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IB Maths Applications & Interpretation · SL and HL · Topic 5

Calculus

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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →

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+ Add to test✓ In your test1IBMI-034-02-r1-01Integration and the trapezoidal rule Paper SLEasy5 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 point	Mark	Notes
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.

+ Add to test✓ In your test2IBMI-034-02-r1-02Integration and the trapezoidal rule Paper SLEasy6 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 point	Mark	Notes
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.

+ Add to test✓ In your test3IBMI-034-00-r1-01Gradients, tangents and rates of change Paper SLEasy6 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 point	Mark	Notes
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.

+ Add to test✓ In your test4IBMI-034-07-r2-02Integration and the trapezoidal rule Paper SLEasy6 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 point	Mark	Notes
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.

+ Add to test✓ In your test5IBMI-034-00-r2-02Gradients, tangents and rates of change Paper SLEasy7 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 point	Mark	Notes
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.

+ Add to test✓ In your test6IBMI-034-01-r2-01Optimisation in context Paper SLEasy5 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 point	Mark	Notes
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.

+ Add to test✓ In your test7IBMI-034-01-r2-02Optimisation in context Paper SLEasy6 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 point	Mark	Notes
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.

+ Add to test✓ In your test8IBMI-034-06-r1-01Optimisation in context Paper SLEasy5 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 point	Mark	Notes
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.

+ Add to test✓ In your test9IBMI-034-07-r1-01Integration and the trapezoidal rule Paper SLEasy5 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 point	Mark	Notes
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.

+ Add to test✓ In your test10IBMI-034-07-r1-02Integration and the trapezoidal rule Paper SLEasy6 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 point	Mark	Notes
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.

+ Add to test✓ In your test11IBMI-034-02-r2-02Integration and the trapezoidal rule Paper SLEasy5 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 point	Mark	Notes
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.

+ Add to test✓ In your test12IBMI-034-05-r1-02Gradients, tangents and rates of change Paper SLEasy7 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 point	Mark	Notes
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.

+ Add to test✓ In your test13IBMI-034-06-r2-02Optimisation in context Paper SLEasy7 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 point	Mark	Notes
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 $2x + y = 80$.

+ Add to test✓ In your test14IBMI-034-05-r2-02Gradients, tangents and rates of change Paper SLEasy7 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 point	Mark	Notes
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.

+ Add to test✓ In your test15IBMI-034-00-r1-02Gradients, tangents and rates of change Paper SLEasy8 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 point	Mark	Notes
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.

+ Add to test✓ In your test16IBMI-034-02-r2-01Integration and the trapezoidal rule Paper SLEasy6 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 point	Mark	Notes
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.

+ Add to test✓ In your test17IBMI-034-00-r2-01Gradients, tangents and rates of change Paper SLEasy5 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 point	Mark	Notes
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.

+ Add to test✓ In your test18IBMI-034-06-r2-01Optimisation in context Paper SLEasy5 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 point	Mark	Notes
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.

+ Add to test✓ In your test19IBMI-034-07-r2-01Integration and the trapezoidal rule Paper SLEasy5 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 point	Mark	Notes
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.

+ Add to test✓ In your test20IBMI-034-01-r1-01Optimisation in context Paper SLEasy5 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 point	Mark	Notes
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.

+ Add to test✓ In your test21IBMI-034-01-r1-02Optimisation in context Paper SLEasy6 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 point	Mark	Notes
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.

+ Add to test✓ In your test22IBMI-034-05-r1-01Gradients, tangents and rates of change Paper SLEasy5 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 point	Mark	Notes
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).

+ Add to test✓ In your test23IBMI-034-01-r1-03Optimisation in context Paper SLMedium13 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 point	Mark	Notes
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.

+ Add to test✓ In your test24IBMI-034-01-r1-05Optimisation in context Paper SLMedium12 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 point	Mark	Notes
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.

+ Add to test✓ In your test25IBMI-034-00-r1-03Gradients, tangents and rates of change Paper SLMedium14 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 point	Mark	Notes
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.

+ Add to test✓ In your test26IBMI-034-00-r1-04Gradients, tangents and rates of change Paper SLMedium7 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 point	Mark	Notes
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.

+ Add to test✓ In your test27IBMI-034-05-r2-03Gradients, tangents and rates of change Paper SLMedium12 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 point	Mark	Notes
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.

+ Add to test✓ In your test28IBMI-034-07-r2-04Integration and the trapezoidal rule Paper SLMedium5 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 point	Mark	Notes
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.

+ Add to test✓ In your test29IBMI-034-00-r2-03Gradients, tangents and rates of change Paper SLMedium12 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 point	Mark	Notes
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.

+ Add to test✓ In your test30IBMI-034-01-r2-04Optimisation in context Paper SLMedium7 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 point	Mark	Notes
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.

+ Add to test✓ In your test31IBMI-034-06-r1-03Optimisation in context Paper SLMedium9 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 point	Mark	Notes
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.

+ Add to test✓ In your test32IBMI-034-06-r1-04Optimisation in context Paper SLMedium10 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 point	Mark	Notes
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.

+ Add to test✓ In your test33IBMI-034-07-r1-03Integration and the trapezoidal rule Paper SLMedium11 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 point	Mark	Notes
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.

+ Add to test✓ In your test34IBMI-034-07-r1-04Integration and the trapezoidal rule Paper SLMedium5 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 point	Mark	Notes
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.

+ Add to test✓ In your test35IBMI-034-07-r1-06Integration and the trapezoidal rule Paper SLMedium8 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 point	Mark	Notes
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.

+ Add to test✓ In your test36IBMI-034-02-r2-03Integration and the trapezoidal rule Paper SLMedium7 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 point	Mark	Notes
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.

+ Add to test✓ In your test37IBMI-034-05-r1-04Gradients, tangents and rates of change Paper SLMedium7 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 point	Mark	Notes
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 $+0.4$.

+ Add to test✓ In your test38IBMI-034-06-r2-03Optimisation in context Paper SLMedium11 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 point	Mark	Notes
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.

+ Add to test✓ In your test39IBMI-034-02-r1-03Integration and the trapezoidal rule Paper SLMedium7 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 point	Mark	Notes
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.

+ Add to test✓ In your test40IBMI-034-06-r2-04Optimisation in context Paper SLMedium12 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 point	Mark	Notes
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.

+ Add to test✓ In your test41IBMI-034-02-r2-04Integration and the trapezoidal rule Paper SLMedium12 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 point	Mark	Notes
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.

+ Add to test✓ In your test42IBMI-034-07-r2-03Integration and the trapezoidal rule Paper SLMedium11 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 point	Mark	Notes
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.

+ Add to test✓ In your test43IBMI-034-00-r2-04Gradients, tangents and rates of change Paper SLMedium7 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 point	Mark	Notes
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.

+ Add to test✓ In your test44IBMI-034-01-r2-03Optimisation in context Paper SLMedium11 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 point	Mark	Notes
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.

+ Add to test✓ In your test45IBMI-034-05-r1-03Gradients, tangents and rates of change Paper SLMedium12 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 point	Mark	Notes
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.

+ Add to test✓ In your test46IBMI-034-02-r1-04Integration and the trapezoidal rule Paper SLMedium11 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 point	Mark	Notes
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.

+ Add to test✓ In your test47IBMI-034-05-r2-06Gradients, tangents and rates of change Paper SLHard11 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 point	Mark	Notes
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.

+ Add to test✓ In your test48IBMI-034-06-r1-06Optimisation in context Paper SLHard12 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 point	Mark	Notes
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.

+ Add to test✓ In your test49IBMI-034-02-r1-06Integration and the trapezoidal rule Paper SLHard13 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 point	Mark	Notes
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.

+ Add to test✓ In your test50IBMI-034-03-r1-01Differential equations and slope fields Paper HLEasy7 marks

A cup of tea is left to cool in a room where the air temperature is 20 °C. The temperature of the tea, T °C, t minutes after it is poured, is modelled by the differential equation
$\frac{dT}{dt} = -k(T-20)$, where k is a positive constant.
When the tea is poured its temperature is 90 °C. Ten minutes later its temperature is 60 °C.

(a)

By solving the differential equation, show that $T = 20 + 70e^{-kt}$.

(3)

(b)

Find the value of k.

(2)

(c)

Find the temperature of the tea 25 minutes after it is poured.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Separates variables: $\int \frac{1}{T-20}\,dT = \int -k\,dt$	✓ 1
$\ln\|T-20\| = -kt + c$	✓ 1
Uses T = 90 when t = 0 to give $T-20 = 70e^{-kt}$, so $T = 20 + 70e^{-kt}$	✓ 1
Part (b)
Substitutes t = 10, T = 60: $40 = 70e^{-10k}$	✓ 1
$k = \frac{1}{10}\ln\frac{7}{4} = 0.0560$ (min⁻¹)	✓ 1
Part (c)
Substitutes t = 25 with their k: $T = 20 + 70e^{-25(0.05596)}$	✓ 1
T = 37.3 °C	✓ 1

Exam tip: Keep the unrounded value of k in your calculator memory for the final part; using k = 0.056 gives 37.3 here, but rounding early often changes the third significant figure.

+ Add to test✓ In your test51IBMI-034-03-r1-02Differential equations and slope fields Paper HLEasy5 marks

The volume of water, y thousand litres, in a storage tank t hours after a pump is switched on is modelled by
$\frac{dy}{dt} = t - y$, for t ≥ 0.
A slope field is to be drawn for this differential equation.

(a)

Find the gradient of the line segment in the slope field at each of the points (1, 3), (2, 2) and (3, 1).

(2)

(b)

Write down the equation of the curve on which every line segment of the slope field is horizontal.

(1)

(c)

Show that the straight line y = t − 1 is itself a solution curve of the differential equation.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Substitutes into t − y for at least one point	✓ 1
Gradients −2, 0 and 2 respectively	✓ 1
Part (b)
y = t	✓ 1
Part (c)
For y = t − 1, $\frac{dy}{dt} = 1$	✓ 1
and t − y = t − (t − 1) = 1, so both sides are equal and the line satisfies the equation	✓ 1

Common error: An isocline is not usually a solution curve. To show a line is a solution you must check that its own gradient equals the value of the differential equation at every point on it.

+ Add to test✓ In your test52IBMI-034-04-r2-01Euler's method and coupled systems Paper HLEasy5 marks

A cup of tea is left to cool in a room. Its temperature, T °C, after t minutes satisfies the differential equation
$\frac{dT}{dt} = -0.08(T - 20)$, with T = 90 when t = 0.
Euler's method with a step length of 1 minute is used to estimate the temperature of the tea.

(a)

Find an estimate for the temperature of the tea when t = 1.

(2)

(b)

Hence find an estimate for the temperature of the tea when t = 3.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses T₁ = T₀ + h × (−0.08(T₀ − 20)) = 90 + 1 × (−0.08 × 70)	✓ 1
T ≈ 84.4 °C	✓ 1
Part (b)
Applies the same iteration to the new value: T₂ = 84.4 − 0.08 × 64.4	✓ 1
T₂ = 79.248	✓ 1
T₃ = 79.248 − 0.08 × 59.248 = 74.508…, so T ≈ 74.5 °C	✓ 1

Exam tip: Set Euler's method up as a table (t, T, dT/dt, new T) on your GDC or on paper. Keep every intermediate value at full calculator accuracy and round only the final answer.

+ Add to test✓ In your test53IBMI-034-08-r2-01Differential equations and slope fields Paper HLEasy5 marks

A cup of soup is left to cool in a kitchen where the air temperature is 20 °C. The temperature of the soup, T °C, t minutes after it is poured is modelled by the differential equation
$\frac{dT}{dt} = -k(T-20)$, where k is a positive constant.
When t = 0 the soup is at 90 °C, and when t = 5 it is at 70 °C.

(a)

By solving the differential equation, show that $T = 20 + 70e^{-kt}$.

(3)

(b)

Find the value of k.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Separates variables: $\int \frac{1}{T-20}\,dT = \int -k\,dt$ [M]	✓ 1
$\ln\|T-20\| = -kt + c$ [A]	✓ 1
$T - 20 = Ae^{-kt}$; uses T = 90 at t = 0 to get A = 70, so $T = 20 + 70e^{-kt}$ [A]	✓ 1
Part (b)
Substitutes t = 5, T = 70: $50 = 70e^{-5k}$ [M]	✓ 1
$k = \frac{1}{5}\ln(1.4) = 0.0673$ min⁻¹ [A]	✓ 1

Exam tip: In a "show that" question every step must appear: write the separated integrals, include the constant of integration, and show how the initial condition fixes it.

+ Add to test✓ In your test54IBMI-034-08-r1-01Differential equations and slope fields Paper HLEasy7 marks

A cup of tea is left to cool in a room where the air temperature is 20°C. The temperature of the tea, T°C, t minutes after it is poured, is modelled by the differential equation
$\frac{dT}{dt} = -0.08(T - 20)$
When the tea is poured, its temperature is 90°C.

(a)

By solving the differential equation, show that $T = 20 + 70e^{-0.08t}$.

(3)

(b)

Find the temperature of the tea 10 minutes after it is poured.

(2)

(c)

The tea is pleasant to drink until its temperature falls to 40°C. Find the time it takes for the tea to cool to 40°C.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: separates variables correctly, $\int \frac{1}{T-20}\,dT = \int -0.08\,dt$	✓ 1
A1: $\ln\|T-20\| = -0.08t + c$	✓ 1
M1: uses T = 90 when t = 0 to get $c = \ln 70$, so $T - 20 = 70e^{-0.08t}$, leading to the given answer	✓ 1
Part (b)
M1: substitutes t = 10, $T = 20 + 70e^{-0.8}$	✓ 1
A1: T = 51.5°C	✓ 1
Part (c)
M1: sets $20 + 70e^{-0.08t} = 40$, or uses a GDC solver or graph intersection	✓ 1
A1: t = 15.7 minutes	✓ 1

Exam tip: When you are told to show a result, every step of the working must be visible. Include the constant of integration and show clearly how the initial condition fixes it.

