Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Medium12 marks

A telecoms company launches a new broadband tariff. Its monthly profit is modelled by
P(x) = −0.4x3 + 5x2 − 8x − 15, for 0 ≤ x ≤ 12,
where x is the number of customers in thousands and P is the profit in thousands of euros (€).

(a)

Write down the value of P(0) and interpret it in context.

(1)
(b)

Find P′(x).

(2)
(c)

Solve P′(x) = 0, giving your answers to 3 significant figures.

(2)
(d)

Show that the greater of the two values found in (c) gives a local maximum of P.

(2)
(e)

Find the numbers of customers, in thousands, for which the company makes a profit. Give your answers to 3 significant figures.

(3)
(f)

Find the maximum monthly profit, in thousands of euros, to 3 significant figures. Use the unrounded value of x.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 P(0) = −15: a loss of €15 000 when there are no customers (fixed costs)✓ 1
Part (b)
M1 attempt to differentiate (at least two terms correct)✓ 1
A1 P′(x) = −1.2x2 + 10x − 8✓ 1
Part (c)
M1 set P′(x) = 0 and solve (formula or technology)✓ 1
A1 x = 0.896 and x = 7.44✓ 1
Part (d)
M1 find P″(x) = −2.4x + 10 and evaluate at x = 7.44 (or consider the sign of P′ either side)✓ 1
R1 P″ = −7.85… < 0 (or P′ changes from positive to negative), so a maximum✓ 1
Part (e)
M1 attempt to solve P(x) > 0 or P(x) = 0 (graph or technology)✓ 1
A1 lower limit 3.41✓ 1
A1 upper limit 10.2, so 3.41 < x < 10.2✓ 1
Part (f)
M1 substitute x = 7.4369… into P(x) (ECF from (c))✓ 1
A1 €37.5 thousand (accept 37 500)✓ 1

Exam tip: To justify a maximum, use either the second derivative being negative or a change of sign of the first derivative; stating 'the graph shows it' is not enough.

305IBMI3S-5-01r32-13Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Medium14 marks

A food-processing plant fills a tank with syrup. The rate of flow into the tank is modelled by
r(t) = 0.8 + 0.4t − 0.03t2, for 0 ≤ t ≤ 12,
where r is in litres per minute and t is the time in minutes since filling began. The tank is empty when t = 0.

(a)

Show that the volume of syrup in the tank after t minutes is V(t) = 0.8t + 0.2t2 − 0.01t3 litres.

(3)
(b)

Find the volume of syrup in the tank after 12 minutes.

(2)
(c)

The tank is considered 'full' at 15 litres. Find the time taken to fill it, to 3 significant figures.

(2)
(d)

Find the greatest rate of flow into the tank, and the time at which it occurs. Give your answers to 3 significant figures.

(3)
(e)

Use the trapezoidal rule with three intervals of equal width to estimate the volume of syrup delivered in the first 12 minutes. Give your answer to 3 significant figures.

(3)
(f)

Explain why your answer to (e) is less than your answer to (b).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to integrate r(t)✓ 1
A1 0.8t + 0.2t2 − 0.01t3 + C✓ 1
R1 V(0) = 0 so C = 0✓ 1
Part (b)
M1 substitute t = 12 into V(t)✓ 1
A1 21.12 litres (accept 21.1)✓ 1
Part (c)
M1 set V(t) = 15 and solve (technology)✓ 1
A1 8.45 minutes✓ 1
Part (d)
M1 set r′(t) = 0 (0.4 − 0.06t = 0) or use technology✓ 1
A1 t = 6.67 minutes✓ 1
A1 2.13 litres per minute✓ 1
Part (e)
M1 interval width 4 and values r(0), r(4), r(8), r(12)✓ 1
A1 values 0.8, 1.92, 2.08, 1.28 used correctly: (4/2)(0.8 + 1.28 + 2(1.92 + 2.08))✓ 1
A1 20.2 litres (20.16)✓ 1
Part (f)
R1 the graph of r is concave down (negative coefficient of t2), so the trapezoidal rule underestimates the area✓ 1

Exam tip: Integrating a rate of flow gives a volume; the constant of integration is fixed by a condition such as 'the tank is empty at t = 0'.

306IBMI3S-5-01r33-12Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 2Medium14 marks

Sulfur dioxide is emitted from a volcanic vent. The emission rate, E tonnes per hour, is modelled by E(t) = 20 + 6t − 0.5t2, where t is the time in hours, 0 ≤ t ≤ 10. Readings are taken every 2 hours.

(a)

Complete the table of values of E.

t0246810
E2030………30

(2)
(b)

Use the trapezoidal rule with five intervals to estimate the total mass of sulfur dioxide emitted in the 10 hours.

(3)
(c)

Use integration to find the exact total mass emitted in the 10 hours.

(3)
(d)

Find the percentage error in the trapezoidal estimate, using the exact value from part (c).

(2)
(e)

Explain why the trapezoidal rule gives an underestimate for this function.

(2)
(f)

Find the mean emission rate over the 10 hours, using the exact total. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 E(4) = 36✓ 1
A1 E(6) = 38 and E(8) = 36✓ 1
Part (b)
M1 trapezoidal rule structure with strip width 2✓ 1
A1 correct substitution ½ × 2 × (20 + 30 + 2(30 + 36 + 38 + 36))✓ 1
A1 330 tonnes✓ 1
Part (c)
M1 ∫010 (20 + 6t − 0.5t2) dt✓ 1
A1 antiderivative 20t + 3t2 − t3/6✓ 1
A1 333⅓ tonnes (333.33)✓ 1
Part (d)
M1 (333.33 − 330)/333.33 × 100✓ 1
A1 1.00%✓ 1
Part (e)
R1 the graph of E is concave (curves downwards, E″ = −1 < 0)✓ 1
R1 so the straight-line tops of the trapezoids lie below the curve✓ 1
Part (f)
M1 divide total by 10✓ 1
A1 33.3 tonnes per hour✓ 1

Exam tip: The trapezoidal rule underestimates the area under a concave curve and overestimates under a convex one; state the curvature when asked to explain the direction of error.

307IBMI3S-5-01r33-14Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 2Medium9 marks

The depth of the seabed, d metres, at a distance x metres from a quay is modelled by d(x) = 0.002x3 − 0.18x2 + 4.8x, for 0 ≤ x ≤ 60.

(a)

Find d′(x).

(2)
(b)

Show that the seabed is level (d′(x) = 0) at x = 20 and at x = 40.

(3)
(c)

Use the second derivative to classify each stationary point.

(2)
(d)

Find the depth of the seabed at the local minimum.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 0.006x2 − 0.36x + 4.8✓ 1
Part (b)
M1 set d′(x) = 0✓ 1
M1 attempt to solve (factorise 0.006(x − 20)(x − 40) or use GDC)✓ 1
A1 x = 20 and x = 40✓ 1
Part (c)
M1 d″(x) = 0.012x − 0.36✓ 1
A1 d″(20) = −0.12 < 0 so local maximum; d″(40) = 0.12 > 0 so local minimum✓ 1
Part (d)
M1 substitute x = 40 into d(x)✓ 1
A1 32 m✓ 1

Exam tip: In a show-that question, solve the equation rather than only substituting; solving also proves there are no other roots.

308IBMI3S-5-02r33-08Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; 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Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Medium10 marks

A conservator makes an open tray from a rectangular aluminium sheet 60 cm by 40 cm. A square of side x cm is cut from each corner and the sides are folded up to form a tray of depth x cm.

(a)

Show that the volume of the tray is V = 4x3 − 200x2 + 2400x cm3.

(2)
(b)

State the possible values of x for this tray.

(1)
(c)

Find dV/dx.

(2)
(d)

Find the value of x that gives the maximum volume. Give your answer to 3 significant figures.

(3)
(e)

Justify that this value gives a maximum, and find the maximum volume to 3 significant figures. Use the unrounded value of x.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 base length 60 − 2x and width 40 − 2x✓ 1
A1 V = x(60 − 2x)(40 − 2x) expanded to 4x3 − 200x2 + 2400x (AG)✓ 1
Part (b)
A1 0 < x < 20✓ 1
Part (c)
M1 attempt to differentiate✓ 1
A1 12x2 − 400x + 2400✓ 1
Part (d)
M1 set dV/dx = 0✓ 1
M1 solve the quadratic (formula or GDC)✓ 1
A1 x = 7.85 (reject 25.5 which is outside the domain)✓ 1
Part (e)
R1 d2V/dx2 = 24x − 400 is negative at x = 7.85 (or sign change of gradient)✓ 1
A1 8450 cm3✓ 1

Common error: A quadratic from dV/dx = 0 often has a root outside the physical domain; always check it against the limits on x.

309IBMI3S-5-01r36-04Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; 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The main cable of a footbridge hangs between two towers 80 m apart. The height of the cable above the deck is modelled by h(x) = 0.005x2 − 0.4x + 12 metres, where x is the horizontal distance in metres from the left tower, 0 ≤ x ≤ 80.

(a)

Find the gradient of the cable when x = 20.

(2)
(b)

Find the lowest height of the cable above the deck.

(3)
(c)

Find the equation of the tangent to the cable at x = 20. Give your answer in the form y = mx + c.

(3)
(d)

This tangent line is extended until it meets the level of the deck (y = 0). Find the value of x at this point.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate: h′(x) = 0.01x − 0.4✓ 1
A1 −0.2✓ 1
Part (b)
M1 set h′(x) = 0✓ 1
A1 x = 40✓ 1
A1 h(40) = 4 m✓ 1
Part (c)
A1 h(20) = 6✓ 1
M1 use y − 6 = −0.2(x − 20)✓ 1
A1 y = −0.2x + 10✓ 1
Part (d)
M1 −0.2x + 10 = 0 (or ECF from (c))✓ 1
A1 x = 50✓ 1

Common error: The gradient at a point is found by substituting x into the derivative, not into the original function.

310IBMI3S-5-01r36-05Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; 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A small roastery produces x kg of coffee per day, 20 ≤ x ≤ 120. The daily cost of production in euros is modelled by C(x) = 0.05x2 + 4x + 180. The average cost per kilogram is A(x) = C(x) ÷ x.

(a)

Show that A(x) = 0.05x + 4 + 180x−1.

(2)
(b)

Find A′(x).

(2)
(c)

Hence find the daily production that minimises the average cost per kilogram, and state this minimum average cost.

(3)
(d)

Justify that this value of x gives a minimum.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 divide each term of C(x) by x✓ 1
A1 0.05x + 4 + 180/x, which equals the given expression✓ 1
Part (b)
M1 attempt to differentiate, including the term in x−1✓ 1
A1 A′(x) = 0.05 − 180x−2✓ 1
Part (c)
M1 set A′(x) = 0✓ 1
A1 x = 60 kg✓ 1
A1 minimum average cost €10 per kg✓ 1
Part (d)
M1 attempt to find A″(x) = 360x−3 or a sign change of A′ either side of 60✓ 1
R1 A″(60) > 0 (or A′ changes from negative to positive), so it is a minimum✓ 1

Exam tip: To minimise a quotient such as average cost, first rewrite it as separate terms so each can be differentiated with the power rule.

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Rainwater drains from a bridge deck into a storage tank. During a storm the inflow rate is I(t) = 12t − t2 litres per minute, 0 ≤ t ≤ 10, where t is the time in minutes after the storm begins. Water leaves the tank through a pump at a constant 20 litres per minute. At t = 0 the tank contains 50 litres.

(a)

Find the two times at which the inflow rate equals the outflow rate.

(3)
(b)

Find the net increase in the volume of water in the tank between t = 2 and t = 10.

(3)
(c)

Find the minimum volume of water in the tank during the storm. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 set 12t − t2 = 20✓ 1
A1 t = 2✓ 1
A1 t = 10✓ 1
Part (b)
M1 attempt to integrate I(t) − 20 from 2 to 10✓ 1
A1 [6t2 − t3/3 − 20t] from 2 to 10✓ 1
A1 85.3 litres (accept 256/3)✓ 1
Part (c)
R1 outflow exceeds inflow for 0 < t < 2, so the minimum is at t = 2✓ 1
M1 volume = 50 + ∫02(I(t) − 20) dt = 50 + (24 − 8/3 − 40)✓ 1
A1 31.3 litres✓ 1

Exam tip: A volume in a tank is the starting amount plus the integral of (inflow − outflow); a minimum occurs where the net rate changes from negative to positive.

312IBMI3S-5-04r36-03Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 2Medium10 marks

A cooling system fails in a server room. The temperature T °C of the room is modelled by T = 20 + 0.5t2 − 0.02t3, for 0 ≤ t ≤ 20, where t is the time in minutes after the failure.

(a)

Find dT/dt.

(2)
(b)

Find the temperature, and the rate at which it is rising, when t = 10.

(2)
(c)

Find the two times at which the temperature is rising at 4 °C per minute.

(3)
(d)

Find the greatest rate at which the temperature rises, and the time when this occurs. Give the rate correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 dT/dt = t − 0.06t2✓ 1
Part (b)
A1 T = 50 °C✓ 1
A1 rate = 4 °C per minute✓ 1
Part (c)
M1 set t − 0.06t2 = 4✓ 1
A1 t = 6.67 minutes (6.666…)✓ 1
A1 t = 10 minutes✓ 1
Part (d)
M1 set the second derivative 1 − 0.12t = 0 (or find the maximum of dT/dt on a GDC)✓ 1
A1 t = 8.33 minutes✓ 1
A1 greatest rate = 4.17 °C per minute✓ 1

Exam tip: A rate of change is a derivative; the fastest rate is a maximum of that derivative, not of the original function.

313IBMI3S-5-04r36-04Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 2Medium10 marks

A cycle route climbs and then descends a hill. Its height h metres above the start is modelled by h = 0.12x − 0.0004x2, where x is the horizontal distance in metres from the start, for 0 ≤ x ≤ 300.

(a)

Find dh/dx.

(2)
(b)

Find the gradient of the route at x = 100, and the angle it makes with the horizontal, correct to 3 significant figures.

(3)
(c)

Find the height of the highest point of the route.

(3)
(d)

A sign warns of a downhill section with gradient −0.04. Find the distance from the start at which the gradient is −0.04.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiate✓ 1
A1 dh/dx = 0.12 − 0.0008x✓ 1
Part (b)
A1 gradient 0.04✓ 1
M1 angle = tan−1(0.04)✓ 1
A1 2.29°✓ 1
Part (c)
M1 set dh/dx = 0✓ 1
A1 x = 150✓ 1
A1 h = 9 m✓ 1
Part (d)
M1 set 0.12 − 0.0008x = −0.04✓ 1
A1 x = 200 m✓ 1

Common error: A negative gradient means the route is descending; do not drop the minus sign when setting the derivative equal to a downhill gradient.

314IBMI3S-5-04r36-10Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 2Medium9 marks

A backup battery in a data centre is charged and discharged. Its net charging rate is modelled by R(t) = −t2 + 8t − 12 kWh per hour, for 0 ≤ t ≤ 8, where t is the time in hours. At t = 0 the battery holds 50 kWh.

(a)

Find the times at which R(t) = 0.

(2)
(b)

State the interval during which the battery is discharging, giving a reason.

(2)
(c)

Find the amount of charge in the battery at t = 2, correct to 3 significant figures.

(3)
(d)

Find the amount of charge in the battery at t = 8. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 solve −t2 + 8t − 12 = 0✓ 1
A1 t = 2 and t = 6✓ 1
Part (b)
A1 0 ≤ t < 2 and 6 < t ≤ 8✓ 1
R1 the net charging rate R is negative on these intervals✓ 1
Part (c)
M1 50 + the integral of R(t) from 0 to 2✓ 1
A1 −t3/3 + 4t2 − 12t evaluated, giving −10.666…✓ 1
A1 39.3 kWh✓ 1
Part (d)
M1 50 + the integral of R(t) from 0 to 8✓ 1
A1 39.3 kWh✓ 1

Exam tip: Starting amount plus the integral of the net rate gives the amount at a later time; do not forget the starting amount.

315IBMI3S-5-01r39-08Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 2Medium13 marks

A marine biology centre is designing an open-topped cylindrical tank of radius r metres and height h metres to hold 20 m3 of water. The base costs €40 per m2 to make and the curved wall costs €25 per m2. There is no lid.

(a)

Show that h = 20/(πr2).

(1)
(b)

Show that the total cost of making the tank is C = 40πr2 + 1000/r euros.

(3)
(c)

Find dC/dr.

(2)
(d)

Find the value of r that minimises the cost, to 3 significant figures.

(2)
(e)

Show that this value of r gives a minimum.

(2)
(f)

Calculate the minimum cost to the nearest euro. Use the unrounded value of r.

(2)
(g)

Find the height of the tank at the minimum cost, to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 V = πr2h = 20 rearranged (AG)✓ 1
Part (b)
M1 base cost 40 × πr2✓ 1
M1 wall cost 25 × 2πrh and substitution of h✓ 1
A1 50πr × 20/(πr2) = 1000/r, so result shown (AG)✓ 1
Part (c)
M1 differentiate at least one term✓ 1
A1 80πr − 1000r−2✓ 1
Part (d)
M1 set dC/dr = 0, so r3 = 1000/(80π)✓ 1
A1 r = 1.58✓ 1
Part (e)
M1 d2C/dr2 = 80π + 2000r−3✓ 1
R1 positive for r > 0, so a minimum (or sign change of the gradient from negative to positive)✓ 1
Part (f)
M1 substitute unrounded r into C✓ 1
A1 €947✓ 1
Part (g)
A1 h = 20/(π × 1.58462) = 2.54 m✓ 1

Common error: A "show that" result must be reached by clear steps from the given information; do not start from the answer.

