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IB Maths Applications & Interpretation · 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 16 October 2026 · new questions added continuouslyFree access until 31 December 2026How every question is checked →

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+ Add to test✓ In your test1IBMI-034-03-r1-01Differential equations and slope fields Paper 1Easy7 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 test2IBMI-034-03-r1-02Differential equations and slope fields Paper 1Easy5 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 test3IBMI-034-04-r2-01Euler's method and coupled systems Paper 1Easy5 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 test4IBMI-034-08-r2-01Differential equations and slope fields Paper 1Easy5 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 test5IBMI-034-08-r1-01Differential equations and slope fields Paper 1Easy7 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 test6IBMI-034-08-r1-02Differential equations and slope fields Paper 1Easy7 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 test7IBMI-034-03-r2-01Differential equations and slope fields Paper 1Easy7 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 test8IBMI-034-04-r1-02Euler's method and coupled systems Paper 1Easy7 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 test9IBMI-034-08-r2-02Differential equations and slope fields Paper 1Easy7 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 test10IBMI-034-04-r2-02Euler's method and coupled systems Paper 1Easy6 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 test11IBMI-034-04-r1-01Euler's method and coupled systems Paper 1Easy6 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 test12IBMI3H-5-10-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The concentration of a drug in the blood, C mg per litre, t hours after an injection is modelled by
C(t) = 12te^−0.4t, t ≥ 0.

(a)

Find $\frac{dC}{dt}$.

(2)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

Find the maximum concentration.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule with the chain rule for the exponential	✓ 1
A1 $12e^{-0.4t}-4.8te^{-0.4t}$	✓ 1
Part (b)
M1 set $\frac{dC}{dt}=0$ and factorise, 12(1 − 0.4t)e^−0.4t = 0	✓ 1
A1 t = 2.5 hours	✓ 1
Part (c)
M1 substitute t = 2.5 into C	✓ 1
A1 11.0 mg per litre	✓ 1

Common error: The exponential factor is never zero, so only the bracket can be zero. Factorise rather than expanding.

+ Add to test✓ In your test13IBMI3H-5-10-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

In a harbour the height of the tide, h metres, t hours after midnight is modelled by
h(t) = 4.5 + 2.1 sin(0.507t), 0 ≤ t ≤ 24,
where the angle is in radians.

(a)

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

(2)

(b)

Find the rate at which the tide is rising at 02:00.

(2)

(c)

Find the greatest rate at which the tide rises.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule, multiplying by 0.507	✓ 1
A1 $1.06\cos(0.507t)$ (1.0647 cos(0.507t))	✓ 1
Part (b)
M1 substitute t = 2 with the calculator in radian mode	✓ 1
A1 0.563 m per hour	✓ 1
Part (c)
A1 1.06 m per hour (the maximum of cosine is 1)	✓ 1

Common error: With trigonometric models in calculus the calculator must be in radian mode, and the inner function's coefficient multiplies the derivative.

+ Add to test✓ In your test14IBMI3H-5-10-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

A tram leaves a station and travels along a straight track. Its velocity, v m s⁻¹, at time t seconds is
v(t) = 1.2t − 0.06t², 0 ≤ t ≤ 20,
and the tram stops when t = 20.

(a)

Find the maximum speed of the tram.

(4)

(b)

Find the distance between the two stops.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a = $\frac{dv}{dt}$ = 1.2 − 0.12t	✓ 1
A1 set a = 0 to get t = 10	✓ 1
A1 substitute into v	✓ 1
A1 6 m s⁻¹	✓ 1
Part (b)
M1 $\int_0^{20}v\,dt$	✓ 1
A1 80 m	✓ 1

Exam tip: Maximum speed occurs when acceleration is zero (the velocity graph's turning point), not when displacement is greatest.

+ Add to test✓ In your test15IBMI3H-5-10b-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

After a patient takes an oral dose, the concentration of a drug in the blood is modelled by C(t) = 5te^−0.4t mg/L, where t is the time in hours after the dose, t ≥ 0.

(a)

Use the product rule to find C′(t).

(3)

(b)

Find the greatest concentration of the drug and the time at which it occurs.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at the product rule	✓ 1
A1 5e^−0.4t	✓ 1
A1 −2te^−0.4t, so C′(t) = e^−0.4t(5 − 2t)	✓ 1
Part (b)
M1 set C′(t) = 0, giving 5 − 2t = 0 and t = 2.5 hours	✓ 1
A1 C = 4.60 mg/L	✓ 1

Common error: The factor e⁻⁰.⁴t is never zero, so only the bracket (5 − 2t) can make the derivative zero.

+ Add to test✓ In your test16IBMI3H-5-10b-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The height of the water in a harbour is modelled by h(t) = 5 + 3 sin(0.51t) metres, where t is the time in hours after midnight, with the argument in radians.

(a)

Find dh/dt.

(2)

(b)

The water level is falling at 1.0 m h⁻¹ twice in the first 12 hours. Find the first time after midnight at which this happens.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the chain rule	✓ 1
A1 1.53 cos(0.51t)	✓ 1
Part (b)
M1 set 1.53 cos(0.51t) = −1	✓ 1
A1 0.51t = 2.283 (calculator in radian mode)	✓ 1
A1 t = 4.48 hours	✓ 1

Common error: Check that the calculator is in radian mode, because the argument 0.51t is an angle in radians.

+ Add to test✓ In your test17IBMI3H-5-10b-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

Chlorine is added to a water tank. The concentration of chlorine in the water is modelled by C(t) = 20t/(t² + 9) mg/L, where t is the time in minutes after dosing, t ≥ 0.

(a)

Use the quotient rule to show that C′(t) = 20(9 − t²)/(t² + 9)².

(3)

(b)

Find the maximum concentration of chlorine.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at the quotient rule	✓ 1
A1 numerator 20(t² + 9) − 20t × 2t	✓ 1
A1 simplify to 20(9 − t²) over (t² + 9)²	✓ 1
Part (b)
M1 set C′(t) = 0, so t = 3 (as t ≥ 0)	✓ 1
A1 C = 3.33 mg/L	✓ 1

Exam tip: In the quotient rule the numerator is (derivative of top × bottom) − (top × derivative of bottom); the order matters.

+ Add to test✓ In your test18IBMI3H-5-10b-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy4 marks

The amount A mg of a drug in a patient's blood satisfies dA/dt = 5 − 0.3A, where t is the time in hours since an infusion began, and A = 0 when t = 0.

(a)

Use Euler's method with a step length of 0.5 hours to estimate the amount of drug in the blood at t = 1.5.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 A_n+1 = A_n + 0.5(5 − 0.3A_n)	✓ 1
A1 at t = 0.5, A = 2.5	✓ 1
A1 at t = 1, A = 4.625	✓ 1
A1 at t = 1.5, A = 6.43 mg	✓ 1

Exam tip: Show each step in a short table (t, A, gradient); it makes slips easy to spot and earns the method mark.

+ Add to test✓ In your test19IBMI3H-5-03r8-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The air pressure P kilopascals (kPa) at a height h metres above sea level on a mountain is modelled by P(h) = 101.3e^−0.000125h, for 0 ≤ h ≤ 8000.

(a)

Find P′(h).

(2)

(b)

Find the rate of change of pressure at a height of 4500 m. Give your answer to three significant figures and include units.

(1)

(c)

A climber's camp is at the height where the pressure is 60 kPa. Find this height, to the nearest metre.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate an exponential using the chain rule	✓ 1
A1 P′(h) = −0.0126625e^−0.000125h	✓ 1
Part (b)
A1 −0.00721 kPa per metre (−0.00722 kPa m⁻¹)	✓ 1
Part (c)
M1 equation 101.3e^−0.000125h = 60	✓ 1
M1 take natural logs, e.g. −0.000125h = ln(60/101.3)	✓ 1
A1 h = 4190 m	✓ 1

Common error: In the chain rule the factor from the exponent (−0.000125) multiplies the original exponential; do not change the power of e.

+ Add to test✓ In your test20IBMI3H-5-03r8-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

Sugar dissolves in a mixing vessel at a rate of r(t) = 6te^−0.5t² grams per minute, where t is the time in minutes after mixing starts.

(a)

Use the substitution u = 0.5t² to show that ∫ 6te^−0.5t² dt = −6e^−0.5t² + c.

(3)

(b)

Find the mass of sugar dissolved in the first 3 minutes. Give your answer in grams to three significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du = t dt (or du/dt = t)	✓ 1
M1 integral becomes ∫ 6e^−u du	✓ 1
A1 = −6e^−u + c = −6e^−0.5t² + c	✓ 1
Part (b)
M1 [−6e^−0.5t²] evaluated between 0 and 3, i.e. 6 − 6e^−4.5	✓ 1
A1 5.93 g	✓ 1

Exam tip: When using a substitution on a definite integral, either change the limits to u-values or return to t before substituting the limits, but never mix the two.

+ Add to test✓ In your test21IBMI3H-5-03r8-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

A bottle of water is taken out of a tent where its temperature is 20 °C into air at −5 °C. Its temperature T °C after t minutes is modelled by dT/dt = −0.08(T + 5), with T = 20 when t = 0.

(a)

Use Euler's method with a step length of 2 minutes to estimate T when t = 6.

(3)

(b)

The exact solution is T = −5 + 25e^−0.08t. Find the exact value of T when t = 6, to four significant figures.

(1)

(c)

Find the amount by which the Euler estimate differs from the exact value, to three significant figures. Use the unrounded values.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 T_n+1 = T_n + 2(−0.08)(T_n + 5)	✓ 1
A1 T(2) = 16 and T(4) = 12.64	✓ 1
A1 T(6) = 9.8176 °C	✓ 1
Part (b)
A1 10.47 °C	✓ 1
Part (c)
A1 0.652 °C (Euler estimate is lower)	✓ 1

Exam tip: Write each Euler step as a short table of t, T and the gradient so that a slip in one line is easy to spot and can be carried forward.

+ Add to test✓ In your test22IBMI3H-5-03r8-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy3 marks

The capacity of a data link, in Mbit per second, is modelled by C(p) = 4ln(1 + 0.25p²), where p is the transmitted power in watts.

(a)

Find dC/dp.

(2)

(b)

Find the rate at which capacity increases with power when p = 5. Give your answer to three significant figures, with units.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule: 4 × (1/(1 + 0.25p²)) × 0.5p	✓ 1
A1 dC/dp = 2p/(1 + 0.25p²)	✓ 1
Part (b)
A1 1.38 Mbit per second per watt	✓ 1

Common error: The derivative of ln(f) is f′/f; forgetting the inner derivative 0.5p is a frequent slip.

+ Add to test✓ In your test23IBMI3H-5-04r8-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy4 marks

The depth of water, D metres, in a harbour channel t hours after midnight is modelled by D(t) = 6 + 2.5sin(0.5t), where 0.5t is in radians.

(a)

Find D′(t).

(2)

(b)

Find the rate at which the depth is changing at 02:00. Give your answer to three significant figures, in metres per hour.

(1)

(c)

State the greatest rate at which the depth can increase.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule applied to sin(0.5t)	✓ 1
A1 D′(t) = 1.25cos(0.5t)	✓ 1
Part (b)
A1 0.675 m per hour	✓ 1
Part (c)
R1 greatest value of cos is 1, so greatest rate = 1.25 m per hour	✓ 1

Common error: Differentiating sin(0.5t) gives 0.5cos(0.5t); the factor 0.5 must not be dropped or applied to the 6.

+ Add to test✓ In your test24IBMI3H-5-04r8-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy4 marks

The number of fish P in a marine reserve t years after monitoring begins satisfies dP/dt = 0.3P(1 − P/800), with P = 200 when t = 0.

(a)

Use Euler's method with a step length of 1 year to estimate the number of fish when t = 2. Give your answer to the nearest whole number.

(3)

(b)

State one limit on the population predicted by the differential equation.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 P_n+1 = P_n + 1 × 0.3P_n(1 − P_n/800)	✓ 1
A1 P = 245 at t = 1	✓ 1
A1 296 fish at t = 2	✓ 1
Part (b)
R1 the population levels off near 800 (where dP/dt = 0)	✓ 1

Exam tip: Keep the unrounded value from one step into the next and round only the final answer.

+ Add to test✓ In your test25IBMI3H-5-04r8-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The profile of a curved roof is modelled by y = 3 + 2cos(0.4x), where x and y are in metres and 0.4x is in radians, for 0 ≤ x ≤ 6.

(a)

Find dy/dx.

(2)

(b)

Find the gradient of the roof where x = 2, to three significant figures.

(1)

(c)

The tangent to the roof at x = 2 meets the y-axis at the point (0, c). Find c to three significant figures, using unrounded values.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule on cos(0.4x)	✓ 1
A1 dy/dx = −0.8sin(0.4x)	✓ 1
Part (b)
A1 −0.574	✓ 1
Part (c)
M1 height at x = 2 is 3 + 2cos 0.8 = 4.393	✓ 1
M1 c = y − (gradient)(2)	✓ 1
A1 c = 5.54	✓ 1

Common error: Set the calculator to radians; in degree mode sin(0.8) and cos(0.8) give completely different values.

+ Add to test✓ In your test26IBMI3H-5-04r8-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The concentration y of krill at time x days in a bay is modelled by dy/dx = x − 2y. A slope field is drawn for this equation.

(a)

Calculate the gradient of the slope field at (1, 0) and at (3, 1).

(2)

(b)

Find the equation of the line along which the slope field segments are horizontal.

(2)

(c)

A solution curve passes through (3, 1). Use its tangent at this point to estimate y when x = 3.4.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 at (1, 0): 1	✓ 1
A1 at (3, 1): 1	✓ 1
Part (b)
M1 x − 2y = 0	✓ 1
A1 y = x/2	✓ 1
Part (c)
M1 y ≈ 1 + 1 × 0.4	✓ 1
A1 y ≈ 1.4	✓ 1

Exam tip: A tangent-line estimate works best close to the known point; further away the curve can drift from the tangent.

+ Add to test✓ In your test27IBMI3H-5-04r11-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy4 marks

In an online game, a player's experience points after $t$ hours of play are modelled by $E(t) = 40\ln(t^2 + 4)$.

(a)

Find $\frac{dE}{dt}$.

(2)

(b)

Find the rate at which experience points are being gained at $t = 3$, in points per hour, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the chain rule with $\frac{d}{dt}(t^2+4) = 2t$	✓ 1
A1 $\frac{dE}{dt} = \frac{80t}{t^2 + 4}$	✓ 1
Part (b)
M1 substitute $t = 3$: $\frac{80(3)}{9 + 4}$	✓ 1
A1 18.5 points per hour	✓ 1

Common error: Differentiating $\ln(t^2+4)$ as $\frac{1}{t^2+4}$ and forgetting the inner derivative $2t$.

+ Add to test✓ In your test28IBMI3H-5-04r11-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy4 marks

The depth of water in a reservoir is modelled by $h = 12 + 3\sin(0.5t)$, where $h$ is in metres, $t$ is the time in days and $t \ge 0$. Angles are in radians.

(a)

Find $\frac{dh}{dt}$ at $t = 2$, giving your answer to 3 significant figures.

(2)

(b)

Find the first time, $t > 0$, at which the depth is not changing, and state the depth then.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{dh}{dt} = 3(0.5)\cos(0.5t)$	✓ 1
A1 $1.5\cos 1 = 0.81$ m per day	✓ 1
Part (b)
M1 set $1.5\cos(0.5t) = 0$ so $0.5t = \frac{\pi}{2}$	✓ 1
A1 $t = \pi = 3.14$ days and depth $= 15$ m	✓ 1

Exam tip: Make sure your calculator is in radian mode whenever the angle is given without degrees.

+ Add to test✓ In your test29IBMI3H-5-04r11-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

A stunt ramp in a game has a profile modelled by $y = x^3 - 4x + 1$ for $0 \le x \le 3$, where $x$ and $y$ are in metres.

(a)

Find $\frac{dy}{dx}$.

(1)

(b)

Find the equation of the tangent to the ramp at the point where $x = 2$.

(3)

(c)

Find the exact $x$-coordinate where this tangent crosses the $x$-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dy}{dx} = 3x^2 - 4$	✓ 1
Part (b)
A1 gradient $= 3(2)^2 - 4 = 8$ and $y = 8 - 8 + 1 = 1$	✓ 1
M1 $y - 1 = 8(x - 2)$	✓ 1
A1 $y = 8x - 15$	✓ 1
Part (c)
M1 set $8x - 15 = 0$	✓ 1
A1 $x = \frac{15}{8}$ ($= 1.875$)	✓ 1

Exam tip: To find a tangent you need two things: the gradient from the derivative and the point from the original function.

+ Add to test✓ In your test30IBMI3H-5-04r11-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy4 marks

The flow rate of water through a pipe was recorded every 2 minutes.
Time (min): 0, 2, 4, 6, 8, 10
Flow rate (L/min): 0, 14, 22, 26, 24, 18

(a)

Use the trapezoidal rule with all six readings to estimate the total volume of water that flowed in the 10 minutes.

(3)

(b)

Suggest one way to obtain a more accurate estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 trapezoidal rule with strip width 2	✓ 1
M1 $\frac{2}{2}\left[0 + 18 + 2(14 + 22 + 26 + 24)\right]$	✓ 1
A1 $190$ litres	✓ 1
Part (b)
R1 record the flow rate more often (smaller time intervals) and use the rule again	✓ 1

Exam tip: With $n$ readings there are $n-1$ strips. Here six readings give five strips of width 2, so the outer values are counted once and the inner values twice.

+ Add to test✓ In your test31IBMI3H-5-03r14-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The atmospheric pressure $P$ kPa at height $h$ metres above sea level on a mountain is modelled by $P = 101.3e^{-0.000125h}$, for $0 \le h \le 6000$.

(a)

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

(2)

(b)

Calculate the rate of change of pressure at a camp at height 3000 m. Give your answer in kPa m⁻¹ to 3 significant figures.

(2)

(c)

Use the value from part (b), unrounded, to estimate the fall in pressure when a climber ascends from 3000 m to 3100 m. Give your answer in kPa to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses the chain rule, or states that the derivative of $e^{kh}$ is $ke^{kh}$	✓ 1
A1 $\frac{dP}{dh}=-0.0126625e^{-0.000125h}$ (accept $-101.3\times0.000125e^{-0.000125h}$)	✓ 1
Part (b)
M1 substitutes $h=3000$ to get $-0.0126625e^{-0.375}$	✓ 1
A1 −0.00870 kPa m⁻¹	✓ 1
Part (c)
M1 $100\times0.0087028\ldots$ (accept $100\times0.00870$)	✓ 1
A1 fall of 0.870 kPa	✓ 1

Exam tip: Check the sign: a negative derivative means pressure falls as height increases, so state the fall as a positive quantity when asked for a decrease.

+ Add to test✓ In your test32IBMI3H-5-03r14-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy8 marks

The voltage $V$ volts across a component in a mountain radio link, $t$ seconds after a transient pulse begins, is modelled by $V = 12te^{-0.4t}$, for $t \ge 0$.

(a)

Use the product rule to show that $\frac{dV}{dt}=12e^{-0.4t}(1-0.4t)$.

(3)

(b)

Find the time at which the voltage is a maximum, and justify that it is a maximum.

(3)

(c)

Find the maximum voltage, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts the product rule with $u=12t$ and $v=e^{-0.4t}$	✓ 1
A1 $u'=12$ and $v'=-0.4e^{-0.4t}$	✓ 1
A1 $\frac{dV}{dt}=12e^{-0.4t}-4.8te^{-0.4t}=12e^{-0.4t}(1-0.4t)$	✓ 1
Part (b)
M1 sets $\frac{dV}{dt}=0$	✓ 1
A1 $t=2.5$ s	✓ 1
R1 since $e^{-0.4t}>0$, the derivative is positive for $t<2.5$ and negative for $t>2.5$, so a maximum (accept second derivative negative)	✓ 1
Part (c)
M1 $V(2.5)=12\times2.5\times e^{-1}$	✓ 1
A1 11.0 V	✓ 1

Common error: Do not forget the chain rule when differentiating $e^{-0.4t}$: the factor −0.4 must appear in $v'$.

+ Add to test✓ In your test33IBMI3H-5-03r14-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

Snow depth at a mountain hut is rising during a storm. The rate of snowfall, in cm per hour, was measured every two hours:

Time t (hours)	0	2	4	6	8
Rate (cm h⁻¹)	1.2	2.0	2.6	2.2	1.4

(a)

Use the trapezoidal rule with four intervals to estimate the total snowfall between $t=0$ and $t=8$.

(3)

(b)

The graph of rate against time is smooth and concave down for $0\le t\le 8$. State, with a reason, whether your estimate is an overestimate or an underestimate.

(2)

(c)

At $t=0$ the snow depth was 40 cm. Use your answer to part (a) to estimate the depth at $t=8$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 interval width $h=2$	✓ 1
M1 $\frac{2}{2}\left[1.2+1.4+2(2.0+2.6+2.2)\right]$	✓ 1
A1 16.2 cm	✓ 1
Part (b)
R1 on a concave-down curve each chord lies below the curve	✓ 1
A1 so the trapezoidal rule gives an underestimate	✓ 1
Part (c)
M1 $40+16.2$ (ECF from (a))	✓ 1
A1 56.2 cm	✓ 1

Common error: The trapezoidal rule is applied to the rate values, so the end values are used once and the interior values are doubled.

+ Add to test✓ In your test34IBMI3H-5-05r17-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

Marine biologists model the density of a plankton bloom beneath a research buoy. The density, P thousand cells per m³, is given by $P(t)=40te^{-0.2t}$, where t is the time in hours after sunrise, $t\ge 0$.

(a)

Find $\frac{dP}{dt}$.

(2)

(b)

Find the time at which the density is a maximum.

(2)

(c)

Find the maximum density. Give your answer to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule on $40t$ and $e^{-0.2t}$	✓ 1
A1 $\frac{dP}{dt}=40e^{-0.2t}(1-0.2t)$ (or equivalent, e.g. $40e^{-0.2t}-8te^{-0.2t}$)	✓ 1
Part (b)
M1 set $\frac{dP}{dt}=0$ (ECF from (a)), noting $e^{-0.2t}\ne 0$	✓ 1
A1 $t=5$ hours	✓ 1
Part (c)
A1 $P=40(5)e^{-1}=73.6$ thousand cells per m³ (ECF from (b))	✓ 1

Exam tip: State that the exponential factor is never zero when solving $\frac{dP}{dt}=0$; it shows why only the bracket can be zero.

+ Add to test✓ In your test35IBMI3H-5-05r17-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

A decorative curved wall has a height profile modelled by $y=5\ln\left(1+0.2x^2\right)$ for $0\le x\le 8$, where x is the horizontal distance and y the height, both in metres.

(a)

Find $\frac{dy}{dx}$.

(2)

(b)

Find the gradient of the wall profile when $x=4$. Give your answer to 3 significant figures.

(1)

(c)

Find the equation of the tangent to the profile at $x=4$. Write it as $y=mx+c$ with m and c to 3 significant figures, using unrounded values in your working.

(2)

(d)

Find the value of x in the given domain at which the gradient equals 1. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the chain rule with $\ln(\cdot)$	✓ 1
A1 $\frac{dy}{dx}=\dfrac{2x}{1+0.2x^2}$	✓ 1
Part (b)
A1 1.90 (from $\frac{8}{4.2}$, ECF from (a))	✓ 1
Part (c)
M1 $y-5\ln 4.2=\frac{8}{4.2}(x-4)$ (point $(4,\,7.175...)$ with the gradient from (b))	✓ 1
A1 $y=1.90x-0.444$	✓ 1
Part (d)
M1 $\frac{2x}{1+0.2x^2}=1$, giving $0.2x^2-2x+1=0$ (ECF from (a))	✓ 1
A1 $x=0.528$ (the other root, 9.47, is outside the domain)	✓ 1

Common error: Forgetting to check the domain: the quadratic has two roots, but only one lies in $0\le x\le 8$.

+ Add to test✓ In your test36IBMI3H-5-05r17-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

A shipping company designs an open-topped wooden crate with a square base of side x m and height h m. The crate must have a volume of 2 m³. The wood costs €18 per m².

(a)

Show that the total outer surface area of the crate, S m², is given by $S=x^2+\dfrac{8}{x}$.

(2)

(b)

Find the value of x that minimises the surface area. Give your answer to 3 significant figures.

(3)

(c)

Find the minimum cost of the wood for one crate, using the unrounded value of x. Give your answer to the nearest cent.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $h=\frac{2}{x^2}$ and $S=x^2+4xh$ (base plus four sides, no top)	✓ 1
A1 $S=x^2+\frac{8}{x}$ (shown)	✓ 1
Part (b)
M1 $\frac{dS}{dx}=2x-\frac{8}{x^2}=0$	✓ 1
A1 $x^3=4$	✓ 1
A1 $x=1.59$ m	✓ 1
Part (c)
A1 €136.07	✓ 1

Exam tip: After finding a stationary point in an optimisation problem, substitute back into the original function (not the derivative) to find the minimum value.

+ Add to test✓ In your test37IBMI3H-5-05r17-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The depth of water in a harbour channel is measured at points across its width. The channel is 30 m wide and the depths, d m, are recorded every 5 m from one bank.
Distance from bank (m): 0, 5, 10, 15, 20, 25, 30
Depth d (m): 0.0, 1.8, 3.1, 3.6, 3.2, 2.0, 0.0

(a)

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

(3)

(b)

Water flows through the channel at an average speed of 0.6 m s⁻¹. Calculate the volume of water passing a point each second, using your answer to (a).

(1)

(c)

Suggest one way to obtain a more accurate estimate of the cross-section area.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{5}{2}\left[\text{first}+\text{last}+2(\text{sum of others})\right]$	✓ 1
A1 $\frac{5}{2}\left[0+0+2(1.8+3.1+3.6+3.2+2.0)\right]$	✓ 1
A1 68.5 m²	✓ 1
Part (b)
A1 $0.6\times 68.5=41.1$ m³ per second (ECF from (a))	✓ 1
Part (c)
R1 take depth measurements at smaller intervals (more strips), since this makes each trapezium follow the bed more closely	✓ 1

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

+ Add to test✓ In your test38IBMI3H-5-04r20-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The height h metres of the valve on a bicycle wheel above the ground is modelled by $h(t)=0.34+0.34\sin(12t)$, where t is the time in seconds after the valve is level with the wheel's centre and rising. Angles are in radians.

(a)

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

(2)

(b)

Find the rate of change of height when t = 0.15, giving your answer to 3 significant figures.

(1)

(c)

Find the first time, t > 0, at which the valve is at its greatest height. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule applied to $\sin(12t)$	✓ 1
A1 $\frac{dh}{dt}=4.08\cos(12t)$	✓ 1
Part (b)
A1 $4.08\cos(1.8)=-0.927$ m s⁻¹ (the valve is moving downwards)	✓ 1
Part (c)
M1 set $\frac{dh}{dt}=0$, so $\cos(12t)=0$ and $12t=\frac{\pi}{2}$	✓ 1
A1 t = 0.131 s	✓ 1

Common error: Calculators must be in radian mode for derivatives of trigonometric models like this one; the factor from the chain rule is the coefficient of t inside the sine.

+ Add to test✓ In your test39IBMI3H-5-04r20-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

A train begins braking at t = 0 seconds. Its speed is modelled by $v(t)=30e^{-0.04t}$ m s⁻¹, for $t\ge0$.

(a)

Find an expression for the acceleration of the train at time t.

(2)

(b)

Find the acceleration when t = 10, to 3 significant figures.

(1)

(c)

Find the distance travelled by the train in the first 60 seconds, to the nearest metre.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiate using the chain rule	✓ 1
A1 $a(t)=-1.2e^{-0.04t}$	✓ 1
Part (b)
A1 $a(10)=-1.2e^{-0.4}=-0.804$ m s⁻²	✓ 1
Part (c)
M1 $\int_0^{60}30e^{-0.04t}\,dt$	✓ 1
A1 $\left[-750e^{-0.04t}\right]_0^{60}=750(1-e^{-2.4})$	✓ 1
A1 682.0 so 682 m	✓ 1

Exam tip: Integrating $e^{kt}$ gives $\frac{1}{k}e^{kt}$; when k is negative, keep track of the sign before substituting the limits.