+ Add to test✓ In your test55IBMI-034-08-r1-02Differential equations and slope fields Paper HLEasy7 marks

A thermometer is placed in an oven whose temperature rises steadily. In suitable units, the thermometer reading θ at time t minutes is modelled by
$\frac{d\theta}{dt} = t - \theta, \quad t \ge 0$
A slope field for this differential equation can be drawn in the t–θ plane.

(a)

Write down the equation of the isocline on which the slope field has gradient zero, and state what this tells you about any solution curve that crosses it.

(2)

(b)

Show that $\theta = t - 1$ is a solution of the differential equation.

(2)

(c)

The general solution is $\theta = t - 1 + Ce^{-t}$. At t = 0 the reading is 2. Find the value of C, and hence find the reading when t = 2.

(2)

(d)

Describe the long-term behaviour of the thermometer reading relative to the line θ = t.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: θ = t	✓ 1
R1: any solution curve that crosses this line has a horizontal tangent (stationary point) there	✓ 1
Part (b)
M1: differentiates to get $\frac{d\theta}{dt} = 1$	✓ 1
A1: substitutes, $t - \theta = t - (t - 1) = 1$, so both sides are equal	✓ 1
Part (c)
A1: C = 3	✓ 1
A1: $\theta = 1 + 3e^{-2} = 1.41$	✓ 1
Part (d)
R1: as t increases, $3e^{-t} \to 0$, so the reading approaches θ = t − 1. In the long run the reading lags the line θ = t by 1 unit	✓ 1

Common error: An isocline is a line along which every slope has the same value. It is not itself a solution curve. Here θ = t has gradient 1, yet every slope drawn on it is 0.

+ Add to test✓ In your test56IBMI-034-03-r2-01Differential equations and slope fields Paper HLEasy7 marks

A cup of tea is left to cool in a room kept at a constant 20 °C. The temperature of the tea, T °C, t minutes after it is poured is modelled by the differential equation
$\frac{dT}{dt} = -k(T-20)$, where k is a positive constant.
When the tea is poured, its temperature is 90 °C.

(a)

By solving the differential equation, show that $T = 20 + 70e^{-kt}$.

(3)

(b)

After 10 minutes the temperature of the tea is 60 °C. Find the value of k.

(2)

(c)

Find the time taken for the temperature of the tea to fall to 30 °C.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Separates variables, e.g. $\int \frac{1}{T-20}\,dT = \int -k\,dt$	✓ 1
$\ln\|T-20\| = -kt + c$	✓ 1
Uses $T=90$ when $t=0$ to obtain $T-20 = 70e^{-kt}$, leading to $T = 20 + 70e^{-kt}$	✓ 1
Part (b)
Substitutes: $60 = 20 + 70e^{-10k}$, giving $e^{-10k} = \frac{4}{7}$	✓ 1
$k = \frac{\ln 1.75}{10} = 0.0560$ (3 s.f.)	✓ 1
Part (c)
Substitutes: $30 = 20 + 70e^{-kt}$, giving $e^{-kt} = \frac{1}{7}$, so $t = \frac{\ln 7}{k} = \frac{10\ln 7}{\ln 1.75}$	✓ 1
$t = 34.8$ minutes (3 s.f.)	✓ 1
Note: the exact value of $k$ gives 34.77…, which rounds to 34.8. Using the rounded $k = 0.0560$ gives 34.7. Accept 34.7 for the final mark only if $k = 0.0560$ is clearly shown.	—

Exam tip: Store the unrounded value of k in your calculator and use it when finding the time to reach 30 °C. Rounding k to 0.0560 first shifts the answer from 34.8 to 34.7 minutes.

+ Add to test✓ In your test57IBMI-034-04-r1-02Euler's method and coupled systems Paper HLEasy7 marks

The number of bacteria in a culture, N thousand, t hours after the start of an experiment, is modelled by
$\frac{dN}{dt} = 0.4N\left(1 - \frac{N}{50}\right)$, with N = 5 when t = 0.

(a)

Use Euler's method with a step length of 0.5 hours to find an estimate for N when t = 0.5.

(2)

(b)

Continue to use Euler's method with the same step length to estimate the number of bacteria when t = 1.5. Give your answer to the nearest bacterium.

(3)

(c)

State how the method could be adapted to give a more accurate estimate.

(1)

(d)

Interpret the value 50 in the context of the model.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$N_1 = 5 + 0.5 \times 0.4 \times 5 \times (1 - \frac{5}{50})$ (M1)	✓ 1
N₁ = 5.9 (A1)	✓ 1
Part (b)
Repeated Euler step, e.g. $N_2 = 5.9 + 0.5 \times 0.4 \times 5.9 \times (1 - \frac{5.9}{50})$ (M1)	✓ 1
N₂ = 6.94076 (A1)	✓ 1
N₃ = 6.94076 + 0.5 × 2.39091… = 8.13622 thousand, so approximately 8136 bacteria (A1)	✓ 1
Part (c)
Use a smaller step length, e.g. 0.1 hours (accept: use a higher-order method such as improved Euler or Runge–Kutta) (A1)	✓ 1
Part (d)
50 thousand is the carrying capacity, the largest population the culture can sustain in the long term, since dN/dt = 0 when N = 50 (A1)	✓ 1

Exam tip: Keep full calculator values between Euler steps. Rounding each step to 3 s.f. lets errors build up and can change the final answer.

+ Add to test✓ In your test58IBMI-034-08-r2-02Differential equations and slope fields Paper HLEasy7 marks

The number of fish, P, in a newly stocked pond t years after stocking is modelled by
$\frac{dP}{dt} = 0.4P - 0.001P^2$, with P = 100 when t = 0.

(a)

Find the non-zero value of P for which the population remains constant.

(2)

(b)

Use Euler's method with a step length of 0.5 years to estimate the number of fish in the pond when t = 1.5.

(4)

(c)

State how your Euler's method estimate in the previous part could be made more accurate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Sets $0.4P - 0.001P^2 = 0$, i.e. $P(0.4 - 0.001P) = 0$ [M]	✓ 1
P = 400 fish (the other solution, P = 0, is excluded) [A]	✓ 1
Part (b)
Uses $P_{n+1} = P_n + 0.5(0.4P_n - 0.001P_n^2)$ [M]	✓ 1
$P_1 = 115$ [A]	✓ 1
$P_2 = 131.3875$ [A]	✓ 1
$P_3 = 149.03\ldots \approx 149$ fish [A]	✓ 1
Part (c)
Use a smaller step length (and so more steps) [R]	✓ 1

Exam tip: Store each Euler value in your calculator rather than rounding it. Rounding at every step lets errors build up and can change the final 3 s.f. answer.

+ Add to test✓ In your test59IBMI-034-04-r2-02Euler's method and coupled systems Paper HLEasy6 marks

On an island, the number of rabbits, x, and the number of foxes, y, after t years are modelled by the coupled differential equations
$\frac{dx}{dt} = 0.4x - 0.01xy$
$\frac{dy}{dt} = -0.3y + 0.005xy$
When t = 0, there are 100 rabbits and 20 foxes. Euler's method with a step length of 0.5 years is used to estimate the populations. Give your estimates to one decimal place where necessary.

(a)

Find estimates for the number of rabbits and the number of foxes when t = 0.5.

(3)

(b)

Find estimates for the number of rabbits and the number of foxes when t = 1.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses x₁ = x₀ + 0.5 × (dx/dt) and y₁ = y₀ + 0.5 × (dy/dt), with dx/dt = 20 and dy/dt = 4 at t = 0	✓ 1
x ≈ 110 rabbits	✓ 1
y ≈ 22 foxes	✓ 1
Part (b)
Evaluates both derivatives at (110, 22): dx/dt = 19.8 and dy/dt = 5.5	✓ 1
x ≈ 110 + 0.5 × 19.8 = 119.9 rabbits	✓ 1
y ≈ 22 + 0.5 × 5.5 = 24.75 ≈ 24.8 foxes	✓ 1

Common error: In a coupled system, both derivatives must be evaluated at the old values of x and y. Do not use the updated x when you work out the new y within the same step.

+ Add to test✓ In your test60IBMI-034-04-r1-01Euler's method and coupled systems Paper HLEasy6 marks

A cup of tea is left to cool in a room. Its temperature, T °C, t minutes after it is poured, satisfies the differential equation
$\frac{dT}{dt} = -0.08(T - 20)$, with T = 90 when t = 0.

(a)

Use Euler's method with a step length of 1 minute to estimate the temperature of the tea after 3 minutes.

(4)

(b)

State, with a reason, whether your answer to part (a) is an overestimate or an underestimate of the true temperature.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses $T_{n+1} = T_n + 1 \times (-0.08(T_n - 20))$ (M1)	✓ 1
T₁ = 90 − 5.6 = 84.4 (A1)	✓ 1
T₂ = 84.4 − 5.152 = 79.248 (A1)	✓ 1
T₃ = 79.248 − 4.73984 = 74.5 °C (3 s.f.) (A1)	✓ 1
Part (b)
Valid reason, for example: $\frac{d^2T}{dt^2} = -0.08\frac{dT}{dt} = 0.0064(T-20) > 0$, so the curve is concave up and each Euler tangent step lies below the curve (R1)	✓ 1
OR: the exact solution $T = 20 + 70e^{-0.08t}$ gives T(3) ≈ 75.1 °C, which is greater than 74.5 °C (R1)	—
Underestimate (A1)	✓ 1

Common error: Each new step must use the gradient calculated from the most recent temperature, not the starting value of 90 °C every time.

+ Add to test✓ In your test61IBMI-034-02-r1-05Integration and the trapezoidal rule Paper HLMedium7 marks

A designer makes a ceramic bowl. The inside surface is formed by rotating the curve $y = 0.1x^2$, for $0 \le y \le 10$, through 360° about the y-axis. All lengths are in centimetres.

(a)

Show that $x^2 = 10y$.

(1)

(b)

Find the volume of liquid the bowl holds when full.

(3)

(c)

Water is poured in until the bowl is exactly half full by volume. Find the depth of the water.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
From $y = 0.1x^2$, multiplying by 10 gives $x^2 = 10y$	✓ 1
Part (b)
$V = \pi\int_0^{10} x^2\,dy$	✓ 1
$= \pi\int_0^{10} 10y\,dy$	✓ 1
= 1570 cm³ (500π)	✓ 1
Part (c)
Volume to depth h: $\pi\int_0^{h} 10y\,dy = 5\pi h^2$	✓ 1
$5\pi h^2 = 250\pi$ (785.4)	✓ 1
h = 7.07 cm	✓ 1

Common error: Rotating about the y-axis needs x² written in terms of y, with y-limits. Half the volume does not mean half the depth, because the bowl widens towards the top.

+ Add to test✓ In your test62IBMI-034-00-r1-05Gradients, tangents and rates of change Paper HLMedium7 marks

The depth of water, $d$ metres, at a harbour entrance $t$ hours after midnight is modelled by
$d(t) = 5 + 2.4\sin(0.5t)$, for $0 \le t \le 12$,
where $0.5t$ is measured in radians.

(a)

Find $d'(t)$.

(2)

(b)

Find $d'(8)$ and interpret your answer in context.

(3)

(c)

Find the value of $t$ at which the depth of water is decreasing most rapidly.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: $1.2$ (from chain rule, $2.4 \times 0.5$)	✓ 1
A1: $\cos(0.5t)$, giving $d'(t) = 1.2\cos(0.5t)$	✓ 1
Part (b)
M1: substituting $t = 8$ with the calculator in radians: $1.2\cos(4)$	✓ 1
A1: $d'(8) = -0.784$	✓ 1
R1: at 08:00 the water depth is decreasing at 0.784 metres per hour	✓ 1
Part (c)
M1: $\cos(0.5t) = -1$, i.e. $0.5t = \pi$, or the GDC minimum of $d'(t)$	✓ 1
A1: $t = 6.28$ hours	✓ 1

Common error: Calculus with trigonometric functions needs the calculator in radian mode. With degrees, $1.2\cos(4)$ gives about 1.20 instead of −0.784.

+ Add to test✓ In your test63IBMI-034-03-r1-05Differential equations and slope fields Paper HLMedium13 marks

A drug is injected into the bloodstream of a patient. The concentration of the drug in the blood is x mg L⁻¹ and in the body tissue is y mg L⁻¹, t hours after injection. The concentrations are modelled by
$\frac{dx}{dt} = -3x + y$
$\frac{dy}{dt} = 2x - 2y$
Initially x = 10 and y = 0.

(a)

Find the eigenvalues and corresponding eigenvectors of the matrix $\begin{pmatrix} -3 & 1 \\ 2 & -2 \end{pmatrix}$.

(4)

(b)

Hence find expressions for x and y in terms of t.

(3)

(c)

Describe the long-term behaviour of the drug concentrations, justifying your answer with reference to the eigenvalues.