316IBMI3S-5-02r42-10Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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Water flows into an empty training pool through a pipe. The flow rate, R litres per second, t seconds after the pipe is opened is modelled by R = 6t − 0.6t2, for 0 ≤ t ≤ 10.

(a)

Find the maximum flow rate and the time at which it occurs.

(3)
(b)

Find the exact volume of water, in litres, that flows into the pool during the first 10 seconds.

(3)
(c)

Use the trapezoidal rule with 5 intervals of equal width to estimate the volume of water that flows in during the first 10 seconds.

(4)
(d)

Find the percentage error in the trapezoidal estimate, taking your answer to part (b) as the exact value.

(2)
(e)

Explain why the trapezoidal rule gives an underestimate in this case.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 dR/dt = 6 − 1.2t = 0✓ 1
A1 t = 5 s✓ 1
A1 R(5) = 15 litres per second✓ 1
Part (b)
M1 ∫010(6t − 0.6t2) dt✓ 1
A1 3t2 − 0.2t3 (or correct value from calculator)✓ 1
A1 100 litres✓ 1
Part (c)
M1 trapezoidal rule with h = 2✓ 1
A1 R(2) = 9.6, R(4) = 14.4, R(6) = 14.4, R(8) = 9.6, with R(0) = R(10) = 0✓ 1
A1 (1/2) × 2 × (0 + 0 + 2(9.6 + 14.4 + 14.4 + 9.6))✓ 1
A1 96 litres✓ 1
Part (d)
M1 |96 − 100| / 100 × 100 (ECF from (b) and (c))✓ 1
A1 4 %✓ 1
Part (e)
A1 the graph of R against t is concave down (a chord joining two points lies below the curve), so each trapezium lies under the curve✓ 1

Exam tip: To explain the direction of a trapezoidal-rule error, describe the shape of the curve (concave up or down) and compare the chord with the curve.

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Optimisation in context; Optimisation in context Paper 2Medium13 marks

A rail operator estimates that the cost of running a locomotive is €900 per hour plus a fuel cost of \(0.0012v^2\) euros per kilometre, where v km h−1 is its constant speed. Let C be the total cost in euros per kilometre travelled, for \(40\le v\le 120\).

(a)

Show that \(C=\frac{900}{v}+0.0012v^2\).

(2)
(b)

Calculate C when \(v=60\).

(2)
(c)

Find \(\frac{dC}{dv}\).

(2)
(d)

Find the speed that minimises the cost per kilometre.

(3)
(e)

Justify that this speed gives a minimum value of C.

(2)
(f)

Using the unrounded value of v from part (d), find the minimum cost per kilometre, and the cost of a 450 km journey at this speed, to the nearest euro.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 time to travel 1 km is \(\frac{1}{v}\) hours, so hourly cost per km is \(\frac{900}{v}\)✓ 1
A1 adds fuel cost per km to obtain the given expression (shown)✓ 1
Part (b)
M1 substitute \(v=60\): \(\frac{900}{60}+0.0012(3600)\)✓ 1
A1 €19.32 per km✓ 1
Part (c)
M1 attempt to differentiate \(900v^{-1}\) giving a term in \(v^{-2}\)✓ 1
A1 \(\frac{dC}{dv}=-\frac{900}{v^2}+0.0024v\)✓ 1
Part (d)
M1 set \(\frac{dC}{dv}=0\)✓ 1
A1 \(v^3=375\,000\)✓ 1
A1 \(v=72.1\) km h−1✓ 1
Part (e)
M1 \(\frac{d^2C}{dv^2}=\frac{1800}{v^3}+0.0024\)✓ 1
R1 this is positive for \(v>0\), so the stationary point is a minimum✓ 1
Part (f)
A1 minimum cost €18.72 per km✓ 1
A1 €8424✓ 1

Exam tip: When the question asks you to justify a minimum, state the sign of the second derivative (or the change in sign of the first) and say what it means.

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Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; 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Optimisation in context; Optimisation in context Paper 2Medium12 marks

A train leaves a station and stops at the next. Its speed is recorded every 30 seconds, as shown. The speed at 120 s is k m s−1.

Time (s)0306090120150180
Speed (m s−1)0122024k180

(a)

Show that the trapezoidal rule gives an estimate for the distance between the stations of \(30(74+k)\) metres.

(3)
(b)

The trapezoidal estimate of the distance is 3000 m. Find the value of k.

(2)
(c)

Use the 3000 m estimate to find the average speed over the whole journey.

(2)
(d)

Find the average acceleration between 60 s and 90 s.

(2)
(e)

State why the trapezoidal rule gives only an estimate of the distance here.

(1)
(f)

Assuming constant deceleration in the last 30 seconds, find the size of the deceleration.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 trapezoidal rule with 6 intervals of width 30✓ 1
A1 \(\frac{30}{2}\left[0+0+2(12+20+24+k+18)\right]\)✓ 1
A1 \(15\times 2(74+k)=30(74+k)\)✓ 1
Part (b)
M1 \(30(74+k)=3000\)✓ 1
A1 \(k=26\)✓ 1
Part (c)
M1 \(\frac{3000}{180}\) (ECF from (b) if the distance used is correct for their k)✓ 1
A1 16.7 m s−1✓ 1
Part (d)
M1 \(\frac{24-20}{30}\)✓ 1
A1 0.133 m s−2✓ 1
Part (e)
R1 the speed is only known at certain times, and is not necessarily linear between readings✓ 1
Part (f)
M1 \(\frac{0-18}{30}\) (or \(\frac{18}{30}\))✓ 1
A1 0.6 m s−2✓ 1

Exam tip: If an unknown appears in the data, keep it as a letter through the trapezoidal-rule bracket, then solve the equation at the end.

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A beekeeper builds an open-topped storage box for hive frames. The box has a square base of side x metres and height h metres, and its volume is 0.9 m3. The surface area of the box (the base and four sides) is A m2.

(a)

Show that A = x2 + 3.6/x.

(3)
(b)

Find dA/dx.

(2)
(c)

Find the value of x for which the surface area is a minimum. Give your answer correct to 3 significant figures.

(3)
(d)

Show that this value of x gives a minimum.

(2)
(e)

Find the minimum surface area. Use the unrounded value of x. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 h = 0.9/x2 from x2h = 0.9✓ 1
M1 A = x2 + 4xh✓ 1
A1 A = x2 + 4x(0.9/x2) = x2 + 3.6/x (AG)✓ 1
Part (b)
M1 attempt to differentiate, including a negative power✓ 1
A1 2x − 3.6x−2✓ 1
Part (c)
M1 set dA/dx = 0✓ 1
A1 2x3 = 3.6, so x3 = 1.8✓ 1
A1 x = 1.22 m✓ 1
Part (d)
M1 d2A/dx2 = 2 + 7.2x−3 (or a sign change of the gradient either side)✓ 1
R1 this is positive for x > 0, so the stationary point is a minimum✓ 1
Part (e)
M1 substitute x = 1.81/3 (= 1.2164…) into A✓ 1
A1 A = 4.44 m2✓ 1

Exam tip: Store the unrounded value from the previous part in your calculator and use it in the next part to avoid accumulated rounding errors.

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Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 2Medium13 marks

A Formula 1 car starts from rest. During its launch, its acceleration is a(t) = 9 − 1.5t m s−2, where t is the time in seconds after the start, for 0 ≤ t ≤ 6.

(a)

Find an expression for the velocity v(t) of the car.

(3)
(b)

Find the velocity of the car when t = 6.

(2)
(c)

Find the distance travelled by the car in the first 6 seconds.

(3)
(d)

Find the mean speed of the car over the first 6 seconds.

(2)
(e)

After t = 6, the car continues at a constant speed of 27 m s−1. Find the total time, from the start, for the car to travel 513 m.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 integrate a(t)✓ 1
A1 v = 9t − 0.75t2 + c✓ 1
A1 c = 0 since v(0) = 0✓ 1
Part (b)
M1 substitute t = 6✓ 1
A1 27 m s−1✓ 1
Part (c)
M1 integrate v(t)✓ 1
A1 s = 4.5t2 − 0.25t3 (with s(0) = 0)✓ 1
A1 108 m✓ 1
Part (d)
M1 108 ÷ 6 (ECF from (c))✓ 1
A1 18 m s−1✓ 1
Part (e)
M1 remaining distance 513 − 108 = 405 m✓ 1
M1 405 ÷ 27 = 15 s✓ 1
A1 total time = 21 s✓ 1

Exam tip: Integrating a rate gives a quantity only if the constant of integration is found from the starting condition.

321IBMI3S-5-03r47-06Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 2Medium13 marks

The cross-section of a curved roof over a market hall is modelled by f(x) = x2(15 − x)/40, for 0 ≤ x ≤ 15, where x is the horizontal distance in metres from the left-hand wall and f(x) is the height of the roof above the ground in metres. The roof meets the ground at both ends.

(a)

Find f′(x).

(2)
(b)

Find the maximum height of the roof above the ground, and the value of x at which it occurs.

(3)
(c)

Find the angle that the roof makes with the horizontal at the point where x = 3. Give your answer to the nearest 0.1°.

(3)
(d)

Find the area of the cross-section between the roof and the ground, correct to 3 significant figures.

(3)
(e)

The end wall of the hall has this shape and is to be covered in cladding costing €85 per square metre. Using the unrounded area from (d), find the cost of the cladding to the nearest euro.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expand to (15x2 − x3)/40 and attempt to differentiate✓ 1
A1 f′(x) = (30x − 3x2)/40✓ 1
Part (b)
M1 set f′(x) = 0✓ 1
A1 x = 10 (reject x = 0, which is the wall)✓ 1
A1 maximum height 12.5 m✓ 1
Part (c)
A1 f′(3) = 1.575✓ 1
M1 angle = tan−1(1.575)✓ 1
A1 57.6°✓ 1
Part (d)
M1 ∫015 f(x) dx✓ 1
A1 antiderivative (5x3 − x4/4)/40✓ 1
A1 105 m2 (105.46875…)✓ 1
Part (e)
M1 85 × 105.46875…✓ 1
A1 €8965✓ 1

Exam tip: When a later part says to use an unrounded value, keep the full calculator display; rounding early can change the final euro.

322IBMI-034-05-r2-06Gradients, tangents and rates of change Paper 2Hard11 marks

On a map, a curved road follows the path
\(y = 0.02x^2 - 0.8x + 10\), for \(0 \le x \le 40\),
where x and y are distances in kilometres east and north of a reference point. The positive y-direction points due north. A town T is located at the point (0, 4). Planners want to build a new straight road from T that joins the curved road at a point P so that the new road is a tangent to the curved road at P, with P having a positive x-coordinate.

(a)

Find \(\frac{dy}{dx}\).

(1)
(b)

Find the coordinates of the point on the curved road where the road runs exactly east–west.

(2)
(c)

Show that the tangent to the curved road at the point where x = a meets the y-axis at \(y = 10 - 0.02a^2\).

(3)
(d)

Hence find the coordinates of P.

(3)
(e)

Find the length of the new road from T to P.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\frac{dy}{dx} = 0.04x - 0.8\)✓ 1
Part (b)
M1 setting \(0.04x - 0.8 = 0\)✓ 1
A1 (20, 2)✓ 1
Part (c)
M1 tangent: \(y = (0.02a^2 - 0.8a + 10) + (0.04a - 0.8)(x - a)\)✓ 1
M1 substituting x = 0: \(y = 0.02a^2 - 0.8a + 10 - 0.04a^2 + 0.8a\)✓ 1
A1 simplifies to \(y = 10 - 0.02a^2\)✓ 1
Part (d)
M1 setting \(10 - 0.02a^2 = 4\)✓ 1
A1 a = √300 = 17.3✓ 1
A1 P(17.3, 2.14)✓ 1
Part (e)
M1 \(\sqrt{17.32^2 + (4 - 2.144)^2}\)✓ 1
A1 17.4 km✓ 1

Common error: T is not on the curve, so its coordinates cannot be substituted into the derivative. Write the tangent at a general point x = a, then make it pass through T.

323IBMI-034-06-r1-06Optimisation in context Paper 2Hard12 marks

A packaging designer makes an open-topped box from a rectangular sheet of card measuring 60 cm by 40 cm. A square of side x cm is cut from each corner and the sides are folded up. The volume of the box is V cm3.

(a)

Show that \(V = 4x^3 - 200x^2 + 2400x\).

(2)
(b)

Explain why \(0 < x < 20\).

(1)
(c)

Use differentiation to find the value of x that maximises the volume of the box, justifying that your answer gives a maximum.

(4)
(d)

Find the maximum volume of the box.

(2)
(e)

The box must hold at least 8000 cm3. Find the range of possible values of x.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Base dimensions \((60 - 2x)\) and \((40 - 2x)\), so \(V = x(60 - 2x)(40 - 2x)\) (M1)✓ 1
Expanding: \(x(2400 - 200x + 4x^2) = 4x^3 - 200x^2 + 2400x\) (A1)✓ 1
Part (b)
x must be positive and \(40 - 2x > 0\) (the shorter side must remain positive), so \(x < 20\) (R1)✓ 1
Part (c)
\(\dfrac{\mathrm{d}V}{\mathrm{d}x} = 12x^2 - 400x + 2400\) (A1)✓ 1
Setting \(12x^2 - 400x + 2400 = 0\) and solving (M1)✓ 1
\(x = 7.85\) cm (7.8475…). The other root, 25.5, is rejected because it lies outside \(0 < x < 20\) (A1)✓ 1
\(\dfrac{\mathrm{d}^2V}{\mathrm{d}x^2} = 24x - 400 = -212 < 0\) at \(x = 7.85\), so this is a maximum. Equivalently, a sign change of \(\dfrac{\mathrm{d}V}{\mathrm{d}x}\) from + to − is acceptable (R1)✓ 1
Part (d)
Substituting \(x = 7.8475…\) into V (M1)✓ 1
\(V = 8450\) cm3 (8450.45…) (A1)✓ 1
Part (e)
Solving \(4x^3 - 200x^2 + 2400x = 8000\) (or \(V \ge 8000\)), e.g. using a GDC (M1)✓ 1
Lower bound \(x = 5.86\) (5.8578…, which equals \(20 - \sqrt{200}\)) (A1)✓ 1
Upper bound \(x = 10\), so \(5.86 \le x \le 10\) (A1)✓ 1

Common error: A GDC may show a third solution of V = 8000 near x = 34.1. Always check solutions against the domain found earlier and discard any that do not make physical sense.

324IBMI-034-02-r1-06Integration and the trapezoidal rule Paper 2Hard13 marks

A plot of land lies between a straight road and a river. The road runs along the x-axis from x = 0 to x = 60, and the plot is bounded at each end by straight fences perpendicular to the road. A surveyor measures the perpendicular distance, y metres, from the road to the riverbank at intervals of 10 m.

x (m): 0, 10, 20, 30, 40, 50, 60
y (m): 20.0, 27.9, 31.6, 33.8, 32.3, 27.1, 20.0

The riverbank is modelled by \(f(x) = 20 + 0.9x - 0.015x^2\), for \(0 \le x \le 60\).

(a)

Use the trapezoidal rule with the surveyor's data to estimate the area of the plot.

(3)
(b)

Use the model to find the area of the plot.

(3)
(c)

Explain why, if the model is accurate, the trapezoidal rule will underestimate the area of the plot.

(2)
(d)

The owner wants to put a new fence perpendicular to the road at x = k, so that the section of the plot with 0 ≤ x ≤ k has an area of 600 m2. Use the model to find k.

(3)
(e)

The land is valued at 85 US dollars (USD) per square metre. Find the value of the plot calculated using the model minus its value calculated using the trapezoidal estimate.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
h = 10✓ 1
\(\frac{10}{2}[20 + 20 + 2(27.9 + 31.6 + 33.8 + 32.3 + 27.1)]\)✓ 1
= 1727 m2 (accept 1730)✓ 1
Part (b)
\(\int_0^{60} (20 + 0.9x - 0.015x^2)\,dx\)✓ 1
\(= [20x + 0.45x^2 - 0.005x^3]_0^{60}\) or GDC evaluation✓ 1
= 1740 m2✓ 1
Part (c)
The coefficient of x2 is negative (f''(x) = −0.03 < 0), so the curve is concave down (arch-shaped)✓ 1
so the straight tops of the trapezia lie below the riverbank and leave out some of the area✓ 1
Part (d)
\(\int_0^{k} f(x)\,dx = 600\)✓ 1
\(20k + 0.45k^2 - 0.005k^3 = 600\), solved using the GDC✓ 1
k = 21.9 m (21.86)✓ 1
Part (e)
(1740 − 1727) × 85✓ 1
= USD 1105 (accept USD 1110; follow through from their answers to (a) and (b), e.g. (1740 − 1730) × 85 = USD 850)✓ 1

Exam tip: Carry the exact trapezoidal value 1727 into later parts. If you round it to 1730 first, the difference in value drops from USD 1105 to USD 850.

325IBMI3S-5-09-12Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 2Hard14 marks

A drinks company makes closed cylindrical cans of volume 330 cm3. A can has radius r cm and height h cm. The company wants to use as little metal as possible.

(a)

Show that h = \(\frac{330}{\pi r^2}\).

(1)
(b)

Show that the total surface area of the can is S = 2πr2 + \(\frac{660}{r}\).

(2)
(c)

Find \(\frac{dS}{dr}\).

(2)
(d)

Find the value of r that minimises S.

(3)
(e)

Find the minimum surface area.