+ Add to test✓ In your test40IBMI3H-5-04r20-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

During a meteor shower, the rate at which meteors are seen from an observatory is $R(t)=4t\cos(t^2)$ hundreds of meteors per hour, for $0\le t\le1.2$, where t is the time in hours after the shower begins.

(a)

Use the substitution $u=t^2$ to show that $\int4t\cos(t^2)\,dt=2\sin(t^2)+c$.

(3)

(b)

Find the total number of meteors expected in the first 1.2 hours, to the nearest meteor.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{du}{dt}=2t$, so $4t\,dt=2\,du$	✓ 1
M1 integral becomes $\int2\cos u\,du$	✓ 1
A1 $2\sin u+c=2\sin(t^2)+c$ (AG)	✓ 1
Part (b)
M1 $\left[2\sin(t^2)\right]_0^{1.2}=2\sin(1.44)$	✓ 1
A1 1.9829 hundred, so 198 meteors	✓ 1

Common error: Angles in $\cos(t^2)$ are in radians; also remember that the answer here is in hundreds of meteors, so convert before stating the number.

+ Add to test✓ In your test41IBMI3H-5-06r21-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The height h metres of a pedal above the ground during a steady ride is modelled by h(t) = 0.45 + 0.2 sin(5t), where t is the time in seconds and the angle is in radians.

(a)

Use the chain rule to find h′(t).

(2)

(b)

Find the rate of change of the height at t = 0.2, to 3 significant figures, and state whether the pedal is rising or falling.

(1)

(c)

Find the first time after t = 0 at which the pedal is momentarily at rest vertically. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates sin(5t) to give 5cos(5t) (chain rule)	✓ 1
A1 h′(t) = cos(5t)	✓ 1
Part (b)
A1 0.540 m s⁻¹, so the pedal is rising (positive rate)	✓ 1
Part (c)
M1 solves cos(5t) = 0	✓ 1
A1 5t = π/2, so t = π/10 = 0.314 s	✓ 1

Common error: Forgetting the factor 5 from the chain rule gives a derivative of 0.2 cos(5t) with the wrong amplitude; check by differentiating the inside function.

+ Add to test✓ In your test42IBMI3H-5-06r21-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

An archaeologist makes an open finds tray from a rectangular sheet of card 30 cm by 20 cm. A square of side x cm is cut from each corner and the sides are folded up.

(a)

Show that the volume of the tray is V = 4x³ − 100x² + 600x cm³.

(2)

(b)

Find the value of x that gives the greatest volume, to 3 significant figures, explaining why the other stationary point is rejected.

(3)

(c)

Find the greatest volume, to the nearest cm³.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 length = 30 − 2x, width = 20 − 2x, height = x	✓ 1
A1 expands x(30 − 2x)(20 − 2x) to 4x³ − 100x² + 600x (AG)	✓ 1
Part (b)
M1 V′ = 12x² − 200x + 600 = 0	✓ 1
A1 x = 3.92 or x = 12.7	✓ 1
R1 x = 12.7 is rejected because the width 20 − 2x would be negative (need 0 < x < 10)	✓ 1
Part (c)
M1 substitutes the unrounded value of x into V	✓ 1
A1 1056 cm³	✓ 1

Common error: Both roots of V′ = 0 must be tested against the physical limits on x before one is chosen.

+ Add to test✓ In your test43IBMI3H-5-06r21-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

A trench is dug with a uniform cross-section. The depth y metres of the trench is measured at horizontal distances x metres from one edge.
x: 0, 1, 2, 3, 4, 5
y: 0, 0.8, 1.3, 1.5, 1.1, 0

(a)

Use the trapezoidal rule with all six values to estimate the area of the cross-section.

(2)

(b)

The trench is 12 m long. Find the volume of soil removed.

(1)

(c)

Soil removal costs €18.50 per m³. Find the cost of removing the soil, to the nearest cent.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 trapezoidal rule with width 1: ½ × 1 × (0 + 0 + 2(0.8 + 1.3 + 1.5 + 1.1))	✓ 1
A1 4.7 m²	✓ 1
Part (b)
A1 4.7 × 12 = 56.4 m³ (ECF from (a))	✓ 1
Part (c)
M1 56.4 × 18.50	✓ 1
A1 €1043.40 (ECF)	✓ 1

Exam tip: State the number of strips and their width before substituting into the trapezoidal rule; it makes the working easy to follow.

+ Add to test✓ In your test44IBMI3H-5-06r21-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

A slope field is drawn for the differential equation dy/dx = x − y.

(a)

Find the gradient of the solution curve at the point (3, 1).

(1)

(b)

Show that y = x − 1 + 2e^−x is the solution of the differential equation that passes through (0, 1).

(3)

(c)

Find the coordinates of the stationary point of this solution curve, to 3 significant figures, and determine its nature.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 2	✓ 1
Part (b)
A1 y(0) = −1 + 2 = 1	✓ 1
M1 dy/dx = 1 − 2e^−x	✓ 1
A1 x − y = 1 − 2e^−x, which equals dy/dx (AG)	✓ 1
Part (c)
M1 sets 1 − 2e^−x = 0	✓ 1
A1 x = ln 2 = 0.693, y = 0.693	✓ 1
R1 d²y/dx² = 2e^−x > 0, so it is a minimum	✓ 1

Exam tip: On the line where the gradient is zero (here y = x) every slope segment is horizontal; solution curves have stationary points where they cross it.

+ Add to test✓ In your test45IBMI3H-5-06r22-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

During roasting, the bean temperature is modelled by T(t) = 210 − 190e^−0.08t °C, where t is the time in minutes after the beans enter the roaster.

(a)

Find the initial bean temperature.

(1)

(b)

Find $\frac{dT}{dt}$.

(2)

(c)

Find the rate of increase of temperature when t = 5, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 20 °C	✓ 1
Part (b)
M1 attempt to differentiate the exponential term using the chain rule	✓ 1
A1 $15.2e^{-0.08t}$	✓ 1
Part (c)
M1 $15.2e^{-0.4}$	✓ 1
A1 10.2 °C min⁻¹	✓ 1

Common error: The derivative of $e^{kt}$ is $ke^{kt}$: remember the factor k, and the sign from −190 × (−0.08).

+ Add to test✓ In your test46IBMI3H-5-06r22-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The depth of a river was measured at equal intervals across a bridge of width 20 m.
Distance from bank (m): 0, 5, 10, 15, 20
Depth (m): 0, 2.1, 3.4, 2.6, 0

(a)

Use the trapezoidal rule with four intervals to estimate the cross-sectional area of the river.

(3)

(b)

The water flows at an average speed of 1.2 m s⁻¹. Find the volume of water passing under the bridge each second.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{5}{2}\{\ldots\}$ with first and last values and twice the middle values	✓ 1
A1 $\frac{5}{2}\left[0+0+2(2.1+3.4+2.6)\right]$	✓ 1
A1 40.5 m²	✓ 1
Part (b)
M1 area × speed, $40.5\times1.2$	✓ 1
A1 48.6 m³ per second (ECF from (a))	✓ 1

Common error: The trapezoidal rule doubles the interior values only: the first and last depths are counted once.

+ Add to test✓ In your test47IBMI3H-5-06r22-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

An inspection drone climbs a bridge pylon. Its velocity is v(t) = 3t² − 18t + 24 m s⁻¹ (positive is upward) for 0 ≤ t ≤ 4, where t is in seconds. At t = 0 the drone is at its start position.

(a)

Find the acceleration at t = 1.

(2)

(b)

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

(2)

(c)

Find the total distance travelled by the drone during the first 4 seconds.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $a=6t-18$	✓ 1
A1 −12 m s⁻²	✓ 1
Part (b)
M1 $3t^2-18t+24=0$	✓ 1
A1 t = 2 and t = 4	✓ 1
Part (c)
M1 displacement $s=t^3-9t^2+24t$	✓ 1
A1 s(2) = 20 and s(4) = 16	✓ 1
A1 distance = 20 + 4 = 24 m	✓ 1

Common error: Total distance is not the same as displacement: split the interval where the velocity changes sign.

+ Add to test✓ In your test48IBMI3H-5-06r22-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

The main cable of a suspension bridge hangs in the shape y = 0.004x² − 0.16x + 12, for 0 ≤ x ≤ 80, where x is the horizontal distance in metres and y the height above the water in metres.

(a)

Find $\frac{dy}{dx}$ and the gradient of the cable at x = 30.

(2)

(b)

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

(2)

(c)

Find the lowest height of the cable above the water.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dy}{dx}=0.008x-0.16$	✓ 1
A1 gradient 0.08	✓ 1
Part (b)
A1 y = 10.8 at x = 30	✓ 1
A1 y = 0.08x + 8.4	✓ 1
Part (c)
M1 $\frac{dy}{dx}=0$	✓ 1
A1 x = 20	✓ 1
A1 10.4 m	✓ 1

Exam tip: The tangent line uses the gradient at the point: substitute x into the derivative, not into the original function.

+ Add to test✓ In your test49IBMI3H-5-06r23-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy8 marks

Calculator

A seismic activity index at a volcano, t days after monitoring begins, is modelled by S(t) = 5 ln(t² + 4), t ≥ 0.

(a)

Find S′(t).

(2)

(b)

Find the rate of increase of the index when t = 3, correct to 3 significant figures.

(2)

(c)

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

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of the chain rule, with derivative of ln u equal to 1/u	✓ 1
A1 S′(t) = 10t/(t² + 4)	✓ 1
Part (b)
M1 substitute t = 3 into S′	✓ 1
A1 30/13 = 2.31 index units per day	✓ 1
Part (c)
M1 attempt to differentiate S′ using the quotient rule	✓ 1
A1 S″(t) = 10(4 − t²)/(t² + 4)²	✓ 1
A1 S″ = 0 gives t = 2 days	✓ 1
A1 greatest rate = 10×2/(2² + 4) = 2.5 index units per day	✓ 1

Common error: The rate of change of S is greatest where S″ = 0, not where S′ = 0; setting S′ = 0 finds a turning point of S.

+ Add to test✓ In your test50IBMI3H-5-06r23-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

Calculator

The power used by a stage lighting rig during a show is P(t) = 3t√(1 + t²) kW, where t is the time in hours since the rig was switched on.

(a)

Use the substitution u = 1 + t² to find ∫ 3t√(1 + t²) dt.

(3)

(b)

Hence find the energy used in the first 2 hours, correct to 3 significant figures.

(2)

(c)

Using the unrounded energy from (b), find the mean power over the first 2 hours, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du = 2t dt, so t dt = ½ du	✓ 1
A1 (3/2) ∫ u^1/2 du	✓ 1
A1 (1 + t²)^3/2 + C	✓ 1
Part (b)
M1 (1 + 2²)^3/2 − (1 + 0)^3/2	✓ 1
A1 5√5 − 1 = 10.2 kWh	✓ 1
Part (c)
M1 divide the energy by 2 hours	✓ 1
A1 5.09 kW	✓ 1

Exam tip: After a substitution, either change the limits to the new variable or return to the original variable before substituting the limits; do not mix the two.

+ Add to test✓ In your test51IBMI3H-5-06r23-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy8 marks

Calculator

The speed v m s⁻¹ of a cargo ship, t seconds after the engines are set to full power, satisfies dv/dt = (100 − v²)/5000. Initially v = 2.

(a)

Use Euler's method with a step length of 50 seconds to estimate the speed of the ship when t = 100. Give your answer to 3 significant figures.

(4)

(b)

Find the speed that the ship approaches over a long time.

(2)

(c)

The second derivative of v with respect to t is negative for 0 < v < 10. State, with a reason, whether the estimate in (a) is an overestimate or an underestimate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 v_n+1 = v_n + 50 × (100 − v_n²)/5000	✓ 1
A1 v₁ = 2.96 at t = 50	✓ 1
M1 second step using v₁	✓ 1
A1 v = 3.87 m s⁻¹	✓ 1
Part (b)
M1 set dv/dt = 0	✓ 1
A1 v = 10 m s⁻¹ (taking the positive root)	✓ 1
Part (c)
R1 the solution curve is concave down, so each tangent line step lies above the curve	✓ 1
A1 overestimate	✓ 1

Exam tip: In Euler's method keep unrounded values from each step until the end, and state the step size and starting values clearly before the first calculation.

+ Add to test✓ In your test52IBMI3H-5-05r24-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The wind speed v m s⁻¹ at a coastal wind farm, t hours after midnight, is modelled by v(t) = 9 + 4cos(πt/12), for 0 ≤ t ≤ 24, where the angle is in radians.

(a)

Find dv/dt.

(2)

(b)

Calculate the rate of change of wind speed at 05:00. Give your answer to 3 significant figures.

(2)

(c)

Find the minimum wind speed predicted by the model and the time at which it occurs.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate cos(πt/12) using the chain rule	✓ 1
A1 dv/dt = −(π/3)sin(πt/12)	✓ 1
Part (b)
M1 substitute t = 5 into dv/dt	✓ 1
A1 −1.01 m s⁻¹ per hour	✓ 1
Part (c)
A1 t = 12 (that is, 12:00)	✓ 1
A1 minimum wind speed 5 m s⁻¹	✓ 1

Common error: Set the calculator to radians for any model written with π; a degrees setting gives a completely different derivative value.

+ Add to test✓ In your test53IBMI3H-5-05r24-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

A library records the rate at which books are borrowed after it opens. For 0.5 ≤ t ≤ 8, where t is the number of hours after opening, the rate is modelled by B(t) = 50 ln(2t) / t books per hour.

(a)

Use the quotient rule to show that B′(t) = 50(1 − ln(2t)) / t².

(3)

(b)

Find the time at which the borrowing rate is greatest. Give your answer to 3 significant figures.

(2)

(c)

Find the greatest borrowing rate. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at the quotient rule with u = 50 ln(2t), v = t	✓ 1
A1 u′ = 50/t	✓ 1
A1 (50 − 50 ln(2t)) / t² = given result	✓ 1
Part (b)
M1 set B′(t) = 0, so ln(2t) = 1	✓ 1
A1 t = e/2 = 1.36 hours	✓ 1
Part (c)
M1 substitute t = e/2 (unrounded) into B(t)	✓ 1
A1 36.8 books per hour	✓ 1

Exam tip: When a derivative is a fraction, a product is zero only when its numerator is zero, so set just the numerator equal to zero.

+ Add to test✓ In your test54IBMI3H-5-05r24-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

A training pool is 25 m long and 10 m wide. The depth of the pool is measured at 5 m intervals along its length, as shown.
Distance from shallow end (m): 0, 5, 10, 15, 20, 25
Depth (m): 1.0, 1.1, 1.4, 1.9, 2.2, 2.0

(a)

Use the trapezoidal rule with all of the data to estimate the area of the vertical cross-section along the length of the pool.

(3)

(b)

Hence estimate the volume of water in the pool when it is full.

(1)

(c)

Water is pumped into the empty pool at a constant rate of 0.03 m³ per second. Find the time needed to fill the pool, in hours.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 correct use of the trapezoidal rule with h = 5	✓ 1
A1 (5/2)[1.0 + 2.0 + 2(1.1 + 1.4 + 1.9 + 2.2)]	✓ 1
A1 40.5 m²	✓ 1
Part (b)
A1 405 m³ (ECF from (a): 10 × area)	✓ 1
Part (c)
M1 405 ÷ 0.03 = 13 500 s (ECF from (b))	✓ 1
A1 3.75 hours	✓ 1

Common error: Give the time in the unit asked for: dividing seconds by 3600 is a separate step that is easily forgotten.

+ Add to test✓ In your test55IBMI3H-5-05r24-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

A library manager models the number of loans per week, y (in hundreds), x weeks after a promotion by the differential equation dy/dx = (y − x)/4. A slope field is to be drawn for this equation.

(a)

Calculate the gradient of the slope field at the points (2, 3) and (4, 1).

(2)

(b)

Find the equation of the isocline along which every gradient is 0.5.

(2)

(c)

Show that y = x + 4 is a solution of the differential equation.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 at (2, 3) the gradient is 0.25	✓ 1
A1 at (4, 1) the gradient is −0.75	✓ 1
Part (b)
M1 (y − x)/4 = 0.5	✓ 1
A1 y = x + 2	✓ 1
Part (c)
M1 dy/dx = 1 and (x + 4 − x)/4 = 1	✓ 1
R1 both sides are equal, so it is a solution	✓ 1

Exam tip: An isocline is a curve of constant gradient, so set the right-hand side of the differential equation equal to a constant and rearrange.

+ Add to test✓ In your test56IBMI3H-5-05r25-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

A shipping company designs a solid float for a mooring buoy. The float is modelled by rotating through 360° about the x-axis the region bounded by $y=2-0.1x^2$, the x-axis, the y-axis and the line x = 3. All lengths are in metres.

(a)

Write down an integral that gives the volume of the float.

(2)

(b)

Find the volume of the float, to 3 significant figures.

(2)

(c)

The float is filled with foam that costs €18.50 per m³. Use the unrounded volume to find the cost of the foam, to the nearest euro.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $V=\pi\int y^2\,dx$ with $y=2-0.1x^2$	✓ 1
A1 $\pi\int_0^3(2-0.1x^2)^2\,dx$	✓ 1
Part (b)
M1 evaluates the integral (GDC or expansion)	✓ 1
A1 27.9 m³ (27.916...)	✓ 1
Part (c)
M1 multiplies the volume by 18.50	✓ 1
A1 €516	✓ 1

Exam tip: The integrand is y², so square the whole expression before integrating; $(2-0.1x^2)^2$ is not $4-0.01x^4$.

+ Add to test✓ In your test57IBMI3H-5-05r25-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

A trapdoor lift in a theatre raises a stage prop from the floor. The lift's upward velocity is $v=1.2t-0.3t^2$ m s⁻¹, for $0\le t\le4$, where t is in seconds.

(a)

Find the acceleration of the lift when t = 1.

(2)

(b)

Find the greatest speed of the lift.

(2)

(c)

Find the distance the lift rises in the 4 seconds.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates: $a=1.2-0.6t$	✓ 1
A1 0.6 m s⁻²	✓ 1
Part (b)
M1 sets $a=0$, giving t = 2	✓ 1
A1 1.2 m s⁻¹	✓ 1
Part (c)
M1 $\int_0^4(1.2t-0.3t^2)\,dt$	✓ 1
A1 3.2 m	✓ 1

Exam tip: Acceleration is the derivative of velocity and distance is the integral of velocity; for greatest speed set the acceleration equal to zero.

+ Add to test✓ In your test58IBMI3H-5-05r25-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The height of the tide in a harbour is modelled by $h=4.5+2.8\sin(0.5t)$ metres, where t is the time in hours after midnight, for $0\le t\le12$.

(a)

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

(2)

(b)

Find the rate at which the tide is rising at 02:00. Give your answer to 3 significant figures.

(2)

(c)

Find the first time after midnight at which the tide is rising at 0.5 m per hour, giving your answer in hours to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule: derivative of $\sin(0.5t)$ is $0.5\cos(0.5t)$	✓ 1
A1 $\frac{dh}{dt}=1.4\cos(0.5t)$	✓ 1
Part (b)
M1 substitutes t = 2 (radians)	✓ 1
A1 0.756 m per hour	✓ 1
Part (c)
M1 $1.4\cos(0.5t)=0.5$	✓ 1
A1 t = 2.41 hours	✓ 1

Common error: Leaving the calculator in degree mode gives wrong values for trigonometric derivatives in a modelling context; use radians.

+ Add to test✓ In your test59IBMI3H-5-04r26-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

During an outbreak, the number of new cases reported per day is modelled by C(t) = 120t e^−0.25t, where t is the number of days since the first report, t ≥ 0.

(a)

Find C′(t).

(3)

(b)

Find the time at which the number of new cases per day is greatest.

(2)

(c)

Find the greatest number of new cases per day, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule with u = 120t and v = e^−0.25t	✓ 1
A1 derivative of e^−0.25t is −0.25e^−0.25t	✓ 1
A1 C′(t) = 120e^−0.25t − 30te^−0.25t	✓ 1
Part (b)
M1 set C′(t) = 0, giving 120 − 30t = 0	✓ 1
A1 t = 4 days	✓ 1
Part (c)
M1 substitute t = 4: 480e⁻¹ = 176.6…	✓ 1
A1 177 cases per day	✓ 1

Common error: When differentiating e⁻⁰.²⁵t, remember the factor −0.25 from the chain rule; leaving it out changes the turning point.

+ Add to test✓ In your test60IBMI3H-5-04r26-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy8 marks

After the power is switched off, the angular speed ω rad s⁻¹ of a wind-turbine rotor at time t seconds satisfies dω/dt = −0.02ω², with ω = 10 when t = 0.

(a)

Solve the differential equation to show that ω = 1/(0.02t + 0.1).

(4)

(b)

Find the time at which the angular speed falls to 4 rad s⁻¹.

(2)

(c)

Find the angular speed after 60 seconds, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 separate variables: ∫ω⁻² dω = ∫−0.02 dt	✓ 1
A1 −1/ω = −0.02t + c	✓ 1
M1 substitute t = 0, ω = 10 to obtain c = −0.1	✓ 1
A1 1/ω = 0.02t + 0.1, so ω = 1/(0.02t + 0.1) (AG)	✓ 1
Part (b)
M1 0.02t + 0.1 = 1/4	✓ 1
A1 t = 7.5 s	✓ 1
Part (c)
M1 ω = 1/(0.02 × 60 + 0.1) = 1/1.3	✓ 1
A1 0.769 rad s⁻¹	✓ 1

Exam tip: In a 'show that' question every step must be written; the given result should appear only as your final line.

+ Add to test✓ In your test61IBMI3H-5-04r26-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

The depth of water in an estuary, used to site a tidal turbine, is modelled by d(t) = 6 + 2.5 sin(0.52t) metres, where t is the time in hours after midnight. Use radians throughout.

(a)

Find an expression for dd/dt.

(2)

(b)

Find the rate at which the depth is changing at 02:00, to 3 significant figures.

(1)

(c)

Find the two times, in hours after midnight, between t = 0 and t = 12 at which the depth is increasing at 0.5 m per hour. Give your answers to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule: multiply by the derivative 0.52 of the inner function	✓ 1
A1 dd/dt = 1.3 cos(0.52t)	✓ 1
Part (b)
A1 1.3 cos(1.04) = 0.658 m per hour	✓ 1
Part (c)
M1 1.3 cos(0.52t) = 0.5	✓ 1
A1 0.52t = 1.177…	✓ 1
A1 t = 2.26	✓ 1
A1 t = 9.82 (from 0.52t = 2π − 1.177…)	✓ 1

Exam tip: Check that your calculator is in radian mode before evaluating trigonometric functions of t; a second solution comes from the symmetry of the cosine curve.

+ Add to test✓ In your test62IBMI3H-5-04r26-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The power output, P MW, of a solar farm was recorded every 2 hours.
Time (hours after midnight): 8, 10, 12, 14, 16
P (MW): 12, 30, 41, 38, 22

(a)

Use the trapezoidal rule with all five values to estimate the energy produced between 08:00 and 16:00, in MWh.

(3)

(b)

The power curve is concave down over this period. State whether the trapezoidal estimate is an overestimate or an underestimate, and explain why.

(2)

(c)

Use your answer to part (a) to find the mean power output over the 8 hours.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of trapezoidal rule with h = 2	✓ 1
M1 (2/2)[12 + 22 + 2(30 + 41 + 38)]	✓ 1
A1 252 MWh	✓ 1
Part (b)
R1 for a concave down curve the straight-line tops of the trapezoids lie below the curve	✓ 1
A1 underestimate	✓ 1
Part (c)
A1 252/8 = 31.5 MW	✓ 1

Common error: The interior values are doubled in the trapezoidal rule; only the first and last values are counted once.

+ Add to test✓ In your test63IBMI3H-5-05r28-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

Calculator

The height of the tide in a marine reserve is modelled by $h(t)=4.2+1.8\cos(0.5t)$, where $h$ is in metres and $t$ is the time in hours after midnight. Angles are in radians.

(a)

Find an expression for $\frac{dh}{dt}$.

(2)

(b)

Find the rate of change of the tide height at $t=2$, and state whether the tide is rising or falling then.

(2)

(c)

Write down the greatest rate at which the tide rises.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate using the chain rule, e.g. a factor of 0.5 or $-\sin(0.5t)$	✓ 1
A1 $\frac{dh}{dt}=-0.9\sin(0.5t)$	✓ 1
Part (b)
M1 substitute $t=2$ into $-0.9\sin(0.5t)$ (calculator in radian mode), i.e. $-0.9\sin 1$	✓ 1
A1 $-0.757$ m h^-1 (negative, so the tide is falling)	✓ 1
Part (c)
A1 0.9 m h^-1	✓ 1

Common error: Leaving the calculator in degree mode gives -0.0157 instead of -0.757; the question states radians.

+ Add to test✓ In your test64IBMI3H-5-05r28-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

Calculator

The number of algae cells per mL in a tank of seawater, $t$ days after it is seeded, is modelled by $P(t)=200te^{-0.25t}$ for $t\ge 0$.

(a)

Show that $P'(t)=200e^{-0.25t}(1-0.25t)$.

(3)

(b)

Find the value of $t$ at which the population is greatest.

(2)

(c)

Find the greatest population, to the nearest whole number of cells per mL.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule with $u=200t$ and $v=e^{-0.25t}$	✓ 1
A1 $P'(t)=200e^{-0.25t}-50te^{-0.25t}$	✓ 1
A1 factorise to reach $200e^{-0.25t}(1-0.25t)$ (AG)	✓ 1
Part (b)
M1 set $P'(t)=0$ (and note $e^{-0.25t}\neq0$)	✓ 1
A1 $t=4$ days	✓ 1
Part (c)
M1 substitute $t=4$ into $P(t)$, i.e. $800e^{-1}$	✓ 1
A1 294 cells per mL	✓ 1

Exam tip: For a 'show that' question, every line of working must be written; the final line must match the given result exactly.

+ Add to test✓ In your test65IBMI3H-5-05r28-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

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A researcher models the number of fish $P$ in a lagoon, $t$ years after monitoring begins, by $\dfrac{dP}{dt}=0.3P\left(1-\dfrac{P}{800}\right)$ with $P(0)=100$. Use Euler's method with a step size of $h=1$.

(a)

Show that the estimate of $P(1)$ is 126.25.

(2)

(b)

Use Euler's method to estimate $P(3)$. Carry unrounded values through the working and give the final answer to 1 decimal place.

(3)

(c)

Explain, using the differential equation, why the model never predicts more than 800 fish if $P(0)=100$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use $P_{n+1}=P_n+h\frac{dP}{dt}$ with $P_0=100$, i.e. $100+0.3\times100\times(1-\frac{100}{800})$	✓ 1
A1 126.25 (AG)	✓ 1
Part (b)
M1 attempt to apply the method again with $P_1=126.25$	✓ 1
A1 $P_2=158.14785\ldots$ (accept 158.1)	✓ 1
A1 $P(3)\approx196.2$	✓ 1
Part (c)
R1 when $P=800$, $\frac{dP}{dt}=0$ so the population stops changing (the population cannot cross this value)	✓ 1

Exam tip: Write the iteration formula once, then show each substitution; a correct formula with a slip earns the method mark.

+ Add to test✓ In your test66IBMI3H-5-05r28-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

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The profile of an arch in a building is modelled by $y=\ln(x^2+1)$, where $x$ and $y$ are in metres.

(a)

Find $\frac{dy}{dx}$.

(2)

(b)

Find the equation of the tangent to the arch at the point where $x=1$. Give your answer in the form $y=mx+c$ with $c$ to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule with $\ln$ and inner function $x^2+1$	✓ 1
A1 $\frac{dy}{dx}=\frac{2x}{x^2+1}$	✓ 1
Part (b)
A1 gradient $=\frac{2}{2}=1$ and point $(1,\ln2)$	✓ 1
M1 use $y-\ln2=1(x-1)$	✓ 1
A1 $y=x-0.307$	✓ 1

Common error: The tangent passes through $(1,\ln 2)$, not $(1,2)$; evaluate $y$ using the curve, not the gradient.