(2)

(d)

Find the time at which the concentration in the tissue is greatest, and find this maximum concentration.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Characteristic equation $(-3-\lambda)(-2-\lambda) - 2 = 0$, i.e. $\lambda^2 + 5\lambda + 4 = 0$	✓ 1
λ = −1 and λ = −4	✓ 1
For λ = −1: eigenvector $\begin{pmatrix}1\\2\end{pmatrix}$	✓ 1
For λ = −4: eigenvector $\begin{pmatrix}1\\-1\end{pmatrix}$	✓ 1
Part (b)
General solution $\begin{pmatrix}x\\y\end{pmatrix} = Ae^{-t}\begin{pmatrix}1\\2\end{pmatrix} + Be^{-4t}\begin{pmatrix}1\\-1\end{pmatrix}$	✓ 1
Uses initial conditions: A + B = 10, 2A − B = 0	✓ 1
$x = \frac{10}{3}e^{-t} + \frac{20}{3}e^{-4t}$, $y = \frac{20}{3}e^{-t} - \frac{20}{3}e^{-4t}$	✓ 1
Part (c)
Both eigenvalues are negative, so the origin is a stable equilibrium and all trajectories approach it	✓ 1
The drug is eventually eliminated from both blood and tissue, with concentrations ultimately decaying in the ratio x : y = 1 : 2 (along the slower eigenvector)	✓ 1
Part (d)
Sets $\frac{dy}{dt} = -\frac{20}{3}e^{-t} + \frac{80}{3}e^{-4t} = 0$	✓ 1
$e^{3t} = 4$, t = 0.462 hours	✓ 1
Substitutes their t into y	✓ 1
y = 3.15 mg L⁻¹	✓ 1

Exam tip: Any non-zero multiple of an eigenvector is acceptable, but use the same vectors consistently when you form the general solution and apply the initial conditions.

+ Add to test✓ In your test64IBMI-034-04-r2-05Euler's method and coupled systems Paper HLMedium11 marks

A drug is injected directly into a patient's bloodstream. The amount of drug in the blood, x mg, and the amount in body tissue, y mg, t hours after the injection are modelled by
$\frac{dx}{dt} = -0.3x + 0.1y$
$\frac{dy}{dt} = 0.2x - 0.2y$
When t = 0, x = 30 and y = 0.

(a)

Find the eigenvalues and corresponding eigenvectors of the matrix $\begin{pmatrix} -0.3 & 0.1 \\ 0.2 & -0.2 \end{pmatrix}$.

(4)

(b)

Hence find expressions for x and y in terms of t.

(3)

(c)

Find the maximum amount of drug in the tissue, and the time at which it occurs.

(3)

(d)

Describe the long-term behaviour of the amount of drug in the patient's body. Justify your answer.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Characteristic equation: (−0.3 − λ)(−0.2 − λ) − 0.02 = 0, i.e. λ² + 0.5λ + 0.04 = 0	✓ 1
λ = −0.1 and λ = −0.4	✓ 1
For λ = −0.1: eigenvector (1, 2) or any multiple of it	✓ 1
For λ = −0.4: eigenvector (1, −1) or any multiple of it	✓ 1
Part (b)
General solution: (x, y) = Ae^−0.1t(1, 2) + Be^−0.4t(1, −1)	✓ 1
Uses the initial conditions: A + B = 30 and 2A − B = 0, so A = 10 and B = 20	✓ 1
x = 10e^−0.1t + 20e^−0.4t, y = 20e^−0.1t − 20e^−0.4t	✓ 1
Part (c)
Sets dy/dt = 0: −2e^−0.1t + 8e^−0.4t = 0	✓ 1
t = (ln 4)/0.3 ≈ 4.62 hours	✓ 1
Maximum y ≈ 9.45 mg	✓ 1
Part (d)
Both eigenvalues are negative, so both exponential terms tend to 0: the amounts in the blood and in the tissue both tend to zero, meaning the drug is eventually eliminated from the body	✓ 1

Exam tip: Check each eigenvector by multiplying it by the matrix. The result should be the eigenvalue times the eigenvector. This quick check avoids carrying an error into the general solution.

+ Add to test✓ In your test65IBMI-034-05-r2-04Gradients, tangents and rates of change Paper HLMedium6 marks

Oil leaking from a damaged tanker spreads out on calm water as a circular slick. The area of the slick increases at a constant rate of 3 m² per minute.

(a)

Find the rate at which the radius of the slick is increasing at the instant when the radius is 12 m.

(4)

(b)

Find the rate at which the circumference of the slick is increasing at this instant.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 using $A = \pi r^2$	✓ 1
A1 $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$	✓ 1
M1 substituting: $3 = 2\pi(12)\frac{dr}{dt}$	✓ 1
A1 $\frac{dr}{dt} = 0.0398$ m per minute	✓ 1
Part (b)
M1 $\frac{dC}{dt} = 2\pi \frac{dr}{dt}$	✓ 1
A1 0.25 m per minute	✓ 1

Common error: Substituting r = 12 into A = πr² before differentiating turns the area into a constant and loses the rate. Differentiate with respect to time first, then substitute the instant's values.

+ Add to test✓ In your test66IBMI-034-07-r2-05Integration and the trapezoidal rule Paper HLMedium7 marks

A remote-controlled cart moves along a straight track. Its velocity, v m s⁻¹, at time t seconds is modelled by

$v(t) = 5e^{-0.2t} - 2, \quad 0 \le t \le 10.$

(a)

Find the time at which the cart changes direction.

(2)

(b)

Find the displacement of the cart from its starting position at t = 10.

(2)

(c)

Find the total distance travelled by the cart during the 10 seconds.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: sets v(t) = 0, 5e^−0.2t = 2	✓ 1
A1: t = 4.58 s (4.58145…)	✓ 1
Part (b)
M1: $\int_0^{10} (5e^{-0.2t} - 2)\,dt$	✓ 1
A1: 1.62 m (1.61662…)	✓ 1
Part (c)
M1: $\int_0^{10} \|v(t)\|\,dt$, or splits the integral at t = 4.58	✓ 1
A1: 5.837 m forwards and 4.220 m backwards (accept either correct value seen)	✓ 1
A1: 10.1 m	✓ 1

Common error: Integrating v(t) directly from 0 to 10 gives displacement, not distance. When the velocity changes sign, integrate |v(t)| or split the interval at the turning time.

+ Add to test✓ In your test67IBMI-034-00-r2-05Gradients, tangents and rates of change Paper HLMedium6 marks

A cup of tea is left to cool in a room. Its temperature, T °C, t minutes after it is poured is modelled by
$T(t) = 22 + 68e^{-0.05t}$, for $t \ge 0$.

(a)

Find $T'(t)$.

(2)

(b)

Find the rate of change of the temperature of the tea when $t = 10$.

(2)

(c)

Find the time at which the tea is cooling at a rate of 1 °C per minute.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses the chain rule for $e^{-0.05t}$	✓ 1
$T'(t) = -3.4e^{-0.05t}$	✓ 1
Part (b)
$T'(10) = -3.4e^{-0.5}$	✓ 1
$-2.06$ °C per minute	✓ 1
Part (c)
Sets $3.4e^{-0.05t} = 1$ (or $T'(t) = -1$)	✓ 1
$t = 24.5$ minutes	✓ 1

Common error: Cooling at 1 °C per minute means T'(t) = −1, not +1. Setting T'(t) = 1 has no solution, because the derivative is always negative.

+ Add to test✓ In your test68IBMI-034-08-r2-04Differential equations and slope fields Paper HLMedium8 marks

Water drains from a cylindrical tank through a small hole in its base. The depth of water, h metres, t minutes after the hole is opened is modelled by
$\frac{dh}{dt} = -0.15\sqrt{h}$.
Initially the depth of water is 2.5 m.

(a)

Solve the differential equation to find an expression for $\sqrt{h}$ in terms of t.

(4)

(b)

Find the depth of water after 10 minutes.

(2)

(c)

Find the time taken for the tank to empty.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Separates variables: $\int h^{-\frac{1}{2}}\,dh = \int -0.15\,dt$ [M]	✓ 1
$2\sqrt{h} = -0.15t + c$ [A]	✓ 1
Uses h = 2.5 at t = 0: $c = 2\sqrt{2.5}$ [M]	✓ 1
$\sqrt{h} = \sqrt{2.5} - 0.075t$ [A]	✓ 1
Part (b)
$\sqrt{h} = \sqrt{2.5} - 0.75 = 0.8311\ldots$ [M]	✓ 1
h = 0.691 m [A]	✓ 1
Part (c)
Sets h = 0: $\sqrt{2.5} - 0.075t = 0$ [M]	✓ 1
t = 21.1 minutes [A]	✓ 1

Common error: Integrating $h^{-1/2}$ gives $2h^{1/2}$, not $\frac{1}{2}h^{1/2}$. Check the result by differentiating it.

+ Add to test✓ In your test69IBMI-034-03-r2-03Differential equations and slope fields Paper HLMedium12 marks

A conservation team reintroduces 50 otters into a river system. The number of otters, P, after t years is modelled by the logistic differential equation
$\frac{dP}{dt} = 0.3P\left(1 - \frac{P}{500}\right)$, with $P(0) = 50$.

(a)

State the carrying capacity of the river system and the population at which the model predicts the otter population grows fastest.

(2)

(b)

Use Euler's method with a step length of 1 year to estimate the number of otters after 3 years.

(4)

(c)

The exact solution of the differential equation is $P = \frac{500}{1 + 9e^{-0.3t}}$. Find the percentage error in the Euler estimate of part (b) at $t = 3$.

(3)

(d)

Explain why Euler's method underestimates the population over the first 3 years.

(2)

(e)

Suggest how the Euler estimate could be made more accurate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Carrying capacity 500 otters	✓ 1
Fastest growth when P = 250	✓ 1
Part (b)
Uses $P_{n+1} = P_n + 1\times 0.3P_n\left(1 - \frac{P_n}{500}\right)$ (M1)	✓ 1
$P_1 = 63.5$ (A1)	✓ 1
$P_2 = 80.1$ (80.13065) (A1)	✓ 1
$P_3 = 100.3$, so about 100 otters (A1)	✓ 1
Part (c)
$P(3) = \frac{500}{1 + 9e^{-0.9}} = 107.316...$ (A1)	✓ 1
Percentage error $= \frac{\|100.317 - 107.316\|}{107.316}\times 100$ (M1)	✓ 1
$= 6.52\%$ (3 s.f.) (A1)	✓ 1
Part (d)
For $P < 250$ the rate of growth is increasing, so the solution curve is concave up (R1)	✓ 1
Each Euler step follows the tangent, which lies below a concave-up curve, so each estimate is too small (R1)	✓ 1
Part (e)
Use a smaller step length, e.g. $h = 0.1$	✓ 1

Exam tip: Store each Euler value at full calculator accuracy and round only at the end; rounding P₁ and P₂ early can change the third significant figure of the final estimate.

+ Add to test✓ In your test70IBMI-034-03-r2-04Differential equations and slope fields Paper HLMedium6 marks

A water butt is emptied through a tap at its base. The volume of water in the butt, V litres, t minutes after the tap is opened is modelled by
$\frac{dV}{dt} = -k\sqrt{V}$, where k is a positive constant.
Initially the butt holds 400 litres. After 10 minutes it holds 225 litres.
By solving the differential equation, find the time taken for the water butt to empty.

(a)

Find the time taken for the water butt to empty.

(6)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Separates variables: $\int V^{-\frac{1}{2}}\,dV = \int -k\,dt$ (M1)	✓ 1
$2\sqrt{V} = -kt + c$ (A1)	✓ 1
Uses $V = 400$ at $t = 0$: $c = 40$ (A1)	✓ 1
Uses $V = 225$ at $t = 10$: $30 = -10k + 40$ (M1)	✓ 1
$k = 1$ (A1)	✓ 1
Empty when $V = 0$: $0 = -t + 40$, so $t = 40$ minutes (A1)	✓ 1

Common error: Integrating 1/√V as ln√V or as √V/(1/2)·V is a frequent slip. Rewrite it as V to the power −1/2 before integrating.

+ Add to test✓ In your test71IBMI-034-04-r1-03Euler's method and coupled systems Paper HLMedium10 marks

On an island, the number of rabbits, x, and the number of foxes, y, t years after a study begins, are modelled by the coupled differential equations
$\frac{dx}{dt} = 0.5x - 0.01xy$
$\frac{dy}{dt} = -0.4y + 0.005xy$
At t = 0, there are 100 rabbits and 40 foxes.

(a)

Find the equilibrium point at which both populations are non-zero.

(3)

(b)

Use Euler's method with a step length of 0.2 years to find estimates for x and y when t = 0.2.

(3)

(c)

Hence estimate the number of rabbits and the number of foxes when t = 0.4.

(3)

(d)

With reference to your answers, describe how the two populations are changing at the start of the study.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Sets both derivatives equal to zero (M1)	✓ 1
$x(0.5 - 0.01y) = 0 \Rightarrow y = 50$ (A1)	✓ 1
$y(-0.4 + 0.005x) = 0 \Rightarrow x = 80$; equilibrium (80, 50) (A1)	✓ 1
Part (b)
At t = 0: dx/dt = 10 and dy/dt = 4 (M1)	✓ 1
x₁ = 100 + 0.2 × 10 = 102 (A1)	✓ 1
y₁ = 40 + 0.2 × 4 = 40.8 (A1)	✓ 1
Part (c)
Uses x₁ = 102, y₁ = 40.8 in both derivatives: 9.384 and 4.488 (M1)	✓ 1
x₂ = 102 + 0.2 × 9.384 = 103.8768 ≈ 104 rabbits (A1)	✓ 1
y₂ = 40.8 + 0.2 × 4.488 = 41.6976 ≈ 41.7 foxes (A1)	✓ 1
Part (d)
Both populations are increasing: there are more rabbits than the equilibrium value of 80, so the foxes have enough food to grow, while there are fewer foxes than 50, so rabbits still increase (R1)	✓ 1

Common error: In a coupled system, the second step must use both updated values. Using the new x with the old y mixes two different time points.

+ Add to test✓ In your test72IBMI-034-04-r1-05Euler's method and coupled systems Paper HLMedium7 marks

A mass on a spring is pulled down and released. Its displacement from its equilibrium position, x metres, t seconds after release, satisfies
$\frac{d^2x}{dt^2} + 0.4\frac{dx}{dt} + 4x = 0$
When t = 0, x = 0.5 and $\frac{dx}{dt} = 0$.