(2)
(f)

Show that this value of r gives a minimum.

(2)
(g)

Find the height of the can with minimum surface area. Compare it with the diameter.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of \(V=\pi r^2h=330\) and rearrange✓ 1
Part (b)
M1 surface area 2πr2 + 2πrh✓ 1
A1 substitute h to reach the given result✓ 1
Part (c)
M1 attempt to differentiate, using 660r−1✓ 1
A1 4πr − 660r−2✓ 1
Part (d)
M1 set \(\frac{dS}{dr}=0\)✓ 1
M1 rearrange to r3 = 165/π✓ 1
A1 r = 3.74 cm✓ 1
Part (e)
M1 substitute r into S✓ 1
A1 264 cm2✓ 1
Part (f)
M1 \(\frac{d^2S}{dr^2}=4\pi+1320r^{-3}\)✓ 1
R1 positive for r > 0, so a minimum✓ 1
Part (g)
M1 h = 330/(πr2) = 7.49 cm✓ 1
A1 the height equals the diameter (2r = 7.49 cm)✓ 1

Exam tip: In a 'show that' part, write every step: a missing substitution loses the final mark even when the result is printed.

326IBMI3S-5-09-13Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 2Hard14 marks

A drone flies along a straight horizontal wire. Its velocity, v m s−1, at time t seconds is given by
v(t) = t2 − 7t + 10,   0 ≤ t ≤ 8.
When t = 0 the drone is at the point A.

(a)

Find the times at which the drone is instantaneously at rest.

(2)
(b)

Find the displacement of the drone from A when t = 8.

(2)
(c)

Find the total distance the drone travels in the first 8 seconds.

(5)
(d)

Hence find the average speed of the drone over the first 8 seconds.

(2)
(e)

Find the speed of the drone at the instant its acceleration is zero.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 set v = 0 and solve✓ 1
A1 t = 2 and t = 5✓ 1
Part (b)
M1 \(\int_0^8 v\,dt\)✓ 1
A1 26.7 m✓ 1
Part (c)
M1 consider the three intervals 0–2, 2–5 and 5–8 separately✓ 1
A1 distance for 0–2 is 8.67 m✓ 1
A1 distance for 2–5 is 4.5 m✓ 1
A1 distance for 5–8 is 22.5 m✓ 1
A1 total 35.7 m✓ 1
Part (d)
M1 total distance ÷ 8✓ 1
A1 4.46 m s−1✓ 1
Part (e)
M1 a = 2t − 7 and set to zero✓ 1
A1 t = 3.5✓ 1
A1 speed 2.25 m s−1 (velocity is −2.25)✓ 1

Common error: Distance and displacement differ when the velocity changes sign. Split the integral at each time the drone is at rest and add the sizes.

327IBMI3S-5-09-14Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 2Hard13 marks

A hiking trail rises and falls. The height of the trail above sea level is h metres at a horizontal distance x km from the start, 0 ≤ x ≤ 20. The gradient of the trail is
\(\frac{dh}{dx}=0.06x^2-0.9x+2.7\),
and the trail starts at a height of 120 m.

(a)

Find an expression for h(x).

(4)
(b)

Find the values of x at which the trail is horizontal.

(3)
(c)

Find the height of the trail at each of these two points. State which is a local maximum and which is a local minimum.

(3)
(d)

A walker claims the highest point of the trail is the local maximum found in (c). Is the walker correct? Justify your answer.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to integrate the gradient✓ 1
A1 0.02x3 − 0.45x2 + 2.7x + c✓ 1
M1 use h(0) = 120✓ 1
A1 h = 0.02x3 − 0.45x2 + 2.7x + 120✓ 1
Part (b)
M1 set \(\frac{dh}{dx}=0\)✓ 1
A1 x = 4.15✓ 1
A1 x = 10.9✓ 1
Part (c)
M1 substitute both values into h✓ 1
A1 local maximum at x = 4.15, height 125 m (124.9)✓ 1
A1 local minimum at x = 10.9, height 122 m (121.9)✓ 1
Part (d)
M1 evaluate h(20) = 154✓ 1
A1 154 m is greater than 125 m✓ 1
R1 not correct: the highest point is the end of the trail at x = 20✓ 1

Common error: A local maximum is not always the greatest value on an interval. Always compare it with the heights at the endpoints.

328IBMI3S-5-09b-12Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 2Hard12 marks

A water company will lay a pipe from a reservoir R to a village V. The reservoir is 4 km from a straight road, at the point P on the road nearest to R. The village is on the road, 10 km from P. The pipe runs in a straight line underground from R to a point Q on the road, where PQ = x km (0 ≤ x ≤ 10), and then along the road from Q to V. Laying the pipe away from the road costs £120 000 per km and laying it along the road costs £80 000 per km. Let C be the total cost in thousands of pounds.

(a)

Show that C(x) = 120√(16 + x2) + 80(10 − x).

(2)
(b)

Calculate the cost when Q is at P and when Q is at V, to the nearest thousand pounds.

(2)
(c)

Use your graphing calculator to find the value of x that minimises the cost, and the minimum cost to the nearest thousand pounds.

(3)
(d)

Find the range of values of x for which the cost is less than £1 200 000.

(3)
(e)

Use your graphing calculator to find dC/dx when x = 2, and interpret your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 Pythagoras gives RQ = √(16 + x2) and QV = 10 − x✓ 1
A1 cost = 120 × RQ + 80 × QV gives the printed expression✓ 1
Part (b)
A1 x = 0: 120 × 4 + 80 × 10 = 1280 thousand✓ 1
A1 x = 10: 120√116 = 1292 thousand✓ 1
Part (c)
M1 graph C(x) and find the minimum✓ 1
A1 x = 3.58 km✓ 1
A1 minimum cost 1158 thousand pounds✓ 1
Part (d)
M1 solve C(x) = 1200 using a graphing calculator✓ 1
A1 x = 1.32 and x = 6.68✓ 1
A1 1.32 < x < 6.68✓ 1
Part (e)
M1 numerical derivative of C at x = 2✓ 1
A1 -26.3: the cost falls by about £26 300 for each extra kilometre that Q moves away from P✓ 1

Common error: State the units of C in the interpretation: the derivative is thousands of pounds per kilometre, not pounds.

329IBMI3S-5-09b-13Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 2Hard11 marks

After an injection, the concentration of a drug in a patient's blood is modelled by C(t) = 12te−0.5t mg/L, where t is the time in hours, 0 ≤ t ≤ 12.

(a)

Find the greatest concentration and the time at which it occurs.

(2)
(b)

The drug is considered effective when the concentration is at least 6 mg/L. Find the interval of time for which this is true.

(2)
(c)

The total exposure is the area under the curve. Find ∫012 C(t) dt.

(2)
(d)

Hence find the mean concentration over the 12 hours.

(2)
(e)

Use the trapezoidal rule with four equal intervals to estimate the integral in part (c), and comment on the accuracy of your estimate.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 find the maximum using a graphing calculator✓ 1
A1 t = 2 hours and C = 8.83 mg/L✓ 1
Part (b)
M1 solve C(t) = 6 using a graphing calculator✓ 1
A1 0.715 ≤ t ≤ 4.31✓ 1
Part (c)
M1 definite integral of C from 0 to 12✓ 1
A1 47.2 mg h/L✓ 1
Part (d)
M1 47.2 ÷ 12✓ 1
A1 3.93 mg/L✓ 1
Part (e)
M1 trapezoidal rule with h = 3✓ 1
A1 39.0 mg h/L✓ 1
R1 an underestimate, because the intervals are too wide: the peak near t = 2 lies between the sampled points and the chord from t = 0 to t = 3 cuts it off✓ 1

Common error: Using a trapezoidal estimate with too few strips on a curve that rises and falls quickly gives a result that is far from the true area.

330IBMI3S-5-02r5-12Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Hard12 marks

A farmer makes an open box for storing seed from a rectangular metal sheet measuring 60 cm by 40 cm. A square of side x cm is cut from each corner and the sides are folded up.

(a)

Show that the volume of the box is V = 4x3 − 200x2 + 2400x cm3.

(2)
(b)

State the range of possible values of x.

(1)
(c)

Find dV/dx.

(2)
(d)

Find the value of x that maximises the volume. Give your answer to 3 significant figures.

(2)
(e)

Show that this value of x gives a maximum.

(2)
(f)

Find the maximum volume of the box, to the nearest cm3.

(1)
(g)

The farmer needs the box to hold at least 8000 cm3. Find the smallest value of x for which the volume is 8000 cm3. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 box dimensions are (60 − 2x), (40 − 2x) and x✓ 1
A1 expansion of x(60 − 2x)(40 − 2x) to the given result✓ 1
Part (b)
A1 0 < x < 20✓ 1
Part (c)
M1 differentiation of each term✓ 1
A1 12x2 − 400x + 2400✓ 1
Part (d)
M1 set dV/dx = 0 and solve (quadratic formula or GDC)✓ 1
A1 x = 7.85 cm; the other root 25.5 is rejected as it is outside the range✓ 1
Part (e)
M1 d2V/dx2 = 24x − 400✓ 1
R1 negative (-211.7) at x = 7.85, so a maximum✓ 1
Part (f)
A1 8450 cm3 (using unrounded x)✓ 1
Part (g)
M1 solve 4x3 − 200x2 + 2400x = 8000 using a GDC✓ 1
A1 x = 5.86 cm✓ 1

Exam tip: After solving a derivative equal to zero, check each root against the range allowed by the context before deciding which one to use.

331IBMI3S-5-02r5-13Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Hard13 marks

The volume envelope of a note on a synthesiser is modelled by f(t) = 5te−t, for 0 ≤ t ≤ 6, where t is the time in seconds after the key is pressed and f(t) is the volume in arbitrary units. Use technology where appropriate.

(a)

Find f(2). Give your answer to 3 significant figures.

(2)
(b)

Find the coordinates of the maximum point of the graph of f, and state what it represents in context.

(3)
(c)

Use technology to find the gradient of the graph at t = 3. Give your answer to 3 significant figures.

(2)
(d)

Use the trapezoidal rule with 6 intervals of equal width to estimate ∫06 f(t) dt. Give your answer to 3 significant figures.

(3)
(e)

Use technology to find the exact value of the integral, to 3 significant figures.

(1)
(f)

Find the percentage error of the trapezoidal estimate, taking the technology value as exact. Use unrounded values and give your answer to 2 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitution of t = 2✓ 1
A1 1.35✓ 1
Part (b)
M1 use of the GDC maximum function✓ 1
A1 t = 1✓ 1
A1 f = 1.84: the loudest the note gets, reached 1 s after the key is pressed✓ 1
Part (c)
M1 use of the GDC derivative at t = 3✓ 1
A1 -0.498✓ 1
Part (d)
M1 width 1 and ordinates f at t = 0, 1, 2, …, 6✓ 1
M1 half the width × (first ordinate + last ordinate + 2 × sum of the five middle ordinates)✓ 1
A1 4.51✓ 1
Part (e)
A1 4.91✓ 1
Part (f)
M1 percentage error = |estimate − exact| ÷ exact × 100 (the estimate is an underestimate)✓ 1
A1 8.2% (ECF from (d) and (e))✓ 1

Exam tip: Percentage error compares with the exact value in the denominator; use unrounded values for both integrals.

332IBMI3S-5-02r5-14Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Hard12 marks

A farmer designs a closed cylindrical silo of volume 50 m3, radius r metres and height h metres. The two circular ends cost €25 per m2 and the curved wall costs €12 per m2.

(a)

Write h in terms of r.

(1)
(b)

Show that the total cost is C = 50πr2 + 1200r−1 euros.

(3)
(c)

Find dC/dr.

(2)
(d)

Find the value of r that minimises the cost. Give your answer to 3 significant figures.

(2)
(e)

Justify that this value of r gives a minimum cost.

(2)
(f)

Find the minimum cost, to the nearest euro.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 πr2h = 50 so h = 50 ÷ (πr2)✓ 1
Part (b)
M1 ends: 2πr2 × 25 = 50πr2✓ 1
M1 wall: 2πrh × 12 with h from (a)✓ 1
A1 2πr × 50/(πr2) × 12 = 1200r−1, giving the given result✓ 1
Part (c)
M1 differentiation of each term✓ 1
A1 100πr − 1200r−2✓ 1
Part (d)
M1 set dC/dr = 0 giving r3 = 12/π✓ 1
A1 r = 1.56 m✓ 1
Part (e)
M1 d2C/dr2 = 100π + 2400r−3✓ 1
R1 positive for all r > 0, so a minimum✓ 1
Part (f)
M1 substitution of the unrounded r into C✓ 1
A1 €1151✓ 1

Common error: The curved surface area is 2πrh, not πr²h; check that each surface is paired with its own cost per square metre.

333IBMI3S-5-02r6-09Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Hard14 marks

A solar heating system uses an open-topped hot-water tank in the shape of a cuboid with a square base of side \(x\) metres and height \(h\) metres. The volume of the tank must be 18 m3. The base material costs £40 per m2 and the material for the four vertical sides costs £25 per m2. Let £\(C\) be the total cost of the material.

(a)

Show that \(h = \frac{18}{x^2}\).

(1)
(b)

Show that \(C = 40x^2 + \frac{1800}{x}\).

(3)
(c)

Find \(\frac{dC}{dx}\).

(2)
(d)

Find the value of \(x\) that minimises the cost, to 3 significant figures.

(3)
(e)

Find the minimum cost, using the unrounded value of \(x\), to the nearest pound.

(2)
(f)

Justify that this value of \(x\) gives a minimum cost.

(2)
(g)

Find the height of the tank when the cost is a minimum, to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(x^2 h = 18\), so \(h = \frac{18}{x^2}\)✓ 1
Part (b)
M1 base cost \(40x^2\)✓ 1
M1 sides cost \(25 \times 4xh\)✓ 1
A1 substitute \(h = \frac{18}{x^2}\) to obtain \(\frac{1800}{x}\)✓ 1
Part (c)
M1 attempt to differentiate, including \(1800x^{-1}\)✓ 1
A1 \(80x - \frac{1800}{x^2}\)✓ 1
Part (d)
M1 set \(\frac{dC}{dx} = 0\)✓ 1
M1 \(x^3 = 22.5\)✓ 1
A1 \(x = 2.82\) m✓ 1
Part (e)
M1 substitute the unrounded \(x\) into \(C\)✓ 1
A1 £956✓ 1
Part (f)
M1 \(\frac{d^2C}{dx^2} = 80 + \frac{3600}{x^3}\)✓ 1
R1 this is positive for \(x > 0\), so the cost is a minimum✓ 1
Part (g)
A1 \(h = \frac{18}{x^2} = 2.26\) m✓ 1

Exam tip: After finding a stationary value, always confirm its nature, either with the second derivative or by checking the sign of the gradient either side.

334IBMI3S-5-02r6-10Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Hard15 marks

The electricity demand on a regional grid, in megawatts (MW), is modelled by \(D(t) = 0.01t^3 - 0.45t^2 + 5.4t + 30\), where \(t\) is the time in hours after midnight, for \(0 \le t \le 24\).

(a)

Find \(D'(t)\).

(2)
(b)

Find the first time at which \(D'(t) = 0\), and the demand at that time, giving both answers to 3 significant figures.

(3)
(c)

Find the equation of the tangent to the graph of \(D\) at \(t = 6\).

(3)
(d)

Find \(\int_0^{24} D(t)\,dt\), which represents the energy demanded in MWh over the day.

(2)
(e)

Use the trapezoidal rule with four intervals of equal width to estimate the same quantity.

(3)
(f)

Find the percentage error of the trapezoidal estimate relative to the value in part (d), to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate each term✓ 1
A1 \(D'(t) = 0.03t^2 - 0.9t + 5.4\)✓ 1
Part (b)
M1 set \(0.03t^2 - 0.9t + 5.4 = 0\) and solve✓ 1
A1 \(t = 8.29\) hours (the other solution, \(t = 21.7\), is a local minimum)✓ 1
A1 \(D = 49.5\) MW✓ 1
Part (c)
A1 \(D(6) = 48.36\)✓ 1
A1 \(D'(6) = 1.08\)✓ 1
M1 use of \(D - 48.36 = 1.08(t - 6)\), giving \(D = 1.08t + 41.88\)✓ 1
Part (d)
M1 attempt to integrate✓ 1
A1 1031.04 MWh✓ 1
Part (e)
M1 correct formula with width 6✓ 1
M1 correct values \(D(0) = 30\), \(D(6) = 48.36\), \(D(12) = 47.28\), \(D(18) = 39.72\), \(D(24) = 38.64\) substituted✓ 1
A1 1018.08 MWh✓ 1
Part (f)
M1 \(\frac{1031.04 - 1018.08}{1031.04} \times 100\)✓ 1
A1 1.26%✓ 1

Exam tip: Substitute the exact function values into the trapezoidal rule, not rounded readings, so that the percentage error you report is meaningful.

335IBMI3S-5-02r6-11Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Hard14 marks

During a vaccination campaign, the number of people vaccinated per day, in thousands, is modelled by \(v(t) = 7.2t - 0.6t^2\), where \(t\) is the number of days since the campaign began, for \(0 \le t \le 12\).

(a)

Find the maximum rate of vaccination and the time at which it occurs.

(4)
(b)

Find \(\int_0^8 v(t)\,dt\) and state what it represents.

(3)
(c)

Find the time at which the total vaccinated first reaches 120 thousand people, to 2 significant figures.

(2)
(d)

Use the trapezoidal rule with four intervals to estimate \(\int_0^8 v(t)\,dt\).