+ Add to test✓ In your test67IBMI3H-5-06r28-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

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The height of the tip of a wind-turbine blade above the ground is $h(t)=80+45\sin(1.2t)$ metres, where $t$ is the time in seconds. Angles are in radians.

(a)

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

(2)

(b)

Write down the maximum rate at which the height of the tip is increasing.

(1)

(c)

Find the first time $t>0$ at which the tip is rising at exactly 20 m s^-1, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule giving a factor 1.2	✓ 1
A1 $\frac{dh}{dt}=54\cos(1.2t)$	✓ 1
Part (b)
A1 54 m s^-1	✓ 1
Part (c)
M1 set $54\cos(1.2t)=20$	✓ 1
A1 $1.2t=\cos^{-1}\left(\frac{20}{54}\right)=1.19\ldots$ (radians)	✓ 1
A1 $t=0.993$ s	✓ 1

Exam tip: Use the smallest positive solution of the inverse cosine for the first time; later solutions come from the periodic behaviour.

+ Add to test✓ In your test68IBMI3H-5-06r28-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

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The rate of visitors entering a library is modelled by $N(t)=120e^{-0.1(t-3)^2}$ people per hour, where $t$ is the time in hours after opening.

(a)

Find $N'(t)$.

(2)

(b)

Hence find the time at which the rate of arrivals is greatest.

(1)

(c)

Find the rate of change of $N$ at $t=5$, to 3 significant figures, and interpret the sign.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule with the exponent $-0.1(t-3)^2$ differentiated	✓ 1
A1 $N'(t)=-24(t-3)e^{-0.1(t-3)^2}$	✓ 1
Part (b)
A1 $N'(t)=0$ when $t=3$ hours	✓ 1
Part (c)
M1 substitute $t=5$: $-24(2)e^{-0.4}$	✓ 1
A1 $-32.2$ people per hour per hour	✓ 1
R1 negative: the arrival rate is falling at that time	✓ 1

Common error: Differentiate the whole exponent: the factor from $(t-3)^2$ is $2(t-3)$, which gives the $-24(t-3)$ term.

+ Add to test✓ In your test69IBMI3H-5-06r28-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

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The chlorine concentration $C$ (mg L^-1) in a pool, $t$ hours after a dosing system starts, satisfies $\dfrac{dC}{dt}=1.2-0.4C$ with $C(0)=0.5$. Use Euler's method with step size $h=0.5$.

(a)

Estimate $C(0.5)$ and $C(1)$.

(3)

(b)

Estimate $C(1.5)$.

(1)

(c)

Find the value that $C$ approaches in the long term according to the model.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use $C_{n+1}=C_n+h(1.2-0.4C_n)$	✓ 1
A1 $C=1.0$ at $t=0.5$	✓ 1
A1 $C=1.4$ at $t=1$	✓ 1
Part (b)
A1 $1.72$ (ECF from (a))	✓ 1
Part (c)
M1 set $\frac{dC}{dt}=0$	✓ 1
A1 $C=3$ mg L^-1	✓ 1

Exam tip: The long-term value of a model like this is where the rate of change is zero; no further iteration is needed.

+ Add to test✓ In your test70IBMI3H-5-06r28-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

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The installed capacity of a wind farm, in GW, $x$ years after a baseline date, is modelled by $f(x)=\dfrac{5}{1+e^{-0.8x}}$.

(a)

Find $f'(x)$.

(3)

(b)

Find the equation of the normal to the curve at $x=0$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 write $f=5(1+e^{-0.8x})^{-1}$ and use the chain rule	✓ 1
A1 $-5(1+e^{-0.8x})^{-2}\times(-0.8e^{-0.8x})$	✓ 1
A1 $f'(x)=\frac{4e^{-0.8x}}{(1+e^{-0.8x})^2}$	✓ 1
Part (b)
A1 $f(0)=2.5$ and $f'(0)=1$	✓ 1
M1 normal gradient is the negative reciprocal, $-1$	✓ 1
A1 $y=2.5-x$	✓ 1

Common error: The normal has gradient $-\frac{1}{f'(x)}$; using the tangent gradient is a frequent slip.

+ Add to test✓ In your test71IBMI3H-5-07r28-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

On a mountaineering expedition, the air pressure P kPa at a height h km above sea level is modelled by P = 101e^−0.12h, for 0 ≤ h ≤ 8.

(a)

Calculate the pressure at a height of 4 km. Give your answer to 3 significant figures.

(2)

(b)

Find dP/dh.

(2)

(c)

Find the value of dP/dh at h = 4, giving your answer to 3 significant figures, and interpret it in context.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute h = 4 into the model	✓ 1
A1 P = 62.5 kPa	✓ 1
Part (b)
M1 differentiate the exponential using the chain rule (multiply by −0.12)	✓ 1
A1 dP/dh = −12.12e^−0.12h	✓ 1
Part (c)
M1 substitute h = 4 into the derivative	✓ 1
A1 −7.50	✓ 1
R1 at 4 km the pressure is falling by about 7.50 kPa for each extra kilometre climbed	✓ 1

Common error: The derivative of ekh is k·ekh; forgetting the factor k (here −0.12) gives a wrong gradient.

+ Add to test✓ In your test72IBMI3H-5-07r28-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

In a food science laboratory, the concentration of a flavour compound in a sauce is modelled by C(t) = 5te^−0.4t mg L⁻¹, where t ≥ 0 is the time in hours after the compound is added.

(a)

Use the product rule to find C′(t).

(2)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

Find the greatest concentration, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule with u = 5t and v = e^−0.4t	✓ 1
A1 C′(t) = 5e^−0.4t − 2te^−0.4t = 5e^−0.4t(1 − 0.4t)	✓ 1
Part (b)
M1 set C′(t) = 0 and note e^−0.4t > 0, so 1 − 0.4t = 0	✓ 1
A1 t = 2.5 hours	✓ 1
Part (c)
M1 substitute t = 2.5: 12.5e⁻¹	✓ 1
A1 4.60 mg L⁻¹	✓ 1

Exam tip: After setting a derivative to zero, say why the exponential factor cannot be zero; it removes a false solution and earns the reasoning mark.

+ Add to test✓ In your test73IBMI3H-5-07r28-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

A mobile network’s cumulative data traffic is modelled by D(t) = 50 ln(t + 1) gigabytes, where t is the time in months since launch.

(a)

Find D′(t).

(2)

(b)

Find the rate of increase of traffic when t = 9.

(1)

(c)

Find the value of t at which the rate of increase is 2 gigabytes per month.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule on ln(t + 1)	✓ 1
A1 D′(t) = 50/(t + 1)	✓ 1
Part (b)
A1 5 gigabytes per month	✓ 1
Part (c)
M1 50/(t + 1) = 2	✓ 1
A1 t = 24 months	✓ 1

Exam tip: Give units with a rate of change: here gigabytes per month, not just gigabytes.

+ Add to test✓ In your test74IBMI3H-5-08r28-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

A marine biologist models the number of plankton cells in a sample as P(t) = 500e^0.04t thousand cells, where t is the time in days.

(a)

Find P′(t).

(2)

(b)

State the rate of growth at the start of the experiment.

(1)

(c)

Find the time at which the rate of growth is 30 thousand cells per day. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiate the exponential (multiply by 0.04)	✓ 1
A1 P′(t) = 20e^0.04t	✓ 1
Part (b)
A1 20 thousand cells per day	✓ 1
Part (c)
M1 20e^0.04t = 30, so e^0.04t = 1.5	✓ 1
M1 take natural logarithms: t = ln 1.5/0.04	✓ 1
A1 t = 10.1 days	✓ 1

Common error: Do not drop the factor 0.04 when differentiating; the derivative is 0.04 × 500e⁰.⁰⁴t, not 500e⁰.⁰⁴t.

+ Add to test✓ In your test75IBMI3H-5-08r28-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy8 marks

An aquarium tank drains through a hole in its base. The depth h metres of water after t seconds satisfies dh/dt = −0.05√h, with h = 1.44 when t = 0.

(a)

Separate the variables and solve to show that h = (1.2 − 0.025t)².

(5)

(b)

Find the depth of water after 20 seconds.

(1)

(c)

Find the time at which the tank is empty.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 separate: ∫ h^−1/2 dh = ∫ −0.05 dt	✓ 1
A1 2√h = −0.05t + c	✓ 1
M1 use h(0) = 1.44: c = 2.4	✓ 1
A1 √h = 1.2 − 0.025t	✓ 1
A1 h = (1.2 − 0.025t)²	✓ 1
Part (b)
A1 0.49 m	✓ 1
Part (c)
M1 set 1.2 − 0.025t = 0	✓ 1
A1 t = 48 s	✓ 1

Exam tip: When you integrate h⁻¹/², the power goes up to ½ and you divide by ½, which gives 2√h.

+ Add to test✓ In your test76IBMI3H-5-08r28-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy8 marks

A marker buoy bobs on waves. Its height above the mean sea level is h(t) = 1.5 sin(0.8t) metres at time t seconds.

(a)

Find the period of the motion. Give your answer to 3 significant figures.

(2)

(b)

Find an expression for the vertical velocity of the buoy.

(2)

(c)

Find the first time t > 0 at which the buoy reaches its greatest height. Give your answer to 3 significant figures.

(2)

(d)

Find the vertical speed of the buoy when t = 1. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 period = 2π/0.8	✓ 1
A1 7.85 s	✓ 1
Part (b)
M1 differentiate with the chain rule (multiply by 0.8)	✓ 1
A1 v = 1.2 cos(0.8t)	✓ 1
Part (c)
M1 v = 0 and sin(0.8t) = 1, so 0.8t = π/2	✓ 1
A1 t = 1.96 s	✓ 1
Part (d)
M1 substitute t = 1 into v, with the calculator in radian mode	✓ 1
A1 0.836 m s⁻¹	✓ 1

Common error: With an angle of 0.8t in radians, using degree mode on the calculator gives a very different value.

+ Add to test✓ In your test77IBMI3H-5-04r29-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

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The height, h metres, of the tip of a robotic arm above a workbench at time t seconds is modelled by h(t) = 0.8 + 0.35 sin(1.5t), where the angle is in radians.

(a)

Find an expression for dh/dt.

(2)

(b)

Find the rate of change of height at t = 2. Give your answer to 3 significant figures and state its units.

(2)

(c)

Find the first positive time at which the tip is momentarily at rest. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate using the chain rule, with a factor 1.5	✓ 1
A1 dh/dt = 0.525 cos(1.5t)	✓ 1
Part (b)
M1 substitutes t = 2 into 0.525 cos(3)	✓ 1
A1 −0.520 m s⁻¹ (−0.5197...)	✓ 1
Part (c)
M1 sets 0.525 cos(1.5t) = 0, so 1.5t = π/2	✓ 1
A1 t = 1.05 s (π/3 = 1.0471...)	✓ 1

Common error: Make sure the calculator is in radian mode, and remember the factor of 1.5 from the chain rule when differentiating sin(1.5t).

+ Add to test✓ In your test78IBMI3H-5-04r29-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

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A fishery is closed for a recovery period. The number of fish in the stock, N, t years after closure is modelled by N(t) = 3000 ln(2t + 1), for t ≥ 0.

(a)

Find N′(t).

(2)

(b)

Find the rate at which the stock is growing when t = 4. Give your answer to the nearest whole number of fish per year.

(2)

(c)

Find the value of t when the stock is growing at 400 fish per year.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate with the chain rule	✓ 1
A1 N′(t) = 6000 / (2t + 1)	✓ 1
Part (b)
M1 substitutes t = 4 into their N′(t)	✓ 1
A1 667 fish per year (666.6...)	✓ 1
Part (c)
M1 sets 6000 / (2t + 1) = 400	✓ 1
A1 t = 7	✓ 1

Exam tip: The derivative of ln(g(x)) is g′(x)/g(x). Here g′(x) = 2, so the 3000 becomes 6000 in the numerator.

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A fishing boat records its catch rate, r tonnes per day, every two days during a 10-day trip.
Day t: 0, 2, 4, 6, 8, 10
Rate r: 4.2, 5.1, 6.3, 6.0, 4.8, 3.6

(a)

Use the trapezoidal rule with all six values to estimate the total catch over the 10 days.

(3)

(b)

Hence estimate the mean catch rate over the 10 days.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses the trapezoidal rule with h = 2	✓ 1
M1 (2/2)[4.2 + 3.6 + 2(5.1 + 6.3 + 6.0 + 4.8)]	✓ 1
A1 52.2 tonnes	✓ 1
Part (b)
M1 divides their total by 10	✓ 1
A1 5.22 tonnes per day (ECF from (a))	✓ 1

Common error: In the trapezoidal rule only the first and last values are counted once. Every middle value is doubled.

+ Add to test✓ In your test80IBMI3H-5-04r29-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

A robotic arm moves along a straight rail. Its displacement from the start, in cm, at time t seconds is s(t) = 2t³ − 15t² + 24t + 3, for 0 ≤ t ≤ 5.

(a)

Find the velocity of the arm when t = 2.

(2)

(b)

Find the two times when the arm is momentarily at rest.

(2)

(c)

Find the acceleration of the arm when t = 4.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 v = ds/dt = 6t² − 30t + 24	✓ 1
A1 v(2) = −12 cm s⁻¹	✓ 1
Part (b)
M1 sets 6t² − 30t + 24 = 0, i.e. 6(t − 1)(t − 4) = 0	✓ 1
A1 t = 1 s and t = 4 s	✓ 1
Part (c)
M1 a = dv/dt = 12t − 30	✓ 1
A1 18 cm s⁻²	✓ 1

Exam tip: Velocity is the derivative of displacement, and acceleration is the derivative of velocity. A negative velocity means motion in the negative direction, not a slower speed.

+ Add to test✓ In your test81IBMI3H-5-04r31-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy4 marks

The density of air, ρ kg m⁻³, at a height h km above sea level is modelled by ρ(h) = 1.225e^−h/8.5, for 0 ≤ h ≤ 12. An aircraft is climbing through this air.

(a)

Find an expression for dρ/dh.

(2)

(b)

Find the value of dρ/dh when h = 3. Give your answer to 3 significant figures.

(1)

(c)

Interpret your answer to (b).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of the chain rule with the exponent −h/8.5	✓ 1
A1 dρ/dh = −(1.225/8.5)e^−h/8.5 (= −0.144e^−h/8.5)	✓ 1
Part (b)
A1 -0.101	✓ 1
Part (c)
R1 at a height of 3 km the air density is decreasing at a rate of about 0.101 kg m⁻³ for each additional kilometre of height	✓ 1

Common error: The derivative of ekh is kekh; forgetting the factor k (here −1/8.5) is the usual slip.

+ Add to test✓ In your test82IBMI3H-5-04r31-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The displacement of the midpoint of a guitar string from its rest position is modelled by y = 0.004 sin(1650t), where y is in metres and t is the time in seconds.

(a)

Find an expression for the velocity of the midpoint, v = dy/dt.

(2)

(b)

Write down the maximum speed of the midpoint.

(1)

(c)

Find the velocity of the midpoint when t = 0.0004. Use radians and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of the chain rule (multiplying by 1650)	✓ 1
A1 v = 6.6 cos(1650t)	✓ 1
Part (b)
A1 6.6 m s⁻¹	✓ 1
Part (c)
M1 substituting t = 0.0004 into the velocity expression, i.e. 6.6 cos(0.66)	✓ 1
A1 5.21 m s⁻¹	✓ 1

Exam tip: Differentiating sin(kt) gives k cos(kt) only when the angle is in radians; set the calculator to radian mode.

+ Add to test✓ In your test83IBMI3H-5-04r31-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The temperature, T °C, inside a greenhouse t hours after the heating is switched on satisfies the differential equation dT/dt = 0.3(25 − T). A slope field is drawn for this equation.

(a)

Calculate the gradient of the slope field at the point (t, T) = (1, 15) and at the point (1, 22).

(2)

(b)

Explain why the gradients along any horizontal line (constant T) in the slope field are all equal.

(1)

(c)

The temperature is 10 °C when t = 0. Describe the shape of the solution curve through the point (0, 10), referring to the line T = 25.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 0.3(25 − 15) = 3	✓ 1
A1 0.3(25 − 22) = 0.9	✓ 1
Part (b)
R1 the right-hand side depends only on T and not on t	✓ 1
Part (c)
A1 the curve is increasing, with the gradient decreasing (concave down)	✓ 1
A1 it approaches the horizontal line T = 25 without crossing it	✓ 1

Exam tip: A slope field only needs the gradient at each point; substitute the coordinates into the right-hand side of the differential equation.

+ Add to test✓ In your test84IBMI3H-5-06r32-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

After an oral dose, the concentration of a drug in the blood is modelled by $C(t)=12te^{-0.5t}$ mg/L, where $t$ is the time in hours after the dose, $t\ge0$.

(a)

Find $\dfrac{dC}{dt}$.

(3)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

Find the greatest concentration, to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule applied to $12t$ and $e^{-0.5t}$	✓ 1
A1 derivative of $e^{-0.5t}$ is $-0.5e^{-0.5t}$	✓ 1
A1 $\frac{dC}{dt}=12e^{-0.5t}(1-0.5t)$	✓ 1
Part (b)
M1 sets $\frac{dC}{dt}=0$ (ECF from (a))	✓ 1
A1 $t=2$ hours	✓ 1
Part (c)
A1 $24e^{-1}=8.83$ mg/L	✓ 1

Exam tip: $e^{-0.5t}$ is never zero, so a stationary point can only come from the other factor.

+ Add to test✓ In your test85IBMI3H-5-06r32-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The depth of water in a reservoir intake channel is modelled by $h(t)=4+1.5\sin(0.5t)$ metres, where $t$ is the time in hours and the angle is in radians.

(a)

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

(2)

(b)

Find the rate at which the depth is changing when $t=2$, to 3 significant figures.

(2)

(c)

State the greatest rate at which the depth can increase.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule	✓ 1
A1 $0.75\cos(0.5t)$	✓ 1
Part (b)
M1 substitutes $t=2$ with the calculator in radian mode (ECF from (a))	✓ 1
A1 $0.405$ m/h	✓ 1
Part (c)
A1 $0.75$ m/h	✓ 1

Common error: Check the calculator is in radian mode; $\cos(1)$ in degrees gives a completely different value.

+ Add to test✓ In your test86IBMI3H-5-06r32-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The concentration $C$ mg/L of a drug in a patient receiving a steady infusion satisfies $\dfrac{dC}{dt}=2-0.3C$, where $t$ is the time in hours and $C=0$ when $t=0$.

(a)

Use Euler's method with a step size of 1 hour to estimate $C$ when $t=3$.

(4)

(b)

The concentration approaches a steady value in the long term. Find this value, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $C_{n+1}=C_n+1\times(2-0.3C_n)$	✓ 1
A1 $C_1=2$	✓ 1
A1 $C_2=3.4$	✓ 1
A1 $C_3=4.38$ mg/L	✓ 1
Part (b)
M1 sets $\frac{dC}{dt}=0$	✓ 1
A1 $6.67$ mg/L	✓ 1

Exam tip: Write each Euler step as a line in a small table; the final answer is the last $y$-value, not the sum of the increments.

+ Add to test✓ In your test87IBMI3H-5-06r32-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

A character in a game moves along a straight track. Its displacement from the start is $s(t)=t^3-6t^2+9t$ metres, where $t\ge0$ is the time in seconds.

(a)

Find an expression for the velocity of the character.

(2)

(b)

Find the times at which the character is at rest.

(2)

(c)

Find the acceleration of the character when $t=3$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates $s$	✓ 1
A1 $v=3t^2-12t+9$	✓ 1
Part (b)
M1 sets $v=0$ (ECF from (a))	✓ 1
A1 $t=1$ and $t=3$ seconds	✓ 1
Part (c)
A1 $a=6t-12=6$ m/s$^2$	✓ 1

Exam tip: "At rest" means velocity is zero; acceleration can still be non-zero at that instant.

+ Add to test✓ In your test88IBMI3H-5-07r32-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The tip of a wind turbine blade rotates, and its height above the ground is modelled by $h(t)=80+45\sin(0.6t)$ metres, where $t$ is the time in seconds after the tip is first level with the hub. Use radians throughout.

(a)

Find an expression for $h'(t)$.

(2)

(b)

Find the rate of change of height when $t=3$, giving your answer to 3 significant figures.

(2)

(c)

Interpret the sign of your answer to part (b).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate using the chain rule, e.g. $45\cos(0.6t)\times 0.6$	✓ 1
A1 $h'(t)=27\cos(0.6t)$	✓ 1
Part (b)
M1 substitute $t=3$: $27\cos(1.8)$	✓ 1
A1 $-6.13$ m s⁻¹	✓ 1
Part (c)
R1 the tip is moving downwards (height decreasing) at this instant	✓ 1

Common error: Remember the chain rule: the factor 0.6 from the inner function must multiply the derivative, and the calculator must be in radian mode.

+ Add to test✓ In your test89IBMI3H-5-07r32-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

A battery store at a wind farm holds $C$ MWh of energy $t$ hours after charging begins. Its charge satisfies the differential equation $\frac{dC}{dt}=0.4(120-C)$, with $C=30$ when $t=0$.

(a)

Use Euler's method with a step length of $h=0.5$ to estimate $C$ when $t=1$.

(3)

(b)

The exact solution is $C=120-90e^{-0.4t}$. Find the exact value of $C$ at $t=1$ to 3 significant figures and state whether Euler's method has over- or under-estimated it.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use $C_{n+1}=C_n+h\,\frac{dC}{dt}$ with $h=0.5$	✓ 1
A1 $C(0.5)\approx 30+0.5\times0.4\times90=48$	✓ 1
A1 $C(t{=}1)\approx 48+0.5\times0.4\times72=62.4$ MWh	✓ 1
Part (b)
A1 $C(t{=}1)=120-90e^{-0.4}=59.7$ MWh	✓ 1
R1 Euler's estimate (62.4) is an overestimate	✓ 1

Exam tip: Write each Euler step as a short calculation showing the gradient used; method marks are awarded for the pattern even if arithmetic slips.

+ Add to test✓ In your test90IBMI3H-5-07r32-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The number of people in a library $t$ hours after it opens is modelled by $N(t)=40te^{-0.25t}$ for $0\le t\le 12$.

(a)

Use the product rule to find $N'(t)$.

(3)

(b)

Find the time at which the library is busiest and the number of people present at that time, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at product rule with $u=40t$, $v=e^{-0.25t}$	✓ 1
A1 $\frac{d}{dt}(e^{-0.25t})=-0.25e^{-0.25t}$ seen	✓ 1
A1 $N'(t)=40e^{-0.25t}(1-0.25t)$	✓ 1
Part (b)
M1 set $N'(t)=0$	✓ 1
A1 $t=4$ hours	✓ 1
A1 $N(4)=160e^{-1}=58.9$ people	✓ 1

Exam tip: Factorise the derivative before setting it to zero; the exponential factor is never zero, so only the bracket gives a solution.

+ Add to test✓ In your test91IBMI3H-5-07r32-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The power output $P$ (MW) of a wind farm is recorded every two hours for eight hours.

Time $t$ (hours)	0	2	4	6	8
$P$ (MW)	3.2	4.1	5.5	4.8	3.9

(a)

Use the trapezoidal rule with four intervals to estimate the energy produced, $\int_0^8 P\,dt$, in MWh.

(3)

(b)

State one way to make the estimate more accurate.

(1)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 trapezoidal rule with $h=2$	✓ 1
A1 $\frac{2}{2}\left[3.2+3.9+2(4.1+5.5+4.8)\right]$	✓ 1
A1 $35.9$ MWh	✓ 1
Part (b)
R1 record the output more frequently (use more, narrower intervals)	✓ 1
Part (c)
M1 energy divided by 8 hours (ECF from (a))	✓ 1
A1 $4.49$ MW	✓ 1

Exam tip: The trapezoidal rule takes the first and last ordinates once and doubles all the others; list them separately before substituting.

+ Add to test✓ In your test92IBMI3H-5-08r32-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The concentration of algae in a marine tank $t$ days after a treatment is modelled by $C(t)=8+20e^{-0.25t}$ mg per litre.

(a)

Find $C'(2)$ to 3 significant figures, and state what it represents.

(3)

(b)

Find the time at which the concentration is falling at exactly 1 mg per litre per day, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule applied to $20e^{-0.25t}$	✓ 1
A1 $C'(t)=-5e^{-0.25t}$	✓ 1
A1 $C'(2)=-3.03$ mg per litre per day: the concentration is falling at this rate after 2 days	✓ 1
Part (b)
M1 $-5e^{-0.25t}=-1$, so $e^{-0.25t}=0.2$	✓ 1
A1 $t=4\ln5=6.44$ days	✓ 1

Common error: The derivative of $e^{kt}$ is $ke^{kt}$; forgetting the factor $k=-0.25$ changes both the value and the sign of the rate.

+ Add to test✓ In your test93IBMI3H-5-08r32-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

A fish population $N$ in a reserve $t$ years after monitoring began satisfies $\frac{dN}{dt}=0.2N\left(1-\frac{N}{500}\right)$, with $N=100$ when $t=0$.

(a)

Find the non-zero population at which $\frac{dN}{dt}=0$.

(2)

(b)

Use Euler's method with step length $h=1$ to estimate $N$ when $t=2$, giving your answer to the nearest whole number.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set $1-\frac{N}{500}=0$	✓ 1
A1 $N=500$	✓ 1
Part (b)
M1 $N_{n+1}=N_n+h\times0.2N_n\left(1-\frac{N_n}{500}\right)$	✓ 1
A1 $N_1=116$	✓ 1
M1 second step: $116+0.2\times116\times\left(1-\frac{116}{500}\right)$	✓ 1
A1 $N_2=134$ fish (from 133.8)	✓ 1

Exam tip: Keep the unrounded value from each Euler step in your calculator and round only the final answer when the question asks for a whole number.

+ Add to test✓ In your test94IBMI3H-5-08r32-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

A forklift moves along a straight warehouse aisle. Its displacement from the loading bay is $s=t^3-6t^2+9t$ metres at time $t$ seconds, for $0\le t\le4$.

(a)

Find an expression for the velocity of the forklift.

(2)

(b)

Find the times when the forklift is instantaneously at rest.

(2)

(c)

Find the greatest distance from the loading bay reached before $t=3$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiate $s$ with respect to $t$	✓ 1
A1 $v=3t^2-12t+9$	✓ 1
Part (b)
M1 solve $3t^2-12t+9=0$, i.e. $3(t-1)(t-3)=0$	✓ 1
A1 $t=1$ and $t=3$	✓ 1
Part (c)
A1 $s(t{=}1)=4$ m	✓ 1

Exam tip: 'At rest' means velocity zero, not displacement zero; differentiate first and then solve $v=0$.

+ Add to test✓ In your test95IBMI3H-5-08r32-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The depth of the sea bed below the surface at a distance $x$ km from the shore is modelled by $h(x)=10x-5x\ln x$ metres, for $1\le x\le5$.

(a)

Use the product rule to find $h'(x)$.

(3)

(b)

Find the distance from the shore at which the sea bed is deepest, giving your answer to 3 significant figures.

(2)

(c)

Find the greatest depth, to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule on $x\ln x$	✓ 1
A1 $\frac{d}{dx}(x\ln x)=\ln x+1$	✓ 1
A1 $h'(x)=5-5\ln x$	✓ 1
Part (b)
M1 set $h'(x)=0$, so $\ln x=1$	✓ 1
A1 $x=e=2.72$ km	✓ 1
Part (c)
A1 $h(e)=5e=13.6$ m	✓ 1

Common error: The derivative of $x\ln x$ is $\ln x+1$, not $\frac{1}{x}$; the product rule must be applied to both factors.