(a)

By letting $v = \frac{dx}{dt}$, write the differential equation as a system of two first-order differential equations.

(2)

(b)

Use Euler's method with a step length of 0.1 seconds to estimate the displacement of the mass when t = 0.3.

(4)

(c)

Interpret the sign of your estimate of v when t = 0.2 in context.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\frac{dx}{dt} = v$ (A1)	✓ 1
$\frac{dv}{dt} = -0.4v - 4x$ (A1)	✓ 1
Part (b)
Uses $x_{n+1} = x_n + 0.1v_n$ and $v_{n+1} = v_n + 0.1(-0.4v_n - 4x_n)$ (M1)	✓ 1
t = 0.1: x = 0.5, v = −0.2 (A1)	✓ 1
t = 0.2: x = 0.48, v = −0.392 (A1)	✓ 1
t = 0.3: x = 0.441 m (0.4408) (A1)	✓ 1
Part (c)
v is negative, so the mass is moving back towards its equilibrium position (displacement decreasing) (R1)	✓ 1

Common error: At the first step v = 0, so x does not change from 0.5 — this is correct, not a mistake. Update x using the old v, not the new one.

+ Add to test✓ In your test73IBMI-034-04-r1-06Euler's method and coupled systems Paper HLMedium9 marks

In a computer strategy game, two armies fight each other. The number of soldiers in army X is x and in army Y is y, t hours after the battle starts. The battle is modelled by
$\frac{dx}{dt} = -0.2y$
$\frac{dy}{dt} = -0.1x$
Initially army X has 500 soldiers and army Y has 400 soldiers. The battle ends when one army has no soldiers left.

(a)

Use Euler's method with a step length of 1 hour to estimate the number of soldiers in each army after 2 hours.

(4)

(b)

Show that $\frac{d}{dt}\left(0.1x^2 - 0.2y^2\right) = 0$.

(2)

(c)

Hence determine which army wins the battle and find the number of soldiers remaining in the winning army.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses $x_{n+1} = x_n - 0.2y_n$, $y_{n+1} = y_n - 0.1x_n$ (M1)	✓ 1
t = 1: x = 420, y = 350 (A1)	✓ 1
t = 2: x = 420 − 70 = 350 (A1)	✓ 1
t = 2: y = 350 − 42 = 308 (A1)	✓ 1
Part (b)
$\frac{d}{dt}(0.1x^2 - 0.2y^2) = 0.2x\frac{dx}{dt} - 0.4y\frac{dy}{dt}$ (M1)	✓ 1
$= 0.2x(-0.2y) - 0.4y(-0.1x) = -0.04xy + 0.04xy = 0$ (A1)	✓ 1
Part (c)
$0.1(500)^2 - 0.2(400)^2 = -7000$, so $0.1x^2 - 0.2y^2 = -7000$ throughout (M1)	✓ 1
The quantity is negative, so y can never reach zero; army Y wins when x = 0 (R1)	✓ 1
$0.2y^2 = 7000 \Rightarrow y = \sqrt{35000} = 187$ soldiers (A1)	✓ 1

Exam tip: A quantity that stays constant lets you find the final state without solving the system: substitute the initial values, then set the losing army's size to zero.

+ Add to test✓ In your test74IBMI-034-06-r1-05Optimisation in context Paper HLMedium11 marks

The concentration of a medicine in a patient's bloodstream, C mg L⁻¹, t hours after an oral dose is taken, is modelled by
$C(t) = 12t\,e^{-0.4t}$, for $t \ge 0$.

(a)

Find $C'(t)$.

(3)

(b)

Show that the maximum concentration occurs when $t = 2.5$.

(2)

(c)

Find the maximum concentration.

(1)

(d)

The medicine is effective while its concentration is at least 6 mg L⁻¹. Find the length of time for which the medicine is effective.

(3)

(e)

Find $C'(5)$ and interpret your answer in context.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Attempt at product rule (M1)	✓ 1
$12e^{-0.4t}$ (A1)	✓ 1
$-4.8t\,e^{-0.4t}$, giving $C'(t) = e^{-0.4t}(12 - 4.8t)$ (A1)	✓ 1
Part (b)
$C'(t) = 0$ and $e^{-0.4t} \ne 0$, so $12 - 4.8t = 0 \Rightarrow t = 2.5$ (M1)	✓ 1
$C'(t) > 0$ for $t < 2.5$ and $C'(t) < 0$ for $t > 2.5$, so this is a maximum (R1)	✓ 1
Part (c)
$C(2.5) = 30e^{-1} = 11.0$ mg L⁻¹ (11.036…) (A1)	✓ 1
Part (d)
Solving $12t\,e^{-0.4t} = 6$ using GDC (M1)	✓ 1
$t = 0.648$ and $t = 6.36$ (6.3566…) (A1)	✓ 1
Time $= 6.3566 - 0.6479 = 5.71$ hours (A1)	✓ 1
Part (e)
$C'(5) = e^{-2}(12 - 24) = -1.62$ (−1.6240…) (A1)	✓ 1
5 hours after the dose, the concentration is decreasing at 1.62 mg L⁻¹ per hour (R1)	✓ 1

Exam tip: An interpretation must mention the quantity, the direction of change, the units and the time, e.g. 'decreasing at 1.62 mg per litre per hour after 5 hours'.

+ Add to test✓ In your test75IBMI-034-05-r1-05Gradients, tangents and rates of change Paper HLMedium6 marks

The concentration, $C$ mg L⁻¹, of a medicine in a patient's bloodstream $t$ hours after an injection is modelled by
$C(t) = 20te^{-0.4t}$, for $t \ge 0$.

(a)

Find $C'(t)$.

(3)

(b)

Find the rate of change of concentration 5 hours after the injection and interpret your answer in context.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Attempt to use the product rule M1	✓ 1
$20e^{-0.4t}$ A1	✓ 1
$-8te^{-0.4t}$, giving $C'(t) = 20e^{-0.4t}(1 - 0.4t)$ A1	✓ 1
Part (b)
Substituting $t = 5$: $20e^{-2}(1 - 2)$ M1	✓ 1
$-2.71$ mg L⁻¹ h⁻¹ A1	✓ 1
After 5 hours the concentration of the medicine is decreasing at 2.71 mg L⁻¹ per hour R1	✓ 1

Exam tip: Store your derivative in the GDC and evaluate it at $t = 5$, or use the numerical derivative feature to check your hand-differentiated expression gives the same value.

+ Add to test✓ In your test76IBMI-034-06-r2-05Optimisation in context Paper HLMedium10 marks

After a patient takes a tablet, the concentration of a drug in their bloodstream, $C$ mg L⁻¹, is modelled by
$C(t) = 12t\,e^{-0.25t}$, for $t \ge 0$,
where $t$ is the time in hours after the tablet is taken.

(a)

Find $C'(t)$.

(3)

(b)

Hence find the time at which the concentration is greatest.

(2)

(c)

The drug works when its concentration is at least 10 mg L⁻¹. Find the length of time for which the drug works.

(3)

(d)

The concentration is safe while it stays below 18 mg L⁻¹. Find the maximum concentration and state, with a reason, whether this dose is safe.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses the product rule M1	✓ 1
$12e^{-0.25t}$ A1	✓ 1
$-3t\,e^{-0.25t}$ A1	✓ 1
$C'(t) = 12e^{-0.25t}(1 - 0.25t)$	—
Part (b)
Sets $C'(t) = 0$; since $e^{-0.25t} \neq 0$, $1 - 0.25t = 0$ M1	✓ 1
$t = 4$ hours A1	✓ 1
Part (c)
Sets $12t\,e^{-0.25t} = 10$ and solves with a GDC M1	✓ 1
$t = 1.10$ and $t = 9.90$ (1.0960…, 9.8991…) A1	✓ 1
Duration $= 8.80$ hours A1	✓ 1
Part (d)
$C(4) = 48e^{-1} = 17.7$ mg L⁻¹ (17.658…) A1	✓ 1
This is the greatest value of $C$ and it is below 18 mg L⁻¹, so the dose is safe R1	✓ 1

Common error: Differentiating $12t\,e^{-0.25t}$ as $12 \times (-0.25)e^{-0.25t}$ leaves out the product rule. Both factors depend on $t$.

+ Add to test✓ In your test77IBMI-034-08-r1-03Differential equations and slope fields Paper HLMedium14 marks

A conservation group stocks a lake with 50 fish. The number of fish, N, t years later is modelled by the logistic differential equation
$\frac{dN}{dt} = 0.4N\left(1 - \frac{N}{500}\right)$

(a)

Write down the carrying capacity of the lake and interpret it in context.

(2)

(b)

Use Euler's method with a step length of 0.5 years to estimate the number of fish after 1 year.

(4)

(c)

The exact solution is $N = \frac{500}{1 + 9e^{-0.4t}}$. Find the exact value of N at t = 1, and hence the percentage error in your Euler estimate.

(3)

(d)

Explain why Euler's method gives an underestimate here.

(1)

(e)

Find the population at which the fish population grows fastest, and use the exact solution to find when this occurs.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: 500 fish	✓ 1
R1: in the long run the fish population approaches 500, the largest population the lake can sustain	✓ 1
Part (b)
M1: uses $N_{n+1} = N_n + 0.5 \times 0.4N_n\left(1 - \frac{N_n}{500}\right)$	✓ 1
A1: $N_1 = 50 + 0.5 \times 18 = 59$	✓ 1
M1: second step, $59 + 0.5 \times 20.8152$	✓ 1
A1: 69.4 fish	✓ 1
Part (c)
A1: N = 71.1 (71.0946…)	✓ 1
M1: $\frac{71.0946 - 69.4076}{71.0946} \times 100$	✓ 1
A1: 2.37%	✓ 1
Part (d)
R1: for N < 250 the solution curve is concave up (the gradient is increasing), so each tangent step lies below the curve	✓ 1
Part (e)
M1: maximises $0.4N - 0.0008N^2$, e.g. by differentiating and setting $0.4 - 0.0016N = 0$, or using the vertex of the quadratic	✓ 1
A1: N = 250	✓ 1
M1: solves $\frac{500}{1 + 9e^{-0.4t}} = 250$, i.e. $9e^{-0.4t} = 1$	✓ 1
A1: t = 5.49 years	✓ 1

Exam tip: Set out Euler's method as a table with columns t, N and dN/dt. Each intermediate value then earns credit, and you can carry the unrounded values from your GDC.

+ Add to test✓ In your test78IBMI-034-01-r1-04Optimisation in context Paper HLMedium8 marks

The concentration of a medication in a patient's blood, $t$ hours after an injection, is modelled by
$C(t) = -0.2t^3 + 1.8t^2$, for $0 \le t \le 9$,
where $C$ is measured in mg L⁻¹.

(a)

Find $C'(t)$.

(2)

(b)

Find the maximum concentration of the medication and the time at which it occurs.

(3)

(c)

Find the time at which the concentration is increasing most rapidly, and state this rate of increase.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate, with at least one term correct	✓ 1
A1 $C'(t) = -0.6t^2 + 3.6t$	✓ 1
Part (b)
M1 setting $-0.6t^2 + 3.6t = 0$, or using a GDC to locate the maximum of $C$	✓ 1
A1 $t = 6$ hours (rejecting $t = 0$; e.g. $C(0) = C(9) = 0$ confirms the maximum)	✓ 1
A1 maximum concentration $C(6) = 21.6$ mg L⁻¹	✓ 1
Part (c)
M1 recognising that the maximum of $C'(t)$ is required, e.g. $C''(t) = -1.2t + 3.6 = 0$ or a GDC maximum of $C'(t)$	✓ 1
A1 $t = 3$ hours	✓ 1
A1 rate $C'(3) = 5.40$ mg L⁻¹ per hour	✓ 1

Common error: The time of fastest increase is where C'(t) is a maximum, not where C(t) is a maximum. Find where C''(t) = 0, then find the rate by substituting that time into C'(t), not into C(t).

+ Add to test✓ In your test79IBMI-034-03-r2-05Differential equations and slope fields Paper HLMedium12 marks

On an island, the population of rabbits, x, and foxes, y, t years after a survey began are modelled by the coupled differential equations
$\frac{dx}{dt} = 0.4x - 0.02xy$
$\frac{dy}{dt} = -0.3y + 0.005xy$
At the start of the survey there were 100 rabbits and 15 foxes.

(a)

Find the equilibrium point at which both populations are non-zero.

(3)

(b)

Find the value of $\frac{dy}{dx}$ at the start of the survey, and interpret its sign in context.

(2)

(c)

Use Euler's method with a step length of 0.5 years to estimate the number of rabbits and the number of foxes after 1 year.

(5)

(d)

Give one reason why the estimates in the previous part may be unreliable, and explain how they could be improved.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Sets $0.4 - 0.02y = 0$ and $-0.3 + 0.005x = 0$ (M1)	✓ 1
$x = 60$ (A1)	✓ 1
$y = 20$ (A1)	✓ 1
Part (b)
$\frac{dy}{dx} = \frac{-4.5 + 7.5}{40 - 30} = \frac{3}{10} = 0.3$ (A1)	✓ 1
It is positive, so initially the two populations change in the same direction: as the number of rabbits increases, the number of foxes also increases (R1)	✓ 1
Part (c)
Uses $x_{n+1} = x_n + 0.5(0.4x_n - 0.02x_ny_n)$ and $y_{n+1} = y_n + 0.5(-0.3y_n + 0.005x_ny_n)$ (M1)	✓ 1
$x_1 = 105$ (A1)	✓ 1
$y_1 = 16.5$ (A1)	✓ 1
$x_2 = 108.675 \approx 109$ rabbits (A1)	✓ 1
$y_2 = 18.356 \approx 18.4$ foxes (A1)	✓ 1
Part (d)
Reason: the step length of 0.5 years is large, and Euler's method uses only the gradient at the start of each step even though the rates change during the step, so errors accumulate (R1)	✓ 1
Improvement: use a smaller step length, which gives a more accurate estimate (R1)	✓ 1

Exam tip: In coupled Euler's method, update x and y using the old values of both. Using the new x to update y in the same step is a common way to lose accuracy marks.