(3)
(e)

Find the percentage error in the estimate from part (d).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(v'(t) = 7.2 - 1.2t\)✓ 1
M1 set \(v'(t) = 0\)✓ 1
A1 \(t = 6\) days✓ 1
A1 \(v(6) = 21.6\) thousand per day✓ 1
Part (b)
M1 attempt to integrate, \(3.6t^2 - 0.2t^3\)✓ 1
A1 128✓ 1
R1 the total number of people vaccinated in the first 8 days, in thousands✓ 1
Part (c)
M1 set \(3.6T^2 - 0.2T^3 = 120\)✓ 1
A1 \(T = 7.6\) days✓ 1
Part (d)
M1 correct formula with width 2✓ 1
M1 values \(v(0) = 0\), \(v(2) = 12\), \(v(4) = 19.2\), \(v(6) = 21.6\), \(v(8) = 19.2\) used✓ 1
A1 124.8✓ 1
Part (e)
M1 \(\frac{128 - 124.8}{128} \times 100\)✓ 1
A1 2.5%✓ 1

Common error: The total vaccinated is the integral of the rate, not the value of the rate; do not substitute a time into \(v(t)\) when asked for a total.

336IBMI3S-5-01r8-07Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Hard14 marks

A company packs chlorine tablets in closed cylindrical tubs of radius r cm and height h cm. Each tub has volume 2000 cm3. The company wishes to minimise the surface area of a tub.

(a)

Show that the total surface area is A = 2πr2 + 4000/r.

(3)
(b)

Find dA/dr.

(2)
(c)

Find the value of r that minimises A, giving your answer to 3 significant figures.

(3)
(d)

Show that this value of r gives a minimum.

(2)
(e)

Find the minimum surface area, giving your answer to 3 significant figures.

(2)
(f)

The material costs £0.0012 per cm2. Using the unrounded minimum area, find the cost of the material for one tub, giving your answer in pounds to 2 decimal places.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 h = 2000/(πr2)✓ 1
M1 A = 2πr2 + 2πrh✓ 1
A1 substitution gives 2πr2 + 4000/r (AG)✓ 1
Part (b)
M1 attempt to differentiate✓ 1
A1 4πr − 4000/r2✓ 1
Part (c)
M1 set dA/dr = 0✓ 1
M1 r3 = 1000/π✓ 1
A1 r = 6.83 cm✓ 1
Part (d)
M1 d2A/dr2 = 4π + 8000/r3✓ 1
R1 positive for r = 6.83, so a minimum✓ 1
Part (e)
M1 substitute the unrounded r into A✓ 1
A1 879 cm2✓ 1
Part (f)
M1 878.755… × 0.0012✓ 1
A1 £1.05✓ 1

Exam tip: Reduce the area to a function of one variable before differentiating, and justify a minimum with the second derivative or a sign change.

337IBMI3S-5-01r8-10Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Hard13 marks

An offshore turbine platform P is 6 km from the nearest point N of a straight shoreline. A substation S is on the shore 14 km from N. A cable runs in a straight line under the sea from P to a point X on the shore, x km from N towards S, then along the shore from X to S. Laying cable under the sea costs €500 000 per km and on land costs €300 000 per km. Let the total cost be C thousand euros.

(a)

Show that C = 500√(36 + x2) + 300(14 − x), for 0 ≤ x ≤ 14.

(2)
(b)

Use your graphic display calculator to find the value of x that minimises C.

(2)
(c)

Find the minimum cost, in euros.

(2)
(d)

The alternative is a cable straight from P to S under the sea. Find its cost, to 3 significant figures, in euros.

(2)
(e)

Using the unrounded cost in (d), find the percentage saving made by the minimum-cost route in (c) compared with the direct sea route. Give your answer to 3 significant figures.

(2)
(f)

The cost of laying cable on land rises to €400 000 per km. Find the new value of x that minimises the total cost.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 sea length = √(36 + x2) and land length = 14 − x✓ 1
A1 combine with costs per km to obtain the given expression (AG)✓ 1
Part (b)
M1 graph or minimum function used on 0 ≤ x ≤ 14✓ 1
A1 x = 4.5✓ 1
Part (c)
M1 substitute x = 4.5✓ 1
A1 C = 6600, so €6 600 000✓ 1
Part (d)
M1 500√(36 + 142)✓ 1
A1 €7 620 000✓ 1
Part (e)
M1 (7615.77… − 6600) ÷ 7615.77…✓ 1
A1 13.3 %✓ 1
Part (f)
M1 C = 500√(36 + x2) + 400(14 − x)✓ 1
M1 graph or minimum function used✓ 1
A1 x = 8✓ 1

Exam tip: For cost-minimisation with a square root, build the cost function carefully, then use the calculator's minimum function on the stated domain.

338IBMI3S-5-01r8-12Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Hard13 marks

A library plans a reading nook whose floor is a rectangle of width 2r metres and length h metres, with a semicircle of radius r attached to one of the 2r sides. The nook is enclosed by a glass wall along the two sides of length h, the other 2r side and the curved edge, totalling 12 m of glass wall.

(a)

Show that h = 6 − r − (πr)/2.

(2)
(b)

Show that the floor area is A = 12r − (2 + π/2)r2.

(3)
(c)

Find the value of r that maximises A, to 3 significant figures.

(2)
(d)

Justify that this value gives a maximum.

(2)
(e)

Find the maximum floor area, to 3 significant figures.

(2)
(f)

Find the percentage of the maximum floor area that is the semicircular part, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 2h + 2r + πr = 12✓ 1
A1 rearrange to the given result (AG)✓ 1
Part (b)
M1 A = 2rh + ½πr2✓ 1
M1 substitute for h✓ 1
A1 12r − 2r2 − πr2 + ½πr2 = 12r − (2 + π/2)r2 (AG)✓ 1
Part (c)
M1 A′ = 12 − (4 + π)r = 0✓ 1
A1 r = 1.68 m✓ 1
Part (d)
M1 A″ = −(4 + π)✓ 1
R1 negative, so a maximum✓ 1
Part (e)
M1 substitute the unrounded r into A✓ 1
A1 10.1 m2✓ 1
Part (f)
M1 ½πr2 ÷ 10.0817… × 100 (unrounded values)✓ 1
A1 44.0 %✓ 1

Exam tip: Use the constraint to eliminate one variable, then differentiate; keep unrounded values in later parts.

339IBMI3S-5-01r13-12Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Hard14 marks

A library atrium window is a rectangle of width 2r metres and height h metres, topped by a semicircle of radius r metres. The total length of the frame (the two vertical sides, the base and the semicircular arc) is 12 m.

(a)

Show that h = 6 − r − (πr)/2.

(2)
(b)

Hence show that the area of the window is A = 12r − (2 + π/2)r2.

(3)
(c)

Find the value of r that maximises the area. Give your answer to 3 significant figures.

(3)
(d)

Show that this value of r gives a maximum area.

(2)
(e)

Find the maximum area of the window. Use the unrounded value of r and give your answer to 3 significant figures.

(2)
(f)

Glass costs €48 per square metre. Using the unrounded area from part (e), find the cost of the glass to the nearest euro.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 frame length: 2h + 2r + πr = 12 (the arc has length πr)✓ 1
A1 rearrange to h = 6 − r − (πr)/2 (AG)✓ 1
Part (b)
M1 A = 2rh + ½πr2✓ 1
M1 substitute the expression for h✓ 1
A1 A = 12r − 2r2 − πr2 + ½πr2 = 12r − (2 + π/2)r2 (AG)✓ 1
Part (c)
M1 dA/dr = 12 − (4 + π)r✓ 1
M1 set derivative equal to zero✓ 1
A1 r = 12/(4 + π) = 1.68 m✓ 1
Part (d)
M1 d2A/dr2 = −(4 + π)✓ 1
R1 second derivative is negative (for all r), so the area is a maximum✓ 1
Part (e)
M1 substitute r = 12/(4 + π) into the area formula (ECF from (c))✓ 1
A1 A = 10.1 m2 (10.08… unrounded)✓ 1
Part (f)
M1 multiply the unrounded area by 48✓ 1
A1 €484✓ 1

Exam tip: Keep unrounded values in your calculator memory between parts to avoid accumulating rounding errors.

340IBMI3S-5-01r13-14Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Hard12 marks

A cycle path beside a river is modelled by h(x) = 0.000001x3 − 0.00045x2 + 0.06x + 12, for 0 ≤ x ≤ 300, where x is the horizontal distance in metres along the path and h is the height in metres above the river.

(a)

Show that h′(x) = 0.000003x2 − 0.0009x + 0.06.

(2)
(b)

Find the values of x at which the path has a horizontal tangent.

(3)
(c)

Find the height of the path at each of these points and state whether each is a local maximum or a local minimum.

(3)
(d)

A safety rule states that the gradient of the path must not exceed 0.05. Find the values of x, in the domain, for which the gradient is greater than 0.05. Give each boundary value to 3 significant figures.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 0.000003x2 − 0.0009x + 0.06 (AG)✓ 1
Part (b)
M1 set h′(x) = 0✓ 1
M1 solve the quadratic (equivalent to x2 − 300x + 20 000 = 0)✓ 1
A1 x = 100 and x = 200✓ 1
Part (c)
A1 h(100) = 14.5 m and h(200) = 14 m✓ 1
M1 gradient changes sign (or the second derivative is examined) at each point✓ 1
A1 x = 100 is a local maximum and x = 200 is a local minimum✓ 1
Part (d)
M1 set 0.000003x2 − 0.0009x + 0.06 = 0.05✓ 1
M1 solve the quadratic to obtain x = 11.56… and x = 288.44…✓ 1
A1 0 ≤ x < 11.6✓ 1
A1 288 < x ≤ 300✓ 1

Common error: A quadratic inequality has an interval, or a pair of intervals, as its answer: check the sign of the quadratic between and outside its roots.

341IBMI3S-5-03r21-12Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 2Hard15 marks

A logistics company sends a van on a delivery of 240 km at a constant speed of v km h-1, where 30 ≤ v ≤ 90. The driver is paid €20 per hour. The fuel cost is (0.00008v2 + 0.06) euros per kilometre. Let T be the total of the driver's pay and the fuel cost for the journey, in euros.

(a)

Show that T(v) = 4800/v + 0.0192v2 + 14.4.

(3)
(b)

Find T′(v).

(2)
(c)

Find the speed that minimises the total cost.

(3)
(d)

Find the minimum total cost.

(2)
(e)

Show that this is a minimum value of T.

(2)
(f)

A customer asks for the delivery to be made at 80 km h-1. Find the percentage increase in the total cost compared with the minimum cost. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 journey time = 240/v hours, so driver's pay = 20 × 240/v = 4800/v✓ 1
M1 fuel cost = 240(0.00008v2 + 0.06)✓ 1
A1 expansion giving 0.0192v2 + 14.4, and adding to the driver's pay✓ 1
Part (b)
M1 attempt to differentiate with 4800/v written as 4800v−1✓ 1
A1 −4800v−2 + 0.0384v✓ 1
Part (c)
M1 set T′(v) = 0✓ 1
A1 v3 = 125000✓ 1
A1 v = 50 km h-1✓ 1
Part (d)
M1 substitute v = 50 into T✓ 1
A1 €158.40 (ECF from (c))✓ 1
Part (e)
M1 T″(v) = 9600v−3 + 0.0384✓ 1
R1 this is positive for all v > 0, so a minimum✓ 1
Part (f)
M1 T(80) = 60 + 122.88 + 14.4✓ 1
A1 €197.28✓ 1
A1 24.5%✓ 1

Exam tip: A percentage increase is the change divided by the original (here the minimum) value, multiplied by 100.

342IBMI3S-5-03r21-13Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 2Hard15 marks

A pedestrian tunnel under a canal has a parabolic cross-section modelled by y = 9 − x2/4 for −6 ≤ x ≤ 6, where y is the height above the floor in metres and x is the horizontal distance in metres from the centre line. The floor is the line y = 0.

(a)

Show that the width of the tunnel floor is 12 m.

(2)
(b)

Find the exact area of the cross-section of the tunnel by integration.

(3)
(c)

The tunnel is 150 m long. Find the volume of the tunnel.

(2)
(d)

Use the trapezoidal rule with four intervals of equal width to estimate the cross-sectional area.

(3)
(e)

Find the percentage error in the estimate from part (d), relative to the exact area.

(2)
(f)

Explain why the trapezoidal rule gives an underestimate in this case.

(1)
(g)

A rectangular delivery pod of width 8 m is to be moved through the tunnel, centred on the centre line. Find the greatest height the pod can have.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 set y = 0, so x2 = 36✓ 1
A1 x = ±6, so width = 12 m✓ 1
Part (b)
M1 set up ∫−66 (9 − x2/4) dx✓ 1
A1 9x − x3/12✓ 1
A1 72 m2✓ 1
Part (c)
M1 72 × 150✓ 1
A1 10800 m3✓ 1
Part (d)
A1 heights 0, 6.75, 9, 6.75, 0 at x = −6, −3, 0, 3, 6✓ 1
M1 use of trapezoidal rule with interval width 3✓ 1
A1 ½ × 3 × [0 + 0 + 2(6.75 + 9 + 6.75)] = 67.5 m2✓ 1
Part (e)
M1 (72 − 67.5)/72 × 100✓ 1
A1 6.25%✓ 1
Part (f)
R1 the curve is concave down, so each chord lies below the curve✓ 1
Part (g)
M1 substitute x = 4 into y✓ 1
A1 5 m✓ 1

Exam tip: For an underestimate or overestimate, link the answer to the shape of the curve: chords of a concave-down curve lie below it.

343IBMI3S-5-03r21-14Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 2Hard13 marks

A fish farm raises sea bass in a pen. After t weeks the number of fish still alive is n(t) = 5000 − 45t and the mean mass of a fish is m(t) = 0.3 + 0.05t kilograms. The total mass of fish in the pen is B(t) = n(t) × m(t) kilograms.

(a)

Show that B(t) = 1500 + 236.5t − 2.25t2.

(2)
(b)

Find B′(t).

(2)
(c)

Find the value of t for which the total mass is greatest. Give your answer to 3 significant figures.

(2)
(d)

The fish can only be harvested at the end of a whole number of weeks. Determine the number of weeks after which the farm should harvest to obtain the greatest total mass.

(3)
(e)

Find the greatest value of t for which the model for the number of fish is valid.

(2)
(f)

The fish are sold for €6.40 per kilogram. Find the income from harvesting after the number of weeks found in part (d).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expansion of (5000 − 45t)(0.3 + 0.05t)✓ 1
A1 1500 + 250t − 13.5t − 2.25t2, giving the result✓ 1
Part (b)
M1 attempt to differentiate (at least one term correct)✓ 1
A1 236.5 − 4.5t✓ 1
Part (c)
M1 set B′(t) = 0✓ 1
A1 t = 52.6✓ 1
Part (d)
M1 evaluate B at the two whole numbers either side of 52.6✓ 1
A1 B(52) = 7714 and B(53) = 7714.25✓ 1
A1 53 weeks, since B(53) is greater✓ 1
Part (e)
M1 set 5000 − 45t = 0✓ 1
A1 t = 111.1 weeks✓ 1
Part (f)
M1 7714.25 × 6.40✓ 1
A1 €49371.20 (ECF from (d))✓ 1

Common error: Giving the non-integer answer from the calculus as the final answer when the context needs a whole number of weeks. Compare the whole-number values either side.

344IBMI3S-5-02r24-10Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Hard13 marks

During a coffee roast the temperature of the beans, T °C, is modelled by T(t) = 20 + 48t − 4t2 + 0.12t3, where t is the time in minutes after the beans enter the drum, for 0 ≤ t ≤ 12.

(a)

Find the rate of change of temperature when t = 2.

(2)
(b)

Find the time at which the temperature is rising at 20 °C per minute. Give your answer to 3 significant figures.

(3)
(c)

Find the least rate of temperature rise during the roast, and the time at which it occurs.

(3)
(d)

The roaster controller predicts the temperature at t = 7 using the tangent to the curve at t = 6. Find the predicted temperature, to 3 significant figures.

(3)
(e)

Determine whether the tangent prediction in part (d) is greater or less than the value given by the model at t = 7. Give a reason.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 T′(t) = 48 − 8t + 0.36t2✓ 1
A1 33.4 °C per minute (33.44)✓ 1
Part (b)
M1 48 − 8t + 0.36t2 = 20✓ 1
A1 solving: t = 4.35 or t = 17.9✓ 1
R1 reject 17.9 as it is outside the domain 0 ≤ t ≤ 12✓ 1
Part (c)
M1 T″(t) = −8 + 0.72t = 0✓ 1
A1 t = 11.1 minutes✓ 1
A1 3.56 °C per minute (compare T′(12) = 3.84 and T′(0) = 48)✓ 1
Part (d)
M1 T(6) = 189.92 and T′(6) = 12.96✓ 1
M1 tangent: T = 189.92 + 12.96(t − 6), i.e. T = 12.96t + 112.16✓ 1
A1 203 °C (202.88)✓ 1
Part (e)
M1 T″(7) = −2.96 < 0 (or T(7) = 201.16)✓ 1
R1 the curve is concave down at this point, so the tangent lies above it: the prediction is greater✓ 1

Exam tip: A tangent line is a good prediction only close to the point of contact; the second derivative tells you which side of the curve the tangent lies on.

345IBMI3S-5-02r24-13Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Hard12 marks

A coffee company designs a closed cylindrical tin of radius r cm and height h cm with volume 1000 cm3. The two circular ends cost €0.012 per cm2 of material. The curved side costs €0.005 per cm2 of material. The total material cost of one tin is C euros.