+ Add to test✓ In your test96IBMI3H-5-06r34-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The height above ground, h metres, of the tip of a wind-turbine blade at time t seconds is modelled by h(t) = 80 + 45 cos(0.6t), where the angle is in radians.

(a)

Find h′(t).

(2)

(b)

Find the rate of change of the height when t = 2. Give your answer to 3 significant figures and state whether the tip is rising or falling.

(2)

(c)

State the maximum rate at which the height of the tip changes.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Use of the chain rule on cos(0.6t)	✓ 1
h′(t) = −27 sin(0.6t)	✓ 1
Part (b)
h′(2) = −27 sin 1.2	✓ 1
= −25.2 m s⁻¹, so the tip is falling	✓ 1
Part (c)
27 m s⁻¹ (the amplitude of h′)	✓ 1

Common error: Make sure the calculator is in radian mode when the model uses radians.

+ Add to test✓ In your test97IBMI3H-5-06r34-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The temperature T °C of a swimming pool satisfies dT/dt = 0.2(30 − T), where t is the time in hours after a heater is switched on. Initially T = 22.

(a)

Use Euler's method with a step length of 1 hour to estimate T after 3 hours.

(3)

(b)

The exact solution is T = 30 − 8e^−0.2t. Find the exact value of T when t = 3, to 3 significant figures, and state whether Euler's method over-estimates or under-estimates it.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
T₁ = 22 + 1 × 0.2(8) = 23.6	✓ 1
T₂ = 23.6 + 0.2(6.4) = 24.88	✓ 1
T₃ = 24.88 + 0.2(5.12) = 25.904, so about 25.9 °C	✓ 1
Part (b)
T(3) = 30 − 8e^−0.6 = 25.6 °C	✓ 1
Euler's estimate is greater, so it over-estimates	✓ 1

Common error: Each Euler step uses the gradient at the start of the step, evaluated with the newly updated value of T, not the original value.

+ Add to test✓ In your test98IBMI3H-5-06r34-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

A swimmer moves along a 25 m pool. Her velocity is v(t) = 2.2 − 0.04t m s⁻¹ for 0 ≤ t ≤ 40, where t is the time in seconds, and she starts at the wall at t = 0.

(a)

Write down her acceleration.

(1)

(b)

Find an expression for her distance s from the starting wall at time t.

(2)

(c)

Find the time she takes to reach the far wall (s = 25). Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
a = v′ = −0.04 m s⁻²	✓ 1
Part (b)
s = ∫(2.2 − 0.04t) dt = 2.2t − 0.02t² + C	✓ 1
C = 0 since s = 0 at t = 0	✓ 1
Part (c)
2.2t − 0.02t² = 25	✓ 1
Solve the quadratic equation	✓ 1
t = 12.9 s (the other root, 97.1 s, is outside the domain)	✓ 1

Common error: Include the constant of integration and use the starting condition to find it before using the position formula.

+ Add to test✓ In your test99IBMI3H-5-06r34-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

The power output P (in MW) of a wind farm is recorded each hour:
Time t (hours): 0, 1, 2, 3, 4, 5, 6
Power P (MW): 2.0, 3.4, 4.8, 5.2, 4.1, 3.0, 2.2

(a)

Use the trapezoidal rule with 6 intervals to estimate the energy produced in the 6 hours, which is the area under the power-time graph, in MWh.

(3)

(b)

Between t = 1 and t = 3 the graph of power against time is concave down. State whether the trapezoidal rule over-estimates or under-estimates the area over this interval, and explain.

(2)

(c)

The electricity is sold at €48 per MWh. Using your answer to part (a), find the estimated income, in euros, to 2 decimal places.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Width of each interval is 1	✓ 1
Trapezoidal rule: (1/2)(2.0 + 2.2) + 3.4 + 4.8 + 5.2 + 4.1 + 3.0	✓ 1
Estimate = 22.6 MWh	✓ 1
Part (b)
Under-estimates	✓ 1
The chords (tops of the trapezoids) lie below a concave down curve	✓ 1
Part (c)
22.6 × 48	✓ 1
€1084.80	✓ 1

Exam tip: The first and last values in the trapezoidal rule are halved; every interior value is counted in full.

+ Add to test✓ In your test100IBMI3H-5-07r34-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

After an injection, the concentration of a drug in a patient's blood is modelled by $C(t) = 12te^{-0.4t}$ mg L⁻¹, where t is the time in hours after the injection, $t \ge 0$.

(a)

Find an expression for $\frac{dC}{dt}$.

(3)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

Find the greatest concentration of the drug, giving your answer to three significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule on $12t \times e^{-0.4t}$	✓ 1
A1 $12e^{-0.4t}$ seen	✓ 1
A1 $\frac{dC}{dt} = 12e^{-0.4t} - 4.8te^{-0.4t}$ (or $12e^{-0.4t}(1-0.4t)$)	✓ 1
Part (b)
M1 set $\frac{dC}{dt}=0$ (since $e^{-0.4t}>0$, $1-0.4t=0$)	✓ 1
A1 $t = 2.5$ hours	✓ 1
Part (c)
M1 substitute $t=2.5$ into $C(t)$, giving $30e^{-1}$ (ECF from (b))	✓ 1
A1 $C = 11.0$ mg L⁻¹ (11.036… before rounding)	✓ 1

Exam tip: Use the unrounded value of t when you substitute back into C; here t = 2.5 is exact so the answer 11.036… rounds cleanly.

+ Add to test✓ In your test101IBMI3H-5-07r34-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy8 marks

In an online game, a player's ranking score after t hours of play in a tournament is modelled by $P(t) = 50\ln(3t^2 + 1)$, for $t \ge 0$.

(a)

Show that $\frac{dP}{dt} = \frac{300t}{3t^2+1}$.

(3)

(b)

Find the rate of change of the score when $t=2$, giving your answer to three significant figures.

(2)

(c)

Use the tangent to the graph of $P$ at $t=2$ to estimate the score when $t = 2.1$. Give your answer to three significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the chain rule (derivative of $\ln u$ is $\frac{u'}{u}$)	✓ 1
A1 $u' = 6t$ seen	✓ 1
A1 $\frac{dP}{dt} = 50 \times \frac{6t}{3t^2+1} = \frac{300t}{3t^2+1}$ (AG)	✓ 1
Part (b)
M1 substitute $t=2$ to get $\frac{600}{13}$	✓ 1
A1 46.2 points per hour	✓ 1
Part (c)
M1 $P(2) = 50\ln 13 = 128.247…$ seen	✓ 1
M1 $P(2.1) \approx P(2) + 0.1 \times \frac{600}{13}$ (use unrounded values)	✓ 1
A1 133 (132.86… before rounding)	✓ 1

Common error: Using the rounded gradient 46.2 and rounded P(2) can shift a three significant figure answer; keep the unrounded values until the end.

+ Add to test✓ In your test102IBMI3H-5-07r34-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The number of active players of an online game, P thousand, t months after launch, satisfies the differential equation $\frac{dP}{dt} = 0.2P\left(1 - \frac{P}{500}\right)$. At launch, $P = 50$. Euler's method with a step length of $h = 1$ month is used to estimate P.

(a)

Show that the estimate of P after one month is 59 thousand.

(2)

(b)

Find the estimate of P after two months, to three significant figures.

(2)

(c)

The true solution is concave up for $P < 250$. State, with a reason, whether Euler's method gives an overestimate or an underestimate of the true value of P at $t = 2$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $P_1 = 50 + 1\times 0.2\times 50\times\left(1-\frac{50}{500}\right)$	✓ 1
A1 $= 50 + 9 = 59$ (AG)	✓ 1
Part (b)
M1 $P_2 = 59 + 1\times 0.2\times 59\times\left(1-\frac{59}{500}\right)$ (use 59 exactly)	✓ 1
A1 69.4 thousand (69.4076… before rounding)	✓ 1
Part (c)
A1 underestimate	✓ 1
R1 each step uses the gradient at the start of the step, which is smaller than the gradient later in the step when the curve is concave up (the tangent lies below the curve)	✓ 1

Exam tip: In Euler's method, use the exact previous value in each step and round only the final answer.

+ Add to test✓ In your test103IBMI3H-5-07r34-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

The depth of water, h metres, in a reservoir that is being filled while also leaking satisfies $\frac{dh}{dt} = 0.8 - 0.1h$, where t is the time in hours. The slope field of this equation is drawn for $t \ge 0$ and $h \ge 0$.

(a)

Calculate the gradient of the slope field at the points where $h=2$ and where $h=10$.

(2)

(b)

Find the depth at which the gradient is zero, and state what this means for the water level.

(2)

(c)

The depth is 2 m at $t=0$. Use Euler's method with a step length of 1 hour to estimate the depth at $t=2$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 gradient $0.6$ at $h=2$	✓ 1
A1 gradient $-0.2$ at $h=10$	✓ 1
Part (b)
M1 solve $0.8 - 0.1h = 0$	✓ 1
A1 $h = 8$ m, at which the depth stays constant (inflow equals leakage)	✓ 1
Part (c)
M1 $h_1 = 2 + 1\times(0.8 - 0.1\times 2) = 2.6$	✓ 1
M1 $h_2 = 2.6 + 1\times(0.8 - 0.1\times 2.6)$ (ECF from $h_1$)	✓ 1
A1 3.14 m	✓ 1

Common error: The gradient depends only on h here, so the slope-field segments are parallel along every horizontal line.

+ Add to test✓ In your test104IBMI3H-5-07r35-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

After a short sprint, the excess blood lactate concentration of an athlete is modelled by $C(t)=8te^{-0.5t}$, where $C$ is in mmol L⁻¹ and $t\ge 0$ is the time in minutes after the sprint ends.

(a)

Find $\frac{dC}{dt}$.

(2)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

Find the greatest concentration, giving your answer correct to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 $\frac{dC}{dt}=8e^{-0.5t}-4te^{-0.5t}$ (or equivalent, e.g. $8e^{-0.5t}(1-0.5t)$)	✓ 1
Part (b)
M1 set $\frac{dC}{dt}=0$	✓ 1
A1 $t=2$ minutes	✓ 1
Part (c)
A1 $C(2)=16e^{-1}=5.89$ mmol L⁻¹	✓ 1

Exam tip: Because $e^{-0.5t}$ is never zero, you can divide it out when solving $\frac{dC}{dt}=0$; the remaining factor gives the turning point directly.

+ Add to test✓ In your test105IBMI3H-5-07r35-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The brightness of a pulsating star is modelled by $F(t)=4.2+0.8\sin\left(\frac{2\pi t}{5.4}\right)$, where $F$ is the brightness in units of the Sun's brightness divided by 10⁴ and $t$ is the time in days. The argument of the sine is in radians.

(a)

Find $\frac{dF}{dt}$.

(2)

(b)

Find the rate at which the brightness is changing when $t=0.9$. Give your answer correct to 3 significant figures.

(2)

(c)

Find the first time, $t>0$, at which the brightness is at a maximum.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the chain rule with a sine function	✓ 1
A1 $\frac{dF}{dt}=0.8\times\frac{2\pi}{5.4}\cos\left(\frac{2\pi t}{5.4}\right)$ (accept $0.931\cos\left(\frac{2\pi t}{5.4}\right)$)	✓ 1
Part (b)
M1 substitute $t=0.9$ into their $\frac{dF}{dt}$	✓ 1
A1 $0.465$ per day (ECF from (a))	✓ 1
Part (c)
M1 set $\frac{dF}{dt}=0$ or use $\sin\left(\frac{2\pi t}{5.4}\right)=1$	✓ 1
A1 $t=1.35$ days	✓ 1

Common error: Calculators must be in radian mode here; the same expression evaluated in degree mode gives a completely different gradient.

+ Add to test✓ In your test106IBMI3H-5-07r35-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The height of a ski run above the valley floor is modelled by $h(x)=60-0.8x+0.004x^2$, for $0\le x\le 100$, where $x$ is the horizontal distance in metres from the top and $h$ is in metres.

(a)

Find the gradient of the run at $x=40$.

(2)

(b)

Find the equation of the tangent to the run at $x=40$.

(2)

(c)

Find the angle below the horizontal at which the run descends at $x=40$. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate, $h'(x)=-0.8+0.008x$	✓ 1
A1 $h'(40)=-0.48$	✓ 1
Part (b)
A1 $h(40)=34.4$	✓ 1
A1 $y-34.4=-0.48(x-40)$, i.e. $y=-0.48x+53.6$	✓ 1
Part (c)
M1 $\tan\theta=0.48$ (use of the gradient)	✓ 1
A1 $25.6^\circ$	✓ 1

Common error: The angle of descent is $rctan$ of the magnitude of the gradient; do not use the gradient value itself as an angle.

+ Add to test✓ In your test107IBMI3H-5-07r35-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

A rower's speed is recorded every 2 seconds at the start of a race.

Time t (s)	0	2	4	6	8	10
Speed v (m s⁻¹)	0	3.1	4.6	5.2	5.5	5.6

(a)

Use the trapezoidal rule with all six values to estimate the distance rowed in the first 10 seconds.

(3)

(b)

Hence find the average speed over the 10 seconds.

(1)

(c)

The speed increases at a decreasing rate. State, with a reason, whether the estimate in (a) is an overestimate or an underestimate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of the trapezoidal rule with strip width 2	✓ 1
A1 $\frac{2}{2}\left[0+5.6+2(3.1+4.6+5.2+5.5)\right]$	✓ 1
A1 $42.4$ m	✓ 1
Part (b)
A1 $\frac{42.4}{10}=4.24$ m s⁻¹ (ECF from (a))	✓ 1
Part (c)
A1 underestimate	✓ 1
R1 the graph is concave down, so each chord (trapezium top) lies below the curve	✓ 1

Exam tip: The distance is the area under a speed–time graph, so the trapezoidal rule applied to speed readings estimates distance, not speed.

+ Add to test✓ In your test108IBMI3H-5-04r39-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The number of bees, P, in a new hive t days after it is established is modelled by the differential equation dP/dt = 0.3P(1 − P/800), with P = 120 when t = 0.

(a)

Use Euler's method with a step length of 1 day to estimate the number of bees when t = 1.

(2)

(b)

Continue the method, using the unrounded value from (a), to estimate the number of bees when t = 2. Give your answer to 1 decimal place.

(2)

(c)

Find the value of P for which the model predicts that the colony stops changing in size, with P > 0.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 P₁ = 120 + 1 × 0.3 × 120(1 − 120/800)	✓ 1
A1 P₁ = 150.6	✓ 1
Part (b)
M1 P₂ = 150.6 + 0.3 × 150.6(1 − 150.6/800)	✓ 1
A1 P₂ = 187.3 (ECF from (a))	✓ 1
Part (c)
A1 P = 800 (solving dP/dt = 0)	✓ 1

Exam tip: In Euler's method, write the new value as old value + h × (gradient at the old point); keep unrounded values between steps.

+ Add to test✓ In your test109IBMI3H-5-04r39-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

A beekeeper makes a honey funnel by rotating the region under the curve y = 1.5e^−0.2x, for 0 ≤ x ≤ 5, through 360° about the x-axis. Lengths are in centimetres.

(a)

Write down an integral for the volume of the funnel's interior.

(2)

(b)

Find the volume of the funnel, in cm³, to 3 significant figures.

(2)

(c)

The funnel is filled to the top with honey of density 1.4 g/cm³. Find the mass of honey, in grams, using the unrounded volume and giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 V = π∫₀⁵ y² dx	✓ 1
A1 V = π∫₀⁵ 2.25e^−0.4x dx	✓ 1
Part (b)
M1 evaluate the integral (analytically or by GDC)	✓ 1
A1 V = 15.3 cm³	✓ 1
Part (c)
M1 mass = density × volume = 1.4 × 15.28…	✓ 1
A1 21.4 g	✓ 1

Common error: The volume formula uses y², not y; square the whole function, including its coefficient, before integrating.

+ Add to test✓ In your test110IBMI3H-5-04r39-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The concentration of a drug in a patient's blood is modelled by C(t) = 40t/(t² + 16) mg/L, where t ≥ 0 is the time in hours after the drug is given.

(a)

Show that C′(t) = 40(16 − t²)/(t² + 16)².

(3)

(b)

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

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule: [40(t² + 16) − 40t × 2t]/(t² + 16)²	✓ 1
A1 numerator 40t² + 640 − 80t²	✓ 1
A1 = 40(16 − t²)/(t² + 16)²	✓ 1
Part (b)
M1 C′(t) = 0, so 16 − t² = 0	✓ 1
A1 t = 4 hours	✓ 1
A1 C = 5 mg/L	✓ 1

Exam tip: Set the numerator of a quotient-rule derivative equal to zero to find stationary points; the denominator is never zero here.

+ Add to test✓ In your test111IBMI3H-5-04r39-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

The temperature inside a beehive is modelled by T(t) = 34 + 3sin(πt/12) °C, where t is the number of hours after midnight, 0 ≤ t ≤ 24. Angles are in radians.

(a)

Find an expression for dT/dt.

(2)

(b)

Find the rate of change of temperature at 14:00. Give your answer to 3 significant figures.

(2)

(c)

Use the tangent to the curve at t = 2 to estimate the temperature at t = 2.5. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule	✓ 1
A1 dT/dt = (π/4)cos(πt/12)	✓ 1
Part (b)
M1 substitute t = 14: (π/4)cos(7π/6)	✓ 1
A1 -0.680 °C per hour	✓ 1
Part (c)
A1 T(2) = 35.5	✓ 1
A1 gradient at t = 2 is (π/4)cos(π/6) = 0.680	✓ 1
A1 estimate = 35.5 + 0.5 × 0.680 = 35.8 °C	✓ 1

Common error: Make sure the calculator is in radian mode when the angle in a model has no degree sign.

+ Add to test✓ In your test112IBMI3H-5-05r40-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

A café owner models her daily profit, P euros, from selling q cups of a seasonal drink by $P(q) = -0.02q^2 + 14q - 300$, for $0 \le q \le 700$.

(a)

Find $\frac{dP}{dq}$.

(2)

(b)

Find the number of cups that must be sold to maximise the daily profit.

(2)

(c)

Show that this value of q gives a maximum, and find the maximum daily profit.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate a term of the polynomial	✓ 1
A1 $\frac{dP}{dq} = -0.04q + 14$	✓ 1
Part (b)
M1 set $\frac{dP}{dq}=0$ i.e. $-0.04q+14=0$	✓ 1
A1 $q = 350$	✓ 1
Part (c)
R1 $\frac{d^2P}{dq^2} = -0.04 < 0$, so a maximum	✓ 1
M1 substitute $q=350$ into $P(q)$	✓ 1
A1 maximum profit €2150	✓ 1

Exam tip: Always justify a maximum with the second derivative or a sign change of the gradient; stating that the gradient is zero is not enough.

+ Add to test✓ In your test113IBMI3H-5-05r40-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

A solar farm records its power output, P megawatts (MW), every two hours during one day.

Time t (hours after 08:00): 0, 2, 4, 6, 8
Power P (MW): 0, 3.0, 5.6, 4.4, 1.2

(a)

Use the trapezoidal rule with four intervals to estimate the total energy produced, in megawatt-hours (MWh), between t = 0 and t = 8.

(3)

(b)

Find the mean power output over these eight hours.

(2)

(c)

The energy is sold at £48 per MWh. Use your answer to part (a) to find the income from this energy, to the nearest pound.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of trapezoidal rule with $h=2$	✓ 1
M1 correct substitution $\frac{2}{2}\left[0+1.2+2(3.0+5.6+4.4)\right]$	✓ 1
A1 27.2 MWh	✓ 1
Part (b)
M1 divide energy by 8 hours: $27.2 \div 8$	✓ 1
A1 3.4 MW, ECF from (a)	✓ 1
Part (c)
M1 $27.2 \times 48$	✓ 1
A1 £1306 (1305.6 rounded)	✓ 1

Exam tip: The trapezoidal rule uses the first and last ordinates once and every middle ordinate twice.

+ Add to test✓ In your test114IBMI3H-5-05r40-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

After a dose of medicine, the concentration in a patient's blood is modelled by $C(t) = \dfrac{6t}{t^2+9}$ mg L⁻¹, where t is the time in hours after the dose, $t \ge 0$.

(a)

Use the quotient rule to show that $C'(t) = \dfrac{54 - 6t^2}{(t^2+9)^2}$.

(3)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

Find the greatest concentration.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at quotient rule	✓ 1
A1 $C'(t) = \dfrac{6(t^2+9) - 6t(2t)}{(t^2+9)^2}$	✓ 1
A1 simplify numerator to $54-6t^2$ (AG)	✓ 1
Part (b)
M1 set numerator $54 - 6t^2 = 0$	✓ 1
A1 $t = 3$ hours	✓ 1
Part (c)
A1 $C(3) = \dfrac{18}{18} = 1$ mg L⁻¹	✓ 1

Exam tip: Set only the numerator of a quotient-rule derivative equal to zero, because the denominator is never zero here.

+ Add to test✓ In your test115IBMI3H-5-05r40-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The number of students, N, infected with a virus at a school is modelled by $\frac{dN}{dt} = 0.2N\left(1 - \frac{N}{500}\right)$, where t is in days. At $t=0$, $N = 50$.

(a)

Find the value of $\frac{dN}{dt}$ when $N=50$.

(1)

(b)

Use Euler's method with a step size of 1 day to estimate N at $t=1$.

(2)

(c)

Continue Euler's method to estimate N at $t=2$, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $0.2(50)(0.9) = 9$ students per day	✓ 1
Part (b)
M1 $N_1 = 50 + 1 \times 9$	✓ 1
A1 $N_1 = 59$	✓ 1
Part (c)
M1 $N_2 = 59 + 0.2(59)\left(1-\frac{59}{500}\right)$	✓ 1
A1 69.4	✓ 1

Common error: Euler's method: new value = old value + step × (rate at the old value). Recalculate the rate at each new point.

+ Add to test✓ In your test116IBMI3H-5-04r42-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy4 marks

Calculator

The rate at which patients arrive at a hospital emergency department is recorded every 2 hours during an 8-hour shift. R is the arrival rate in patients per hour, t hours after the start of the shift.

The values recorded are: t = 0, 2, 4, 6, 8 with R = 12, 18, 25, 21, 14 respectively.

(a)

Use the trapezoidal rule with all five values to estimate the integral of R from t = 0 to t = 8.

(3)

(b)

Interpret the value found in (a) in the context of the question.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 trapezoidal rule with interval width 2	✓ 1
A1 ½ × 2 × [12 + 14 + 2(18 + 25 + 21)]	✓ 1
A1 154	✓ 1
Part (b)
R1 estimated total number of patients arriving during the 8-hour shift	✓ 1

Exam tip: In the trapezoidal rule the first and last values are counted once and every middle value is counted twice.

+ Add to test✓ In your test117IBMI3H-5-04r42-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

Calculator

A metal instrument tray at a hospital is removed from a steriliser at 100 °C. Its temperature T °C after t minutes is modelled by T = 22 + 78e^−0.12t, t ≥ 0.

(a)

Find an expression for dT/dt.

(2)

(b)

Find the rate of change of temperature when t = 5, giving your answer to 3 significant figures.

(1)

(c)

Find the time at which the tray reaches 40 °C, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule applied to e^−0.12t	✓ 1
A1 dT/dt = −9.36e^−0.12t	✓ 1
Part (b)
A1 -5.14 °C per minute (the tray is cooling)	✓ 1
Part (c)
M1 22 + 78e^−0.12t = 40, i.e. e^−0.12t = 18/78	✓ 1
A1 t = 12.2 minutes	✓ 1

Common error: The derivative of e⁻⁰.¹²t includes the factor −0.12 from the chain rule, which is easy to forget.

+ Add to test✓ In your test118IBMI3H-5-04r42-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

Calculator

A pump delivers a drug to a patient at a rate of r(t) = 8te^−0.5t² millilitres per minute for 0 ≤ t ≤ 3, where t is in minutes. The volume delivered in the first 3 minutes is ∫₀³ 8te^−0.5t² dt.

(a)

Use the substitution u = 0.5t² to show that this volume equals 8∫₀^4.5 e^−u du.

(3)

(b)

Hence find the volume delivered, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du = t dt	✓ 1
A1 limits changed: t = 0 gives u = 0 and t = 3 gives u = 4.5	✓ 1
A1 integrand becomes 8e^−u (AG)	✓ 1
Part (b)
M1 8[−e^−u]₀^4.5 = 8(1 − e^−4.5)	✓ 1
A1 7.91 mL	✓ 1

Exam tip: When you substitute in a definite integral, change the limits to the new variable so that you do not need to substitute back.

+ Add to test✓ In your test119IBMI3H-5-06r43-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy4 marks

During an outbreak, the number of infected people I at time t days satisfies the differential equation dI/dt = 0.3I − 0.0004I². Initially I = 50.

(a)

Find the rate at which the number of infected people is increasing at t = 0.

(1)

(b)

Use Euler's method with a step size of 1 day to estimate I when t = 2. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 14 people per day	✓ 1
Part (b)
M1 attempt to use I_n+1 = I_n + 1 × (0.3I_n − 0.0004I_n²)	✓ 1
A1 I₁ = 64	✓ 1
A1 I₂ = 81.6	✓ 1

Common error: In Euler's method the gradient must be recalculated at each new point using the latest value of I, not reused from the first step.

+ Add to test✓ In your test120IBMI3H-5-06r43-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The energy stored in a solar battery, Q kWh, at time t hours satisfies dQ/dt = 8 − 0.2Q. A slope field is drawn for this equation.

(a)

Calculate the gradient of the slope-field segment at the point (t, Q) = (2, 10).

(1)

(b)

Explain why the segment at (5, 10) has the same gradient as the one at (2, 10).

(1)

(c)

Find the value of Q at which the slope-field segments are horizontal.

(2)

(d)

Describe the behaviour of the solution curve that starts at Q = 10.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 6	✓ 1
Part (b)
R1 the expression for dQ/dt does not contain t, so the gradient depends only on Q	✓ 1
Part (c)
M1 set 8 − 0.2Q = 0	✓ 1
A1 Q = 40 kWh	✓ 1
Part (d)
A1 Q increases with decreasing gradient, approaching 40 kWh without crossing it	✓ 1

Exam tip: When the differential equation has no t on the right-hand side, all segments along a horizontal line of the slope field are parallel.

+ Add to test✓ In your test121IBMI3H-5-06r43-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

A wind turbine's power output was recorded every 2 hours.
Time t (hours): 0, 2, 4, 6, 8
Power P (kW): 120, 180, 260, 240, 150

(a)

Use the trapezoidal rule with 4 intervals to estimate the energy produced in the 8 hours, in kWh.

(3)

(b)

Hence find the mean power over the 8 hours, to 3 significant figures.

(1)

(c)

Suggest how the estimate of the energy could be made more accurate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the trapezoidal rule with width 2	✓ 1
A1 (2/2)[120 + 150 + 2(180 + 260 + 240)]	✓ 1
A1 1630 kWh	✓ 1
Part (b)
A1 1630/8 = 204 kW	✓ 1
Part (c)
R1 record the power at shorter time intervals (more, narrower trapezoids)	✓ 1

Common error: The width of each trapezoid is 2 hours, so the factor in front of the bracket is half of 2, which equals 1.

+ Add to test✓ In your test122IBMI3H-5-06r43-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

A trolley in a retail warehouse moves along a straight rail for 4 seconds with velocity v = 5 sin(πt/4) m s⁻¹, for 0 ≤ t ≤ 4.

(a)

Write down the time at which the speed is greatest.

(1)

(b)

Find the acceleration at t = 1, to 3 significant figures.

(2)

(c)

Find the distance travelled by the trolley in the 4 seconds, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 t = 2 s	✓ 1
Part (b)
M1 attempt to differentiate using the chain rule: (5π/4) cos(πt/4)	✓ 1
A1 2.78 m s⁻²	✓ 1
Part (c)
M1 attempt to integrate v from 0 to 4	✓ 1
A1 antiderivative −(20/π) cos(πt/4)	✓ 1
A1 12.7 m	✓ 1

Common error: Acceleration is the derivative of velocity, but distance is the integral of speed; do not swap these.