+ Add to test✓ In your test80IBMI-034-02-r2-05Integration and the trapezoidal rule Paper HLMedium5 marks

The inside of a plant pot is modelled by rotating the curve
$r = 6 + 0.02h^2$, for $0 \le h \le 20$,
through 360° about the h-axis, where r is the internal radius and h is the height above the base, both in cm.

(a)

Find the internal volume of the pot.

(3)

(b)

A gardener needs a pot that holds at least 5 litres of soil. Determine whether this pot is suitable, justifying your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses volume = $\pi \int r^2\,dh$	✓ 1
$\pi \int_0^{20} (6 + 0.02h^2)^2\,dh$	✓ 1
5080 cm³ (5076.8…, $1616\pi$)	✓ 1
Part (b)
Converts using 1000 cm³ = 1 litre: 5.08 litres (accept 5.077 litres)	✓ 1
Yes, the pot is suitable because 5.08 litres > 5 litres	✓ 1

Common error: The radius must be squared before integrating. Using $\pi \int r\,dh$ gives an answer with the wrong units and loses the method mark.

+ Add to test✓ In your test81IBMI-034-03-r1-04Differential equations and slope fields Paper HLMedium6 marks

The depth of water, h metres, in a reservoir t months after the start of a dry season is modelled by
$\frac{dh}{dt} = 2 - 0.25th$, for t ≥ 0, h > 0.
A slope field for this differential equation is being studied.

(a)

Write down the gradient of every line segment on the line t = 0.

(1)

(b)

Find the gradient of the line segment at the point (6, 3).

(1)

(c)

Find the equation of the curve along which $\frac{dh}{dt} = 0$.

(1)

(d)

One solution curve has a stationary point when t = 4. Find $\frac{d^2h}{dt^2}$ at this point and hence show that the depth of water is a maximum there.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
2	✓ 1
Part (b)
2 − 0.25(6)(3) = −2.5	✓ 1
Part (c)
$2 - 0.25th = 0 \Rightarrow h = \frac{8}{t}$ (for t > 0)	✓ 1
Part (d)
At the stationary point $2 - 0.25(4)h = 0$, so $h = 2$ (or uses $h = \frac{8}{t}$ at $t = 4$)	✓ 1
Differentiates using the product rule: $\frac{d^2h}{dt^2} = -0.25\left(h + t\frac{dh}{dt}\right)$	✓ 1
At (4, 2) with $\frac{dh}{dt}=0$: $\frac{d^2h}{dt^2} = -0.5 < 0$, so the depth has a maximum of 2 m there	✓ 1

Common error: When differentiating 0.25th with respect to t, remember that h is a function of t. The product rule is needed, and treating h as a constant loses the t(dh/dt) term.

+ Add to test✓ In your test82IBMI-034-08-r1-04Differential equations and slope fields Paper HLMedium6 marks

Water drains from a tank through a small hole in its base. The depth of water, h metres, t minutes after the hole is opened satisfies
$\frac{dh}{dt} = -0.6\sqrt{h}$
Initially the depth of water is 4 m.

(a)

Solve the differential equation to find h in terms of t.

(4)

(b)

Find the time taken for the tank to empty.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: separates variables, $\int h^{-1/2}\,dh = \int -0.6\,dt$	✓ 1
A1: $2\sqrt{h} = -0.6t + c$	✓ 1
M1: substitutes h = 4 when t = 0, giving c = 4	✓ 1
A1: $h = (2 - 0.3t)^2$	✓ 1
(The model is valid for $0 \le t \le \frac{20}{3}$; after that time h = 0.)	—
Part (b)
M1: sets h = 0, so $2 - 0.3t = 0$	✓ 1
A1: $t = \frac{20}{3} \approx 6.67$ minutes	✓ 1

Common error: The integral of $h^{-1/2}$ is $2h^{1/2}$, not $\frac{1}{2}h^{1/2}$. Losing the factor of 2 gives the wrong constant and the wrong emptying time.

+ Add to test✓ In your test83IBMI-034-04-r2-03Euler's method and coupled systems Paper HLMedium10 marks

A mass is attached to a spring and moves vertically in a viscous liquid. Its displacement from equilibrium, x metres, at time t seconds satisfies
$\frac{d^2x}{dt^2} + 0.4\frac{dx}{dt} + 4x = 0$
When t = 0, x = 0.5 and the mass is at rest.

(a)

By letting $v = \frac{dx}{dt}$, show that the motion can be described by the system
$\frac{dx}{dt} = v$, $\frac{dv}{dt} = -4x - 0.4v$.

(2)

(b)

Use Euler's method with a step length of 0.1 seconds to estimate the displacement of the mass when t = 0.3.

(5)

(c)

Find an estimate for the velocity of the mass when t = 0.3. Hence state, with a reason, whether the mass is moving towards or away from its equilibrium position at this time.

(2)

(d)

State one way in which your Euler's method estimates of the displacement and velocity at t = 0.3 could be made more accurate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
States that v = dx/dt, so dv/dt = d²x/dt²	✓ 1
Substitutes and rearranges the equation to give dv/dt = −4x − 0.4v (AG)	✓ 1
Part (b)
Uses x_n+1 = x_n + 0.1v_n	✓ 1
Uses v_n+1 = v_n + 0.1(−4x_n − 0.4v_n)	✓ 1
At t = 0.1: x = 0.5, v = −0.2	✓ 1
At t = 0.2: x = 0.48, v = −0.392	✓ 1
At t = 0.3: x = 0.4408, so x ≈ 0.441 m	✓ 1
Part (c)
v ≈ −0.568 m s⁻¹	✓ 1
x is positive and v is negative, so the displacement is decreasing and the mass is moving towards its equilibrium position	✓ 1
Part (d)
Use a smaller step length (with more steps)	✓ 1

Exam tip: A second-order equation is turned into a system by introducing v = dx/dt. You then need two update formulas, one for x and one for v. Apply both at every step, and use the old values of x and v in each formula.

+ Add to test✓ In your test84IBMI-034-08-r2-05Differential equations and slope fields Paper HLMedium12 marks

On an island, x is the number of rabbits in hundreds and y is the number of foxes in tens, t years after a study begins. The populations are modelled by the coupled differential equations
$\frac{dx}{dt} = 0.3x - 0.01xy$
$\frac{dy}{dt} = -0.2y + 0.005xy$
When t = 0, x = 50 and y = 20.

(a)

Find the equilibrium point at which both populations are non-zero.

(3)

(b)

Find the value of $\frac{dy}{dx}$ when t = 0, and describe what is happening to both populations at that moment.

(3)

(c)

Use Euler's method with a step length of 0.5 years to estimate the values of x and y when t = 1.

(4)

(d)

Give one reason why the estimates in the previous part may not be reliable, and suggest how they could be improved.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Sets both derivatives to zero with x, y ≠ 0: $0.3 - 0.01y = 0$ and $-0.2 + 0.005x = 0$ [M]	✓ 1
y = 30 [A]	✓ 1
x = 40 [A]	✓ 1
Part (b)
Uses $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-4+5}{15-10}$ [M]	✓ 1
$\frac{dy}{dx} = 0.2$ [A]	✓ 1
dx/dt = 5 > 0 and dy/dt = 1 > 0, so both rabbits and foxes are increasing [R]	✓ 1
Part (c)
Uses $x_{n+1} = x_n + 0.5\,\frac{dx}{dt}$ and $y_{n+1} = y_n + 0.5\,\frac{dy}{dt}$ [M]	✓ 1
At t = 0.5: $x = 52.5$, $y = 20.5$ [A]	✓ 1
At t = 1: $x = 55.0$ (54.99375) [A]	✓ 1
At t = 1: $y = 21.1$ (21.140625) [A]	✓ 1
Part (d)
A step of 0.5 years is large and the rates of change vary within each step, so errors may build up [R]	✓ 1
Use a smaller step length, or a more accurate numerical method [R]	✓ 1

Exam tip: In a coupled Euler calculation, work out both new values from the old values of x and y before moving to the next step. Do not use the new x when updating y.

+ Add to test✓ In your test85IBMI-034-08-r2-03Differential equations and slope fields Paper HLMedium9 marks

A conservation group reintroduces deer into a forest. The deer population, P hundred, t years after reintroduction is modelled by
$\frac{dP}{dt} = 0.1P\left(1 - \frac{P}{50}\right)$.
A slope field for this differential equation is drawn for $t \ge 0$ and $0 \le P \le 70$.

(a)

Find the gradient of the slope field at any point where P = 20.

(2)

(b)

Find the values of P for which the slope field is horizontal, and interpret these values in context.

(3)

(c)

Describe the long-term behaviour of the deer population if initially (i) P = 10, (ii) P = 60.

(2)

(d)

Find the maximum rate at which the deer population can grow.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Substitutes: $0.1 \times 20 \times (1 - 0.4)$ [M]	✓ 1
= 1.2 (hundred deer per year) [A]	✓ 1
Part (b)
P = 0 and P = 50 [A]	✓ 1
P = 0: with no deer present the population stays at zero, so there is no growth [R]	✓ 1
P = 50: the population of 5000 deer stays constant; this is the largest population the forest can sustain (the carrying capacity) [R]	✓ 1
Part (c)
(i) dP/dt > 0 for 0 < P < 50, so the population increases and approaches 50 hundred (5000 deer) [R]	✓ 1
(ii) dP/dt < 0 for P > 50, so the population decreases and approaches 50 hundred (5000 deer) [R]	✓ 1
Part (d)
dP/dt is a quadratic in P with its maximum at P = 25 (from the vertex, or by solving $0.1 - 0.004P = 0$) [M]	✓ 1
Maximum rate = $0.1 \times 25 \times 0.5 = 1.25$ hundred deer per year [A]	✓ 1

Common error: A horizontal slope field means dP/dt = 0, not P = 0. Set the whole right-hand side equal to zero and give every root, including P = 0. Then interpret each root separately.

+ Add to test✓ In your test86IBMI-034-01-r2-05Optimisation in context Paper HLMedium10 marks

After a patient takes a tablet, the concentration of a drug in the bloodstream, C mg L⁻¹, is modelled by
$C(t) = 12t\,e^{-0.4t}$, $t \ge 0$,
where t is the time in hours after the tablet is taken.

(a)

Show that $C'(t) = 12e^{-0.4t}(1 - 0.4t)$.

(2)

(b)

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

(3)

(c)

Find the time at which the concentration is decreasing most rapidly.

(3)

(d)

Find the rate of change of concentration at this time. Interpret your answer in context.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses the product rule: $12e^{-0.4t} + 12t(-0.4e^{-0.4t})$ M1	✓ 1
Factorises to $12e^{-0.4t}(1 - 0.4t)$ A1	✓ 1
Part (b)
$e^{-0.4t} > 0$, so $C'(t) = 0$ when $1 - 0.4t = 0$ (or uses a GDC to find the maximum of C) M1	✓ 1
t = 2.5 hours A1	✓ 1
C(2.5) = 30e⁻¹ = 11.0 mg L⁻¹ A1	✓ 1
Part (c)
Recognises that this occurs at the minimum of C′(t), i.e. where $C''(t) = 0$ M1	✓ 1
$C''(t) = 12e^{-0.4t}(0.16t - 0.8)$, or a GDC is used to find the minimum point of the graph of C′(t) A1	✓ 1
t = 5 hours A1	✓ 1
Part (d)
C′(5) = −12e⁻² = −1.62 mg L⁻¹ h⁻¹ A1	✓ 1
Five hours after the tablet is taken, the concentration is falling at 1.62 mg L⁻¹ per hour, the fastest rate of decrease R1	✓ 1

Common error: The most rapid decrease is not where C′(t) = 0. It is where C′ itself has a minimum, so solve C″(t) = 0 or find the minimum of C′ on a GDC.

+ Add to test✓ In your test87IBMI-034-04-r2-04Euler's method and coupled systems Paper HLMedium7 marks

A patient swallows a tablet. The concentration of the drug in the patient's bloodstream, C mg L⁻¹, t hours after the tablet is taken is modelled by
$\frac{dC}{dt} = 5e^{-0.5t} - 0.3C$, with C = 0 when t = 0.
Euler's method with a step length of 0.5 hours is used to estimate C.

(a)

Find an estimate for the concentration when t = 0.5.

(2)

(b)

Find an estimate for the concentration when t = 1.5.

(3)

(c)

The exact solution of the differential equation is $C = 25(e^{-0.3t} - e^{-0.5t})$. Find the percentage error in your Euler estimate of the concentration when t = 1.5.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
C₁ = 0 + 0.5 × (5e⁰ − 0.3 × 0)	✓ 1
C ≈ 2.5 mg L⁻¹	✓ 1
Part (b)
Uses C_n+1 = C_n + 0.5(5e^−0.5t_n − 0.3C_n) with t updated at each step	✓ 1
At t = 1, C ≈ 4.072	✓ 1
At t = 1.5, C ≈ 4.98 mg L⁻¹	✓ 1
Part (c)
Exact value at t = 1.5 is 4.1315…; percentage error = \|4.9775 − 4.1315\| ÷ 4.1315 × 100	✓ 1
Percentage error ≈ 20.5%	✓ 1

Common error: When the derivative depends on t as well as C, remember to update t at every step. Keeping t = 0 in the exponential term gives the wrong gradient from the second step onwards.