(a)

Show that h = 1000 ÷ (πr2).

(1)
(b)

Show that C = 0.024πr2 + 10 ÷ r.

(3)
(c)

Find dC/dr.

(2)
(d)

Find the value of r that minimises C. Give your answer to 3 significant figures.

(2)
(e)

Show that this value of r gives a minimum cost.

(2)
(f)

Find the minimum cost of material for one tin. Use the unrounded value of r and give your answer to the nearest cent.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
R1 volume πr2h = 1000, rearranged✓ 1
Part (b)
M1 ends: 2πr2 × 0.012 = 0.024πr2✓ 1
M1 side: 2πrh × 0.005 with h from part (a), giving 0.01πr × 1000 ÷ (πr2)✓ 1
A1 10 ÷ r, hence the given expression✓ 1
Part (c)
A1 0.048πr✓ 1
A1 −10r−2 (or −10 ÷ r2)✓ 1
Part (d)
M1 0.048πr − 10 ÷ r2 = 0, i.e. r3 = 10 ÷ (0.048π)✓ 1
A1 r = 4.05 cm✓ 1
Part (e)
A1 d2C/dr2 = 0.048π + 20 ÷ r3✓ 1
R1 positive for r > 0, so a minimum✓ 1
Part (f)
M1 substituting r = 4.0476… into C✓ 1
A1 €3.71✓ 1

Common error: A closed tin has two ends and one curved side; counting only one end, or giving the ends and side the same price, changes the optimal radius.

346IBMI3S-5-02r24-14Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Hard9 marks

A small rover released from a space probe moves in a straight line. Its velocity is v(t) = t2 − 8t + 12 m s−1, where t is the time in seconds, for 0 ≤ t ≤ 8. Positive velocity is away from the probe.

(a)

Find the times at which the rover is momentarily at rest.

(2)
(b)

Find the displacement of the rover from its starting position at t = 8. Give your answer to 3 significant figures.

(3)
(c)

Find the total distance travelled by the rover between t = 0 and t = 8.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 t2 − 8t + 12 = 0✓ 1
A1 t = 2 and t = 6✓ 1
Part (b)
M1 ∫08(t2 − 8t + 12) dt✓ 1
A1 [t3 ÷ 3 − 4t2 + 12t]08✓ 1
A1 10.7 m away from the probe (32/3)✓ 1
Part (c)
M1 splitting the integral at t = 2 and t = 6 (the velocity changes sign)✓ 1
A1 ∫02 = 32/3 and ∫68 = 32/3✓ 1
A1 ∫26 = −32/3 (the rover moves back towards the probe)✓ 1
A1 total distance = 32/3 + 32/3 + 32/3 = 32 m✓ 1

Common error: Displacement is the integral of velocity, but distance travelled is the integral of speed; where velocity is negative, take the positive size of that part of the area.

347IBMI3S-5-01r27-07Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Hard12 marks

The total number of people infected with a seasonal illness in a region is modelled by N(t) = −0.02t3 + 1.2t2 + 5t + 10, for 0 ≤ t ≤ 40, where t is the time in days after the first report.

(a)

Write down the number of people infected when t = 0.

(1)
(b)

Find N′(t).

(2)
(c)

Find N″(t) and hence find the time at which the number of infected people is increasing most rapidly.

(3)
(d)

Find the greatest rate of increase of the number of infected people.

(2)
(e)

Justify that the rate found in part (d) is the greatest rate of increase over the whole domain.

(2)
(f)

Find the mean rate of increase of the number of infected people over the first 20 days.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 10 people✓ 1
Part (b)
M1 differentiating at least two terms correctly✓ 1
A1 −0.06t2 + 2.4t + 5✓ 1
Part (c)
A1 N″(t) = −0.12t + 2.4✓ 1
M1 setting N″(t) = 0✓ 1
A1 t = 20 days✓ 1
Part (d)
M1 substituting t = 20 into N′(t)✓ 1
A1 29 people per day✓ 1
Part (e)
R1 N″ changes from positive to negative at t = 20 (so N′ has a maximum there)✓ 1
R1 at the ends of the domain N′(0) = 5 and N′(40) = 5, both less than 29✓ 1
Part (f)
M1 (N(20) − N(0)) ÷ 20 with N(20) = 430✓ 1
A1 21 people per day✓ 1

Common error: The point where N′(t) is greatest is found by setting the second derivative to zero, not the first. Setting N′ = 0 would find where the total stops changing.

348IBMI3S-5-01r27-11Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Hard12 marks

A solar water heater uses a closed cylindrical storage tank of radius r metres and height h metres. The tank holds exactly 0.5 m3 of water. The total outer surface area of the tank is A m2, and the manufacturer wants to minimise the amount of metal used.

(a)

Express h in terms of r.

(2)
(b)

Hence show that A = 2πr2 + 1/r.

(2)
(c)

Find dA/dr.

(2)
(d)

Find the value of r that minimises A, correct to 3 significant figures.

(2)
(e)

Show that this value of r gives a minimum surface area.

(2)
(f)

Find the minimum surface area, correct to 3 significant figures. Use the unrounded value of r.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 πr2h = 0.5✓ 1
A1 h = 0.5 ÷ (πr2)✓ 1
Part (b)
M1 A = 2πr2 + 2πrh✓ 1
A1 substituting for h to obtain 2πr2 + 1/r✓ 1
Part (c)
M1 differentiating at least one term correctly✓ 1
A1 4πr − 1/r2✓ 1
Part (d)
M1 setting dA/dr = 0, giving r3 = 1/(4π)✓ 1
A1 r = 0.430 m✓ 1
Part (e)
M1 d2A/dr2 = 4π + 2/r3✓ 1
R1 this is positive for all r > 0, so the value is a minimum✓ 1
Part (f)
M1 substituting r = 0.43013… into A✓ 1
A1 3.49 m2✓ 1

Exam tip: When a later part says to use the unrounded value, store the full GDC value in memory rather than retyping the rounded answer.

349IBMI3S-5-02r27-10Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Hard12 marks

The width, w metres, of the waterline of a ship at a distance x metres from its bow is modelled by w(x) = 0.008x(100 − x), for 0 ≤ x ≤ 100.

Distance x (m): 0, 25, 50, 75, 100
Width w (m): 0, a, 20, b, 0

(a)

Find the values of a and b.

(2)
(b)

Use the trapezoidal rule with four intervals to estimate the area of the waterline surface of the ship.

(3)
(c)

Find the exact area of the waterline surface by integration. Give your answer to the nearest m2.

(3)
(d)

Find the percentage error of the estimate in part (b), using the exact area 1333.33… m2. Give your answer correct to 3 significant figures.

(2)
(e)

Loading 500 tonnes of cargo makes the ship sink deeper. Estimate the increase in draught, in metres, by assuming the waterline area stays constant at 1333.33… m2. The density of seawater is 1025 kg m−3. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 a = 15✓ 1
A1 b = 15✓ 1
Part (b)
M1 using the trapezoidal rule with width 25✓ 1
A1 (25/2)[0 + 0 + 2(15 + 20 + 15)]✓ 1
A1 1250 m2✓ 1
Part (c)
M1 integral of w between 0 and 100✓ 1
A1 antiderivative 0.4x2 − 0.008x3/3✓ 1
A1 1333 m2✓ 1
Part (d)
M1 (1333.33 − 1250) ÷ 1333.33 × 100✓ 1
A1 6.25%✓ 1
Part (e)
M1 volume displaced = 500 000 ÷ 1025 = 487.8 m3✓ 1
A1 increase in draught = volume ÷ area = 0.366 m✓ 1

Exam tip: A table of values only gives an estimate of an area; comparing it with the exact integral shows how the estimate improves as more intervals are used.

350IBMI3S-5-03r27-05Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 2Hard13 marks

A pharmacy packs liquid medicine in closed cylindrical plastic vials of volume 400 cm3. A vial has radius r cm and height h cm. The curved surface costs £0.002 per cm2 to make and each circular end costs £0.005 per cm2.

(a)

Show that h = 400/(πr2).

(1)
(b)

Show that the cost of making one vial, £C, is C = 0.01πr2 + 1.6/r.

(3)
(c)

Find dC/dr and hence find the value of r that minimises the cost. Give your answer to 3 significant figures.

(4)
(d)

Find the height of this vial, and the cost of making it. Use the unrounded value of r from (c).

(2)
(e)

The vials must fit on a shelf, so h cannot exceed 12 cm. Find the minimum possible cost of a vial in this case, giving your answer to 3 significant figures. Justify that this is the minimum.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 volume πr2h = 400 rearranged to give the result✓ 1
Part (b)
M1 C = 2 × 0.005 × πr2 + 0.002 × 2πrh✓ 1
M1 substitute h = 400/(πr2)✓ 1
A1 0.002 × 2πr × 400/(πr2) = 1.6/r, so C = 0.01πr2 + 1.6/r✓ 1
Part (c)
M1 attempt to differentiate✓ 1
A1 dC/dr = 0.02πr − 1.6r−2✓ 1
M1 set dC/dr = 0✓ 1
A1 r = 2.94 cm✓ 1
Part (d)
A1 h = 14.7 cm✓ 1
A1 C = £0.816✓ 1
Part (e)
M1 h = 12 gives r = √(400/(12π)) = 3.257...✓ 1
R1 C decreases for r < 2.94 and increases for r > 2.94, so for r ≥ 3.257 the lowest cost is at r = 3.257✓ 1
A1 £0.825✓ 1

Common error: After finding a stationary point, check any constraint given in the question. The unconstrained minimum may not be allowed.

351IBMI3S-5-03r27-10Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 2Hard12 marks

After an oral dose, the concentration of a drug in the blood is modelled by C(t) = at3 + bt2, where C is in mg/L, t is in hours, 0 ≤ t ≤ 6, and a and b are constants. The concentration reaches its maximum value of 8 mg/L at t = 4.

(a)

Find the values of a and b.

(5)
(b)

Find the length of time for which the concentration is greater than 6 mg/L. Give your answer to 3 significant figures.

(2)
(c)

The total exposure to the drug is the area under the graph of C against t from t = 0 to t = 6. Find its value, in mg h/L.

(3)
(d)

Hence find the mean concentration over the 6 hours.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 C′(t) = 3at2 + 2bt✓ 1
M1 C′(4) = 0, so 48a + 8b = 0✓ 1
M1 C(4) = 8, so 64a + 16b = 8✓ 1
A1 a = −0.25✓ 1
A1 b = 1.5✓ 1
Part (b)
M1 solve −0.25t3 + 1.5t2 = 6, giving t = 2.69... and t = 5.06...✓ 1
A1 2.37 hours✓ 1
Part (c)
M1 attempt to integrate C(t) from 0 to 6✓ 1
A1 −0.0625t4 + 0.5t3✓ 1
A1 27 mg h/L✓ 1
Part (d)
M1 27 ÷ 6✓ 1
A1 4.5 mg/L (ECF from (c))✓ 1

Exam tip: When a question gives a maximum, it supplies two conditions: the derivative is zero there, and the function takes the stated value.

352IBMI3S-5-03r27-14Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context Paper 2Hard12 marks

The surface area of a reservoir at depth h metres is A thousand m2. Measurements give:

h (m): 0, 1, 3, 6, 10
A (thousand m2): 0, 1.5, 7.5, 24, 60

The volume of water in the reservoir between depth 0 and depth 10 m, in thousand m3, is the area under the graph of A against h. The values in the table fit the model A(h) = 0.5h2 + h.

(a)

Use the trapezoidal rule, with a separate trapezium for each interval between consecutive depths in the table, to estimate the volume.

(4)
(b)

Use integration of the model to find the volume. Give your answer to 4 significant figures.

(3)
(c)

Find the percentage error of the estimate in (a), using the exact value from integration. Give your answer to 3 significant figures.

(2)
(d)

Explain why the trapezoidal rule gives an overestimate here.

(1)
(e)

The operator requires the estimate to be within 2% of the value found in (b). Decide whether the estimate in (a) meets this requirement, giving your reason.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 area of each strip = (width/2)(sum of parallel sides)✓ 1
A1 at least two correct strip areas from 0.75, 9, 47.25, 168✓ 1
A1 all four strip areas correct✓ 1
A1 225 thousand m3✓ 1
Part (b)
M1 ∫010 (0.5h2 + h) dh✓ 1
A1 h3/6 + h2/2✓ 1
A1 216.7 thousand m3✓ 1
Part (c)
M1 (225 − 650/3)/(650/3) × 100✓ 1
A1 3.85 %✓ 1
Part (d)
R1 the gradient of A, A′(h) = h + 1, increases with h, so the graph curves upwards and every chord lies above the curve✓ 1
Part (e)
M1 comparison of the percentage difference with 2%✓ 1
R1 the estimate is about 3.85% too large, which is more than 2%, so it does not meet the requirement (ECF from (a) and (b))✓ 1

Common error: Percentage error is measured relative to the exact value, not relative to the estimate.

353IBMI3S-5-02r30-03Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 2Hard12 marks

A data centre needs an open-topped water tank with a square base of side x metres and height h metres. The tank must hold exactly 32 m3 of water. The tank is made from metal panels, which cost €85 per square metre.

(a)

Show that \(h = \frac{32}{x^2}\).

(1)
(b)

Show that the total area A m2 of panel used is \(A = x^2 + \frac{128}{x}\).

(2)
(c)

Find \(\frac{dA}{dx}\).

(2)
(d)

Find the value of x that gives the least area of panels.

(3)
(e)

Show that this value of x gives a minimum.

(2)
(f)

Find the least possible cost of the panels.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 volume = x2h = 32, so h = 32/x2✓ 1
Part (b)
M1 area = base + four sides = x2 + 4xh✓ 1
A1 substituting h: 4x × 32/x2 = 128/x✓ 1
Part (c)
M1 attempt to differentiate, with 128x−1 giving a term in x−2✓ 1
A1 \(2x - 128x^{-2}\)✓ 1
Part (d)
M1 set \(\frac{dA}{dx} = 0\)✓ 1
A1 \(x^3 = 64\)✓ 1
A1 x = 4 m✓ 1
Part (e)
M1 \(\frac{d^2A}{dx^2} = 2 + 256x^{-3}\)✓ 1
R1 at x = 4 this is 6, which is positive, so the area is a minimum (accept a sign change of the gradient either side of x = 4)✓ 1
Part (f)
M1 least area = 42 + 128/4 = 48 m2✓ 1
A1 48 × 85 = €4080✓ 1
ECF from (d).—

Common error: Forgetting that the tank has no lid. Count only the base and four sides in the area formula.

354IBMI3S-5-02r30-09Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 2Hard14 marks

A company sells cloud-storage plans. When the monthly price is €p, the number of customers per month is \(N = 600 - 10p\), for 20 ≤ p ≤ 55. Each customer costs the company €12 per month to serve, and there are fixed costs of €1500 per month.

(a)

Find the number of customers when the price is €20.

(1)
(b)

Show that the monthly profit €P is given by \(P = -10p^2 + 720p - 8700\).

(3)
(c)

Find \(\frac{dP}{dp}\).

(2)
(d)

Find the price that gives the greatest profit if the company may choose any price in the given range.

(2)
(e)

A regulator limits the price to a maximum of €30. Find the greatest possible monthly profit under this rule, justifying your answer.

(4)
(f)

At a price of €30, find the company's total monthly revenue.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 600 − 200 = 400✓ 1
Part (b)
M1 revenue = p(600 − 10p)✓ 1
M1 costs = 12(600 − 10p) + 1500✓ 1
A1 P = p(600 − 10p) − 12(600 − 10p) − 1500, which simplifies to the given expression✓ 1
Part (c)
M1 attempt to differentiate✓ 1
A1 −20p + 720✓ 1
Part (d)
M1 set \(\frac{dP}{dp} = 0\)✓ 1
A1 p = €36✓ 1
ECF from (c).—
Part (e)
M1 30 < 36 (the unconstrained optimum is not allowed)✓ 1
R1 dP/dp > 0 for p < 36, so profit increases up to the cap and the best allowed price is €30✓ 1
A1 P(30) = −9000 + 21600 − 8700✓ 1
A1 €3900✓ 1
Part (f)
M1 30 × (600 − 10 × 30)✓ 1
A1 €9000✓ 1

Common error: Giving the unconstrained optimum when the question limits the allowed values. Always check the answer against the stated range.

355IBMI3S-5-02r30-13Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 2Hard13 marks

A burial mound has a cross-section modelled by the curve \(y = 3 - 0.12x^2\) for −5 ≤ x ≤ 5, where x and y are in metres and the x-axis is the ground. The mound's cross-section is estimated using the trapezoidal rule with 5 intervals of equal width, at x = −5, −3, −1, 1, 3, 5.

(a)

Find the values of y at x = −3 and x = −1.

(2)
(b)

Use the trapezoidal rule to estimate the area of the cross-section.

(3)
(c)

Find the exact area of the cross-section by integration.

(3)
(d)

Find the percentage error in the estimate from (b), taking the exact area as the true value.

(2)
(e)

Explain why the trapezoidal rule gives an underestimate for this curve.