+ Add to test✓ In your test123IBMI3H-5-05r45-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

The partial pressure of oxygen in the air at altitude h metres on a mountain is modelled by p(h) = 21.2e^−0.000118h kPa, for 0 ≤ h ≤ 9000.

(a)

Find dp/dh and hence calculate its value when h = 5000. Give your answer to 3 significant figures.

(3)

(b)

A climber notices that the rate of change of pressure with altitude is −0.0010 kPa per metre. Find the altitude at which this occurs, giving your answer to the nearest metre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate using the chain rule (factor of −0.000118 appears)	✓ 1
A1 dp/dh = −0.0025016e^−0.000118h (accept −21.2×0.000118e^−0.000118h)	✓ 1
A1 −0.00139 kPa per metre	✓ 1
Part (b)
M1 sets 0.0025016e^−0.000118h = 0.0010 and attempts to solve using logarithms (e.g. h = ln(2.5016)/0.000118)	✓ 1
A1 h = 7771 m	✓ 1

Common error: Differentiating e⁻⁰.⁰⁰⁰¹¹⁸h but forgetting to multiply by the constant −0.000118 from the chain rule.

+ Add to test✓ In your test124IBMI3H-5-05r45-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The download speed S megabits per second received by a phone at a distance d metres from a wireless router is modelled by S(d) = 200/(1 + 0.05d²), for d ≥ 0.

(a)

Use the quotient rule to find an expression for S′(d).

(2)

(b)

Find S′(8) to 3 significant figures and interpret your answer in context.

(2)

(c)

Find the equation of the tangent to the graph of S at the point where d = 8. Give the coefficients to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at quotient rule (or chain rule on 200(1 + 0.05d²)⁻¹)	✓ 1
A1 S′(d) = −20d/(1 + 0.05d²)²	✓ 1
Part (b)
A1 S′(8) = −9.07	✓ 1
R1 at 8 m from the router the download speed is falling at about 9.07 megabits per second for each extra metre of distance	✓ 1
Part (c)
M1 S(8) = 47.6 (47.619…) and substitutes into S − S(8) = S′(8)(d − 8), using unrounded values	✓ 1
A1 S = −9.07d + 120	✓ 1

Common error: Writing the numerator of the quotient rule as u′v + uv′ or reversing the order of the subtraction, which changes the sign of the derivative.

+ Add to test✓ In your test125IBMI3H-5-05r45-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

A mountaineer’s oxygen cylinder has pressure P bar after t minutes of use. The pressure satisfies the differential equation dP/dt = −0.08P with P = 300 when t = 0.

(a)

Use Euler’s method with a step length of 0.5 to estimate P when t = 1.

(3)

(b)

The exact solution is P = 300e^−0.08t. Find the percentage error in your estimate from part (a), giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses P_n+1 = P_n + 0.5 × (−0.08P_n)	✓ 1
A1 P(0.5) ≈ 288	✓ 1
A1 P ≈ 276.48 when t = 1	✓ 1
Part (b)
M1 percentage error = \|exact − estimate\|/\|exact\| × 100, with exact value 276.93…	✓ 1
A1 0.164 %	✓ 1

Common error: Using the original value 300 in every step instead of updating P after each step of Euler’s method.

+ Add to test✓ In your test126IBMI3H-5-05r45-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

The number of users B, in thousands, connected to a new telecoms network after t months satisfies the differential equation dB/dt = 0.3B(1 − B/80). A slope field is drawn for this equation.

(a)

Calculate the gradient of the slope field at the point (t, B) = (4, 20).

(2)

(b)

State the values of B along which every line segment of the slope field is horizontal.

(1)

(c)

Find the value of B at which the line segments of the slope field are steepest, and find the gradient there.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes B = 20 into the equation	✓ 1
A1 gradient = 0.3 × 20 × (1 − 0.25) = 4.5	✓ 1
Part (b)
A1 B = 0 and B = 80	✓ 1
Part (c)
M1 expands the right-hand side to 0.3B − 0.00375B² and sets its derivative with respect to B equal to zero	✓ 1
A1 B = 40	✓ 1
A1 maximum gradient = 6 thousand users per month	✓ 1

Exam tip: In an autonomous equation the gradient depends only on B, so the line segments along any horizontal line of the slope field are parallel.

+ Add to test✓ In your test127IBMI3H-5-07r45-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

Calculator

Ash from a volcanic vent drifts over a monitoring station. The concentration of ash, $C$ micrograms per cubic metre, is modelled by $C(t)=12te^{-0.4t}$, where $t$ is the time in hours after the eruption, $t\ge 0$.

(a)

Find $C'(t)$.

(3)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

Find the greatest concentration. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule, with the chain rule on $e^{-0.4t}$	✓ 1
A1 $12e^{-0.4t}$ seen as one part of the derivative	✓ 1
A1 $C'(t)=12e^{-0.4t}-4.8te^{-0.4t}$ (accept $12e^{-0.4t}(1-0.4t)$)	✓ 1
Part (b)
M1 set $C'(t)=0$ (ECF from (a))	✓ 1
A1 $t=2.5$ hours (from $1-0.4t=0$)	✓ 1
Part (c)
M1 substitute $t=2.5$ into $C(t)$, i.e. $30e^{-1}$	✓ 1
A1 $11.0$ micrograms per cubic metre (11.036...)	✓ 1

Exam tip: After finding a stationary point, substitute back into the original function, not into the derivative, to get the maximum value.

+ Add to test✓ In your test128IBMI3H-5-07r45-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy7 marks

Calculator

A harbour tender approaches a quay. Its speed in metres per second is $v(t)=0.03t^2-0.9t+6$ for $0\le t\le 10$, where $t$ is in seconds. The tender stops at the quay when $t=10$.

(a)

Find the acceleration of the tender when $t=5$.

(3)

(b)

Find the distance from the quay at $t=0$.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiate $v(t)$	✓ 1
A1 $a(t)=0.06t-0.9$	✓ 1
A1 $a(5)=-0.6$ m s$^{-2}$	✓ 1
Part (b)
M1 distance $=\int_0^{10}v\,dt$	✓ 1
A1 $0.01t^3-0.45t^2+6t$	✓ 1
M1 substitute limits $10$ and $0$	✓ 1
A1 $25$ m	✓ 1

Exam tip: Distance travelled is the integral of speed, and acceleration is the derivative of speed. Check the sign of the acceleration against the story: here the tender is slowing down.

+ Add to test✓ In your test129IBMI3H-5-07r45-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

Calculator

Lava crosses a road and the rate of flow, $r$ cubic metres per minute, is measured every two minutes.
At $t=0, 2, 4, 6, 8$ the rates are $12.0,\ 15.5,\ 18.0,\ 16.0,\ 11.5$ m$^3$/min respectively.

(a)

Use the trapezoidal rule with four intervals to estimate the volume of lava that crossed the road in the 8 minutes.

(3)

(b)

Clearing the lava costs €3.40 per cubic metre. Find the cost of clearing the estimated volume.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of $\frac{h}{2}\left[\text{first}+\text{last}+2(\text{rest})\right]$ with $h=2$	✓ 1
A1 $\frac{2}{2}\left[12.0+11.5+2(15.5+18.0+16.0)\right]$	✓ 1
A1 $122.5$ m$^3$	✓ 1
Part (b)
M1 $122.5\times3.40$	✓ 1
A1 €416.50	✓ 1

Common error: In the trapezoidal rule the first and last values are used once and every middle value is doubled.

+ Add to test✓ In your test130IBMI3H-5-06r46-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

Calculator

The weekly sales of a fashion item t weeks after launch are modelled by $S(t) = 120te^{-0.25t}$ items per week, for $t \ge 0$.

(a)

Find $S'(t)$.

(3)

(b)

Find the time at which sales per week are greatest.

(2)

(c)

Find the greatest number of items sold in one week, correct to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at the product rule	✓ 1
A1 $S'(t) = 120e^{-0.25t} - 30te^{-0.25t}$	✓ 1
A1 equivalent simplified form $30e^{-0.25t}(4 - t)$	✓ 1
Part (b)
M1 set $S'(t) = 0$	✓ 1
A1 $t = 4$ weeks	✓ 1
Part (c)
A1 $S(4) = 480e^{-1} = 177$ items	✓ 1

Exam tip: A stationary point is where the derivative is zero; confirm it is a maximum by checking the sign of the derivative either side or using the context.

+ Add to test✓ In your test131IBMI3H-5-06r46-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy5 marks

Calculator

The horizontal position of the tip of a robot arm is $x(t) = 1.2\sin(3t)$ metres, where $t$ is in seconds. Work in radians.

(a)

Find an expression for the velocity $v = \frac{dx}{dt}$.

(2)

(b)

Find the velocity at $t = 0.5$ s, correct to 3 significant figures.

(1)

(c)

Find the first time $t > 0$ at which the arm tip is momentarily at rest, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule with factor 3	✓ 1
A1 $v = 3.6\cos(3t)$	✓ 1
Part (b)
A1 $0.255$ m s^-1	✓ 1
Part (c)
M1 solve $\cos(3t) = 0$, so $3t = \frac{\pi}{2}$	✓ 1
A1 $t = 0.524$ s	✓ 1

Exam tip: Set your calculator to radian mode for any trigonometric function of time or distance, unless degrees are stated.

+ Add to test✓ In your test132IBMI3H-5-06r46-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy6 marks

Calculator

After a feed is added to a fish tank, the concentration of a nutrient is $c(x) = \frac{50x}{x^2 + 4}$ mg per litre, where $x$ is the time in hours, $x \ge 0$.

(a)

Use the quotient rule to show that $c'(x) = \frac{50(4 - x^2)}{(x^2+4)^2}$.

(3)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

State the greatest concentration.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule with $u = 50x$, $v = x^2 + 4$	✓ 1
A1 $\frac{50(x^2+4) - 50x(2x)}{(x^2+4)^2}$	✓ 1
A1 simplifies to $\frac{50(4 - x^2)}{(x^2+4)^2}$	✓ 1
Part (b)
M1 solve $c'(x) = 0$	✓ 1
A1 $x = 2$ hours	✓ 1
Part (c)
A1 12.5 mg per litre	✓ 1

Exam tip: In a 'show that' question every line of the working must be written; finish with the exact expression you were asked to reach.

+ Add to test✓ In your test133IBMI3H-5-06r46-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Easy4 marks

Calculator

The rate at which a trawler brings fish aboard is recorded every 2 hours.
Time (h): 0, 2, 4, 6, 8, 10
Rate (kg per hour): 0, 14, 26, 30, 22, 9

(a)

Use the trapezoidal rule with all the data to estimate the total mass of fish brought aboard in the 10 hours.

(2)

(b)

Hence find the mean rate over the 10 hours, in kg per hour, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{2}{2}\left[0 + 9 + 2(14 + 26 + 30 + 22)\right]$	✓ 1
A1 193 kg	✓ 1
Part (b)
M1 $\frac{193}{10}$ (ECF from (a))	✓ 1
A1 19.3 kg per hour	✓ 1

Exam tip: The trapezoidal rule gives an approximation to the area under the rate graph, which here represents total mass.

+ Add to test✓ In your test134IBMI-034-02-r1-05Integration and the trapezoidal rule Paper 1Medium7 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 test135IBMI-034-00-r1-05Gradients, tangents and rates of change Paper 1Medium7 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 test136IBMI-034-05-r2-04Gradients, tangents and rates of change Paper 1Medium6 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 test137IBMI-034-07-r2-05Integration and the trapezoidal rule Paper 1Medium7 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 test138IBMI-034-00-r2-05Gradients, tangents and rates of change Paper 1Medium6 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 test139IBMI-034-08-r2-04Differential equations and slope fields Paper 1Medium8 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 test140IBMI-034-03-r2-04Differential equations and slope fields Paper 1Medium6 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 test141IBMI-034-04-r1-05Euler's method and coupled systems Paper 1Medium7 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 test142IBMI-034-05-r1-05Gradients, tangents and rates of change Paper 1Medium6 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 test143IBMI-034-01-r1-04Optimisation in context Paper 1Medium8 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 test144IBMI-034-02-r2-05Integration and the trapezoidal rule Paper 1Medium5 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 test145IBMI-034-03-r1-04Differential equations and slope fields Paper 1Medium6 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 test146IBMI-034-08-r1-04Differential equations and slope fields Paper 1Medium6 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 test147IBMI-034-04-r2-04Euler's method and coupled systems Paper 1Medium7 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 test148IBMI-034-08-r1-06Differential equations and slope fields Paper 1Medium7 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 test149IBMI3H-5-10-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A researcher models the activity level, A units, of an enzyme t hours after it is added to a solution by
A(t) = $\frac{5\ln t}{t}$, t ≥ 1.

(a)

Show that $\frac{dA}{dt}=\frac{5(1-\ln t)}{t^2}$.

(3)

(b)

Find the exact time at which the activity is greatest.

(2)

(c)

Find the maximum activity. Justify that it is a maximum.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the quotient rule	✓ 1
A1 $\frac{5\cdot\frac{1}{t}\cdot t-5\ln t\cdot 1}{t^2}$	✓ 1
A1 simplify to reach the given result	✓ 1
Part (b)
M1 set $\frac{dA}{dt}=0$, giving ln t = 1	✓ 1
A1 t = e hours	✓ 1
Part (c)
A1 $\frac{5}{e}=1.84$ units	✓ 1
R1 the derivative changes from positive to negative either side of t = e	✓ 1

Exam tip: In the quotient rule, subtract in the order (u′v − uv′), and keep the denominator squared.

+ Add to test✓ In your test150IBMI3H-5-10-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Water flows into a storage tank at a rate of Q(t) = 3t$\sqrt{t^2+4}$ cubic metres per minute, where t is the time in minutes, 0 ≤ t ≤ 3.

(a)

Use the substitution u = t² + 4 to show that $\int 3t\sqrt{t^2+4}\,dt=(t^2+4)^{3/2}+c$.

(3)

(b)

Hence find the volume of water that enters the tank in the first 3 minutes.

(2)

(c)

Find the mean rate of flow during these 3 minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $du=2t\,dt$ and rewrite the integral in terms of u	✓ 1
A1 $\frac{3}{2}\int u^{1/2}\,du$	✓ 1
A1 $u^{3/2}+c$ and substitute back	✓ 1
Part (b)
M1 $[(t^2+4)^{3/2}]_0^3=13^{3/2}-8$	✓ 1
A1 38.9 m³	✓ 1
Part (c)
M1 38.9 ÷ 3 (ECF)	✓ 1
A1 13.0 m³ per minute	✓ 1

Common error: After substituting, either change the limits to u-values or return to t before using the original limits. Do not mix them.

+ Add to test✓ In your test151IBMI3H-5-10-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A survey drone starts from rest at a point O and flies in a straight line. Its acceleration is a(t) = 1.5 − 0.3t m s⁻² for 0 ≤ t ≤ 10, where t is in seconds.

(a)

Find an expression for the velocity of the drone at time t.

(3)

(b)

Show that the drone is at rest when t = 10.

(1)

(c)

Find the distance of the drone from O when t = 10.

(2)

(d)

Find the average velocity over the 10 seconds.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate a	✓ 1
A1 1.5t − 0.15t² + c	✓ 1
A1 c = 0 since v(0) = 0	✓ 1
Part (b)
R1 v(10) = 15 − 15 = 0	✓ 1
Part (c)
M1 integrate v from 0 to 10	✓ 1
A1 25 m	✓ 1
Part (d)
A1 2.5 m s⁻¹	✓ 1

Common error: Integrating acceleration gives velocity plus a constant; use the starting condition to find that constant before integrating again.

+ Add to test✓ In your test152IBMI3H-5-10-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The surface of a ramp is modelled by the curve y = 0.5x² + 1, x ≥ 0, with x and y in metres. A straight cable runs from the point (0, −1) and just touches the ramp at the point where x = a.

(a)

Write down $\frac{dy}{dx}$.

(1)

(b)

Show that the equation of the cable is y = ax − 0.5a² + 1.

(2)

(c)

Find the value of a.

(3)

(d)

State the gradient of the cable.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 x	✓ 1
Part (b)
M1 use y − (0.5a² + 1) = a(x − a)	✓ 1
A1 rearrange to reach the given result	✓ 1
Part (c)
M1 substitute (0, −1) into the equation of the cable	✓ 1
A1 a² = 4	✓ 1
A1 a = 2 (since x ≥ 0)	✓ 1
Part (d)
A1 2	✓ 1

Exam tip: For a tangent through an external point, let the point of contact be a general x = a, write the tangent in terms of a, then force it through the given point.

+ Add to test✓ In your test153IBMI3H-5-10b-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

The current in a device is modelled by i(t) = 4te^−0.2t² amperes, where t is the time in seconds, t ≥ 0.

(a)

Use the substitution u = 0.2t² to find ∫ 4te^−0.2t² dt.

(3)

(b)

Hence find the total charge, in coulombs, that passes through the device in the first 3 seconds.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du = 0.4t dt, so 4t dt = 10 du	✓ 1
A1 ∫ 10e^−u du	✓ 1
A1 −10e^−0.2t² + c	✓ 1
Part (b)
M1 substitute the limits 0 and 3	✓ 1
A1 10(1 − e^−1.8)	✓ 1
A1 8.35 C	✓ 1

Common error: When using a substitution on a definite integral, either change the limits to u-values or return to t before substituting; never mix the two.

+ Add to test✓ In your test154IBMI3H-5-10b-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

A decorative glass vase is made by rotating the region bounded by the curve y = 2 + sin(0.5x), the x-axis, the y-axis and the line x = 10 through 360° about the x-axis. Lengths are in centimetres, and the vase is solid-walled with negligible thickness.

(a)

Write down an integral for the volume of the vase.

(3)

(b)

Find the volume of water, in mL, that the vase holds when full.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume = π∫y² dx	✓ 1
A1 integrand (2 + sin(0.5x))²	✓ 1
A1 limits 0 and 10: V = π∫₀¹⁰ (2 + sin(0.5x))² dx	✓ 1
Part (b)
M1 evaluate the integral using a graphing calculator, giving 160 cm³	✓ 1
A1 160 mL (since 1 cm³ = 1 mL)	✓ 1

Common error: The radius is squared inside the integral: π∫y² dx is not π(∫y dx)².

+ Add to test✓ In your test155IBMI3H-5-10b-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A delivery drone moves along a straight track with velocity v(t) = 6te^−0.5t m s⁻¹, for 0 ≤ t ≤ 10, where t is the time in seconds. The drone starts at the origin.

(a)

Find the acceleration of the drone at t = 1.

(3)

(b)

Find the maximum speed of the drone.

(2)

(c)

Find the distance travelled in the first 5 seconds.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at the product rule	✓ 1
A1 a = 6e^−0.5t − 3te^−0.5t	✓ 1
A1 acceleration at t = 1 is 1.82 m s⁻²	✓ 1
Part (b)
M1 set a = 0, so t = 2	✓ 1
A1 maximum speed 4.41 m s⁻¹	✓ 1
Part (c)
M1 ∫₀⁵ v dt (velocity is never negative here)	✓ 1
A1 17.1 m	✓ 1

Exam tip: Maximum speed occurs where the acceleration is zero; distance is the area under the velocity curve (use the modulus if the velocity changes sign).

+ Add to test✓ In your test156IBMI3H-5-10b-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

A games studio models the number of active players P (in thousands) of a new title t weeks after launch by dP/dt = 0.25P(4 − P). A slope field is to be drawn for this equation in the (t, P) plane.

(a)

Calculate the gradient of the slope field at the points (2, 1) and (2, 5).

(2)

(b)

Find the values of P for which every segment of the slope field is horizontal.

(2)

(c)

A solution curve passes through (0, 1). Describe its shape for t > 0 and its long-term behaviour.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute P = 1 and P = 5 into 0.25P(4 − P)	✓ 1
A1 gradients 0.75 and −1.25	✓ 1
Part (b)
M1 set dP/dt = 0	✓ 1
A1 P = 0 and P = 4	✓ 1
Part (c)
A1 the curve is increasing, since the gradient is positive for 0 < P < 4	✓ 1
A1 the gradient shrinks towards zero as P nears 4, so the number of players levels off near 4 thousand and never exceeds it	✓ 1

Exam tip: The equation contains no t, so the segments are identical along every horizontal line P = constant.

+ Add to test✓ In your test157IBMI3H-5-10b-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A rainwater tank is an open-topped rectangular box with a square base of side x metres and height h metres. Its volume must be 32 m³. The base costs £60 per m² to build and the four vertical sides cost £40 per m². Let £C be the total cost of the materials.

(a)

Show that C = 60x² + 5120/x.

(2)

(b)

Find dC/dx.

(1)

(c)

Find the value of x that minimises the cost.

(2)

(d)

Find the minimum cost.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 h = 32/x² and C = 60x² + 40 × 4xh	✓ 1
A1 160x × 32/x² = 5120/x, giving the printed result	✓ 1
Part (b)
A1 120x − 5120/x²	✓ 1
Part (c)
M1 set dC/dx = 0, so x³ = 5120/120	✓ 1
A1 x = 3.49 m	✓ 1
Part (d)
M1 substitute x into C	✓ 1
A1 £2197.85	✓ 1

Exam tip: Eliminate one variable using the fixed volume before differentiating; the cost must be a function of a single variable.

+ Add to test✓ In your test158IBMI3H-5-10b-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

A cyclist's speed is recorded every 4 seconds while accelerating from rest, as shown.

t (s): 0, 4, 8, 12, 16, 20
v (m s⁻¹): 0, 3.8, 6.2, 7.5, 8.0, 8.2

(a)

Use the trapezoidal rule to estimate the distance travelled in the 20 seconds.

(3)

(b)

Hence find the mean speed over the 20 seconds.

(1)

(c)

The speed curve is increasing but its gradient is decreasing throughout. State whether the true distance is greater or less than your estimate in part (a), giving a reason.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 trapezoidal rule with h = 4	✓ 1
A1 (4/2)[0 + 8.2 + 2(3.8 + 6.2 + 7.5 + 8.0)]	✓ 1
A1 118.4 m	✓ 1
Part (b)
A1 118.4 ÷ 20 = 5.92 m s⁻¹	✓ 1
Part (c)
R1 greater, because the curve is concave down so each chord lies below the curve and the trapezia under-estimate the area	✓ 1

Common error: A trapezoidal estimate is an under-estimate for a concave-down curve and an over-estimate for a concave-up curve; link the sign of the second derivative to your answer.

+ Add to test✓ In your test159IBMI3H-5-03r8-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

During the maturing of a sauce, the concentration of a flavour compound is modelled by C(t) = 12te^−0.4t mg per litre, where t is the time in hours after cooking, t ≥ 0.

(a)

Use the product rule to find C′(t).

(3)

(b)

Hence show that the concentration is greatest when t = 2.5.

(2)

(c)

Find the greatest concentration, in mg per litre, to three significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule applied to 12t and e^−0.4t	✓ 1
A1 derivative of e^−0.4t is −0.4e^−0.4t	✓ 1
A1 C′(t) = 12e^−0.4t(1 − 0.4t)	✓ 1
Part (b)
M1 set C′(t) = 0, noting e^−0.4t ≠ 0	✓ 1
R1 1 − 0.4t = 0 so t = 2.5; C′ changes from positive to negative either side of 2.5, so a maximum	✓ 1
Part (c)
A1 11.0	✓ 1

Exam tip: After setting a derivative to zero, factorise first; an exponential factor is never zero, which removes it from the equation.

+ Add to test✓ In your test160IBMI3H-5-03r8-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The amplitude of a radio signal at a distance x kilometres from a mast is modelled by A(x) = cos(0.4x)/(1 + x), for 0 ≤ x ≤ 9, where 0.4x is in radians.

(a)

Find A′(x).

(3)

(b)

Find the value of x, to three significant figures, at which the amplitude has a stationary point.

(2)

(c)

Find the amplitude at this stationary point, to three significant figures. Use the unrounded value of x from part (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule with numerator cos(0.4x) and denominator 1 + x	✓ 1
A1 derivative of numerator −0.4sin(0.4x)	✓ 1
A1 A′(x) = [−0.4(1 + x)sin(0.4x) − cos(0.4x)]/(1 + x)²	✓ 1
Part (b)
M1 set the numerator of A′(x) equal to zero	✓ 1
A1 x = 7.11	✓ 1
Part (c)
M1 substitute x into A(x)	✓ 1
A1 A = −0.118 (ECF from (b))	✓ 1

Common error: Calculator trigonometry must be in radian mode here because the angle 0.4x is given in radians.

+ Add to test✓ In your test161IBMI3H-5-03r8-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

A mould for a jelly is made by rotating the region bounded by the curve y = 2 + sin(x/2), the x-axis, the y-axis and the line x = 6 through 360° about the x-axis. Lengths are in centimetres and x/2 is in radians.

(a)

Write down an integral for the volume of the jelly.

(2)

(b)

Find the volume, in cm³, to three significant figures.

(1)

(c)

The jelly has density 1.05 g cm⁻³. Find its mass to the nearest gram, using the unrounded volume.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume = π∫y² dx	✓ 1
A1 V = π∫₀⁶ (2 + sin(x/2))² dx	✓ 1
Part (b)
A1 135 cm³	✓ 1
Part (c)
M1 mass = 1.05 × volume	✓ 1
A1 142 g	✓ 1

Common error: Square the whole expression 2 + sin(x/2) before integrating; squaring each term separately loses the middle term 4sin(x/2).

+ Add to test✓ In your test162IBMI3H-5-03r8-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

A telecoms engineer models the data buffer level y gigabits (Gb) at time t seconds by dy/dt = 0.5(t − y). A slope field is drawn for this equation.

(a)

Calculate the gradient of the slope field at each of the points (t, y) = (0, 3), (2, 0) and (4, 4).

(3)

(b)

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

(1)

(c)

A solution curve passes through (2, 0). Use the tangent at this point to estimate y when t = 2.4.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 at (0, 3): −1.5	✓ 1
A1 at (2, 0): 1	✓ 1
A1 at (4, 4): 0	✓ 1
Part (b)
A1 y = t	✓ 1
Part (c)
M1 y ≈ 0 + (gradient) × 0.4 using gradient 1	✓ 1
A1 y ≈ 0.4 Gb	✓ 1

Exam tip: Horizontal segments occur where the right-hand side of the differential equation equals zero, so set it to zero to find that isocline.

+ Add to test✓ In your test163IBMI3H-5-03r8-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

A rope hanging from a ledge swings so that its horizontal displacement x metres from the vertical at time t seconds satisfies d²x/dt² = −4x − 0.5(dx/dt). At t = 0, x = 1 and dx/dt = 0.

(a)

By letting y = dx/dt, write the equation as a pair of first-order differential equations.

(2)

(b)

Use Euler's method with step length 0.2 seconds to estimate x when t = 0.4.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 dx/dt = y	✓ 1
A1 dy/dt = −4x − 0.5y	✓ 1
Part (b)
M1 first step: x(0.2) = 1 + 0.2(0) = 1 and y(0.2) = 0 + 0.2(−4) = −0.8	✓ 1
M1 second step uses dy/dt = −4 × 1 − 0.5 × (−0.8) = −3.6	✓ 1
A1 x(0.4) = 1 + 0.2(−0.8) = 0.84 m	✓ 1

Exam tip: Introduce the new variable y = dx/dt explicitly; a second-order equation needs two starting values, one for x and one for y.

+ Add to test✓ In your test164IBMI3H-5-03r8-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A pie is removed from an oven at 90 °C into a kitchen at 20 °C. Its temperature T °C after t minutes satisfies dT/dt = −k(T − 20), where k is a positive constant.

(a)

Solve the differential equation to show that T = 20 + 70e^−kt.

(4)

(b)

After 10 minutes the temperature is 60 °C. Find the value of k, to three significant figures.