+ Add to test✓ In your test88IBMI-034-08-r1-06Differential equations and slope fields Paper HLMedium7 marks

A skydiver falls from rest. Their velocity, v m s⁻¹, t seconds after jumping is modelled by
$\frac{dv}{dt} = 10 - 0.05v^2$

(a)

Find the terminal velocity of the skydiver.

(2)

(b)

For v ≥ 0, describe how the gradients in the slope field of this differential equation vary with v. Hence explain why the skydiver's velocity increases towards the terminal velocity but never exceeds it.

(2)

(c)

Use Euler's method with a step length of 0.5 s to estimate the skydiver's velocity after 1 second.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: sets $10 - 0.05v^2 = 0$	✓ 1
A1: $v = 10\sqrt{2} \approx 14.1$ m s⁻¹	✓ 1
Part (b)
A1: gradients depend only on v. They are positive for 0 ≤ v < 14.1, decreasing from 10 at v = 0 to zero at v = 14.1, and negative for v > 14.1	✓ 1
R1: starting from v = 0 the gradient is positive, so v increases. The gradient tends to zero as v approaches 14.1, and v = 14.1 is an equilibrium solution, so the solution curve approaches it asymptotically without crossing it	✓ 1
Part (c)
M1: uses $v_{n+1} = v_n + 0.5(10 - 0.05v_n^2)$ with $v_0 = 0$	✓ 1
A1: $v_1 = 0 + 0.5 \times 10 = 5$	✓ 1
A1: $v_2 = 5 + 0.5 \times 8.75 = 9.375 \approx 9.38$ m s⁻¹ (accept 9.375)	✓ 1

Common error: In Euler's method, evaluate the gradient for each step at the start of that step. Do not use the new value of v.

+ Add to test✓ In your test89IBMI-034-01-r1-06Optimisation in context Paper HLHard14 marks

The population $P$ of fish in a managed lake, $t$ years after monitoring begins, is modelled by
$\frac{dP}{dt} = 0.4P - 0.00008P^2 - H$,
where $H$ is the number of fish removed by fishing each year. The expression $f(P) = 0.4P - 0.00008P^2$ is the natural growth rate of the population.

(a)

By differentiating $f(P)$, show that the natural growth rate is greatest when $P = 2500$.

(3)

(b)

Hence find the largest value of $H$ for which the population can remain constant, and interpret this value in context.

(2)

(c)

The lake managers set $H = 420$. Find the two population sizes at which the population remains constant.

(3)

(d)

With $H = 420$, the initial population is 1800 fish. Use Euler's method with a step length of 0.5 years to estimate the population after 1.5 years. Give your answer to the nearest whole fish.

(4)

(e)

Describe the long-term behaviour of the population predicted by the model if the initial population were 1400 fish instead of 1800, with $H = 420$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $f'(P) = 0.4 - 0.00016P$	✓ 1
A1 setting $f'(P) = 0$ gives $P = 2500$	✓ 1
R1 $f''(P) = -0.00016 < 0$, so this is a maximum	✓ 1
Part (b)
A1 $f(2500) = 1000 - 500 = 500$, so $H = 500$	✓ 1
R1 500 fish per year is the greatest number that can be removed each year while keeping the population steady (maximum sustainable catch)	✓ 1
Part (c)
M1 setting $0.4P - 0.00008P^2 - 420 = 0$	✓ 1
A1 $P = 1500$	✓ 1
A1 $P = 3500$	✓ 1
Part (d)
M1 correct use of $P_{n+1} = P_n + 0.5(0.4P_n - 0.00008P_n^2 - 420)$	✓ 1
A1 $P_1 = 1820.4$	✓ 1
A1 $P_2 = 1841.93$	✓ 1
A1 $P_3 = 1865$ fish	✓ 1
Part (e)
R1 for $P < 1500$, $\frac{dP}{dt} < 0$ (e.g. at $P = 1400$, $\frac{dP}{dt} = -16.8$)	✓ 1
R1 so the population would keep decreasing and the fish would die out	✓ 1

Exam tip: Write each Euler step in a table with t, P and dP/dt, and keep full calculator accuracy between steps. Only round the final population.

+ Add to test✓ In your test90IBMI-034-00-r1-06Gradients, tangents and rates of change Paper HLHard11 marks

Water flows at a constant rate of 0.05 m³ min⁻¹ into a tank shaped as an inverted right circular cone with its vertex at the bottom. When the depth of water is $h$ metres, the radius of the water surface is $r = 0.4h$ metres.
The volume of a cone is $V = \frac{1}{3}\pi r^2 h$.

(a)

Show that the volume of water in the tank is $V = \dfrac{4\pi}{75}h^3$.

(2)

(b)

Find $\dfrac{dV}{dh}$.

(1)

(c)

Find the rate at which the depth of the water is increasing when $h = 1.5$ m.

(3)

(d)

Find the rate at which the area of the water surface is increasing when $h = 1.5$ m.

(3)

(e)

Explain why the rate at which the depth increases becomes smaller as the tank fills.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: substituting $r = 0.4h$: $V = \frac{1}{3}\pi(0.4h)^2 h$	✓ 1
A1: $= \frac{0.16}{3}\pi h^3 = \frac{4\pi}{75}h^3$ (AG)	✓ 1
Part (b)
A1: $\frac{dV}{dh} = \frac{4\pi}{25}h^2$ (accept $0.16\pi h^2$)	✓ 1
Part (c)
M1: using $\frac{dh}{dt} = \frac{dV}{dt} \div \frac{dV}{dh}$	✓ 1
A1: $\frac{dV}{dh} = 0.16\pi(1.5)^2 = 1.13$ (0.36π)	✓ 1
A1: $\frac{dh}{dt} = \frac{0.05}{1.130\ldots} = 0.0442$ m min⁻¹	✓ 1
Part (d)
M1: $A = \pi r^2 = 0.16\pi h^2$	✓ 1
M1: $\frac{dA}{dt} = 0.32\pi h \times \frac{dh}{dt}$	✓ 1
A1: $\frac{dA}{dt} = 0.32\pi(1.5)(0.04420\ldots) = 0.0667$ m² min⁻¹	✓ 1
Part (e)
R1: $\frac{dV}{dh} = 0.16\pi h^2$ increases as $h$ increases, because the cone widens and the water surface gets larger	✓ 1
R1: the inflow rate is constant, so $\frac{dh}{dt} = \frac{0.05}{0.16\pi h^2}$ decreases as $h$ increases	✓ 1

Exam tip: In related-rates problems, write the chain rule in full, for example $\frac{dA}{dt} = \frac{dA}{dh} \times \frac{dh}{dt}$. This shows the method clearly and helps avoid dividing where you should multiply.

+ Add to test✓ In your test91IBMI-034-04-r2-06Euler's method and coupled systems Paper HLHard14 marks

Two species of fish compete for food in a lake. Their populations, in thousands, after t years are x and y. They are modelled by
$\frac{dx}{dt} = 0.5x(4 - x - y)$
$\frac{dy}{dt} = 0.2y(6 - 2x - y)$
Initially there are 1000 fish of each species.

(a)

Find all the equilibrium points of the system.

(4)

(b)

At a particular time, the populations are x = 3 and y = 3. Find the values of $\frac{dx}{dt}$ and $\frac{dy}{dt}$ at this time, and describe how the populations are changing.

(3)

(c)

The trajectory starting from (1, 1) approaches the point (2, 2). Interpret this in context.

(2)

(d)

Use Euler's method with a step length of 0.1 years to estimate both populations when t = 0.3.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Sets both derivatives to zero: x = 0 or x + y = 4, and y = 0 or 2x + y = 6	✓ 1
(0, 0)	✓ 1
(4, 0) and (0, 6)	✓ 1
(2, 2)	✓ 1
Part (b)
dx/dt = 0.5 × 3 × (−2) = −3	✓ 1
dy/dt = 0.2 × 3 × (−3) = −1.8	✓ 1
Both values are negative, so both populations are decreasing; species x is falling faster, at 3000 fish per year compared with 1800 fish per year	✓ 1
Part (c)
In the long term, the populations tend to 2000 fish of each species	✓ 1
So the two species coexist: neither species dies out	✓ 1
Part (d)
Uses x_n+1 = x_n + 0.1 × 0.5x_n(4 − x_n − y_n) and y_n+1 = y_n + 0.1 × 0.2y_n(6 − 2x_n − y_n)	✓ 1
At t = 0.1: x = 1.1, y = 1.06	✓ 1
At t = 0.2: x = 1.2012, y = 1.11809	✓ 1
At t = 0.3: x ≈ 1.30, i.e. about 1300 fish	✓ 1
At t = 0.3: y ≈ 1.17, i.e. about 1170 fish	✓ 1

Common error: The variables are measured in thousands. Give answers in context by converting back to numbers of fish. Do not round intermediate Euler values, because rounding errors build up quickly over several steps.

+ Add to test✓ In your test92IBMI-034-08-r2-06Differential equations and slope fields Paper HLHard14 marks

Two competing species of beetle live in a woodland. Their populations, x thousand and y thousand, t years after monitoring begins are modelled by
$\frac{dx}{dt} = 0.3x - 0.4y$
$\frac{dy}{dt} = -0.2x + 0.1y$
When t = 0, x = 5 and y = 4.

(a)

Find the eigenvalues of the matrix $\begin{pmatrix} 0.3 & -0.4 \\ -0.2 & 0.1 \end{pmatrix}$.

(3)

(b)

Find an eigenvector corresponding to each eigenvalue.

(2)

(c)

Hence find expressions for x and y in terms of t.

(4)

(d)

Classify the equilibrium point at the origin, giving a reason.

(2)

(e)

According to the model, species y dies out. Find the time at which this happens and the population of species x at that time.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses $\det\begin{pmatrix} 0.3-\lambda & -0.4 \\ -0.2 & 0.1-\lambda \end{pmatrix} = 0$ [M]	✓ 1
$\lambda^2 - 0.4\lambda - 0.05 = 0$ [A]	✓ 1
λ = 0.5 and λ = −0.1 [A]	✓ 1
Part (b)
λ = 0.5: $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$ (or any non-zero multiple) [A]	✓ 1
λ = −0.1: $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ (or any non-zero multiple) [A]	✓ 1
Part (c)
General solution $\begin{pmatrix} x \\ y \end{pmatrix} = Ae^{0.5t}\begin{pmatrix} 2 \\ -1 \end{pmatrix} + Be^{-0.1t}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ [M]	✓ 1
Uses initial values: $2A + B = 5$, $-A + B = 4$ [M]	✓ 1
$A = \frac{1}{3}$, $B = \frac{13}{3}$ [A]	✓ 1
$x = \frac{2}{3}e^{0.5t} + \frac{13}{3}e^{-0.1t}$, $y = -\frac{1}{3}e^{0.5t} + \frac{13}{3}e^{-0.1t}$ [A]	✓ 1
Part (d)
The eigenvalues are real and of opposite sign [R]	✓ 1
so the origin is a saddle point, which is unstable [A]	✓ 1
Part (e)
Sets y = 0: $e^{0.6t} = 13$ [M]	✓ 1
t = 4.27 years [A]	✓ 1
x = 8.48 thousand beetles [A]	✓ 1

Exam tip: Each eigenvalue must be paired with its own eigenvector in the general solution. Check this by substituting t = 0 and confirming that you recover the initial populations.

+ Add to test✓ In your test93IBMI-034-03-r2-06Differential equations and slope fields Paper HLHard14 marks

Two connected tanks contain salt solution. The mass of salt in tank A is x kg and the mass of salt in tank B is y kg, t minutes after mixing begins. The system is modelled by
$\frac{dx}{dt} = -0.2x + 0.1y$
$\frac{dy}{dt} = 0.2x - 0.2y$
Initially tank A contains 20 kg of salt and tank B contains pure water.

(a)

Find the eigenvalues of the matrix $\begin{pmatrix} -0.2 & 0.1 \\ 0.2 & -0.2 \end{pmatrix}$.

(3)

(b)

Find a corresponding eigenvector for each eigenvalue.

(2)

(c)

Hence find expressions for x and y in terms of t.

(3)

(d)

Classify the equilibrium point at the origin and describe what happens to the salt in the long term.

(2)

(e)

Find the maximum mass of salt in tank B and the time at which it occurs.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Characteristic equation $(-0.2-\lambda)^2 - 0.02 = 0$ (M1)	✓ 1
$\lambda_1 = -0.2 + 0.1\sqrt{2} \approx -0.0586$ (A1)	✓ 1
$\lambda_2 = -0.2 - 0.1\sqrt{2} \approx -0.341$ (A1)	✓ 1
Part (b)
For $\lambda_1$: $\begin{pmatrix} 1 \\ \sqrt{2} \end{pmatrix}$ (A1)	✓ 1
For $\lambda_2$: $\begin{pmatrix} 1 \\ -\sqrt{2} \end{pmatrix}$ (A1)	✓ 1
Part (c)
General solution $\begin{pmatrix} x \\ y \end{pmatrix} = Ae^{\lambda_1 t}\begin{pmatrix} 1 \\ \sqrt{2} \end{pmatrix} + Be^{\lambda_2 t}\begin{pmatrix} 1 \\ -\sqrt{2} \end{pmatrix}$ (M1)	✓ 1
Uses initial conditions: $A + B = 20$, $\sqrt{2}(A - B) = 0$, so $A = B = 10$ (A1)	✓ 1
$x = 10(e^{\lambda_1 t} + e^{\lambda_2 t})$, $y = 10\sqrt{2}(e^{\lambda_1 t} - e^{\lambda_2 t})$ (A1)	✓ 1
Part (d)
Both eigenvalues are real and negative, so the origin is a stable node (R1)	✓ 1
The mass of salt in both tanks tends to zero as t increases (all the salt eventually leaves the system) (A1)	✓ 1
Part (e)
Sets $\frac{dy}{dt} = 0$: $\lambda_1 e^{\lambda_1 t} = \lambda_2 e^{\lambda_2 t}$ (M1)	✓ 1
$t = \frac{\ln(\lambda_2/\lambda_1)}{\lambda_1 - \lambda_2}$ (M1)	✓ 1
$t = 6.23$ minutes (A1)	✓ 1
Maximum mass $y = 8.13$ kg (A1)	✓ 1

Exam tip: Keep the eigenvalues in exact surd form or stored at full accuracy; using −0.06 and −0.34 noticeably shifts both the time and the maximum mass.