(2)
(f)

The mound is 18 m long. Using the exact area, find its volume.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 y(−3) = 1.92✓ 1
A1 y(−1) = 2.88✓ 1
Part (b)
M1 use of the trapezoidal rule with width 2✓ 1
A1 \(\frac{2}{2}\left[0 + 0 + 2(1.92 + 2.88 + 2.88 + 1.92)\right]\)✓ 1
A1 19.2 m2✓ 1
Part (c)
M1 \(\int_{-5}^{5}(3 - 0.12x^2)\,dx\)✓ 1
M1 antiderivative \(3x - 0.04x^3\)✓ 1
A1 20 m2✓ 1
Part (d)
M1 |19.2 − 20| ÷ 20 × 100✓ 1
A1 4.00%✓ 1
ECF from (b) and (c).—
Part (e)
R1 the curve is concave (it bends downwards)✓ 1
R1 so each chord joining two points lies below the curve, and the trapezia cover less area than the region✓ 1
Part (f)
A1 20 × 18 = 360 m3✓ 1
ECF from (c).—

Common error: Using the trapezoidal estimate in the final volume when the question asks for the exact area. Read which value each part tells you to use.

356IBMI3S-5-01r31-10Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Hard14 marks

Calculator

A hatchery builds an open-topped concrete tank with a square base of side x metres and height h metres. The tank must hold exactly 32 m3 of water.

(a)

Show that h = 32/x2.

(1)
(b)

Show that the total inner surface area, S m2, of the base and four sides is S = x2 + 128/x.

(2)
(c)

Find dS/dx.

(2)
(d)

Find the value of x that minimises the surface area.

(3)
(e)

Justify that this value of x gives a minimum.

(2)
(f)

Find the minimum surface area.

(2)
(g)

Concrete lining costs €36.50 per m2. Find the minimum cost of lining the tank.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 volume = x2h = 32, so h = 32/x2 (AG)✓ 1
Part (b)
M1 S = x2 + 4xh✓ 1
A1 substituting h: 4x × 32/x2 = 128/x (AG)✓ 1
Part (c)
A1 2x✓ 1
A1 − 128x−2✓ 1
Part (d)
M1 setting dS/dx = 0✓ 1
M1 2x3 = 128 (or x3 = 64)✓ 1
A1 x = 4✓ 1
Part (e)
M1 d2S/dx2 = 2 + 256x−3✓ 1
R1 at x = 4 this is 6 > 0, so a minimum (accept a sign change of the gradient from negative to positive)✓ 1
Part (f)
M1 substituting x = 4: 16 + 32✓ 1
A1 48 m2✓ 1
Part (g)
M1 48 × 36.50 (ECF from (f))✓ 1
A1 €1752 (accept €1752.00)✓ 1

Common error: Eliminate the second variable using the constraint before differentiating; differentiating an expression that still contains both x and h will not give the correct stationary point.

357IBMI3S-5-01r31-14Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context Paper 2Hard14 marks

Calculator

A delivery robot travels along a straight corridor. Its speed, v m s−1, t seconds after leaving the charging point is modelled by v(t) = 6t − t2, for 0 ≤ t ≤ 6.

(a)

Find the acceleration of the robot when t = 1.

(2)
(b)

Find the greatest speed of the robot.

(3)
(c)

Find the distance travelled by the robot in the first 6 seconds.

(4)
(d)

Find the average speed of the robot over the first 6 seconds.

(1)
(e)

Use the trapezoidal rule with 3 strips of equal width to estimate the distance travelled in the first 6 seconds.

(3)
(f)

Explain why the trapezoidal rule gives an underestimate of the distance travelled in this case.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 a(t) = v′(t) = 6 − 2t✓ 1
A1 at t = 1, a = 4 m s−2✓ 1
Part (b)
M1 setting 6 − 2t = 0 (ECF from (a))✓ 1
A1 t = 3✓ 1
A1 v(3) = 9 m s−1✓ 1
Part (c)
M1 ∫06 (6t − t2) dt✓ 1
A1 3t2 − t3/3✓ 1
M1 substituting limits: 108 − 72✓ 1
A1 36 m✓ 1
Part (d)
A1 36 ÷ 6 = 6 m s−1 (ECF from (c))✓ 1
Part (e)
M1 trapezoidal rule with h = 2 and v values at t = 0, 2, 4, 6✓ 1
A1 v(2) = 8 and v(4) = 8, v(0) = v(6) = 0: (2/2)[0 + 0 + 2(8 + 8)]✓ 1
A1 32 m✓ 1
Part (f)
R1 the graph of v is concave down (a curve bending downwards), so the straight-line tops of the trapezoids lie below the curve✓ 1

Common error: Distance travelled is the area under the speed–time graph, found by integrating the speed, not by differentiating it.

358IBMI3S-5-01r32-07Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Hard14 marks

The cross-section of a snow slope is modelled by the curve with equation h(x) = ax2 + bx + c, where x is the horizontal distance and h is the height above sea level, both measured in units of 100 m. The gradient of the slope is −0.8 where x = 1.5 and is 1.2 where x = 4. The point (1.5, 0.9) lies on the curve.

(a)

Write down an expression for h′(x).

(1)
(b)

Find the values of a, b and c.

(5)
(c)

Find the coordinates of the lowest point of the curve.

(3)
(d)

Find the equation of the tangent to the curve at the point where x = 4. Give your answer in the form y = mx + c.

(3)
(e)

This tangent crosses the line h = 0 at the point Q. Find the x-coordinate of Q, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 h′(x) = 2ax + b✓ 1
Part (b)
M1 form two equations from the gradient conditions: 3a + b = −0.8 and 8a + b = 1.2✓ 1
A1 a = 0.4✓ 1
A1 b = −2✓ 1
M1 substitute the point (1.5, 0.9) into h(x) with their a, b to find c✓ 1
A1 c = 3✓ 1
Part (c)
M1 set h′(x) = 0 (or use the vertex formula)✓ 1
A1 x = 2.5✓ 1
A1 h = 0.5, so (2.5, 0.5)✓ 1
Part (d)
A1 h(4) = 1.4✓ 1
M1 use of y − 1.4 = 1.2(x − 4) or equivalent✓ 1
A1 y = 1.2x − 3.4✓ 1
Part (e)
M1 set 1.2x − 3.4 = 0 (ECF from (d))✓ 1
A1 x = 2.83✓ 1

Exam tip: Gradient conditions give equations in the unknown coefficients; use each given piece of information once and check your final curve reproduces all of them.

359IBMI3S-5-01r32-10Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Hard14 marks

A bakery makes open baking trays by cutting a square of side x cm from each corner of a rectangular metal sheet measuring 40 cm by 25 cm, then folding up the sides.

(a)

Show that the volume of the tray is V(x) = 4x3 − 130x2 + 1000x cm3.

(2)
(b)

Write down the set of possible values of x.

(1)
(c)

Find V′(x).

(2)
(d)

Find the value of x that gives the maximum volume, and justify your choice.

(3)
(e)

Find the maximum volume of the tray.

(2)
(f)

The bakery needs trays with a volume of at least 2000 cm3. Find the range of values of x for which this is true. Give your answers to 3 significant figures.

(3)
(g)

Express the maximum volume in litres. (1 litre = 1000 cm3.)

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 base dimensions (40 − 2x) and (25 − 2x) and height x: V = x(40 − 2x)(25 − 2x)✓ 1
A1 expansion leading to the given expression✓ 1
Part (b)
A1 0 < x < 12.5✓ 1
Part (c)
M1 attempt to differentiate (at least two terms correct)✓ 1
A1 V′(x) = 12x2 − 260x + 1000✓ 1
Part (d)
M1 set V′(x) = 0 and solve✓ 1
A1 x = 5 or x = 16.7 (16.67)✓ 1
R1 reject x = 16.7 because it is outside the possible values, so x = 5✓ 1
Part (e)
M1 substitute x = 5 into V(x) (ECF from (d))✓ 1
A1 2250 cm3✓ 1
Part (f)
M1 attempt to solve V(x) = 2000 or V(x) ≥ 2000 (graph or technology)✓ 1
A1 x = 3.20✓ 1
A1 x = 7.01, so 3.20 ≤ x ≤ 7.01✓ 1
Part (g)
A1 2.25 litres (ECF from (e))✓ 1

Exam tip: When an optimisation problem comes from a physical shape, state the domain of the variable and reject any solution outside it.

360IBMI3S-5-01r33-02Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 2Hard12 marks

A volcano observatory plans an open-topped storage box for delicate instruments. The box is a cuboid with a square base of side x m and height h m, and its volume must be 48 m3. The base is made of heavy shielding costing €30 per m2. The four vertical sides cost €20 per m2. There is no lid.

(a)

Show that h = 48/x2, and hence that the total cost, C euros, is given by C = 30x2 + 3840/x.

(3)
(b)

Find dC/dx.

(2)
(c)

Find the value of x that gives the minimum cost.

(3)
(d)

Show that this value of x gives a minimum cost.

(2)
(e)

Find the minimum cost of the box and the height of the box at this cost.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 volume equation x2h = 48 and rearranging to h = 48/x2✓ 1
M1 cost = 30x2 + 20 × 4xh✓ 1
A1 substitution of h to reach C = 30x2 + 3840/x (80 × 48 = 3840) (AG)✓ 1
Part (b)
M1 attempt to differentiate (at least one term correct, e.g. 3840x−1 → −3840x−2)✓ 1
A1 60x − 3840/x2✓ 1
Part (c)
M1 set dC/dx = 0✓ 1
M1 60x3 = 3840, so x3 = 64 (or use GDC to solve)✓ 1
A1 x = 4✓ 1
Part (d)
M1 find d2C/dx2 = 60 + 7680/x3, or test the sign of the gradient either side of x = 4✓ 1
R1 value is positive at x = 4 (or gradient changes from negative to positive), so a minimum✓ 1
Part (e)
A1 C = 30 × 16 + 3840/4 = €1440✓ 1
A1 h = 48/16 = 3 m✓ 1

Exam tip: After finding a stationary point in an optimisation problem, always justify that it is a minimum or maximum; one line using the second derivative or a sign change earns the mark.

361IBMI3S-5-01r33-06Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 2Hard13 marks

A theatre prints a rectangular poster. The printed region has area 720 cm2, with width x cm and height y cm. There is a blank margin of 5 cm on each side (left and right) and of 4 cm at the top and at the bottom. The area of paper, A cm2, is to be as small as possible.

(a)

Show that A = 800 + 8x + 7200/x.

(4)
(b)

Find dA/dx.

(2)
(c)

Find the value of x for which A is a minimum, given that x > 0.

(3)
(d)

Use the second derivative to show that this value of x gives a minimum.

(2)
(e)

Find the dimensions of the poster and its least area.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 poster width x + 10 and height y + 8✓ 1
M1 y = 720/x seen✓ 1
M1 expansion (x + 10)(720/x + 8) = 720 + 8x + 7200/x + 80✓ 1
A1 A = 800 + 8x + 7200/x (AG)✓ 1
Part (b)
M1 attempt to differentiate✓ 1
A1 8 − 7200/x2✓ 1
Part (c)
M1 set dA/dx = 0✓ 1
M1 x2 = 900✓ 1
A1 x = 30✓ 1
Part (d)
M1 d2A/dx2 = 14400/x3✓ 1
R1 positive when x = 30, so a minimum✓ 1
Part (e)
A1 poster is 40 cm wide and 32 cm high (y = 24)✓ 1
A1 least area = 1280 cm2✓ 1

Common error: The dimensions of the poster include the margins; do not report x and y (the printed region) as the size of the paper.

362IBMI3S-5-02r33-03Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Hard13 marks

A company is building a data centre site on a rectangular plot of length x m and width y m. The plot is divided into three equal compounds by two internal fences, each of length y m, parallel to the sides of length y. Perimeter fencing costs €60 per metre and internal fencing costs €40 per metre. The budget for all the fencing is exactly €18 000.

(a)

Show that 3x + 5y = 450.

(3)
(b)

Hence show that the total area, A m2, of the plot is A = 150y − (5/3)y2.

(2)
(c)

Find dA/dy.

(2)
(d)

Find the value of y that gives the greatest area.

(2)
(e)

Justify that this is a maximum.

(2)
(f)

Find the value of x and the greatest area of the plot.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 perimeter cost 60(2x + 2y)✓ 1
M1 internal cost 40 × 2y and total set equal to 18 000: 120x + 200y = 18 000✓ 1
A1 divide by 40 to reach 3x + 5y = 450 (AG)✓ 1
Part (b)
M1 x = 150 − (5/3)y✓ 1
A1 A = xy = 150y − (5/3)y2 (AG)✓ 1
Part (c)
M1 attempt to differentiate✓ 1
A1 150 − (10/3)y✓ 1
Part (d)
M1 set dA/dy = 0✓ 1
A1 y = 45✓ 1
Part (e)
M1 d2A/dy2 = −10/3 (or sign change of the gradient)✓ 1
R1 negative (for all y), so a maximum✓ 1
Part (f)
A1 x = 150 − 75 = 75 m✓ 1
A1 greatest area 75 × 45 = 3375 m2✓ 1

Exam tip: In a show-that question every line of working is marked, so state each cost separately before simplifying.

363IBMI3S-5-02r33-12Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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The power demand of a data centre, P kW, over a 24-hour day is modelled by P(t) = 380 + 12t − 0.5t2, where t is the time in hours, 0 ≤ t ≤ 24.

(a)

Complete the table of values.

t06121824
P380………380

(2)
(b)

Use the trapezoidal rule with four intervals to estimate the energy used in the 24 hours, in kWh.

(3)
(c)

Use integration to find the exact energy used in the 24 hours.

(3)
(d)

Find the percentage error of the estimate in part (b), using the exact value from part (c). Give your answer to 3 significant figures.

(2)
(e)

Explain why the estimate in part (b) is smaller than the exact value.

(1)
(f)

The energy price is €0.12 per kWh. Find the cost of the exact energy used.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 P(6) = 434 and P(18) = 434✓ 1
A1 P(12) = 452✓ 1
Part (b)
M1 trapezoidal rule structure with strip width 6✓ 1
A1 ½ × 6 × (380 + 380 + 2(434 + 452 + 434))✓ 1
A1 10 200 kWh✓ 1
Part (c)
M1 ∫024 (380 + 12t − 0.5t2) dt✓ 1
A1 antiderivative 380t + 6t2 − t3/6✓ 1
A1 10 272 kWh✓ 1
Part (d)
M1 (10 272 − 10 200)/10 272 × 100✓ 1
A1 0.701%✓ 1
Part (e)
R1 P″(t) = −1 < 0, so the curve is concave; the chords used by the trapezoidal rule lie below the curve✓ 1
Part (f)
M1 10 272 × 0.12✓ 1
A1 €1232.64✓ 1

Exam tip: A fully labelled table of values with correct strip width is the quickest way to make a trapezoidal rule calculation checkable.

364IBMI3S-5-02r33-14Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Hard12 marks

The height of a road above sea level, h metres, at a distance x km along a cycling route is modelled by h(x) = ax2 + bx + 40, for 0 ≤ x ≤ 5, where a and b are constants. The road is 42 m above sea level at x = 2, and its gradient at that point is 5 m per km.

(a)

Write down two equations for a and b.

(2)
(b)

Find the values of a and b.

(2)
(c)

Find the minimum height of the road for 0 ≤ x ≤ 5. Give your answer to 3 significant figures.

(3)
(d)

A rider finds the road too steep when its gradient exceeds 12 m per km. Find the value of x at which the gradient first reaches 12 m per km.

(2)
(e)

Find the mean height of the road over the first 4 km. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 4a + 2b + 40 = 42, i.e. 4a + 2b = 2✓ 1
A1 h′(x) = 2ax + b, so 4a + b = 5✓ 1
Part (b)
M1 attempt to solve the simultaneous equations (or GDC)✓ 1
A1 a = 2 and b = −3✓ 1
Part (c)
M1 set h′(x) = 4x − 3 = 0✓ 1
A1 x = 0.75✓ 1
A1 h = 38.875 = 38.9 m✓ 1
Part (d)
M1 set 4x − 3 = 12✓ 1
A1 x = 3.75 km✓ 1
Part (e)
M1 ∫04 (2x2 − 3x + 40) dx ÷ 4✓ 1
A1 integral = 178.67 (178⅔)✓ 1
A1 mean height 44.7 m✓ 1

Exam tip: A condition on gradient gives an equation in the derivative, while a condition on height gives an equation in the original function; set up one equation of each kind.

365IBMI3S-5-01r36-06Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; 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A space agency designs a closed cylindrical fuel tank with internal volume 12 m3. The tank has radius r metres and height h metres. The total outer surface area of the tank is A m2. The tank is made from a material costing £85 per square metre.
Give numerical answers to 3 significant figures unless stated otherwise, and use unrounded values in later parts.

(a)

Write down an expression for h in terms of r.

(2)
(b)

Show that A = 2πr2 + 24/r.

(3)
(c)

Find dA/dr.

(2)
(d)

Find the value of r that minimises the surface area.

(3)
(e)

Show that this value of r gives a minimum.

(2)
(f)

Find the minimum surface area.

(2)
(g)

Find the cost of the material for the tank, to the nearest £, using the unrounded value of the minimum surface area.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 12 = πr2h✓ 1
A1 h = 12/(πr2)✓ 1
Part (b)
M1 A = 2πr2 + 2πrh✓ 1
M1 substitute h from (a)✓ 1
A1 2πr2 + 24/r (AG)✓ 1
Part (c)
M1 attempt to differentiate✓ 1
A1 4πr − 24r−2✓ 1
Part (d)
M1 set dA/dr = 0✓ 1
A1 r3 = 6/π✓ 1
A1 r = 1.24 m✓ 1
Part (e)
M1 d2A/dr2 = 4π + 48r−3✓ 1
R1 this is positive for r > 0, so the stationary point is a minimum✓ 1
Part (f)
M1 substitute the unrounded value of r into A✓ 1
A1 29.0 m2✓ 1
Part (g)
M1 29.0158… × 85✓ 1
A1 £2466✓ 1

Exam tip: Always check the second derivative or a sign change before claiming a stationary point is a minimum; the question often awards a mark for it.