(2)

(c)

Find the time taken for the pie to reach 30 °C, to the nearest tenth of a minute. Use the unrounded value of k.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 separate variables: ∫1/(T − 20) dT = ∫−k dt	✓ 1
A1 ln\|T − 20\| = −kt + c	✓ 1
M1 T − 20 = Ae^−kt	✓ 1
A1 use T = 90 at t = 0 to get A = 70	✓ 1
Part (b)
M1 60 = 20 + 70e^−10k	✓ 1
A1 k = 0.0560	✓ 1
Part (c)
M1 30 = 20 + 70e^−kt, so e^−kt = 1/7	✓ 1
A1 t = 34.8 minutes	✓ 1

Common error: Keep the modulus or absolute value when integrating 1/(T − 20); drop it only after deciding that T − 20 is positive here.

+ Add to test✓ In your test165IBMI3H-5-03r8-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

A delivery drone flies along a straight line towards a mountain hut. Its velocity, in m s⁻¹, is v(t) = 5sin(0.4t) + 1 for 0 ≤ t ≤ 8, where t is in seconds and angles are in radians.

(a)

Find the acceleration of the drone when t = 3. Give your answer to three significant figures.

(2)

(b)

Find the distance travelled by the drone in the first 5 seconds. Give your answer to three significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a = v′(t) = 2cos(0.4t)	✓ 1
A1 a(3) = 0.725 m s⁻²	✓ 1
Part (b)
R1 v > 0 for 0 ≤ t ≤ 5, so distance = displacement	✓ 1
M1 ∫₀⁵ (5sin(0.4t) + 1) dt	✓ 1
A1 [−12.5cos(0.4t) + t]₀⁵ = 12.5(1 − cos 2) + 5	✓ 1
A1 22.7 m	✓ 1

Exam tip: Distance equals the integral of speed; check that the velocity does not change sign on the interval, or split the integral where it does.

+ Add to test✓ In your test166IBMI3H-5-04r8-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

An architect designs a rectangular window whose base lies along the ground and whose top two corners touch an arch with equation y = 6e^−0.1x², where x and y are in metres and the window is symmetric about the y-axis, with x > 0 at its right-hand corners.

(a)

Show that the area of the window is A = 12xe^−0.1x².

(1)

(b)

Find dA/dx.

(3)

(c)

Find the value of x that gives the largest window area.

(2)

(d)

Find the largest area, in m², to three significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R1 width = 2x and height = 6e^−0.1x², so A = 2x × 6e^−0.1x²	✓ 1
Part (b)
M1 product rule	✓ 1
A1 derivative of e^−0.1x² is −0.2xe^−0.1x²	✓ 1
A1 dA/dx = 12e^−0.1x²(1 − 0.2x²)	✓ 1
Part (c)
M1 set dA/dx = 0 so 1 − 0.2x² = 0	✓ 1
A1 x = √5 ≈ 2.24 m	✓ 1
Part (d)
A1 16.3 m²	✓ 1

Exam tip: State why your stationary point is a maximum (for example, the sign of the gradient changes from positive to negative) when the question says largest.

+ Add to test✓ In your test167IBMI3H-5-04r8-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

A courier company models the cost per drop-off, in pounds, of a delivery route with x drop-off points by C(x) = 40e^0.2x/x, for x > 0.

(a)

Use the quotient rule to show that C′(x) = 40e^0.2x(0.2x − 1)/x².

(3)

(b)

Find the number of drop-off points that minimises the cost per drop-off.

(2)

(c)

Find the minimum cost per drop-off, in pounds, to the nearest penny.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule with numerator 40e^0.2x and denominator x	✓ 1
A1 numerator of derivative 8e^0.2xx − 40e^0.2x	✓ 1
A1 factorise to 40e^0.2x(0.2x − 1) over x²	✓ 1
Part (b)
M1 0.2x − 1 = 0	✓ 1
A1 x = 5	✓ 1
Part (c)
A1 £21.75	✓ 1

Common error: Do not set the whole of C′(x) equal to zero without simplifying; the exponential and the x² can never be zero for x > 0.

+ Add to test✓ In your test168IBMI3H-5-04r8-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

The roof of a circular pavilion is formed by rotating the curve y = 4 − 0.25x², for 0 ≤ x ≤ 4, through 360° about the y-axis. Lengths are in metres, and the floor is the plane y = 0.

(a)

Show that x² = 16 − 4y.

(1)

(b)

Find the volume of air enclosed under the roof, giving your answer to three significant figures.

(3)

(c)

A ventilation system replaces 15% of the air in the pavilion each hour. Find the volume of air replaced in one hour, to three significant figures, using the exact volume from part (b).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 rearranging 0.25x² = 4 − y and multiplying by 4	✓ 1
Part (b)
M1 volume = π∫x² dy between y = 0 and y = 4	✓ 1
M1 π[16y − 2y²]₀⁴	✓ 1
A1 32π ≈ 100.5, so 101 m³	✓ 1
Part (c)
A1 0.15 × 32π = 15.1 m³	✓ 1

Exam tip: When rotating about the y-axis, the limits are y-values; here they run from the floor at y = 0 to the top at y = 4.

+ Add to test✓ In your test169IBMI3H-5-04r8-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

Water drains from a ship's ballast tank so that its depth h metres at time t minutes satisfies dh/dt = −0.4√h. Initially the depth is 10 m.

(a)

Solve the differential equation to show that √h = √10 − 0.2t.

(4)

(b)

Find the time taken for the depth to fall to 3 m. Give your answer in minutes to three significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 separate variables: ∫h^−1/2 dh = ∫−0.4 dt	✓ 1
A1 2√h = −0.4t + c	✓ 1
M1 use h = 10 at t = 0 to get c = 2√10	✓ 1
A1 √h = √10 − 0.2t	✓ 1
Part (b)
M1 √3 = √10 − 0.2t	✓ 1
A1 t = 5(√10 − √3) = 7.15 minutes	✓ 1

Common error: Integrate h⁻¹/² to get 2h¹/²; writing h¹/² forgets the division by the new power 1/2.

+ Add to test✓ In your test170IBMI3H-5-04r8-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

An autonomous trolley in a warehouse moves in a straight line with velocity v(t) = 3t/(1 + t²) m s⁻¹ for t ≥ 0, where t is in seconds.

(a)

Show that the greatest velocity occurs when t = 1 and state its value.

(3)

(b)

Use the substitution u = 1 + t² to find the distance travelled in the first 4 seconds. Give your answer in metres to three significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule: v′ = (3(1 + t²) − 6t²)/(1 + t²)²	✓ 1
A1 v′ = 3(1 − t²)/(1 + t²)² = 0 gives t = 1 (for t ≥ 0)	✓ 1
A1 greatest velocity 1.5 m s⁻¹	✓ 1
Part (b)
M1 du = 2t dt, with limits u = 1 to u = 17	✓ 1
A1 ∫ 1.5/u du	✓ 1
A1 1.5[ln u]₁¹⁷ = 1.5ln 17	✓ 1
A1 4.25 m	✓ 1

Exam tip: If you substitute, change the limits to u-values at the same time; the integral is then evaluated without returning to t.

+ Add to test✓ In your test171IBMI3H-5-04r8-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

The sorting efficiency E (parcels per second) of a conveyor system with belt speed x m s⁻¹ is modelled by E(x) = 5ln(2x)/x, for x > 0.

(a)

Find E′(x).

(3)

(b)

Show that the maximum efficiency occurs when x = e/2.

(2)

(c)

Find the maximum efficiency, to three significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule	✓ 1
A1 derivative of 5ln(2x) is 5/x	✓ 1
A1 E′(x) = 5(1 − ln(2x))/x²	✓ 1
Part (b)
M1 set E′(x) = 0 so ln(2x) = 1	✓ 1
R1 2x = e so x = e/2; E′ changes from positive to negative	✓ 1
Part (c)
A1 3.68 parcels per second	✓ 1

Exam tip: ln(2x) = 1 means 2x = e; undo a logarithm by exponentiating both sides.

+ Add to test✓ In your test172IBMI3H-5-04r8-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

A rotating louvre panel on a building produces power P(t) = 3cos(0.5t)e^sin(0.5t) kilowatts, where t is the time in hours after sunrise and angles are in radians.

(a)

Use the substitution u = sin(0.5t) to find ∫ P(t) dt.

(3)

(b)

Find the energy produced in the first 3 hours, in kilowatt-hours, to three significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du = 0.5cos(0.5t) dt	✓ 1
M1 integral becomes ∫ 6e^u du	✓ 1
A1 6e^sin(0.5t) + c	✓ 1
Part (b)
M1 6[e^sin(0.5t)]₀³ = 6(e^{sin 1.5} − 1)	✓ 1
A1 10.3 kWh	✓ 1

Common error: Changing from du to dt gives 3cos(0.5t) dt = 6 du; forgetting this factor of 2 halves the answer.

+ Add to test✓ In your test173IBMI3H-5-04r11-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A water company is designing an open-topped cylindrical tank (a base and a curved side, no lid) of radius $r$ m and height $h$ m that must hold exactly $40\ \text{m}^3$. The company wants to use as little sheet metal as possible, so the surface area $A$ must be as small as possible.

(a)

Express $h$ in terms of $r$.

(1)

(b)

Show that $A = \pi r^2 + \frac{80}{r}$.

(2)

(c)

Find the value of $r$ for which $A$ is stationary, to 4 significant figures.

(3)

(d)

Use the second derivative to show that this value of $r$ gives a minimum.

(1)

(e)

Find the minimum surface area, to 4 significant figures. Use the unrounded value of $r$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\pi r^2 h = 40$ so $h = \frac{40}{\pi r^2}$	✓ 1
Part (b)
M1 $A = \pi r^2 + 2\pi r h$	✓ 1
A1 substitute $h$: $2\pi r\cdot\frac{40}{\pi r^2} = \frac{80}{r}$, hence the result	✓ 1
Part (c)
M1 $\frac{dA}{dr} = 2\pi r - \frac{80}{r^2}$	✓ 1
M1 set equal to 0 so $r^3 = \frac{40}{\pi}$	✓ 1
A1 $r = 2.335$ m	✓ 1
Part (d)
R1 $\frac{d^2A}{dr^2} = 2\pi + \frac{160}{r^3} > 0$ for $r > 0$, so a minimum	✓ 1
Part (e)
A1 $A = 51.39$ m²	✓ 1

Common error: Differentiating before eliminating $h$. The area formula must be a function of one variable, $r$, before you differentiate.

+ Add to test✓ In your test174IBMI3H-5-04r11-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

The concentration of a drug in a patient's blood, $t$ hours after an injection, is modelled by $C(t) = \frac{20t}{t^2 + 4}$ mg/L, for $t \ge 0$.

(a)

Use the quotient rule to show that $C'(t) = \frac{20(4 - t^2)}{(t^2 + 4)^2}$.

(3)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

Find the greatest concentration.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $C'(t) = \frac{20(t^2 + 4) - 20t(2t)}{(t^2 + 4)^2}$	✓ 1
M1 expand the numerator: $20t^2 + 80 - 40t^2$	✓ 1
A1 $= \frac{80 - 20t^2}{(t^2+4)^2} = \frac{20(4 - t^2)}{(t^2+4)^2}$	✓ 1
Part (b)
M1 set $C'(t) = 0$ so $t^2 = 4$	✓ 1
A1 $t = 2$ hours (as $t \ge 0$)	✓ 1
Part (c)
A1 $C(2) = \frac{40}{8} = 5$ mg/L	✓ 1

Exam tip: In a 'show that' question every line of working must be written down; the final line must match the given expression exactly.

+ Add to test✓ In your test175IBMI3H-5-04r11-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A drone in a game moves along a straight track. Its velocity at time $t$ seconds is $v = te^{-0.5t}$ m/s, for $t \ge 0$.

(a)

Find the acceleration at $t = 1$, to 3 significant figures.

(3)

(b)

Find the time at which the velocity of the drone is greatest.

(2)

(c)

The drone starts at the origin. Find its displacement from the origin at $t = 5$, in metres, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule: $a = e^{-0.5t} + t(-0.5)e^{-0.5t}$	✓ 1
A1 $a = e^{-0.5t}(1 - 0.5t)$	✓ 1
A1 $a = 0.303$ m/s² at $t = 1$	✓ 1
Part (b)
M1 set $a = 0$ so $1 - 0.5t = 0$	✓ 1
A1 $t = 2$ s	✓ 1
Part (c)
M1 $\int_0^5 te^{-0.5t}\,dt$ (evaluated on a calculator)	✓ 1
A1 2.85 m	✓ 1

Exam tip: Velocity is greatest when the acceleration is zero and changes from positive to negative. Check the sign of $a$ either side of your answer.

+ Add to test✓ In your test176IBMI3H-5-04r11-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

The flow rate through a valve in a treatment plant is $r(t) = \frac{12t}{t^2 + 1}$ thousand litres per hour, where $t$ is the time in hours.

(a)

Use the substitution $u = t^2 + 1$ to show that $\int r(t)\,dt = 6\ln(t^2 + 1) + c$.

(3)

(b)

Find the volume that flows through the valve between $t = 0$ and $t = 3$, in thousand litres, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{du}{dt} = 2t$ so $12t\,dt = 6\,du$	✓ 1
A1 $\int\frac{6}{u}\,du = 6\ln\|u\| + c$	✓ 1
A1 return to $t$ (note $u > 0$): $6\ln(t^2+1) + c$	✓ 1
Part (b)
M1 $6\ln 10 - 6\ln 1$	✓ 1
A1 13.8 thousand litres	✓ 1

Common error: Forgetting to change the integrand completely into $u$. The $12t\,dt$ must be replaced by $6\,du$ so that no $t$ remains.

+ Add to test✓ In your test177IBMI3H-5-04r11-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

The volume of water in a reservoir, in thousand cubic metres, is modelled by $V(t) = 0.5t^3 - 6t^2 + 18t + 40$, where $t$ is the time in days and $0 \le t \le 7$.

(a)

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

(1)

(b)

Find the values of $t$ for which the volume is not changing.

(2)

(c)

Use the second derivative to classify each stationary point.

(2)

(d)

Find the volume of water at the local minimum.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dV}{dt} = 1.5t^2 - 12t + 18$	✓ 1
Part (b)
M1 set $1.5t^2 - 12t + 18 = 0$, i.e. $t^2 - 8t + 12 = 0$	✓ 1
A1 $t = 2$ and $t = 6$	✓ 1
Part (c)
M1 $\frac{d^2V}{dt^2} = 3t - 12$	✓ 1
A1 at $t = 2$ the value is $-6 < 0$ (local maximum); at $t = 6$ it is $6 > 0$ (local minimum)	✓ 1
Part (d)
A1 $V = 0.5(216) - 6(36) + 18(6) + 40 = 40$ thousand cubic metres	✓ 1

Common error: Stating the value of $t$ as the maximum or minimum. The turning point's $t$-value and its $V$-value answer different questions.

+ Add to test✓ In your test178IBMI3H-5-03r14-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The concentration $C$ mg L⁻¹ of a flavour compound released into a sauce, $t$ minutes after it is added, is modelled by $C=\dfrac{15t}{t^2+6}$, for $t\ge 0$.

(a)

Use the quotient rule to show that $\frac{dC}{dt}=\frac{15(6-t^2)}{(t^2+6)^2}$.

(3)

(b)

Find the exact time at which the concentration is greatest.

(2)

(c)

Find the greatest concentration, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts the quotient rule with $u=15t$, $v=t^2+6$	✓ 1
A1 $\frac{dC}{dt}=\frac{15(t^2+6)-15t\times2t}{(t^2+6)^2}$	✓ 1
A1 $=\frac{15t^2+90-30t^2}{(t^2+6)^2}=\frac{15(6-t^2)}{(t^2+6)^2}$	✓ 1
Part (b)
M1 sets $\frac{dC}{dt}=0$, so $6-t^2=0$	✓ 1
A1 $t=\sqrt{6}$ minutes (reject negative root)	✓ 1
Part (c)
M1 $C=\frac{15\sqrt{6}}{6+6}$	✓ 1
A1 3.06 mg L⁻¹	✓ 1

Exam tip: In a quotient-rule question, expand the numerator carefully before factorising; a sign slip in $-15t\times 2t$ is easy to make.

+ Add to test✓ In your test179IBMI3H-5-03r14-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

A relay in a mountain hut transmits data at a rate of $r(t)=6te^{-0.25t^2}$ megabits per second, for $0 \le t \le 4$, where $t$ is in seconds.

(a)

Use the substitution $u=0.25t^2$ to show that the total data transmitted in the first 4 seconds is $12\int_0^4 e^{-u}\,du$.

(3)

(b)

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

(2)

(c)

State the units of your answer to part (b) and what it represents.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{du}{dt}=0.5t$, so $t\,dt=2\,du$	✓ 1
A1 new limits $u=0$ and $u=4$	✓ 1
A1 $\int_0^4 6te^{-0.25t^2}dt=\int_0^4 12e^{-u}du$	✓ 1
Part (b)
M1 $12\left[-e^{-u}\right]_0^4=12(1-e^{-4})$	✓ 1
A1 11.8	✓ 1
Part (c)
A1 megabits; the total data transmitted in the first 4 seconds	✓ 1

Common error: When you substitute, change the limits to $u$-values (or return to $t$ before substituting limits); mixing the two gives a wrong value.

+ Add to test✓ In your test180IBMI3H-5-03r14-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

A chocolate mould is formed by rotating the region bounded by the curve $y=3e^{-0.15x}$, the $x$-axis, the $y$-axis and the line $x=6$ through $360^\circ$ about the $x$-axis. Lengths are in cm.

(a)

Write down an integral for the volume of the mould and show that it can be written as $9\pi\int_0^6 e^{-0.3x}\,dx$.

(2)

(b)

Calculate the volume of the mould, giving your answer to 3 significant figures.

(2)

(c)

The chocolate has density 1.3 g cm⁻³. Using the unrounded volume from part (b), calculate the mass of chocolate needed to fill the mould, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $V=\pi\int_0^6 \left(3e^{-0.15x}\right)^2dx$	✓ 1
A1 $(3e^{-0.15x})^2=9e^{-0.3x}$, so $V=9\pi\int_0^6e^{-0.3x}dx$	✓ 1
Part (b)
M1 $9\pi\left[\frac{e^{-0.3x}}{-0.3}\right]_0^6=30\pi(1-e^{-1.8})$	✓ 1
A1 78.7 cm³	✓ 1
Part (c)
M1 $1.3\times78.668\ldots$ (accept $1.3\times78.7$ for method)	✓ 1
A1 102 g	✓ 1

Exam tip: The radius is the function value $y$, so square the whole expression $3e^{-0.15x}$, including the 3.

+ Add to test✓ In your test181IBMI3H-5-03r14-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A supply trolley runs along an inclined rail beside a mountain hut. Its displacement $s$ metres from a marker, $t$ seconds after release, is $s=0.2t^3-1.5t^2+2.4t$, for $0\le t\le5$.

(a)

Find an expression for the velocity of the trolley and hence the times at which it is instantaneously at rest.

(3)

(b)

Find the acceleration of the trolley when $t=4$.

(2)

(c)

Find the total distance travelled by the trolley from $t=0$ to $t=5$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates: $v=0.6t^2-3t+2.4$	✓ 1
M1 sets $v=0$, so $0.6(t-1)(t-4)=0$	✓ 1
A1 $t=1$ s and $t=4$ s	✓ 1
Part (b)
M1 $a=1.2t-3$	✓ 1
A1 1.8 m s⁻²	✓ 1
Part (c)
M1 finds the displacements at $t=1$, $t=4$ and $t=5$: 1.1, −1.6 and −0.5	✓ 1
M1 adds the magnitudes of the three separate displacements: $1.1+2.7+1.1$	✓ 1
A1 4.9 m	✓ 1

Common error: Total distance is not $s(5)-s(0)$; the trolley changes direction, so find the distance between each pair of turning points and add them.

+ Add to test✓ In your test182IBMI3H-5-03r14-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The yield $Y$ litres of a fermenting mixture, $t$ hours after the start, is modelled by $Y=12\ln(1+2t)-t$, for $t\ge0$.

(a)

Find $\frac{dY}{dt}$.

(2)

(b)

Find the time at which the yield is greatest.

(2)

(c)

Show that this time gives a maximum.

(1)

(d)

Find the greatest yield, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule on $\ln(1+2t)$	✓ 1
A1 $\frac{dY}{dt}=\frac{24}{1+2t}-1$	✓ 1
Part (b)
M1 sets $\frac{24}{1+2t}=1$	✓ 1
A1 $t=11.5$ hours	✓ 1
Part (c)
R1 $\frac{d^2Y}{dt^2}=-\frac{48}{(1+2t)^2}<0$ (accept sign change of $\frac{dY}{dt}$ from positive to negative)	✓ 1
Part (d)
M1 $Y(11.5)=12\ln24-11.5$	✓ 1
A1 26.6 litres	✓ 1

Common error: The derivative of $\ln(1+2t)$ is $\frac{2}{1+2t}$, not $\frac{1}{1+2t}$; the inner derivative must be included.

+ Add to test✓ In your test183IBMI3H-5-05r17-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

An atrium has a glass roof. The light level on the atrium floor, L lux, is modelled by $L(t)=\dfrac{900\sin\left(\frac{\pi t}{12}\right)}{1+0.1t}$, where t is the time in hours after sunrise, $0\le t\le 12$. The argument of the sine is in radians.

(a)

Find an expression for $L'(t)$.

(3)

(b)

Find $L'(3)$ and interpret your answer in context. Give $L'(3)$ to 3 significant figures.

(2)

(c)

Find the time at which the light level is greatest. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the quotient rule (or product rule with $(1+0.1t)^{-1}$)	✓ 1
A1 derivative of numerator $900\cdot\frac{\pi}{12}\cos\left(\frac{\pi t}{12}\right)$ and derivative of denominator $0.1$ seen	✓ 1
A1 $L'(t)=\dfrac{900\cdot\frac{\pi}{12}\cos\left(\frac{\pi t}{12}\right)(1+0.1t)-90\sin\left(\frac{\pi t}{12}\right)}{(1+0.1t)^2}$ (or equivalent)	✓ 1
Part (b)
A1 $L'(3)=90.5$	✓ 1
R1 at 3 hours after sunrise the light level is increasing at 90.5 lux per hour	✓ 1
Part (c)
M1 solve $L'(t)=0$ (GDC) or find the maximum of $L(t)$	✓ 1
A1 $t=5.05$ hours	✓ 1

Common error: Putting the calculator in degree mode. Here the argument is in radians, so the derivative of the sine brings a factor of $\frac{\pi}{12}$ only in radian mode.

+ Add to test✓ In your test184IBMI3H-5-05r17-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A forklift truck moves along a straight aisle in a warehouse. Its velocity, v m s⁻¹, is modelled by $v(t)=0.9te^{-0.3t}$ for $0\le t\le 10$, where t is the time in seconds after it starts from rest.

(a)

Find an expression for the acceleration of the truck.

(2)

(b)

Find the time at which the speed of the truck is greatest. Give your answer to 3 significant figures.

(2)

(c)

Find the distance travelled by the truck in the first 10 seconds. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate using the product rule	✓ 1
A1 $a(t)=0.9e^{-0.3t}(1-0.3t)$ (or equivalent)	✓ 1
Part (b)
M1 set $a(t)=0$ (ECF from (a))	✓ 1
A1 $t=3.33$ s	✓ 1
Part (c)
M1 distance $=\int v\,dt$ (velocity is never negative on this interval)	✓ 1
A1 correct limits 0 and 10 used	✓ 1
A1 8.01 m	✓ 1

Exam tip: Before integrating velocity to find distance, check whether the velocity changes sign. Here $v\ge 0$, so displacement and distance are equal.

+ Add to test✓ In your test185IBMI3H-5-05r17-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The roll angle, θ radians, of a ship in a swell satisfies the second-order differential equation $\dfrac{d^2\theta}{dt^2}+0.2\dfrac{d\theta}{dt}+0.8\theta=0$, where t is the time in seconds. At $t=0$ the ship has $\theta=0.30$ and is momentarily at rest.

(a)

Let $\omega=\dfrac{d\theta}{dt}$. Write the equation as a system of two first-order differential equations.

(2)

(b)

Use Euler's method with step length 0.5 to estimate θ and ω when $t=0.5$, $t=1$ and $t=1.5$. Give θ at $t=1.5$ as your final answer.

(4)

(c)

Using your answer to (b), state whether the ship has passed through the upright position ($\theta=0$) by $t=1.5$. Justify your answer.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{d\theta}{dt}=\omega$	✓ 1
A1 $\frac{d\omega}{dt}=-0.8\theta-0.2\omega$	✓ 1
Part (b)
M1 $\theta_{n+1}=\theta_n+0.5\omega_n$ and $\omega_{n+1}=\omega_n+0.5(-0.8\theta_n-0.2\omega_n)$ used from $(0.30,\,0)$	✓ 1
A1 $t=0.5$: $\theta=0.30$, $\omega=-0.12$	✓ 1
A1 $t=1$: $\theta=0.24$, $\omega=-0.228$	✓ 1
A1 $t=1.5$: $\theta=0.126$ (and $\omega=-0.3012$)	✓ 1
Part (c)
R1 No, since θ is still positive (0.126 > 0), so the ship has not yet crossed θ = 0 (ECF from (b))	✓ 1

Common error: Updating θ with the newly calculated ω from the same step; in Euler's method both new values come from the old pair $(\theta_n,\omega_n)$.

+ Add to test✓ In your test186IBMI3H-5-05r17-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The side profile of a curved canopy is modelled by $y=9\cos\left(\frac{\pi x}{30}\right)\sin^2\left(\frac{\pi x}{30}\right)$ for $0\le x\le 15$, where x and y are in metres and the argument of each trigonometric function is in radians. The canopy is the region between this curve and the x-axis.

(a)

Use the substitution $u=\sin\left(\frac{\pi x}{30}\right)$ to show that the area of this region is exactly $\dfrac{90}{\pi}$ m².

(5)

(b)

The canopy is a solid with this uniform cross-section and a length of 40 m. Find its volume to the nearest m³.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{du}{dx}=\frac{\pi}{30}\cos\left(\frac{\pi x}{30}\right)$	✓ 1
A1 integrand becomes $9\cdot\frac{30}{\pi}u^2\,du$	✓ 1
A1 limits changed: $x=0\to u=0$, $x=15\to u=1$	✓ 1
M1 $\frac{270}{\pi}\left[\frac{u^3}{3}\right]_0^1$	✓ 1
A1 $\frac{90}{\pi}$	✓ 1
Part (b)
M1 volume = area × length $=\frac{90}{\pi}\times 40$	✓ 1
A1 1146 m³	✓ 1

Exam tip: When using a substitution in a definite integral, change the limits to values of u at the start; you then never need to return to x.

+ Add to test✓ In your test187IBMI3H-5-05r17-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A remotely operated vehicle dives from a survey ship. Its depth below the surface is s metres at time t minutes, where $s(t)=0.5t^3-6t^2+18t$ for $0\le t\le 8$.

(a)

Find an expression for the velocity of the vehicle, $v=\dfrac{ds}{dt}$.

(2)

(b)

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

(2)

(c)

Find the total distance travelled by the vehicle in the 8 minutes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 $v=1.5t^2-12t+18$	✓ 1
Part (b)
M1 set $v=0$, giving $t^2-8t+12=0$	✓ 1
A1 $t=2$ and $t=6$	✓ 1
Part (c)
M1 split the motion at the times found in (b) (or integrate $\|v\|$)	✓ 1
A1 depths $s(2)=16$, $s(6)=0$, $s(8)=16$	✓ 1
A1 total distance $=16+16+16=48$ m	✓ 1

Common error: Giving the final depth (16 m) as the distance travelled. Distance counts every change of direction, so it must be found in sections.

+ Add to test✓ In your test188IBMI3H-5-04r20-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

After an athlete takes a supplement, its concentration in the blood is modelled by $C(t)=t^2e^{-0.5t}$ mg per litre, where t is the time in hours after taking it, $t\ge0$.

(a)

Find $C'(t)$.