+ Add to test✓ In your test94IBMI-034-02-r2-06Integration and the trapezoidal rule Paper HLHard14 marks

A cyclist starts from rest. Her speed, v m s⁻¹, is recorded every 4 seconds.

t (s): 0, 4, 8, 12, 16, 20
v (m s⁻¹): 0, 5.4, 8.4, 10.0, 10.9, 11.4

A coach models her speed by $v(t) = 12(1 - e^{-0.15t})$, for $t \ge 0$.

(a)

Use the trapezoidal rule with the recorded data to estimate the distance she travels in the first 20 seconds.

(3)

(b)

Use the model to find the distance she travels in the first 20 seconds.

(3)

(c)

Find the percentage difference between the trapezoidal estimate and the value from the model, taking the model value as accurate.

(2)

(d)

With reference to the shape of the model, explain why the trapezoidal rule gives a smaller distance than the model.

(2)

(e)

Use the model to find the time at which the cyclist has travelled 400 m.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
h = 4	✓ 1
$\frac{4}{2}[0 + 11.4 + 2(5.4 + 8.4 + 10.0 + 10.9)]$	✓ 1
162 m (161.6)	✓ 1
Part (b)
Distance as integral of speed	✓ 1
$\int_0^{20} 12(1 - e^{-0.15t})\,dt$	✓ 1
164 m (163.98…)	✓ 1
Part (c)
$\frac{163.98 - 161.6}{163.98} \times 100$	✓ 1
1.45 %	✓ 1
Part (d)
The speed–time curve is concave down, because her acceleration decreases over time	✓ 1
The straight tops of the trapezoids lie below the curve, so the rule underestimates the distance	✓ 1
Part (e)
Sets up $\int_0^{T} 12(1 - e^{-0.15t})\,dt = 400$	✓ 1
Correct equation, e.g. $12\left[T + \frac{e^{-0.15T} - 1}{0.15}\right] = 400$	✓ 1
Attempts to solve with GDC (graph or solver)	✓ 1
T = 40.0 s (39.98…)	✓ 1

Exam tip: When the unknown is the upper limit of an integral, set up the equation clearly and then solve it on the GDC. A correctly written equation earns marks even if the numerical solution goes wrong.

+ Add to test✓ In your test95IBMI-034-08-r1-05Differential equations and slope fields Paper HLHard13 marks

An ecologist studies two interacting species near an equilibrium. The deviations of their populations from equilibrium, x and y (in hundreds), at time t years are modelled by
$\frac{dx}{dt} = x + 4y, \quad \frac{dy}{dt} = x + y$
At t = 0, x = 6 and y = 1.

(a)

Find the eigenvalues of the matrix $\begin{pmatrix}1 & 4\\ 1 & 1\end{pmatrix}$.

(3)

(b)

Find a corresponding eigenvector for each eigenvalue.

(2)

(c)

Hence find expressions for x and y in terms of t.

(3)

(d)

Classify the equilibrium point at the origin, giving a reason.

(2)

(e)

Describe the long-term behaviour of the trajectory in the phase plane.

(2)

(f)

Find the value of x when t = 0.5.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: forms the characteristic equation $(1-\lambda)^2 - 4 = 0$	✓ 1
A1: λ = 3	✓ 1
A1: λ = −1	✓ 1
Part (b)
A1: for λ = 3, $\begin{pmatrix}2\\1\end{pmatrix}$ or any non-zero multiple	✓ 1
A1: for λ = −1, $\begin{pmatrix}2\\-1\end{pmatrix}$ or any non-zero multiple	✓ 1
Part (c)
M1: writes the general solution $\begin{pmatrix}x\\y\end{pmatrix} = Ae^{3t}\begin{pmatrix}2\\1\end{pmatrix} + Be^{-t}\begin{pmatrix}2\\-1\end{pmatrix}$	✓ 1
M1: uses the initial conditions, 2A + 2B = 6 and A − B = 1	✓ 1
A1: $x = 4e^{3t} + 2e^{-t}$, $y = 2e^{3t} - e^{-t}$	✓ 1
Part (d)
A1: saddle point (unstable)	✓ 1
R1: the eigenvalues are real with opposite signs	✓ 1
Part (e)
M1: the $e^{3t}$ terms dominate as t increases	✓ 1
A1: $\frac{y}{x} \to \frac{1}{2}$, so the trajectory moves away from the origin and approaches the line $y = \frac{x}{2}$. Both deviations grow without bound	✓ 1
Part (f)
A1: $x = 4e^{1.5} + 2e^{-0.5} = 19.1$	✓ 1

Exam tip: Check your eigenvalues using trace = sum of eigenvalues (here 2 = 3 + (−1)) and determinant = product of eigenvalues (here −3 = 3 × (−1)).

+ Add to test✓ In your test96IBMI-034-05-r1-06Gradients, tangents and rates of change Paper HLHard14 marks

A water storage tank is in the shape of an inverted right circular cone with its vertex at the bottom. The tank has a top radius of 3 m and a depth of 6 m. Water is pumped into the tank at a constant rate of 0.5 m³ per minute. When the depth of water is $h$ m, the radius of the water surface is $r$ m and the volume of water is $V$ m³.

(a)

Show that $V = \frac{\pi h^3}{12}$.

(2)

(b)

Find $\frac{dV}{dh}$.

(1)

(c)

Find the rate at which the depth of water is increasing when $h = 2$.

(3)

(d)

Find the rate at which the area of the water surface is increasing when $h = 2$.

(4)

(e)

Find the rate at which the depth of water is increasing at the instant the tank is half full by volume.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Similar triangles: $\frac{r}{h} = \frac{3}{6}$, so $r = \frac{h}{2}$ M1	✓ 1
$V = \frac{1}{3}\pi\left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}$ A1	✓ 1
Part (b)
$\frac{dV}{dh} = \frac{\pi h^2}{4}$ A1	✓ 1
Part (c)
At $h = 2$: $\frac{dV}{dh} = \frac{\pi (2)^2}{4} = \pi$ A1	✓ 1
$\frac{dh}{dt} = \frac{dV}{dt} \div \frac{dV}{dh} = \frac{0.5}{\pi}$ M1	✓ 1
$0.159$ m per minute A1	✓ 1
Part (d)
$A = \pi r^2 = \frac{\pi h^2}{4}$ M1	✓ 1
$\frac{dA}{dh} = \frac{\pi h}{2}$, which equals $\pi$ at $h = 2$ A1	✓ 1
$\frac{dA}{dt} = \frac{dA}{dh} \times \frac{dh}{dt} = \pi \times \frac{0.5}{\pi}$ M1	✓ 1
$0.5$ m² per minute A1	✓ 1
Part (e)
Full volume $= \frac{\pi (6)^3}{12} = 18\pi$, so half volume $= 9\pi$ M1	✓ 1
$\frac{\pi h^3}{12} = 9\pi \Rightarrow h = \sqrt[3]{108} = 4.76$ m A1	✓ 1
$\frac{dh}{dt} = \frac{0.5}{\pi (4.762)^2/4}$ M1	✓ 1
$0.0281$ m per minute A1	✓ 1

Common error: Assuming the tank is half full when the depth is 3 m. Because the cone is narrow at the bottom, half the volume is reached at a depth of about 4.76 m, so the volume must be used to find $h$.

+ Add to test✓ In your test97IBMI-034-00-r2-06Gradients, tangents and rates of change Paper HLHard14 marks

A water tank is in the shape of an inverted right circular cone with its vertex at the bottom. The tank has height 5 m and the radius of its top is 2 m. Water is pumped into the empty tank at a constant rate of 0.3 m³ per minute. When the depth of water is h metres, the radius of the water surface is r metres and the volume of water is V m³.

(a)

Show that $V = \frac{4\pi h^3}{75}$.

(2)

(b)

Find $\frac{dV}{dh}$.

(1)

(c)

Find the rate at which the depth of water is increasing when $h = 2$.

(3)

(d)

The area of the water surface is $A = \pi r^2$. Find the rate at which this area is increasing when $h = 2$.

(3)

(e)

Find the time taken to fill the tank completely.

(2)

(f)

The tank is emptied and refilled at the same rate. A crack now lets water leak out at a rate of $0.08h$ m³ per minute. Find the depth of water at which the net rate of change of volume would be zero. Hence explain whether the tank will fill.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses similar triangles: $\frac{r}{h} = \frac{2}{5}$, so $r = 0.4h$	✓ 1
$V = \frac{1}{3}\pi(0.4h)^2h = \frac{0.16\pi h^3}{3} = \frac{4\pi h^3}{75}$	✓ 1
Part (b)
$\frac{dV}{dh} = \frac{12\pi h^2}{75} = \frac{4\pi h^2}{25}$	✓ 1
Part (c)
Uses the chain rule: $\frac{dh}{dt} = \frac{dV}{dt} \div \frac{dV}{dh}$	✓ 1
$\frac{0.3}{4\pi(2)^2/25}$	✓ 1
0.149 m per minute	✓ 1
Part (d)
$A = 0.16\pi h^2$	✓ 1
$\frac{dA}{dt} = 0.32\pi h \frac{dh}{dt}$	✓ 1
$0.32\pi(2)(0.1492...) = 0.300$ m² per minute	✓ 1
Part (e)
$V(5) = \frac{4\pi(125)}{75} = 20.94$ m³	✓ 1
$\frac{20.94}{0.3} = 69.8$ minutes	✓ 1
Part (f)
$\frac{dV}{dt} = 0.3 - 0.08h$; setting $0.3 - 0.08h = 0$	✓ 1
$h = 3.75$ m	✓ 1
For $h < 3.75$ we have $\frac{dV}{dt} > 0$ but tending to zero, so the level rises towards 3.75 m; since $3.75 < 5$, the tank never fills	✓ 1

Exam tip: In related-rates problems, write the chain rule out in full, for example dh/dt = (dV/dt) ÷ (dV/dh). This makes your method clear even if a later value is wrong.

+ Add to test✓ In your test98IBMI-034-03-r1-06Differential equations and slope fields Paper HLHard14 marks

The vertical displacement, x cm, of a car body from its rest position t seconds after passing over a bump is modelled by
$\frac{d^2x}{dt^2} + 0.4\frac{dx}{dt} + 4x = 0$.
At t = 0 the displacement is 5 cm and the car body is momentarily at rest.

(a)

By letting $v = \frac{dx}{dt}$, write the differential equation as a system of two first-order differential equations.

(2)

(b)

Find the eigenvalues of the matrix associated with this system.

(3)

(c)

Describe the phase portrait of the system near the origin and interpret its behaviour in context.

(3)

(d)

Use Euler's method with a step length of 0.1 seconds to estimate the displacement of the car body when t = 0.3.

(4)

(e)

Using the eigenvalues you found earlier, find the time taken for one complete oscillation of the car body.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\frac{dx}{dt} = v$	✓ 1
$\frac{dv}{dt} = -4x - 0.4v$	✓ 1
Part (b)
Matrix $\begin{pmatrix}0 & 1\\ -4 & -0.4\end{pmatrix}$	✓ 1
Characteristic equation $\lambda^2 + 0.4\lambda + 4 = 0$	✓ 1
λ = −0.2 ± 1.99i	✓ 1
Part (c)
Complex eigenvalues give trajectories that spiral	✓ 1
The real part is negative, so the spiral is inward and the origin is stable	✓ 1
The car body oscillates up and down with decreasing amplitude and comes to rest at its rest position	✓ 1
Part (d)
Uses $x_0 = 5$, $v_0 = 0$ with $x_{n+1} = x_n + 0.1v_n$ and $v_{n+1} = v_n + 0.1(-4x_n - 0.4v_n)$	✓ 1
t = 0.1: x = 5, v = −2	✓ 1
t = 0.2: x = 4.8, v = −3.92	✓ 1
t = 0.3: x ≈ 4.41 cm	✓ 1
Part (e)
Period $= \frac{2\pi}{\text{imaginary part}} = \frac{2\pi}{1.98997}$	✓ 1
3.16 s	✓ 1

Common error: In the coupled Euler method, both new values must be calculated from the old x and v. Updating x first and then using the new x to find v gives a different, incorrect sequence.

+ Add to test✓ In your test99IBMI-034-04-r1-04Euler's method and coupled systems Paper HLHard16 marks

Two adjoining rooms in a building lose heat to the outside and exchange heat with each other. Let x and y be the temperatures, in °C above the outdoor temperature, of room A and room B respectively, t hours after the heating is switched off. The temperatures are modelled by
$\frac{dx}{dt} = -0.4x + 0.2y$
$\frac{dy}{dt} = 0.2x - 0.4y$
At t = 0, x = 20 and y = 8.

(a)

Find the eigenvalues of the matrix $\begin{pmatrix} -0.4 & 0.2 \\ 0.2 & -0.4 \end{pmatrix}$.

(3)

(b)

Find a corresponding eigenvector for each eigenvalue.

(2)

(c)

Hence find expressions for x and y in terms of t.

(3)

(d)

Use Euler's method with a step length of 0.5 hours to estimate the value of x when t = 1.