366IBMI3S-5-01r36-08Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; 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integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; 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After an engine burn, the speed of a space probe relative to a moon is modelled by v(t) = −0.06t2 + 1.2t km s−1, where t is the time in seconds, 0 ≤ t ≤ 20. The speed is zero at both t = 0 and t = 20.

(a)

Show that the maximum speed occurs when t = 10, and find this maximum speed.

(3)
(b)

Find the distance travelled by the probe between t = 0 and t = 20 by integration.

(3)
(c)

Complete the table of values and use the trapezoidal rule with 4 equal intervals to estimate the distance.
t (s): 0, 5, 10, 15, 20
v (km/s): 0, ___, 6, ___, 0

(3)
(d)

Find the percentage error in the estimate from (c), using 80 km as the exact value.

(2)
(e)

Explain why the trapezoidal rule gives an underestimate in this case.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 v′(t) = −0.12t + 1.2 = 0✓ 1
A1 t = 10✓ 1
A1 v(10) = 6 km s−1✓ 1
Part (b)
M1 attempt to integrate v(t) from 0 to 20✓ 1
A1 [−0.02t3 + 0.6t2]✓ 1
A1 80 km✓ 1
Part (c)
A1 v(5) = 4.5 and v(15) = 4.5✓ 1
M1 (5/2)[0 + 0 + 2(4.5 + 6 + 4.5)]✓ 1
A1 75 km✓ 1
Part (d)
M1 |80 − 75|/80 × 100 (or ECF from (c))✓ 1
A1 6.25%✓ 1
Part (e)
R1 the graph of v is concave down (v″ < 0)✓ 1
R1 so each chord lies below the curve and the trapezia have less area than the region under the curve✓ 1

Exam tip: To check an estimate, compare it with the exact integral and express the difference as a percentage of the exact value.

367IBMI3S-5-01r36-12Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; 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A rectangular metal sheet measuring 40 cm by 25 cm is used to make an open-topped box for a bridge sensor. A square of side x cm is cut from each corner and the sides are folded up.

(a)

Show that the volume of the box is V = 4x3 − 130x2 + 1000x cm3.

(2)
(b)

Write down the set of possible values of x.

(1)
(c)

Find dV/dx.

(2)
(d)

Find the value of x that gives the maximum volume.

(3)
(e)

Justify that this value gives a maximum, and find the maximum volume.

(3)
(f)

Find the area of sheet metal in the box of maximum volume.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 base dimensions (40 − 2x) and (25 − 2x), height x✓ 1
M1 V = x(40 − 2x)(25 − 2x) and expand to the given form✓ 1
Part (b)
A1 0 < x < 12.5✓ 1
Part (c)
M1 attempt to differentiate✓ 1
A1 12x2 − 260x + 1000✓ 1
Part (d)
M1 set dV/dx = 0✓ 1
A1 x = 5 or x = 16.7✓ 1
R1 reject 16.7 as it is outside 0 < x < 12.5, so x = 5✓ 1
Part (e)
M1 d2V/dx2 = 24x − 260 (or sign change of V′)✓ 1
R1 negative at x = 5, so a maximum✓ 1
A1 V(5) = 2250 cm3✓ 1
Part (f)
M1 area of box = sheet area − 4x2 = 1000 − 4(5)2✓ 1
A1 900 cm2✓ 1

Common error: Stationary points outside the allowed domain must be rejected; always state why a root is not valid.

368IBMI3S-5-04r36-05Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 2Hard13 marks

A cyclist rides towards a junction, stops, and then moves off again. Her speed v m s−1 is modelled by v = 0.02t3 − 0.6t2 + 4.5t, for 0 ≤ t ≤ 18, where t is the time in seconds.

(a)

Find the acceleration at t = 2.

(2)
(b)

Find the times at which her acceleration is zero.

(3)
(c)

Find her speed at each of the times found in (b). Hence state the maximum speed in the interval 0 ≤ t ≤ 18, and when she stops.

(3)
(d)

State the interval of time during which she is slowing down, giving a reason.

(2)
(e)

Find the distance she travels in the first 15 seconds. Give your answer correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 differentiate to find a = dv/dt = 0.06t2 − 1.2t + 4.5✓ 1
A1 a = 2.34 m s−2✓ 1
Part (b)
M1 set 0.06t2 − 1.2t + 4.5 = 0✓ 1
A1 t = 5 s✓ 1
A1 t = 15 s✓ 1
Part (c)
A1 v(5) = 10 m s−1✓ 1
A1 v(15) = 0 m s−1, so she stops at t = 15 s✓ 1
R1 compare with the end values v(0) = 0 and v(18) = 3.24, so the maximum speed is 10 m s−1✓ 1
Part (d)
A1 5 < t < 15✓ 1
R1 the acceleration is negative on this interval (dv/dt < 0)✓ 1
Part (e)
M1 integrate speed between 0 and 15✓ 1
A1 0.005t4 − 0.2t3 + 2.25t2 evaluated at 15✓ 1
A1 84.4 m✓ 1

Exam tip: Zero acceleration marks a maximum or minimum of speed; check the end points of the domain before naming the overall maximum.

369IBMI3S-5-04r36-09Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 2Hard15 marks

Two cyclists, A and B, start together from the same point and ride along the same straight track. For 0 ≤ t ≤ 30, where t is in seconds, their speeds in m s−1 are modelled by vA = 8 and vB = 1.2t − 0.04t2.

(a)

Find the two times at which the cyclists have the same speed.

(3)
(b)

Find the greatest speed reached by cyclist B, and the time at which it occurs.

(3)
(c)

Find the distance travelled by cyclist A in the first 20 seconds.

(1)
(d)

Find the distance travelled by cyclist B in the first 20 seconds. Give your answer correct to 3 significant figures.

(3)
(e)

Hence find how far cyclist B is behind cyclist A at t = 20. Use the unrounded distance from (d), and give your answer correct to 3 significant figures.

(2)
(f)

Find the distance by which cyclist B closes the gap on cyclist A between t = 10 and t = 20. Give your answer correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 set 1.2t − 0.04t2 = 8✓ 1
A1 t = 10 s✓ 1
A1 t = 20 s✓ 1
Part (b)
M1 differentiate vB and set equal to zero, or use a GDC maximum✓ 1
A1 t = 15 s✓ 1
A1 vB = 9 m s−1✓ 1
Part (c)
A1 8 × 20 = 160 m✓ 1
Part (d)
M1 integrate vB from 0 to 20✓ 1
A1 0.6t2 − t3/75 evaluated at 20✓ 1
A1 133 m✓ 1
Part (e)
M1 160 − 133.33…✓ 1
A1 26.7 m✓ 1
Part (f)
M1 integrate (vB − vA) from 10 to 20✓ 1
A1 [0.6t2 − t3/75 − 8t] evaluated between 10 and 20✓ 1
A1 6.67 m✓ 1

Common error: The area between two speed–time graphs is the change in the gap between the riders, not the gap itself.

370IBMI3S-5-04r36-13Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 2Hard14 marks

A company runs data centres. When it installs n thousand servers, its annual profit is modelled by P(n) = −0.01n3 + 0.6n2 − 5n − 20 million euros, for 0 ≤ n ≤ 40.

(a)

Find dP/dn.

(2)
(b)

Find the two values of n for which dP/dn = 0, correct to 3 significant figures.

(3)
(c)

State the interval of n for which the profit is increasing as n increases.

(1)
(d)

Find the value of P at n = 0, at n = 40, and at the larger stationary point, using unrounded values of n where needed.

(3)
(e)

Hence state the number of servers that gives the greatest annual profit, and the value of that profit in million euros correct to 3 significant figures.

(3)
(f)

Find the least value of n for which the company makes a profit. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 dP/dn = −0.03n2 + 1.2n − 5✓ 1
Part (b)
M1 solve −0.03n2 + 1.2n − 5 = 0✓ 1
A1 n = 4.72✓ 1
A1 n = 35.3✓ 1
Part (c)
A1 4.72 < n < 35.3 (ECF from (b))✓ 1
Part (d)
A1 P(0) = −20✓ 1
A1 P(40) = 100✓ 1
A1 P(35.275…) = 111.28… million euros✓ 1
Part (e)
R1 compare the three values from (d); the greatest is at the larger stationary point (not at the end n = 40)✓ 1
A1 n = 35.3 thousand servers✓ 1
A1 greatest profit 111 million euros✓ 1
Part (f)
M1 solve P(n) = 0 (graph, GDC solver) for the first positive root✓ 1
A1 n = 14.0 thousand servers✓ 1

Exam tip: On a closed domain, the greatest value can occur at a stationary point or at an end point, so evaluate the function at all of them.

371IBMI3S-5-01r39-12Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 2Hard14 marks

An architect designs a window in the shape of a rectangle of width 2r metres and height h metres, topped by a semicircle of radius r metres. The total length of the outer frame (the two sides, the base and the semicircular arc) is 12 m.

(a)

Show that h = 6 − (2 + π)r/2.

(2)
(b)

Show that the area of the window is A = 12r − (2 + π/2)r2.

(3)
(c)

Find dA/dr.

(2)
(d)

Find the value of r for which the area is a maximum, to 3 significant figures.

(2)
(e)

Find the maximum area of the window, to 3 significant figures. Use the unrounded r.

(2)
(f)

Show that this value of r gives a maximum.

(2)
(g)

Find the height h of the rectangular part at this maximum, to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 frame length 2h + 2r + πr = 12✓ 1
A1 rearrange to the given result (AG)✓ 1
Part (b)
M1 area = 2rh + ½πr2✓ 1
M1 substitute h from (a)✓ 1
A1 simplify to the given result (AG)✓ 1
Part (c)
M1 differentiate✓ 1
A1 12 − (4 + π)r✓ 1
Part (d)
M1 set dA/dr = 0✓ 1
A1 r = 1.68✓ 1
Part (e)
M1 substitute unrounded r into A✓ 1
A1 10.1 m2✓ 1
Part (f)
M1 d2A/dr2 = −(4 + π)✓ 1
R1 negative, so a maximum✓ 1
Part (g)
A1 h = 1.68 m✓ 1

Common error: The curved edge of the semicircle contributes πr to the frame length, not 2πr or πr²; the diameter is not part of the frame.

372IBMI3S-5-01r39-14Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule Paper 2Hard14 marks

A courier plans a 240 km delivery at a constant speed of v km/h, where 40 ≤ v ≤ 90. The van uses fuel costing 0.0004v2 − 0.04v + 1.5 euros per kilometre. The driver is paid €18 for each hour of the journey.

(a)

Show that the total cost of the journey is T(v) = 0.096v2 − 9.6v + 360 + 4320/v euros.

(3)
(b)

Find T′(v).

(3)
(c)

Use your GDC to find the speed, in km/h to 3 significant figures, at which the cost is a minimum.

(2)
(d)

Calculate the minimum cost of the journey to the nearest euro, using the unrounded speed.

(2)
(e)

Show that this speed gives a minimum cost.

(2)
(f)

Find T′(75) to 3 significant figures and interpret its value.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 fuel cost = 240(0.0004v2 − 0.04v + 1.5)✓ 1
M1 time = 240/v hours so driver cost = 18 × 240/v✓ 1
A1 sum simplified to the given result (AG)✓ 1
Part (b)
M1 attempt to differentiate✓ 1
A1 0.192v − 9.6✓ 1
A1 −4320v−2✓ 1
Part (c)
M1 solve T′(v) = 0 (or find the minimum of T)✓ 1
A1 v = 56.9 km/h✓ 1
Part (d)
M1 substitute unrounded v into T✓ 1
A1 €200✓ 1
Part (e)
M1 T″(v) = 0.192 + 8640v−3✓ 1
R1 positive for v > 0, so a minimum✓ 1
Part (f)
A1 T′(75) = 4.03✓ 1
R1 at 75 km/h the cost increases by about €4.03 for each extra 1 km/h of speed✓ 1

Common error: Remember that the driver's cost depends on the time taken, 240/v, which falls as the speed rises; forgetting this makes the cost model one-sided.

373IBMI3S-5-02r41-10Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; 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integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 2Hard12 marks

A space probe fires a thruster for 20 seconds. Its acceleration is \(a = 12 - 0.6t\) m s−2 for \(0\le t\le20\), where t is the time in seconds after the burn begins. The speed of the probe at \(t=0\) is 50 m s−1.

(a)

Find the acceleration at \(t=15\).

(1)
(b)

Show that the speed of the probe is \(v = 50 + 12t - 0.3t^2\).

(3)
(c)

Find the speed at the end of the burn.

(1)
(d)

Find the distance travelled by the probe during the burn.

(3)
(e)

Use the trapezoidal rule with 4 equal intervals to estimate the distance travelled in the burn.

(3)
(f)

Explain why the trapezoidal estimate is an underestimate.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(12-0.6\times15=3\) m s−2✓ 1
Part (b)
M1 \(\int(12-0.6t)\,dt = 12t-0.3t^2+c\)✓ 1
M1 use \(v=50\) at \(t=0\) to find \(c\)✓ 1
A1 \(v=50+12t-0.3t^2\) (AG)✓ 1
Part (c)
A1 \(v(20)=50+240-120=170\) m s−1✓ 1
Part (d)
M1 \(\int_0^{20}(50+12t-0.3t^2)\,dt\)✓ 1
A1 \(\left[50t+6t^2-0.1t^3\right]_0^{20}\) or GDC evaluation✓ 1
A1 2600 m✓ 1
Part (e)
M1 speeds at \(t=0,5,10,15,20\): \(50,\,102.5,\,140,\,162.5,\,170\)✓ 1
M1 \(\frac{5}{2}\left(50+170+2(102.5+140+162.5)\right)\)✓ 1
A1 2575 m✓ 1
Part (f)
R1 the graph of v against t is concave down (a decreasing gradient), so each trapezium lies below the curve✓ 1

Exam tip: For a 'show that' question, include the constant of integration and the condition used to find it.

374IBMI3S-5-02r41-11Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; 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Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 2Hard14 marks

The profile of a tunnel arch is modelled by \(y = 20 - 0.05x^2\) for \(-20\le x\le20\), where x and y are in metres, with the ground at \(y=0\). A rectangular opening for vehicles is cut in the arch. It has its base on the ground, is symmetrical about the y-axis, has width \(2x\) and its top corners lie on the arch. The area of the rectangle is A m2.

(a)

Show that \(A = 40x - 0.1x^3\).

(2)
(b)

Find \(\frac{dA}{dx}\).

(2)
(c)

Find the value of x that gives the maximum area. Give your answer to 3 significant figures.

(2)
(d)

Find the maximum area of the rectangle. Give your answer to 3 significant figures.

(2)
(e)

Justify that this value of x gives a maximum.

(2)
(f)

Find the maximum area of the rectangle as a percentage of the total area under the arch. Give your answer to 3 significant figures.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 width \(2x\) and height \(20-0.05x^2\)✓ 1
A1 \(A=2x(20-0.05x^2)=40x-0.1x^3\) (AG)✓ 1
Part (b)
M1 differentiate term by term✓ 1
A1 \(40-0.3x^2\)✓ 1
Part (c)
M1 \(40-0.3x^2=0\)✓ 1
A1 \(x=11.5\) m✓ 1
Part (d)
M1 substitute the unrounded value of x into A✓ 1
A1 308 m2✓ 1
Part (e)
M1 consider the sign of \(\frac{dA}{dx}\) either side of \(x=11.5\), or find \(\frac{d^2A}{dx^2}=-0.6x\)✓ 1
R1 gradient changes from positive to negative (or second derivative is negative), so a maximum✓ 1
Part (f)
M1 \(\int_{-20}^{20}(20-0.05x^2)\,dx\)✓ 1
A1 \(533\) m2 (533.33…)✓ 1
M1 \(\frac{307.9\ldots}{533.3\ldots}\times100\) (ECF from (d))✓ 1
A1 57.7 %✓ 1

Common error: Remember the rectangle's width is \(2x\) (both sides of the axis), not \(x\).

375IBMI3S-5-02r41-12Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; 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Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule Paper 2Hard14 marks

A coffee roastery's delivery van uses fuel at a rate of \(L = 0.0005v^2 + \frac{512}{v}\) litres per 100 km when travelling at a steady speed of v km h−1, for \(40\le v\le130\).

(a)

Find the fuel used per 100 km at 60 km h−1. Give your answer to 3 significant figures.

(1)
(b)

Find \(\frac{dL}{dv}\).

(2)
(c)

Find the value of \(\frac{dL}{dv}\) at \(v=60\) and interpret its meaning. Give the value to 3 significant figures.

(2)
(d)

Find the speed that minimises the fuel used per 100 km.

(3)
(e)

Find the minimum fuel used per 100 km.

(1)
(f)

Show that \(v=80\) gives a minimum.