(2)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

Find the greatest concentration, and justify that it is a maximum.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule with $u=t^2$ and $v=e^{-0.5t}$	✓ 1
A1 $C'(t)=2te^{-0.5t}-0.5t^2e^{-0.5t}=te^{-0.5t}(2-0.5t)$	✓ 1
Part (b)
M1 set $C'(t)=0$ with $t>0$, so $2-0.5t=0$	✓ 1
A1 t = 4 hours	✓ 1
Part (c)
M1 substitute t = 4 into $C(t)$	✓ 1
A1 $16e^{-2}=2.165$ so 2.17 mg per litre	✓ 1
R1 $C'(t)>0$ for $04$, so the concentration is a maximum at t = 4	✓ 1

Exam tip: Factorise the derivative after the product rule; the exponential factor is never zero, so only the remaining bracket and the factor t need to be considered.

+ Add to test✓ In your test189IBMI3H-5-04r20-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

The outline of a rugby ball is modelled by the curve $y=0.07\cos(11.2x)$ for $-0.14\le x\le0.14$, where x and y are in metres. The ball's surface is formed by rotating this curve through 360° about the x-axis.

(a)

Write down an integral that gives the volume of the ball.

(2)

(b)

Use your calculator to find the volume in m³, to 3 significant figures.

(2)

(c)

A ball is acceptable if its volume is between 2.0 and 2.3 litres. Decide whether this ball is acceptable. (1 litre = 0.001 m³)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $V=\pi\int y^2\,dx$	✓ 1
A1 $V=\pi\int_{-0.14}^{0.14}\left(0.07\cos(11.2x)\right)^2dx$	✓ 1
Part (b)
M1 evaluate the definite integral	✓ 1
A1 0.002159 m³, so 2.16×10⁻³ m³	✓ 1
Part (c)
M1 convert: 2.16 litres (volume × 1000)	✓ 1
A1 between 2.0 and 2.3, so the ball is acceptable	✓ 1

Exam tip: Square the whole expression for y, including the constant 0.07, before integrating; enter the full integrand in the calculator with brackets.

+ Add to test✓ In your test190IBMI3H-5-04r20-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Freight wagons move between two rail depots, X and Y. The numbers of wagons at the depots at time t days are x and y, and they satisfy the coupled system $\frac{dx}{dt}=-0.2x+0.1y,\quad\frac{dy}{dt}=0.2x-0.1y.$ Initially there are 80 wagons at X and 40 wagons at Y.

(a)

Use Euler's method with a step length of 0.5 days to estimate the numbers of wagons at each depot when t = 1.

(4)

(b)

Show that the total number of wagons, $x+y$, is constant.

(1)

(c)

Find the long-term number of wagons at depot Y, where both rates of change are zero.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $x_{n+1}=x_n+0.5(-0.2x_n+0.1y_n)$ and $y_{n+1}=y_n+0.5(0.2x_n-0.1y_n)$	✓ 1
A1 $x_1=74$, $y_1=46$	✓ 1
M1 apply the formulas again from $(x_1,y_1)$	✓ 1
A1 $x_2=68.9$, $y_2=51.1$, so about 68.9 wagons at X and 51.1 at Y	✓ 1
Part (b)
A1 $\frac{d(x+y)}{dt}=(-0.2x+0.1y)+(0.2x-0.1y)=0$	✓ 1
Part (c)
M1 $-0.2x+0.1y=0$ gives $y=2x$, and $x+y=120$	✓ 1
A1 $3x=120$, so $x=40$ and $y=80$ wagons	✓ 1

Exam tip: At an equilibrium of a coupled system, set both derivatives to zero; the two equations here are not independent, so the conserved total supplies the second condition.

+ Add to test✓ In your test191IBMI3H-5-04r20-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The vertical displacement x metres of the tip of a springboard from its resting position, t seconds after release, satisfies $\frac{d^2x}{dt^2}=-4x-0.5\frac{dx}{dt}$. The board is released from rest with $x=0.3$.

(a)

By letting $y=\frac{dx}{dt}$, write the equation as a pair of first-order equations.

(2)

(b)

Use Euler's method with a step length of 0.1 to estimate x and y when t = 0.3. Give exact values.

(4)

(c)

Interpret the sign of your value of y at t = 0.3.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{dx}{dt}=y$	✓ 1
A1 $\frac{dy}{dt}=-4x-0.5y$	✓ 1
Part (b)
M1 $x_{n+1}=x_n+0.1y_n$ and $y_{n+1}=y_n+0.1(-4x_n-0.5y_n)$ with $x_0=0.3$, $y_0=0$	✓ 1
A1 $x_1=0.3$, $y_1=-0.12$	✓ 1
A1 $x_2=0.288$, $y_2=-0.234$	✓ 1
A1 $x_3=0.2646$, $y_3=-0.3375$	✓ 1
Part (c)
R1 y is negative, so the tip is moving downwards (back towards the resting position) at 0.3375 m s⁻¹	✓ 1

Common error: Both variables must be updated using the values from the previous step; using the new value of x inside the update for y gives a different (incorrect) sequence.

+ Add to test✓ In your test192IBMI3H-5-04r20-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The membership y (in hundreds) of a running club after x years satisfies $\frac{dy}{dx}=0.2y\left(1-\frac{y}{50}\right)$. A slope field is drawn for this equation for $y\ge0$.

(a)

Find the gradient of the line segments in the slope field at the points (0, 10) and (0, 40).

(2)

(b)

Describe two features of the slope field, and state what they show about the long-term membership.

(2)

(c)

Find the greatest value of the gradient in the slope field for $0\le y\le50$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute y = 10 and y = 40 into the equation	✓ 1
A1 both gradients equal 1.6	✓ 1
Part (b)
A1 line segments are horizontal along y = 0 and y = 50, and are the same along any vertical line (they depend only on y)	✓ 1
A1 for \(0y = 50)	✓ 1
Part (c)
M1 differentiate $0.2y-0.004y^2$ with respect to y and set equal to zero: $0.2-0.008y=0$	✓ 1
A1 y = 25	✓ 1
A1 greatest gradient 2.5	✓ 1

Exam tip: A slope field for $\frac{dy}{dx}=f(y)$ has identical segments along every horizontal line, so you only need to study the function of y to describe its behaviour.

+ Add to test✓ In your test193IBMI3H-5-04r20-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The height of a railway line above sea level is modelled by $h(x)=0.5x^3-4x^2+9x+20$ metres, where x is the distance in km from a station, for $0\le x\le5$.

(a)

Find $h'(x)$.

(2)

(b)

Find the values of x for which $h'(x)=0$, to 3 significant figures.

(2)

(c)

Find the height of the local maximum, to 3 significant figures, and justify that it is a maximum. Use the unrounded value of x.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiate each term	✓ 1
A1 $h'(x)=1.5x^2-8x+9$	✓ 1
Part (b)
M1 solve $1.5x^2-8x+9=0$	✓ 1
A1 x = 1.61 and x = 3.72	✓ 1
Part (c)
M1 substitute the smaller root $x=\frac{8-\sqrt{10}}{3}$ into $h(x)$	✓ 1
A1 26.208 m, so 26.2 m	✓ 1
R1 $h''(x)=3x-8<0$ at this x, so it is a maximum	✓ 1

Exam tip: The smaller of the two stationary points is the maximum for a positive cubic; confirm this with the sign of the second derivative rather than assuming it.

+ Add to test✓ In your test194IBMI3H-5-04r20-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A sprint cyclist's speed is recorded every 5 seconds during the first 30 seconds of a time trial.
Time (s): 0, 5, 10, 15, 20, 25, 30
Speed (m s⁻¹): 0, 4.6, 7.4, 9.0, 9.8, 10.1, 10.2

(a)

Use the trapezoidal rule with all the data to estimate the distance travelled in the 30 seconds.

(3)

(b)

The speed–time curve is concave down throughout (its gradient is always decreasing). State, with a reason, whether your estimate is an overestimate or an underestimate.

(2)

(c)

Hence find the average speed over the 30 seconds, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{5}{2}\left[\ldots\right]$ with first and last values once	✓ 1
M1 interior values doubled: $2(4.6+7.4+9.0+9.8+10.1)=81.8$	✓ 1
A1 230 m	✓ 1
Part (b)
R1 underestimate	✓ 1
R1 for a concave-down curve each trapezium lies below the curve, so the area is too small	✓ 1
Part (c)
M1 distance ÷ time = 230 ÷ 30	✓ 1
A1 7.67 m s⁻¹	✓ 1

Exam tip: In the trapezoidal rule only the first and last ordinates are counted once; every interior ordinate is counted twice.

+ Add to test✓ In your test195IBMI3H-5-06r21-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A cyclist accelerates from rest in a sprint. Her speed v m s⁻¹ at time t seconds is modelled by v(t) = 8te^−0.3t, t ≥ 0.

(a)

Use the product rule to show that v′(t) = 8e^−0.3t(1 − 0.3t).

(3)

(b)

Find the initial acceleration.

(1)

(c)

Find the time at which her speed is greatest, to 3 significant figures.

(2)

(d)

Hence find her greatest speed, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule with u = 8t and w = e^−0.3t	✓ 1
A1 v′ = 8e^−0.3t + 8t(−0.3)e^−0.3t	✓ 1
A1 factorises to 8e^−0.3t(1 − 0.3t) (AG)	✓ 1
Part (b)
A1 v′(0) = 8 m s⁻²	✓ 1
Part (c)
M1 sets v′(t) = 0 and notes e^−0.3t ≠ 0	✓ 1
A1 t = 10/3 = 3.33 s	✓ 1
Part (d)
M1 substitutes t = 10/3 into v(t)	✓ 1
A1 9.81 m s⁻¹	✓ 1

Exam tip: Once an expression is factorised, a product equals zero only if a factor is zero; an exponential factor is never zero, so only the bracket gives a solution.

+ Add to test✓ In your test196IBMI3H-5-06r21-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

The load c gigabits per second (Gbps) on a data centre network link t hours after 18:00 is modelled by c(t) = 12t/(t² + 4), t ≥ 0.

(a)

Use the quotient rule to show that c′(t) = 12(4 − t²)/(t² + 4)².

(2)

(b)

Find the time at which the load is greatest and justify that it is a maximum.

(3)

(c)

Find the greatest load.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule: ((t² + 4)(12) − 12t(2t))/(t² + 4)²	✓ 1
A1 numerator 12t² + 48 − 24t² = 48 − 12t² = 12(4 − t²) (AG)	✓ 1
Part (b)
M1 sets c′(t) = 0	✓ 1
A1 t = 2 (reject t = −2)	✓ 1
R1 c′ changes from positive to negative at t = 2 (or a second derivative test)	✓ 1
Part (c)
A1 c(2) = 24/8 = 3.00 Gbps	✓ 1

Exam tip: Always say why a stationary point is a maximum: a sign change of the gradient is enough, and in this context only t ≥ 0 is allowed.

+ Add to test✓ In your test197IBMI3H-5-06r21-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The power drawn by a server cluster is modelled by P(t) = t√(t² + 9) kW, 0 ≤ t ≤ 4, where t is the time in hours. The energy used in the four hours is the integral of P with respect to t.

(a)

Use the substitution u = t² + 9 to show that the energy in kWh is ½∫₉²⁵ u^1/2 du.

(3)

(b)

Hence find the exact energy used.

(2)

(c)

Find the mean power over the four hours, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du = 2t dt, so t dt = ½ du	✓ 1
M1 changes limits: t = 0 gives u = 9 and t = 4 gives u = 25	✓ 1
A1 ½∫₉²⁵ u^1/2 du (AG)	✓ 1
Part (b)
M1 ½ × [⅔u^3/2] = ⅓[u^3/2]₉²⁵	✓ 1
A1 (125 − 27)/3 = 98/3 kWh	✓ 1
Part (c)
M1 (98/3) ÷ 4	✓ 1
A1 8.17 kW	✓ 1

Common error: When changing variable in a definite integral, either change the limits to u values or return to t before substituting the original limits; mixing the two is a frequent slip.

+ Add to test✓ In your test198IBMI3H-5-06r21-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The inside of a ceramic vase is formed by rotating the curve y = 4 + 2 sin(0.3x), 0 ≤ x ≤ 20, through 360° about the x-axis. Lengths are in centimetres and 0.3x is in radians.

(a)

Write down an integral for the volume of the vase and use your calculator to evaluate it, to 3 significant figures.

(3)

(b)

Find the greatest radius of the vase and the first value of x at which it occurs, each to 3 significant figures.

(2)

(c)

A scale model is made with every length halved. Find the volume of the model, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 V = π∫₀²⁰ (4 + 2 sin(0.3x))² dx	✓ 1
A1 correct limits and integrand	✓ 1
A1 V = 1140 cm³ (1143.27…)	✓ 1
Part (b)
A1 greatest radius = 6.00 cm	✓ 1
A1 0.3x = π/2, so x = 5.24 cm	✓ 1
Part (c)
M1 volume scale factor (½)³ = 1/8	✓ 1
A1 143 cm³	✓ 1

Exam tip: The volume integral uses the square of the function, so square the whole expression including both terms before entering it.

+ Add to test✓ In your test199IBMI3H-5-06r21-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

The compression x cm of a bicycle suspension fork satisfies d²x/dt² = −4x − 0.5(dx/dt), where t is in seconds. At t = 0, x = 1 and dx/dt = 0.

(a)

Let v = dx/dt. Write the second-order equation as a pair of first-order equations.

(2)

(b)

Use Euler's method with step length 0.1 to estimate x and v at t = 0.3.

(5)

(c)

State which term in the equation represents damping.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 dx/dt = v	✓ 1
A1 dv/dt = −4x − 0.5v	✓ 1
Part (b)
M1 x_new = x + 0.1v and v_new = v + 0.1(−4x − 0.5v)	✓ 1
A1 t = 0.1: x = 1, v = −0.4	✓ 1
A1 t = 0.2: x = 0.96, v = −0.78	✓ 1
A1 x = 0.882 at t = 0.3	✓ 1
A1 v = −1.125 at t = 0.3	✓ 1
Part (c)
R1 −0.5(dx/dt), because it opposes the velocity	✓ 1

Common error: Use the old values of x and v in every Euler step for both variables; updating x first and then using it in the v step gives a different, incorrect scheme.

+ Add to test✓ In your test200IBMI3H-5-06r22-01Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

A space probe fires its thrusters for a short manoeuvre. Its speed gain relative to its initial speed is modelled by v(t) = 12te^−0.25t m s⁻¹, where t is the time in seconds after ignition, for t ≥ 0.

(a)

Find an expression for the acceleration $a(t)=\frac{dv}{dt}$.

(3)

(b)

Find the time at which the acceleration is zero.

(1)

(c)

Find the maximum value of v, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 $12e^{-0.25t}-3te^{-0.25t}$	✓ 1
A1 $a=12e^{-0.25t}(1-0.25t)$ or equivalent	✓ 1
Part (b)
A1 t = 4 s	✓ 1
Part (c)
M1 $v(4)=48e^{-1}$	✓ 1
A1 17.7 m s⁻¹	✓ 1

Exam tip: A maximum of v occurs where the derivative of v (the acceleration) is zero and changes sign from positive to negative.

+ Add to test✓ In your test201IBMI3H-5-06r22-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The rate at which a probe's thruster uses fuel is modelled by r(t) = 8te^−t²/2 kg per hour, where t is the time in hours after ignition.

(a)

Use the substitution $u=-\frac{t^2}{2}$ to find $\int 8te^{-t^2/2}\,dt$.

(3)

(b)

Find the total mass of fuel used in the first 3 hours, to 3 significant figures.

(2)

(c)

The total fuel used approaches a limit as t increases. Find the time at which 99 % of this limit has been used, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $du=-t\,dt$	✓ 1
A1 $\int -8e^{u}\,du$	✓ 1
A1 $-8e^{-t^2/2}+C$	✓ 1
Part (b)
M1 $8\left(1-e^{-4.5}\right)$	✓ 1
A1 7.91 kg	✓ 1
Part (c)
M1 limit is 8 kg, so $8\left(1-e^{-t^2/2}\right)=0.99\times8$	✓ 1
A1 3.03 hours	✓ 1

Exam tip: For a definite integral by substitution, either change the limits to u-values or return to t before substituting the original limits.

+ Add to test✓ In your test202IBMI3H-5-06r22-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A coffee-grinder funnel is modelled by rotating the curve y = 3e^−0.1x, for 0 ≤ x ≤ 12, through 360° about the x-axis. Lengths are in centimetres.

(a)

Write down an integral for the volume of the funnel.

(2)

(b)

Evaluate the volume, to 3 significant figures.

(3)

(c)

Roasted beans have a bulk density of 0.38 g cm⁻³. Using the unrounded volume, find the mass of beans that fills the funnel, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\pi\int y^2\,dx$	✓ 1
A1 $\pi\int_0^{12}9e^{-0.2x}\,dx$	✓ 1
Part (b)
M1 antiderivative $-45e^{-0.2x}$	✓ 1
A1 $45\pi\left(1-e^{-2.4}\right)$	✓ 1
A1 129 cm³	✓ 1
Part (c)
M1 $0.38\times45\pi\left(1-e^{-2.4}\right)$	✓ 1
A1 48.8 g	✓ 1

Common error: The volume uses $\pi y^2$: square the whole expression for y, including the 3, giving 9 and $e^{-0.2x}$.

+ Add to test✓ In your test203IBMI3H-5-06r22-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A probe entering a thin atmosphere decelerates under drag. Its speed v km s⁻¹ at time t seconds satisfies $\frac{dv}{dt}=-0.004v^2$, with v = 8 when t = 0.

(a)

Separate the variables and integrate to show that $\frac{1}{v}=0.004t+C$.

(3)

(b)

Use the initial condition to find C, and hence write v in terms of t.

(2)

(c)

Find the speed when t = 30, to 3 significant figures.

(1)

(d)

Find the time at which the speed is 3 km s⁻¹, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\int\frac{1}{v^2}\,dv=\int-0.004\,dt$	✓ 1
A1 $-\frac{1}{v}$ on the left	✓ 1
A1 $-\frac{1}{v}=-0.004t+k$, so $\frac{1}{v}=0.004t+C$ (AG)	✓ 1
Part (b)
A1 $C=\frac{1}{8}=0.125$	✓ 1
A1 $v=\frac{1}{0.125+0.004t}$	✓ 1
Part (c)
A1 4.08 km s⁻¹	✓ 1
Part (d)
M1 $\frac{1}{3}=0.125+0.004t$	✓ 1
A1 52.1 s	✓ 1

Exam tip: Keep the constant of integration on one side only, and find it from the initial condition before substituting any other values.

+ Add to test✓ In your test204IBMI3H-5-06r22-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

During a roast, the flavour index of the beans is modelled by $F(t)=\frac{45t}{t^2+30}$, where t is the time in minutes, for t > 0.

(a)

Use the quotient rule to show that $F'(t)=\frac{45(30-t^2)}{(t^2+30)^2}$.

(3)

(b)

Find the time at which the flavour index is greatest, to 3 significant figures.

(2)

(c)

Justify that this is a maximum.

(1)

(d)

Find the greatest flavour index, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at quotient rule $\frac{vu'-uv'}{v^2}$	✓ 1
A1 $\frac{45(t^2+30)-45t(2t)}{(t^2+30)^2}$	✓ 1
A1 simplification to $\frac{45(30-t^2)}{(t^2+30)^2}$ (AG)	✓ 1
Part (b)
M1 $F'(t)=0$	✓ 1
A1 t = 5.48 min	✓ 1
Part (c)
R1 F′ changes from positive to negative either side of $t=\sqrt{30}$	✓ 1
Part (d)
M1 $F(\sqrt{30})=\frac{45\sqrt{30}}{60}$	✓ 1
A1 4.11	✓ 1

Common error: Subtract in the correct order in the quotient rule: vu′ − uv′, with the square of the denominator below.

+ Add to test✓ In your test205IBMI3H-5-06r22-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A coffee storage tin has no lid. It is a cuboid with a square base of side x cm and height h cm, and its volume is 1500 cm³.

(a)

Show that the outer surface area is $A=x^2+\frac{6000}{x}$.

(2)

(b)

Find $\frac{dA}{dx}$.

(2)

(c)

Find the value of x that minimises A, to 3 significant figures.

(2)

(d)

Find the minimum surface area, to 3 significant figures, and justify that it is a minimum.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $h=\frac{1500}{x^2}$	✓ 1
A1 $A=x^2+4xh=x^2+\frac{6000}{x}$ (AG)	✓ 1
Part (b)
A1 $2x$	✓ 1
A1 $-6000x^{-2}$	✓ 1
Part (c)
M1 $2x=\frac{6000}{x^2}$, i.e. $x^3=3000$	✓ 1
A1 x = 14.4 cm	✓ 1
Part (d)
A1 624 cm²	✓ 1
R1 $\frac{d^2A}{dx^2}=2+\frac{12000}{x^3}>0$	✓ 1

Exam tip: For an optimisation problem, write the quantity in terms of one variable first, using the fixed constraint to eliminate the other.

+ Add to test✓ In your test206IBMI3H-5-06r23-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

Calculator

Tremor power recorded at a volcano is modelled by P(t) = 8t/(t² + 9) units, where t is in hours and t ≥ 0. The total energy released between t = 0 and t = T is the integral of P over that interval.

(a)

Use the substitution u = t² + 9 to show that the total energy released in the first 4 hours is 4 ln(25/9).

(3)

(b)

Find the mean power over the first 4 hours, correct to 3 significant figures.

(2)

(c)

Find the value of T for which the total energy released is 6 units, correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du = 2t dt, so the integral becomes ∫ 4/u du	✓ 1
A1 new limits u = 9 to u = 25	✓ 1
A1 4[ln u] from 9 to 25 = 4 ln(25/9)	✓ 1
Part (b)
M1 4 ln(25/9) ÷ 4 = ln(25/9)	✓ 1
A1 1.02 units	✓ 1
Part (c)
M1 4 ln((T² + 9)/9) = 6	✓ 1
M1 (T² + 9)/9 = e^1.5	✓ 1
A1 T = 5.60 hours	✓ 1

Exam tip: An integrand of the form f′(t)/f(t) integrates to ln|f(t)|; check for it before attempting a more complicated method.

+ Add to test✓ In your test207IBMI3H-5-06r23-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

Calculator

The half-width y of a ship's waterline, measured at points along its length x, is given in metres in the table.

x: 0, 10, 20, 30, 40, 50, 60
y: 0, 3.2, 4.6, 5.1, 4.4, 2.5, 0

(a)

Use the trapezoidal rule with 6 intervals to estimate the area enclosed between the waterline on one side of the ship and the centre line.

(4)

(b)

The second differences of the y-values are all negative. Explain why the trapezoidal rule gives an underestimate of the true area.

(2)

(c)

A computer model gives the true area as 207 m². Find the percentage error in the estimate from (a), correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 strip width h = 10	✓ 1
M1 ½ × 10 × [first + last + 2(sum of the others)]	✓ 1
A1 correct substitution 5 × [0 + 0 + 2(3.2 + 4.6 + 5.1 + 4.4 + 2.5)]	✓ 1
A1 198 m²	✓ 1
Part (b)
R1 negative second differences indicate that the curve is concave down	✓ 1
R1 the chords lie below the curve, so the trapezia have less area than the region	✓ 1
Part (c)
M1 (207 − 198)/207 × 100	✓ 1
A1 4.35%	✓ 1

Exam tip: State which formula you are using for percentage error: the difference between the estimate and the true value, divided by the true value, multiplied by 100.

+ Add to test✓ In your test208IBMI3H-5-05r24-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

After a pool is treated, the concentration of chlorine, C mg per litre, is modelled by C(t) = 12te^−0.4t, where t ≥ 0 is the time in hours since the treatment.

(a)

Show that C′(t) = 12e^−0.4t(1 − 0.4t).

(2)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

Show that this value of t gives a maximum.

(2)

(d)

Find the maximum concentration. Give your answer to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule: 12e^−0.4t + 12t(−0.4)e^−0.4t	✓ 1
A1 factorise to obtain the given result	✓ 1
Part (b)
M1 set C′(t) = 0, noting e^−0.4t ≠ 0, so 1 − 0.4t = 0	✓ 1
A1 t = 2.5 hours	✓ 1
Part (c)
M1 C″(t) = 12e^−0.4t(0.16t − 0.8)	✓ 1
R1 C″(2.5) = −1.77 < 0, hence a maximum (accept a sign change of C′ from positive to negative)	✓ 1
Part (d)
A1 11.0 mg per litre	✓ 1

Exam tip: Check a stationary point is a maximum by a sign test on the second derivative, and state the sign explicitly so the reasoning mark is earned.

+ Add to test✓ In your test209IBMI3H-5-05r24-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

The power output of a small wind farm is modelled by P(t) = 6te^−0.1t² megawatts (MW), where t is the number of hours after 06:00, for 0 ≤ t ≤ 5. The energy produced in MWh between t = 0 and t = 5 is the integral of P(t) from 0 to 5.

(a)

Use the substitution u = 0.1t² to show that the energy produced equals 30 ∫₀^2.5 e^−u du.

(3)

(b)

Hence find the energy produced between 06:00 and 11:00. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du = 0.2t dt, so 6t dt = 30 du	✓ 1
A1 new limits u = 0 and u = 2.5	✓ 1
A1 30 ∫₀^2.5 e^−u du	✓ 1
Part (b)
M1 30[−e^−u]₀^2.5 = 30(1 − e^−2.5)	✓ 1
A1 27.5 MWh	✓ 1

Exam tip: When you change variable in a definite integral, convert the limits at the same time; then you never need to substitute back.

+ Add to test✓ In your test210IBMI3H-5-05r25-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The biomass y (thousands of tonnes) of a fish stock in a shipping lane's fishery t years after monitoring began is modelled by $\frac{dy}{dt}=0.2y(5-y)$. A slope field is drawn for this differential equation. Initially y = 1.

(a)

Find the gradient of the slope field at y = 1 and at y = 6.

(2)

(b)

State the values of y at which the line segments of the slope field are horizontal, and describe the long-term behaviour of the solution that starts at y = 1.

(2)

(c)

Use Euler's method with step length 1 to estimate y when t = 3. Use unrounded values throughout and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 gradient 0.8 at y = 1	✓ 1
A1 gradient −1.2 at y = 6	✓ 1
Part (b)
A1 y = 0 and y = 5	✓ 1
R1 the biomass increases and levels off, approaching 5 (thousand tonnes)	✓ 1
Part (c)
M1 uses $y_{n+1}=y_n+1\times 0.2y_n(5-y_n)$ starting from $y_0=1$	✓ 1
A1 $y_1=1.8$ and $y_2=2.952$	✓ 1
A1 $y_3=4.16$ (4.1612...)	✓ 1

Common error: Horizontal segments occur where the right-hand side is zero; students sometimes also include points where y is large, which give a steep negative gradient instead.

+ Add to test✓ In your test211IBMI3H-5-05r25-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A fishery models the biomass of krill, x, and of the fish that feed on them, y (both in thousands of tonnes), t years after a survey, by $\frac{dx}{dt}=0.5x-0.02xy,\quad \frac{dy}{dt}=0.01xy-0.3y.$ At the survey, x = 40 and y = 10.

(a)

Calculate $\frac{dx}{dt}$ and $\frac{dy}{dt}$ at t = 0.

(2)

(b)

Use Euler's method with step length 0.5 to estimate x and y when t = 1. Give each answer to 3 significant figures.

(4)

(c)

For y > 0, find the value of x at which the fish biomass is momentarily constant.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dx}{dt}=12$	✓ 1
A1 $\frac{dy}{dt}=1$	✓ 1
Part (b)
M1 uses $x_{n+1}=x_n+0.5\frac{dx}{dt}$ and $y_{n+1}=y_n+0.5\frac{dy}{dt}$	✓ 1
A1 x(0.5) = 46 and y(0.5) = 10.5	✓ 1
M1 applies a second step using these values (ECF)	✓ 1
A1 at t = 1: x ≈ 52.7 and y ≈ 11.3 (52.67 and 11.34 unrounded)	✓ 1
Part (c)
A1 setting $0.01x-0.3=0$ gives x = 30	✓ 1

Exam tip: For coupled equations, update both variables using the values from the previous step; never use a new x in the y update of the same step.