(3)

(e)

Use your expression for x in terms of t to find the value of x when t = 1, and comment on the accuracy of the Euler estimate.

(3)

(f)

Describe the long-term behaviour of the temperatures of the two rooms, justifying your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$(-0.4 - \lambda)^2 - 0.04 = 0$ (M1)	✓ 1
λ = −0.2 (A1)	✓ 1
λ = −0.6 (A1)	✓ 1
Part (b)
For λ = −0.2: $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ (or any non-zero multiple) (A1)	✓ 1
For λ = −0.6: $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ (or any non-zero multiple) (A1)	✓ 1
Part (c)
General solution $\begin{pmatrix} x \\ y \end{pmatrix} = Ae^{-0.2t}\begin{pmatrix} 1 \\ 1 \end{pmatrix} + Be^{-0.6t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ (M1)	✓ 1
A + B = 20 and A − B = 8, giving A = 14, B = 6 (A1)	✓ 1
$x = 14e^{-0.2t} + 6e^{-0.6t}$, $y = 14e^{-0.2t} - 6e^{-0.6t}$ (A1)	✓ 1
Part (d)
At t = 0: dx/dt = −6.4 and dy/dt = 0.8 (M1)	✓ 1
At t = 0.5: x = 16.8 and y = 8.4 (A1)	✓ 1
At t = 0.5: dx/dt = −5.04, so at t = 1: x = 16.8 − 2.52 = 14.3 (14.28) (A1)	✓ 1
Part (e)
Substitutes t = 1: $14e^{-0.2} + 6e^{-0.6}$ (M1)	✓ 1
x = 14.8 °C above the outdoor temperature (14.755…) (A1)	✓ 1
The Euler estimate is lower than the true value by about 0.48 °C (about 3%), so it underestimates; a smaller step length would reduce the error (R1)	✓ 1
Part (f)
Both eigenvalues are negative, so both exponential terms tend to zero and both rooms approach the outdoor temperature (R1)	✓ 1
The $e^{-0.6t}$ term decays faster, so for large t, x ≈ y ≈ $14e^{-0.2t}$: the two rooms have nearly equal temperatures and cool at the same rate (R1)	✓ 1

Exam tip: Check your particular solution by substituting t = 0. It must return x = 20 and y = 8 before you use it to judge the Euler estimate.

+ Add to test✓ In your test100IBMI-034-07-r2-06Integration and the trapezoidal rule Paper HLHard13 marks

A glassmaker designs a vase of height 24 cm. The vase is formed by rotating the curve

$r(h) = 4 + 2\sin\left(\frac{\pi h}{24}\right), \quad 0 \le h \le 24,$

through 2π radians about the h-axis, where r is the internal radius in cm at a height h cm above the base. The internal capacity of the vase is $V = \pi\int_0^{24} r^2\,dh$.

(a)

Find the maximum internal radius of the vase and the height at which it occurs.

(2)

(b)

Use the trapezoidal rule with four strips of equal width to estimate the capacity of the vase.

(4)

(c)

Use integration to find the capacity of the vase, giving your answer correct to 3 significant figures.

(3)

(d)

Without further integration, write down the depth of water needed for the vase to be exactly half full. Justify your answer.

(2)

(e)

Find the percentage error in your trapezoidal estimate of the capacity, taking your answer from integration as the true value.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: maximum radius 6 cm	✓ 1
A1: at h = 12 cm	✓ 1
Part (b)
M1: strip width 6, with r² values 16, 29.31, 36, 29.31, 16	✓ 1
M1: applies the trapezoidal rule to r² (not r) and multiplies by π	✓ 1
M1: correct substitution, $\pi \times \frac{6}{2}\left[16+16+2(29.3137+36+29.3137)\right]$	✓ 1
A1: 2090 cm³ (accept 2085 or 2085.28…)	✓ 1
Part (c)
M1: $\pi\int_0^{24} \left(4 + 2\sin\left(\frac{\pi h}{24}\right)\right)^2 dh$	✓ 1
A1: 2125.17… (accept the exact form 432π + 768)	✓ 1
A1: 2130 cm³	✓ 1
Part (d)
A1: 12 cm	✓ 1
R1: the radius function is symmetric about h = 12 (r(12 − k) = r(12 + k)), so the volume below h = 12 equals the volume above it	✓ 1
Part (e)
M1: $\frac{2125.17 - 2085.28}{2125.17} \times 100$	✓ 1
A1: 1.88% (accept 1.87% to 1.88%)	✓ 1

Common error: For a volume of revolution, the trapezoidal rule must be applied to πr² (cross-sectional areas), not to r. Squaring the trapezoidal estimate of ∫r dh does not give the volume. Use unrounded values when you calculate the percentage error.

+ Add to test✓ In your test101IBMI-034-07-r1-05Integration and the trapezoidal rule Paper HLHard10 marks

A designer models the inside of a glass vase by rotating the curve
$r(x) = 4 + 2\sin\left(\frac{x}{5}\right)$, for $0 \le x \le 20$,
through 2π radians about the x-axis. Here x is the distance in cm along the vase's axis and r is the internal radius in cm. The sine function is in radians.

(a)

Write down an integral for the internal volume of the vase.

(2)

(b)

Find the internal volume of the vase.

(2)

(c)

Use the trapezoidal rule with 4 strips of equal width to estimate $\int_0^{20} r^2\,dx$. Hence estimate the internal volume of the vase.

(4)

(d)

Using unrounded values, find the percentage error in your trapezoidal-rule estimate of the volume, compared with the volume found by integration.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses the volume of revolution formula $\pi\int y^2\,dx$ (M1)	✓ 1
$V = \pi\int_0^{20}\left(4 + 2\sin\frac{x}{5}\right)^2 dx$, with the correct limits and integrand (A1)	✓ 1
Part (b)
Evaluates their integral using a GDC (M1)	✓ 1
1530 cm³ (1531.04…) (A1)	✓ 1
Part (c)
Uses h = 5 with values of r² at x = 0, 5, 10, 15, 20: 16, 32.30, 33.86, 18.34, 6.18 (A1)	✓ 1
Correct substitution: $\frac{5}{2}[16 + 6.182 + 2(32.30 + 33.86 + 18.34)]$ (M1)	✓ 1
Integral ≈ 478 (477.90…) (A1)	✓ 1
Volume ≈ π × 477.90… = 1500 cm³ (1501.38…) (A1)	✓ 1
Part (d)
Correct substitution: $\frac{1531.04 - 1501.38}{1531.04} \times 100$ (M1)	✓ 1
1.94% (accept 1.9% to 2.0% if rounded volumes are used) (A1)	✓ 1

Common error: The trapezoidal rule must be applied to r², not to r. Squaring an estimate of the area under r does not give the volume. Also check that your GDC is in radian mode before evaluating the sine values.

+ Add to test✓ In your test102IBMI-034-01-r2-06Optimisation in context Paper HLHard13 marks

An offshore gas platform, P, is 3 km from the nearest point, A, on a straight coastline. A processing plant, R, is on the coast 8 km from A. A pipeline will run in a straight line under the sea from P to a point Q on the coast, and then along the coast from Q to R. Q lies between A and R, with AQ = x km.
Laying pipe under the sea costs $1.5 million per km. Laying pipe on land costs $0.9 million per km.

(a)

Show that the total cost, C million dollars, is given by $C(x) = 1.5\sqrt{9 + x^2} + 0.9(8 - x)$, for $0 \le x \le 8$.

(2)

(b)

Find $C'(x)$.

(2)

(c)

Find the value of x that minimises the cost.

(3)

(d)

Find the minimum cost. Justify that it is the minimum on the domain $0 \le x \le 8$.

(3)

(e)

Land costs rise to $k million per km, while the undersea cost is unchanged. Find the exact value of k at or above which the cheapest option is to lay the whole pipeline under the sea directly from P to R. Give your answer to 4 significant figures as well.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses Pythagoras: PQ = $\sqrt{9 + x^2}$ M1	✓ 1
States QR = 8 − x and multiplies each length by its cost per km to give the result A1	✓ 1
Part (b)
Uses the chain rule to get $\frac{1.5x}{\sqrt{9 + x^2}}$ A1	✓ 1
Differentiates the land term to get −0.9 A1	✓ 1
Part (c)
Sets $\frac{1.5x}{\sqrt{9+x^2}} = 0.9$ M1	✓ 1
Squares both sides: 2.25x² = 0.81(9 + x²), so 1.44x² = 7.29 A1	✓ 1
x = 2.25 km A1	✓ 1
Part (d)
C(2.25) = 1.5(3.75) + 0.9(5.75) = $10.8 million A1	✓ 1
Evaluates the endpoints: C(0) = 11.7 and C(8) ≈ 12.8 M1	✓ 1
States that both endpoint costs exceed 10.8 and that there is only one stationary point, so x = 2.25 gives the minimum R1	✓ 1
Part (e)
States $C'(x) = \frac{1.5x}{\sqrt{9+x^2}} - k$ and that the first term increases with x, so the greatest value of C′ on the domain is C′(8) M1	✓ 1
The whole pipe goes under the sea when C′(x) ≤ 0 throughout 0 ≤ x ≤ 8, i.e. C′(8) ≤ 0, so C is decreasing and is least at x = 8. This gives k ≥ $\frac{1.5 \times 8}{\sqrt{73}}$ M1	✓ 1
k = $\frac{12}{\sqrt{73}}$ ≈ 1.404 million dollars per km A1	✓ 1

Exam tip: On a closed domain, evaluate the function at both endpoints as well as at the stationary point. For a threshold value, give the exact form: rounding it down can give a value that does not meet the condition.

+ Add to test✓ In your test103IBMI-034-06-r2-06Optimisation in context Paper HLHard14 marks

The number of fish, $P$, in a managed lake is modelled by the differential equation
$\frac{dP}{dt} = 0.4P\left(1 - \frac{P}{5000}\right)$,
where $t$ is the time in years.

(a)

Find the population size at which the population grows fastest, and find this maximum growth rate.

(4)

(b)

The owners plan to remove fish at a constant rate of $H$ fish per year, so that $\frac{dP}{dt} = 0.4P\left(1 - \frac{P}{5000}\right) - H$. Explain why the population cannot survive in the long term if $H > 500$, and state the largest possible sustainable value of $H$.

(2)

(c)

The owners choose $H = 420$. Find the two equilibrium populations. The current population is 3000. Describe what the model predicts about the population in the long term.

(4)

(d)

With $H = 420$ and $P = 3000$ when $t = 0$, use Euler's method with a step length of 0.5 years to estimate the population after 1 year.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Treats the growth rate as a function of $P$, $f(P) = 0.4P - 0.00008P^2$, and uses a valid method to maximise it (differentiation, the symmetry of the quadratic, or a GDC) M1	✓ 1
$f'(P) = 0.4 - 0.00016P = 0$, or the roots of $f$ are $P = 0$ and $P = 5000$ so the vertex is at their midpoint A1	✓ 1
$P = 2500$ fish A1	✓ 1
Maximum rate $= 0.4(2500)(0.5) = 500$ fish per year A1	✓ 1
Part (b)
The natural growth rate is never more than 500, so if $H > 500$ then $\frac{dP}{dt} < 0$ for every $P$ and the population falls until it dies out R1	✓ 1
Largest sustainable harvest $H = 500$ fish per year A1	✓ 1
Part (c)
Sets $0.4P - 0.00008P^2 - 420 = 0$ M1	✓ 1
$P = 1500$ A1	✓ 1
$P = 3500$ A1	✓ 1
$\frac{dP}{dt} > 0$ for $1500 < P < 3500$, so starting from 3000 the population increases towards 3500 fish R1	✓ 1
Part (d)
Uses $P_{n+1} = P_n + 0.5\left[0.4P_n\left(1 - \frac{P_n}{5000}\right) - 420\right]$ M1	✓ 1
At $t = 0.5$: $P_1 = 3000 + 0.5(60) = 3030$ A1	✓ 1
At $t = 1$: $P_2 = 3030 + 0.5(57.528)$ M1	✓ 1
$P_2 = 3058.76… \approx 3060$ fish A1	✓ 1

Exam tip: To find the population that gives the fastest growth, maximise $\frac{dP}{dt}$ as a function of $P$, not $t$. Because $\frac{dP}{dt}$ is a quadratic in $P$, its maximum lies midway between its roots.

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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.	—

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
Part (a)
Separates variables: \(\int \frac{1}{T-20}\,dT = \int -k\,dt\)	✓ 1
\(\ln\|T-20\| = -kt + c\)	✓ 1
Uses T = 90 when t = 0 to give \(T-20 = 70e^{-kt}\), so \(T = 20 + 70e^{-kt}\)	✓ 1
Part (b)
Substitutes t = 10, T = 60: \(40 = 70e^{-10k}\)	✓ 1
\(k = \frac{1}{10}\ln\frac{7}{4} = 0.0560\) (min⁻¹)	✓ 1
Part (c)
Substitutes t = 25 with their k: \(T = 20 + 70e^{-25(0.05596)}\)	✓ 1
T = 37.3 °C	✓ 1

Marking point	Mark	Notes
Part (a)
M1: separates variables correctly, \(\int \frac{1}{T-20}\,dT = \int -0.08\,dt\)	✓ 1
A1: \(\ln\|T-20\| = -0.08t + c\)	✓ 1
M1: uses T = 90 when t = 0 to get \(c = \ln 70\), so \(T - 20 = 70e^{-0.08t}\), leading to the given answer	✓ 1
Part (b)
M1: substitutes t = 10, \(T = 20 + 70e^{-0.8}\)	✓ 1
A1: T = 51.5°C	✓ 1
Part (c)
M1: sets \(20 + 70e^{-0.08t} = 40\), or uses a GDC solver or graph intersection	✓ 1
A1: t = 15.7 minutes	✓ 1