(2)
(g)

The van makes a 450 km delivery. Fuel costs €1.65 per litre. Find how much is saved by driving at 80 km h−1 rather than 100 km h−1. Give your answer to the nearest cent.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(L=0.0005\times3600+\frac{512}{60}=10.3\) litres per 100 km✓ 1
Part (b)
M1 write \(\frac{512}{v}=512v^{-1}\)✓ 1
A1 \(\frac{dL}{dv}=0.001v-512v^{-2}\)✓ 1
Part (c)
A1 \(0.06-\frac{512}{3600}=-0.0822\)✓ 1
R1 at 60 km h−1 fuel use per 100 km is decreasing by about 0.0822 litres for each 1 km h−1 increase in speed✓ 1
Part (d)
M1 \(0.001v-\frac{512}{v^2}=0\)✓ 1
M1 \(v^3=512000\)✓ 1
A1 \(v=80\) km h−1✓ 1
Part (e)
A1 \(L(80)=3.2+6.4=9.6\) litres per 100 km✓ 1
Part (f)
M1 sign of \(\frac{dL}{dv}\) either side of 80 (negative at 60, positive at 100), or a second derivative test✓ 1
R1 gradient changes from negative to positive, so a minimum✓ 1
Part (g)
M1 \(L(100)=5+5.12=10.12\) litres per 100 km✓ 1
M1 \(4.5\times1.65\times(10.12-9.6)\)✓ 1
A1 €3.86✓ 1

Common error: The journey is 4.5 lots of 100 km, so multiply the fuel rate by 4.5, not by 450.

376IBMI3S-5-02r42-12Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Hard13 marks

The number of people inside a library t hours after it opens is modelled by N(t) = t3 − 15t2 + 63t + 20, for 0 ≤ t ≤ 10.

(a)

Find N′(t).

(2)
(b)

Find the values of t at which the number of people in the library is neither increasing nor decreasing.

(3)
(c)

Find the number of people in the library at each of the times found in part (b), and state which of these is a local maximum and which is a local minimum.

(3)
(d)

Find the rate of change of the number of people in the library when t = 10, stating the units.

(2)
(e)

The number of people is decreasing fastest at time t = k. Find the value of k.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate, at least two terms correct✓ 1
A1 3t2 − 30t + 63✓ 1
Part (b)
M1 3t2 − 30t + 63 = 0 (or t2 − 10t + 21 = 0)✓ 1
A1 t = 3✓ 1
A1 t = 7✓ 1
Part (c)
A1 N(3) = 101✓ 1
A1 N(7) = 69✓ 1
A1 local maximum of 101 people at t = 3 and local minimum of 69 people at t = 7 (supported by the sign of N′ either side or by a graph)✓ 1
Part (d)
M1 N′(10) = 3(100) − 300 + 63✓ 1
A1 63 people per hour✓ 1
Part (e)
M1 N″(t) = 6t − 30✓ 1
M1 setting N″(t) = 0 (the minimum of N′ occurs where its gradient is zero)✓ 1
A1 k = 5 hours (N′(5) = -12 people per hour)✓ 1

Common error: A zero of the first derivative marks a turning point of N, but the fastest rate of decrease is at the turning point of N′, which is where the second derivative is zero.

377IBMI3S-5-01r44-10Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; 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Optimisation in context; Optimisation in context Paper 2Hard15 marks

A closed cylindrical storage tank at a rail depot has radius r metres and height h metres. Its total external surface area is 150 m2.

(a)

Find an expression for h in terms of r.

(2)
(b)

Hence show that the volume of the tank is \(V=75r-\pi r^3\).

(2)
(c)

Find \(\frac{dV}{dr}\).

(2)
(d)

Find the value of r that maximises the volume.

(3)
(e)

Using the unrounded value of r, find the maximum volume.

(2)
(f)

Show that at the maximum volume, \(h=2r\).

(2)
(g)

Justify that the stationary value is a maximum.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(2\pi r^2+2\pi rh=150\)✓ 1
A1 \(h=\frac{75-\pi r^2}{\pi r}\)✓ 1
Part (b)
M1 substitute into \(V=\pi r^2h\)✓ 1
A1 \(V=r(75-\pi r^2)=75r-\pi r^3\) (shown)✓ 1
Part (c)
M1 attempt to differentiate✓ 1
A1 \(75-3\pi r^2\)✓ 1
Part (d)
M1 set \(\frac{dV}{dr}=0\)✓ 1
A1 \(r^2=\frac{25}{\pi}\)✓ 1
A1 \(r=2.82\) m✓ 1
Part (e)
M1 substitute \(r\) into \(V=75r-\pi r^3\)✓ 1
A1 \(V=141\) m3✓ 1
Part (f)
M1 use \(75=3\pi r^2\) in the expression for h✓ 1
A1 \(h=\frac{3\pi r^2-\pi r^2}{\pi r}=2r\) (shown)✓ 1
Part (g)
M1 \(\frac{d^2V}{dr^2}=-6\pi r\)✓ 1
R1 negative for \(r>0\), so a maximum✓ 1

Exam tip: With a constraint, eliminate one variable first so the quantity to optimise depends on a single variable.

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Optimisation in context; Optimisation in context Paper 2Hard13 marks

During a 20-minute training run, the lactate concentration in an athlete's blood, L mmol l−1, is modelled by \(L(t)=0.01t^3-0.3t^2+2.4t+1.5\), where t is in minutes and \(0\le t\le 20\).

(a)

Find \(L'(t)\).

(2)
(b)

Find the rate of change of lactate concentration at \(t=2\).

(1)
(c)

Find the values of t at which the concentration has stationary values.

(3)
(d)

Find the lactate concentration at each stationary point.

(2)
(e)

Determine the nature of each stationary point.

(2)
(f)

Find the greatest rate of decrease of lactate concentration, and state when it occurs.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 \(0.03t^2-0.6t+2.4\)✓ 1
Part (b)
A1 \(L'(2)=1.32\) mmol l−1 per minute✓ 1
Part (c)
M1 set \(L'(t)=0\)✓ 1
A1 \(t=5.53\)✓ 1
A1 \(t=14.5\)✓ 1
Part (d)
A1 \(L(5.53)=7.29\)✓ 1
A1 \(L(14.5)=3.71\)✓ 1
Part (e)
M1 \(L''(t)=0.06t-0.6\)✓ 1
R1 \(L''(5.53)<0\) so a maximum, and \(L''(14.5)>0\) so a minimum✓ 1
Part (f)
M1 set \(L''(t)=0\)✓ 1
A1 \(t=10\) minutes✓ 1
A1 \(L'(10)=-0.6\), so the greatest rate of decrease is 0.6 mmol l−1 per minute✓ 1

Common error: The steepest decrease is where the gradient is most negative, which is a stationary point of the derivative, not of the original function.

379IBMI3S-5-01r45-07Integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; 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integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context Paper 2Hard14 marks

A hospital orders a closed cylindrical bin for clinical waste. The bin has base radius r cm, height h cm and volume 50 000 cm3. The base and lid are made of plastic costing €0.02 per cm2. The curved side is made of plastic costing €0.011 per cm2. The total cost of the plastic is €C.

(a)

Show that h = 50 000/(πr2).

(2)
(b)

Show that C = 0.04πr2 + 1100/r.

(3)
(c)

Find dC/dr.

(2)
(d)

Find the value of r that minimises the cost. Give your answer correct to 3 significant figures.

(3)
(e)

Show that this value of r gives a minimum cost.

(2)
(f)

Find the minimum cost of the plastic. Use the unrounded value of r. Give your answer to the nearest cent.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 volume = πr2h = 50 000✓ 1
A1 rearrange to h = 50 000/(πr2) (AG)✓ 1
Part (b)
M1 cost of base and lid = 0.02 × 2πr2 = 0.04πr2✓ 1
M1 cost of side = 0.011 × 2πrh✓ 1
A1 substitute h: 0.011 × 2πr × 50 000/(πr2) = 1100/r, giving the result (AG)✓ 1
Part (c)
M1 attempt to differentiate, including 1100r−1✓ 1
A1 0.08πr − 1100r−2✓ 1
Part (d)
M1 set dC/dr = 0✓ 1
A1 r3 = 1100/(0.08π) = 4376.8…✓ 1
A1 r = 16.4 cm✓ 1
Part (e)
M1 d2C/dr2 = 0.08π + 2200r−3 (or a sign change of the gradient)✓ 1
R1 this is positive for r > 0, so the stationary point is a minimum✓ 1
Part (f)
M1 substitute r = (1100/(0.08π))1/3 = 16.3575… into C✓ 1
A1 €100.87✓ 1

Exam tip: In a 'show that' question, every step must be visible: write the substitution of h explicitly before simplifying.

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The mass of honey stored in a beehive is modelled by M(d) = 0.002d3 − 0.15d2 + 3.2d + 5 kilograms, where d is the number of days after the start of the season, for 0 ≤ d ≤ 50.

(a)

Find M′(d).

(2)
(b)

Find the rate at which the mass of honey is changing when d = 10.

(2)
(c)

Find the values of d for which the mass of honey is neither increasing nor decreasing. Give your answers correct to 3 significant figures.

(3)
(d)

State which of these two values of d gives a local maximum and which gives a local minimum, justifying your answer.

(2)
(e)

Find the greatest mass of honey the hive holds in the first 20 days. Use the unrounded value of d. Give your answer correct to 3 significant figures.

(2)
(f)

Write down the interval(s) of d, for 0 ≤ d ≤ 50, on which the mass of honey is increasing.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 0.006d2 − 0.3d + 3.2✓ 1
Part (b)
M1 substitute d = 10 into M′(d)✓ 1
A1 0.8 kg per day✓ 1
Part (c)
M1 set M′(d) = 0✓ 1
A1 d = 15.4✓ 1
A1 d = 34.6✓ 1
Part (d)
A1 local maximum at d = 15.4 and local minimum at d = 34.6✓ 1
R1 M′ changes from positive to negative at the first value and from negative to positive at the second (or M″ < 0 and M″ > 0 respectively)✓ 1
Part (e)
M1 substitute d = (0.3 − √0.0132)/0.012 = 15.4257… into M(d)✓ 1
A1 26.0 kg✓ 1
Part (f)
A1 0 ≤ d < 15.4✓ 1
A1 34.6 < d ≤ 50✓ 1

Common error: A zero gradient only locates a stationary point; you must show the sign of the gradient on each side (or the sign of the second derivative) to classify it.

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Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context Paper 2Hard14 marks

In an online game, the rate at which a player earns in-game coins is modelled by f(t) = 2 + 0.3t2 coins per minute, where t is the time in minutes since the player logged in, for 0 ≤ t ≤ 6. The total number of coins earned in the first 6 minutes is the area under the graph of f from t = 0 to t = 6.

(a)

Complete the table of values: find f(0), f(2), f(4) and f(6).

(2)
(b)

Use the trapezoidal rule with 3 intervals of equal width to estimate the total number of coins earned in the first 6 minutes.

(3)
(c)

Find the exact total number of coins earned in the first 6 minutes by integration.

(2)
(d)

Find the percentage error of the trapezoidal estimate in part (b), to 3 significant figures.

(2)
(e)

Explain, without further calculation, why the trapezoidal rule gives an overestimate here.

(2)
(f)

For n intervals of equal width, the overestimate in the trapezoidal rule for this function is 10.8 / n2 coins. Find the smallest n for which the percentage error is less than 0.5%.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 f(0) = 2 and f(2) = 3.2✓ 1
A1 f(4) = 6.8 and f(6) = 12.8✓ 1
Part (b)
M1 trapezoidal rule with strip width 2✓ 1
A1 correct substitution: ½ × 2 × [2 + 12.8 + 2(3.2 + 6.8)]✓ 1
A1 estimate = 34.8 coins✓ 1
Part (c)
M1 integrates: 2t + 0.1t3✓ 1
A1 value = 12 + 21.6 = 33.6 coins✓ 1
Part (d)
M1 percentage error = (34.8 − 33.6) / 33.6 × 100✓ 1
A1 3.57%✓ 1
Part (e)
R1 the graph of f is curved upwards (f″ = 0.6 > 0, concave up)✓ 1
R1 so each chord joining two points on the curve lies above the curve, and each trapezium has more area than the region under the curve✓ 1
Part (f)
M1 requires 10.8 / n2 < 0.005 × 33.6 = 0.168✓ 1
M1 n2 > 64.28…, so n > 8.02…✓ 1
A1 n = 9 (checking: n = 8 gives 0.16875, which is a 0.502% error)✓ 1

Exam tip: For a curve that bends upwards, chords lie above it, so the trapezoidal rule overestimates; for a curve that bends downwards it underestimates.

382IBMI3S-5-03r47-07Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; 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Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; 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A fish farm needs an open-topped cylindrical tank of volume 12 m3. The tank has base radius r metres and height h metres. The base costs €40 per square metre to build and the curved side costs €25 per square metre to build. Let €C be the total cost of building the tank.

(a)

Show that C = 40πr2 + 600/r.

(3)
(b)

Find dC/dr.

(2)
(c)

Find the value of r that minimises C, correct to 3 significant figures.

(3)
(d)

Using the unrounded value of r, find the minimum cost, to the nearest euro.

(2)
(e)

Justify that this value of r gives a minimum cost.

(2)
(f)

Planning rules limit the height of the tank to 2.0 m. Find the height of the tank for the minimum cost design, using the unrounded r, correct to 3 significant figures, and state whether the design meets the rule.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 h = 12/(πr2)✓ 1
M1 C = 40πr2 + 25 × 2πrh✓ 1
A1 curved area 2πrh = 24/r, so cost 600/r (AG)✓ 1
Part (b)
M1 attempt to differentiate✓ 1
A1 80πr − 600/r2✓ 1
Part (c)
M1 set dC/dr = 0✓ 1
A1 r3 = 600/(80π) = 2.387…✓ 1
A1 r = 1.34 m✓ 1
Part (d)
M1 substitute r = 1.3365… into C✓ 1
A1 €673✓ 1
Part (e)
M1 d2C/dr2 = 80π + 1200/r3✓ 1
R1 positive for all r > 0, so a minimum✓ 1
Part (f)
A1 h = 12/(π × 1.3365…2) = 2.14 m✓ 1
R1 2.14 > 2.0, so the design does not meet the rule✓ 1

Exam tip: Eliminate one variable using the constraint before differentiating, and keep the unrounded optimum for any later calculation.

383IBMI3S-5-03r47-09Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; 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Optimisation in context; Integration and the trapezoidal rule; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; 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A haulage firm models the running cost of a lorry, in euros per kilometre, by C(v) = 0.00004v2 + 24/v, where v km h−1 is the constant speed, for 30 ≤ v ≤ 100. The first term models fuel and the second term models the driver's wages of €24 per hour.

(a)

Find the cost per kilometre when v = 50.

(2)
(b)

Find C′(v).

(2)
(c)

Find the speed that minimises the cost per kilometre, correct to 3 significant figures.

(3)
(d)

Using the unrounded speed from (c), find the minimum cost per kilometre, correct to 3 significant figures.

(2)
(e)

The lorry makes a journey of 420 km at the speed from (c). Using the unrounded cost per kilometre, find the total cost to the nearest euro.

(2)
(f)

A speed limit of 60 km h−1 applies for the whole of another route. Find the lowest possible cost per kilometre on that route, and justify your answer.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitute v = 50✓ 1
A1 €0.58 per km✓ 1
Part (b)
M1 attempt to differentiate✓ 1
A1 0.00008v − 24/v2✓ 1
Part (c)
M1 set C′(v) = 0✓ 1
A1 v3 = 300000✓ 1
A1 v = 66.9 km h−1✓ 1
Part (d)
M1 substitute v = 66.943… into C✓ 1
A1 €0.538 per km✓ 1
Part (e)
M1 420 × 0.53776…✓ 1
A1 €226✓ 1
Part (f)
M1 C(60) = 0.00004 × 602 + 24/60✓ 1
A1 €0.544 per km✓ 1
R1 C′ < 0 for v < 66.9, so C decreases up to 60 and the lowest cost occurs at the limit✓ 1

Exam tip: A constraint such as a speed limit can move the optimum to the edge of the domain; always compare the stationary point with the allowed range.

384IBMI3S-5-03r47-12Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Optimisation in context; Gradients, tangents and rates of change; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; Optimisation in context; Optimisation in context; Optimisation in context; Gradients, tangents and rates of change; optimisation in context; integration and the trapezoidal rule; Integration and the trapezoidal rule; Gradients, tangents and rates of change; 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The cross-section of a breakwater is modelled by the curve y = x3 − 6x2 + 9x + 2, for 0 ≤ x ≤ 4, where x is the horizontal distance in metres and y is the height in metres above the sea bed.

(a)

Find dy/dx.

(2)
(b)

Find the coordinates of the two points on the curve where the gradient is zero.

(4)
(c)

Find the equation of the tangent to the curve at the point where x = 2. Give your answer in the form y = mx + c.

(3)
(d)

Find the equation of the normal to the curve at the point where x = 2.

(2)
(e)

The tangent and the normal from (d) and (c) each cross the line y = 0. Find the area of the triangle formed by these two crossing points and the point (2, 4), correct to 3 significant figures.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to differentiate✓ 1
A1 3x2 − 12x + 9✓ 1
Part (b)
M1 set 3x2 − 12x + 9 = 0✓ 1
A1 x = 1 and x = 3✓ 1
A1 (1, 6)✓ 1
A1 (3, 2)✓ 1
Part (c)
A1 y(2) = 4 and gradient −3✓ 1
M1 y − 4 = −3(x − 2)✓ 1
A1 y = −3x + 10✓ 1
Part (d)
M1 normal gradient = −1/(−3) = 1/3✓ 1
A1 y − 4 = (1/3)(x − 2), so y = x/3 + 10/3✓ 1
Part (e)
A1 tangent crosses y = 0 at x = 10/3✓ 1
A1 normal crosses y = 0 at x = −10✓ 1
M1 area = ½ × base × height with base 40/3 and height 4 (ECF)✓ 1
A1 26.7 m2 (80/3)✓ 1

Exam tip: The normal is perpendicular to the tangent: its gradient is the negative reciprocal of the tangent gradient.