+ Add to test✓ In your test212IBMI3H-5-05r25-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The inner surface of a volcanic crater is modelled by rotating the curve $y=0.002x^2$, for $x\ge0$, through 360° about the y-axis, where x and y are in metres and y is the height above the lowest point of the crater floor. Molten lava collects in the crater and rises to a height of 12 m.

(a)

Show that $x^2=500y$.

(2)

(b)

Find the volume of lava in the crater, in m³, to 3 significant figures.

(3)

(c)

Lava enters the crater at a constant 1500 m³ per hour, starting from an empty crater. Use the unrounded volume from part (b) to find how long it takes to reach a height of 12 m, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearranges $y=0.002x^2$ to $x^2=\frac{y}{0.002}$	✓ 1
A1 $x^2=500y$ (AG)	✓ 1
Part (b)
M1 $V=\pi\int x^2\,dy$ with limits 0 and 12	✓ 1
A1 $\pi\int_0^{12}500y\,dy$	✓ 1
A1 $V=113000$ m³ (113097...)	✓ 1
Part (c)
M1 divides volume by 1500	✓ 1
A1 75.4 hours	✓ 1

Common error: When rotating about the y-axis, integrate $\pi x^2$ with respect to y; students often square y and integrate with respect to x.

+ Add to test✓ In your test213IBMI3H-5-05r25-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Ash falls on a village downwind of a volcano. The rate of ash deposition is $R(t)=12te^{-0.5t^2}$ kg per m² per hour, where t is the time in hours since the eruption began, for $0\le t\le3$.

(a)

Find $\frac{dR}{dt}$ and hence the time at which the rate of deposition is greatest.

(3)

(b)

Use the substitution $u=0.5t^2$ to find the total mass of ash deposited per m² in the first 3 hours. Give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule with chain rule	✓ 1
A1 $\frac{dR}{dt}=12e^{-0.5t^2}(1-t^2)$	✓ 1
A1 t = 1 hour	✓ 1
Part (b)
M1 $du=t\,dt$	✓ 1
M1 new limits $u=0$ to $u=4.5$	✓ 1
A1 $12\int_0^{4.5}e^{-u}\,du$	✓ 1
A1 11.9 kg per m² (11.867...)	✓ 1

Exam tip: After a substitution, change the limits as well as the variable, then you never need to substitute back.

+ Add to test✓ In your test214IBMI3H-5-05r25-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A ballast tank on a ship drains through a valve in its base. The depth of water is h metres at time t minutes, where $\frac{dh}{dt}=-0.05\sqrt{h}$. Initially the depth is 9 m.

(a)

Solve the differential equation to show that $h=(3-0.025t)^2$.

(4)

(b)

Find the time taken for the tank to empty.

(1)

(c)

Find the time at which the depth is 4.5 m, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 separates variables: $\int h^{-1/2}\,dh=\int-0.05\,dt$	✓ 1
A1 $2\sqrt{h}=-0.05t+c$	✓ 1
M1 uses $h=9$ at $t=0$ to find $c=6$	✓ 1
A1 $h=(3-0.025t)^2$ (AG)	✓ 1
Part (b)
A1 120 minutes	✓ 1
Part (c)
M1 $(3-0.025t)^2=4.5$	✓ 1
A1 35.1 minutes	✓ 1

Common error: Do not forget the constant of integration, and find it using the initial depth before rearranging for h.

+ Add to test✓ In your test215IBMI3H-5-05r25-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A shipping firm makes an open-topped crate with a square base of side x metres and height h metres. The crate must have a volume of 2 m³. The base material costs €20 per m² and the side material costs €12 per m².

(a)

Express h in terms of x.

(1)

(b)

Show that the total cost, in euros, is $C=20x^2+\frac{96}{x}$.

(2)

(c)

Find the value of x that minimises the cost, to 3 significant figures.

(3)

(d)

Justify that this value gives a minimum.

(1)

(e)

Find the minimum cost, to the nearest cent (use the unrounded value of x).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $h=\frac{2}{x^2}$	✓ 1
Part (b)
M1 $C=20x^2+12\times4xh$	✓ 1
A1 substitutes $h=\frac{2}{x^2}$ to obtain $20x^2+\frac{96}{x}$ (AG)	✓ 1
Part (c)
M1 $\frac{dC}{dx}=40x-\frac{96}{x^2}$	✓ 1
M1 sets the derivative to zero: $x^3=2.4$	✓ 1
A1 x = 1.34	✓ 1
Part (d)
R1 $\frac{d^2C}{dx^2}=40+\frac{192}{x^3}>0$ for $x>0$, so a minimum	✓ 1
Part (e)
A1 €107.55	✓ 1

Exam tip: Write the quantity to be optimised in terms of a single variable before differentiating; the volume condition is what eliminates the second variable.

+ Add to test✓ In your test216IBMI3H-5-05r25-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A sensor near a volcanic vent records the concentration of sulfur dioxide, C ppm, at time t hours after it is switched on, where $C=\frac{20\ln t}{t}$ for $t\ge1$.

(a)

Use the quotient rule to show that $\frac{dC}{dt}=\frac{20(1-\ln t)}{t^2}$.

(3)

(b)

Find the exact time at which the concentration is greatest.

(2)

(c)

Find the greatest concentration, to 3 significant figures.

(1)

(d)

Find the rate of change of concentration when t = 10, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 applies the quotient rule with $u=20\ln t$ and $v=t$	✓ 1
A1 $\frac{u'v-uv'}{v^2}=\frac{\frac{20}{t}\times t-20\ln t}{t^2}$	✓ 1
A1 $\frac{20(1-\ln t)}{t^2}$ (AG)	✓ 1
Part (b)
M1 sets $\frac{dC}{dt}=0$, so $\ln t=1$	✓ 1
A1 t = e hours	✓ 1
Part (c)
A1 $\frac{20}{e}=7.36$ ppm	✓ 1
Part (d)
M1 substitutes t = 10 into $\frac{dC}{dt}$	✓ 1
A1 −0.261 ppm per hour	✓ 1

Exam tip: Simplify the numerator of a quotient-rule derivative fully; here the factor of t cancels and the stationary point is then easy to find.

+ Add to test✓ In your test217IBMI3H-5-04r26-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A retailer models its weekly profit, P thousand pounds (£), when £x thousand is spent on advertising by
P(x) = 30 ln x / x, for x ≥ 1.

(a)

Use the quotient rule to show that P′(x) = 30(1 − ln x) / x².

(3)

(b)

Find the exact value of x for which P′(x) = 0.

(2)

(c)

Show that this value of x gives a maximum profit and find the maximum profit, in thousands of pounds, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at the quotient rule with u = 30 ln x, v = x	✓ 1
A1 u′ = 30/x and v′ = 1 substituted: (30 − 30 ln x)/x²	✓ 1
A1 factorises to AG: P′(x) = 30(1 − ln x)/x²	✓ 1
Part (b)
M1 set 1 − ln x = 0	✓ 1
A1 x = e	✓ 1
Part (c)
M1 consider the sign of P′ either side of x = e, e.g. P′(2) = 2.30 > 0 and P′(3) = −0.328 < 0	✓ 1
R1 gradient changes from positive to negative so a maximum	✓ 1
A1 P(e) = 30/e = 11.0 (thousand pounds)	✓ 1

Exam tip: To justify a maximum you can use the sign of the first derivative either side, or the sign of the second derivative; a statement of the sign values is needed, not just the word 'maximum'.

+ Add to test✓ In your test218IBMI3H-5-04r26-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

The number of people, N, infected with a virus in a small town t days after the first test is modelled by
dN/dt = 0.3N(1 − N/500), with N = 20 when t = 0.
Use Euler's method with a step length of 1 day. Do not round intermediate values.

(a)

Estimate the number of infected people after 2 days, giving your answer to 3 significant figures.

(4)

(b)

For 0 < N < 250 the solution curve is concave up. State, with a reason, whether the Euler estimate for N₂ is likely to be an underestimate or an overestimate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of N_n+1 = N_n + 1 × 0.3N_n(1 − N_n/500)	✓ 1
A1 N₁ = 25.76	✓ 1
M1 second step using N₁ = 25.76	✓ 1
A1 N₂ = 33.1	✓ 1
Part (b)
R1 each step follows the tangent at the start of the interval and the gradient is increasing along the true curve	✓ 1
A1 the Euler estimate is an underestimate	✓ 1

Common error: Carry unrounded values from one step to the next and round only the final answer, as the question instructs.

+ Add to test✓ In your test219IBMI3H-5-04r26-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

In an outbreak, S is the number of susceptible people and I the number of infected people after t days. The model is
dS/dt = −0.002SI, dI/dt = 0.002SI − 0.1I,
with S = 500 and I = 10 when t = 0. Use Euler's method with step length 1 day and do not round intermediate values.

(a)

Interpret the term −0.1I in the equation for dI/dt.

(1)

(b)

Find estimates for S₁ and I₁.

(3)

(c)

Hence estimate S₂ and I₂, giving S₂ to the nearest whole number and I₂ to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 each day, 10% of the infected people leave the infected group (e.g. recover)	✓ 1
Part (b)
M1 use S_n+1 = S_n + h dS/dt and I_n+1 = I_n + h dI/dt with h = 1	✓ 1
A1 S₁ = 490	✓ 1
A1 I₁ = 19	✓ 1
Part (c)
M1 second step using S = 490 and I = 19	✓ 1
A1 S₂ = 471.38… ≈ 471	✓ 1
A1 I₂ = 19 + 0.002 × 490 × 19 − 0.1 × 19 = 35.72 ≈ 35.7	✓ 1

Common error: Both new values must be calculated from the old values of S and I; do not use the updated S when finding the new I in the same step.

+ Add to test✓ In your test220IBMI3H-5-04r26-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

A candle maker produces a candle by rotating the region between the curve y = 2 + 0.5 sin x, the x-axis and the lines x = 0 and x = 6 through 360° about the x-axis. Lengths are in centimetres.

(a)

Write down an integral for the volume of the candle.

(2)

(b)

Evaluate the volume, to 3 significant figures. Use radians.

(2)

(c)

The wax costs £0.012 per cm³. Using the unrounded volume, find the cost of the wax, to the nearest penny.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of V = π∫y² dx	✓ 1
A1 V = π∫₀⁶ (2 + 0.5 sin x)² dx	✓ 1
Part (b)
M1 evaluate the integral using technology	✓ 1
A1 78.1 cm³	✓ 1
Part (c)
M1 0.012 × 78.11…	✓ 1
A1 £0.94	✓ 1

Common error: Calculators must be in radian mode for this integral; degree mode gives a very different volume.

+ Add to test✓ In your test221IBMI3H-5-05r28-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

Calculator

A harmless dye is released into a tidal pool. The concentration of dye at a sensor is modelled by $C(t)=\dfrac{5t}{t^2+4}$ mg L^-1, where $t\ge0$ is the time in hours after release.

(a)

Use the quotient rule to show that $C'(t)=\dfrac{5(4-t^2)}{(t^2+4)^2}$.

(3)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

Find the greatest concentration of dye.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at quotient rule with $u=5t$ and $v=t^2+4$	✓ 1
A1 $C'(t)=\frac{5(t^2+4)-5t(2t)}{(t^2+4)^2}$	✓ 1
A1 simplify numerator $20-5t^2=5(4-t^2)$ (AG)	✓ 1
Part (b)
M1 set $C'(t)=0$, i.e. $4-t^2=0$	✓ 1
A1 $t=2$ hours (reject $t=-2$)	✓ 1
Part (c)
A1 $C(2)=1.25$ mg L^-1	✓ 1

Common error: Forgetting the minus sign in the quotient rule numerator ($vu'-uv'$) gives a numerator of $5t^2+20$, which never equals zero.

+ Add to test✓ In your test222IBMI3H-5-05r28-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Calculator

The temperature $T$ (°C) of water in a marine-biology tank, $t$ minutes after a heater is switched on, satisfies $\dfrac{dT}{dt}=0.2(24-T)$. Initially $T=10$ and $T<24$ throughout.

(a)

Solve the differential equation to show that $T=24-14e^{-0.2t}$.

(5)

(b)

Find the time taken for the water to reach 20°C, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 separate variables: $\int\frac{dT}{24-T}=\int0.2\,dt$	✓ 1
A1 $-\ln(24-T)$ on the left (accept $-\ln\|24-T\|$)	✓ 1
A1 $=0.2t+c$	✓ 1
M1 substitute $t=0,\ T=10$ to find the constant, $c=-\ln14$	✓ 1
A1 rearrange to $T=24-14e^{-0.2t}$ (AG)	✓ 1
Part (b)
M1 solve $24-14e^{-0.2t}=20$, i.e. $e^{-0.2t}=\frac{4}{14}$	✓ 1
A1 $t=6.26$ minutes	✓ 1

Exam tip: When integrating $\frac{1}{24-T}$, remember the negative sign from the chain rule; check your final form against $T(0)=10$.

+ Add to test✓ In your test223IBMI3H-5-05r28-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

No calculator

Two connected lagoons contain amounts $x(t)$ and $y(t)$ (kg) of a dissolved nutrient, where $t$ is in days. The amounts satisfy
$\frac{dx}{dt}=-0.5x+0.2y,\qquad \frac{dy}{dt}=0.5x-0.2y.$
Initially $x=100$ and $y=0$.

(a)

Write the system in the form $\frac{d\mathbf{X}}{dt}=\mathbf{A}\mathbf{X}$, stating $\mathbf{A}$.

(1)

(b)

Find the eigenvalues of $\mathbf{A}$.

(3)

(c)

Show that $x+y$ is constant. Hence find the long-term value of $x$, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\mathbf{A}=\begin{pmatrix}-0.5&0.2\\0.5&-0.2\end{pmatrix}$	✓ 1
Part (b)
M1 set $\det(\mathbf{A}-\lambda\mathbf{I})=0$	✓ 1
A1 $(-0.5-\lambda)(-0.2-\lambda)-0.1=0$, giving $\lambda^2+0.7\lambda=0$	✓ 1
A1 $\lambda=0$ and $\lambda=-0.7$	✓ 1
Part (c)
R1 $\frac{d}{dt}(x+y)=0$, so $x+y=100$ for all $t$	✓ 1
M1 equilibrium: $-0.5x+0.2y=0$, so $y=2.5x$ (the eigenvector for $\lambda=0$)	✓ 1
A1 $x+2.5x=100$, so $x=28.6$ kg	✓ 1

Exam tip: An eigenvalue of 0 means there is a steady state; the other eigenvalue being negative tells you the system settles to it.

+ Add to test✓ In your test224IBMI3H-5-05r28-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Calculator

The roof of a museum is a dome. In a vertical cross-section, the edge of the dome is the curve $y=5-0.2x^2$ for $0\le x\le5$, where $x$ and $y$ are in metres. The dome is formed by rotating this curve through $360^\circ$ about the $y$-axis.

(a)

Write $x^2$ in terms of $y$.

(1)

(b)

Find the volume enclosed by the dome and the ground ($y=0$), in m³, to 3 significant figures.

(4)

(c)

Insulating foam costs £14 per m³. Using the unrounded volume, find the cost of filling the dome, to the nearest £.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $x^2=25-5y$	✓ 1
Part (b)
M1 use $V=\pi\int x^2\,dy$	✓ 1
A1 limits $y=0$ to $y=5$ with integrand $25-5y$	✓ 1
A1 $\pi[25y-2.5y^2]_0^5=62.5\pi$	✓ 1
A1 196 m³	✓ 1
Part (c)
M1 multiply $62.5\pi$ by 14	✓ 1
A1 £2749	✓ 1

Common error: When rotating about the y-axis, the integral needs $x^2$ expressed in terms of $y$ with limits in $y$, not in $x$.

+ Add to test✓ In your test225IBMI3H-5-05r28-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

Calculator

The load per metre on a beam, in kN m^-1, at distance $x$ metres from one end is $w(x)=xe^{0.25x^2}$ for $0\le x\le4$.

(a)

Use the substitution $u=0.25x^2$ to find the total load $\int_0^4 xe^{0.25x^2}\,dx$ in kN, to 3 significant figures.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{du}{dx}=0.5x$, so $x\,dx=2\,du$	✓ 1
A1 integral becomes $2\int e^u\,du$	✓ 1
A1 limits change: $x=0\to u=0$, $x=4\to u=4$	✓ 1
A1 $2(e^4-1)$	✓ 1
A1 107 kN	✓ 1

Exam tip: Change the limits with the substitution, or convert back to $x$ before substituting; mixing the two is a common loss of marks.

+ Add to test✓ In your test226IBMI3H-5-05r28-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Calculator

An automated trolley in a warehouse starts from rest. Its velocity is $v(t)=2\sin(0.4t)$ m s^-1 for $0\le t\le\frac{\pi}{0.4}$, where $t$ is in seconds. It then stops.

(a)

Find an expression for the acceleration of the trolley.

(2)

(b)

Find the acceleration when $t=3$, to 3 significant figures.

(2)

(c)

Find the total distance travelled by the trolley.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiate $v$	✓ 1
A1 $a=0.8\cos(0.4t)$ m s^-2	✓ 1
Part (b)
M1 substitute $t=3$ (radians)	✓ 1
A1 0.290 m s^-2	✓ 1
Part (c)
M1 $\int_0^{\pi/0.4}2\sin(0.4t)\,dt$	✓ 1
A1 $\left[-5\cos(0.4t)\right]$ evaluated between the limits	✓ 1
A1 10 m	✓ 1

Exam tip: Distance travelled is the integral of speed; here the velocity stays non-negative over the interval, so integrating $v$ gives distance.

+ Add to test✓ In your test227IBMI3H-5-06r28-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

Calculator

A pool is drained. The depth $h$ metres, $t$ minutes after draining begins, satisfies $\dfrac{dh}{dt}=-0.2\sqrt{h}$ with $h(0)=9$.

(a)

Solve the differential equation to show that $h=(3-0.1t)^2$.

(5)

(b)

Find the time at which the pool is empty.

(1)

(c)

Find the time at which the depth is 4.5 m, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 separate variables: $\int h^{-1/2}\,dh=\int-0.2\,dt$	✓ 1
A1 $2\sqrt{h}$	✓ 1
A1 $=-0.2t+c$	✓ 1
M1 use $h=9$ at $t=0$ to get $c=6$	✓ 1
A1 $\sqrt{h}=3-0.1t$, so $h=(3-0.1t)^2$ (AG)	✓ 1
Part (b)
A1 30 minutes	✓ 1
Part (c)
M1 solve $(3-0.1t)^2=4.5$ with $3-0.1t=\sqrt{4.5}$ (positive root)	✓ 1
A1 8.79 minutes	✓ 1

Common error: When taking the square root of $h=(3-0.1t)^2$, use the positive root since depth cannot be negative and $3-0.1t>0$ before the pool is empty.

+ Add to test✓ In your test228IBMI3H-5-06r28-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

Calculator

A wind-turbine tower is modelled as a solid whose radius at height $h$ metres above the base is $r=2.2e^{-0.005h}$ metres, for $0\le h\le80$. The volume is found by rotating the curve about the $h$-axis.

(a)

Show that the volume of the tower is $V=4.84\pi\int_0^{80}e^{-0.01h}\,dh$.

(3)

(b)

Calculate the volume, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use $V=\pi\int r^2\,dh$	✓ 1
A1 limits 0 and 80	✓ 1
A1 $r^2=4.84e^{-0.01h}$ (AG)	✓ 1
Part (b)
M1 integrate: $4.84\pi\left[-100e^{-0.01h}\right]_0^{80}$	✓ 1
A1 837 m³	✓ 1

Common error: Squaring the radius means squaring both the coefficient and the exponent: $(2.2e^{-0.005h})^2=4.84e^{-0.01h}$.

+ Add to test✓ In your test229IBMI3H-5-06r28-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

Calculator

Books are returned at a library desk at a rate of $B(t)=\dfrac{40t}{t^2+9}$ books per minute, where $t$ is the time in minutes since the desk opened, for $0\le t\le6$.

(a)

Use the substitution $u=t^2+9$ to show that the total number of books returned in the first 6 minutes is $20\ln5$.

(4)

(b)

Find the mean rate of returns over the first 6 minutes, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{du}{dt}=2t$, so $t\,dt=\frac12du$	✓ 1
A1 integral becomes $20\int\frac{1}{u}\,du$	✓ 1
A1 new limits $u=9$ to $u=45$	✓ 1
A1 $20\ln45-20\ln9=20\ln5$ (AG)	✓ 1
Part (b)
M1 divide the total by 6	✓ 1
A1 5.36 books per minute	✓ 1

Exam tip: The mean value of a rate over an interval is the integral divided by the interval length.

+ Add to test✓ In your test230IBMI3H-5-06r28-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

Calculator

A swimmer pushes off the wall of a 50 m pool. Her velocity is $v(t)=2.1-0.9e^{-0.5t}$ m s^-1, where $t$ is in seconds.

(a)

Find her acceleration as a function of $t$, and its value at $t=0$.

(2)

(b)

State the value that her velocity approaches as $t\to\infty$.

(1)

(c)

Find the time at which her velocity is 2.0 m s^-1, to 3 significant figures.

(2)

(d)

Find the distance she swims in the first 10 seconds, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiate $v$	✓ 1
A1 $a=0.45e^{-0.5t}$, so $a(0)=0.45$ m s^-2	✓ 1
Part (b)
A1 2.1 m s^-1	✓ 1
Part (c)
M1 solve $0.9e^{-0.5t}=0.1$	✓ 1
A1 $t=2\ln9=4.39$ s	✓ 1
Part (d)
M1 $\int_0^{10}v\,dt$	✓ 1
A1 $\left[2.1t+1.8e^{-0.5t}\right]_0^{10}$	✓ 1
A1 19.2 m	✓ 1

Exam tip: For a velocity that tends to a limit, take the limit of the exponential term (it tends to zero) to read off the long-term value.

+ Add to test✓ In your test231IBMI3H-5-06r28-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

Calculator

The number of members $y$ (thousands) using a library app, $x$ years after launch, is modelled by $\dfrac{dy}{dx}=0.1y(4-y)$. A slope field is drawn for this model.

(a)

Find the gradient of the slope field at the point $(x,1)$ for any $x$.

(1)

(b)

Find the values of $y$ for which the gradient is zero for all $x$.

(2)

(c)

Describe the long-term behaviour of the solution with (i) $y(0)=1$ and (ii) $y(0)=5$.

(2)

(d)

Find the value of $y$ at which the gradient of the solution curve is greatest, and that greatest gradient.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 0.3	✓ 1
Part (b)
M1 set $0.1y(4-y)=0$	✓ 1
A1 $y=0$ and $y=4$	✓ 1
Part (c)
A1 (i) $y$ increases and approaches 4	✓ 1
A1 (ii) $y$ decreases and approaches 4	✓ 1
Part (d)
M1 differentiate $0.4y-0.1y^2$ with respect to $y$ and set to zero	✓ 1
A1 $y=2$	✓ 1
A1 greatest gradient 0.4 thousand members per year	✓ 1

Exam tip: Horizontal segments at constant $y$ values are equilibria; check the sign of the gradient either side to see whether solutions move towards or away from them.

+ Add to test✓ In your test232IBMI3H-5-07r28-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A telecoms engineer models the signal quality index S received at a distance d km from a mast by S(d) = 80d / (d² + 25), for d > 0.

(a)

Use the quotient rule to show that S′(d) = 80(25 − d²) / (d² + 25)².

(3)

(b)

Find the distance at which the signal quality index is greatest.

(2)

(c)

Find the greatest value of S and justify that it is a maximum.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule with u = 80d, v = d² + 25	✓ 1
A1 S′ = [80(d² + 25) − 80d·2d] / (d² + 25)²	✓ 1
A1 expand numerator: 80d² + 2000 − 160d² = 80(25 − d²), giving the printed result	✓ 1
Part (b)
M1 set S′(d) = 0, so 25 − d² = 0	✓ 1
A1 d = 5 km (reject d = −5 as d > 0)	✓ 1
Part (c)
M1 substitute d = 5 into S	✓ 1
A1 S = 8	✓ 1
R1 S′ > 0 for 0 < d < 5 and S′ < 0 for d > 5, so S changes from increasing to decreasing	✓ 1

Exam tip: The denominator of the quotient-rule derivative is squared and always positive, so only the numerator decides the sign of the gradient.

+ Add to test✓ In your test233IBMI3H-5-07r28-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

During a snowstorm on a high ridge, snow falls at a rate of r(t) = 6te^−t²/8 cm per hour, where t is the time in hours since the storm began, for 0 ≤ t ≤ 4.

(a)

Find the time at which the rate of snowfall is greatest, and the greatest rate, giving the rate to 3 significant figures.

(3)

(b)

Use the substitution u = t²/8 to find the total depth of snow that falls during the first 4 hours. Give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiate with the product and chain rules and set r′(t) = 6e^−t²/8(1 − t²/4) = 0	✓ 1
A1 t = 2 hours	✓ 1
A1 r(2) = 12e^−0.5 = 7.28 cm per hour	✓ 1
Part (b)
M1 du = (t/4) dt, so 6t dt = 24 du	✓ 1
M1 new limits u = 0 to u = 2, giving ∫₀² 24e^−u du	✓ 1
A1 [−24e^−u]₀² = 24(1 − e⁻²)	✓ 1
A1 20.8 cm	✓ 1

Exam tip: When you substitute in a definite integral, change the limits to the new variable at the same time, so you never need to convert back.

+ Add to test✓ In your test234IBMI3H-5-07r28-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A chocolate mould is made by rotating the curve y = 3 sin(x/4), for 0 ≤ x ≤ 4π, through 360° about the x-axis. Lengths are in cm.

(a)

Write down an integral for the volume V of the mould.

(1)

(b)

Find V, giving your answer to 3 significant figures.

(4)

(c)

Chocolate has density 1.30 g cm⁻³. Use the unrounded volume to find the mass of chocolate needed to fill the mould, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 V = π∫₀^4π 9 sin²(x/4) dx	✓ 1
Part (b)
M1 use sin²(x/4) = ½(1 − cos(x/2))	✓ 1
M1 integrate: 9/2 [x − 2 sin(x/2)]₀^4π	✓ 1
A1 the integral equals 18π, so V = 18π²	✓ 1
A1 V = 178 cm³	✓ 1
Part (c)
M1 mass = 1.30 × 18π²	✓ 1
A1 231 g	✓ 1

Common error: Square the whole function before integrating: (3 sin(x/4))² = 9 sin²(x/4), not 3 sin²(x/4).

+ Add to test✓ In your test235IBMI3H-5-07r28-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A cable-car pod on an alpine route moves from rest along a straight cable. For 0 ≤ t ≤ 60 its velocity is v = 1.8 sin(πt/60) m s⁻¹, where t is in seconds.

(a)

State the greatest speed of the pod and the time at which it is reached.

(2)

(b)

Find the acceleration of the pod when t = 15. Give your answer to 3 significant figures.

(3)

(c)

Find the distance travelled by the pod in the 60 seconds. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 1.8 m s⁻¹	✓ 1
A1 t = 30 s	✓ 1
Part (b)
M1 differentiate using the chain rule	✓ 1
A1 a = (1.8π/60) cos(πt/60)	✓ 1
A1 a = 0.0666 m s⁻²	✓ 1
Part (c)
M1 distance = ∫₀⁶⁰ 1.8 sin(πt/60) dt	✓ 1
A1 [−(108/π) cos(πt/60)]₀⁶⁰ = 216/π	✓ 1
A1 68.8 m	✓ 1

Common error: Set the calculator to radians for trigonometric functions of πt/60; degree mode gives a completely different value.

+ Add to test✓ In your test236IBMI3H-5-07r28-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Traffic through a telecoms exchange has a data rate of r(t) = 12t / (3t² + 1) gigabytes per hour, where t is the time in hours since midnight, for t ≥ 0.

(a)

Use the substitution u = 3t² + 1 to find the total traffic from t = 0 to t = 2. Give your answer to 3 significant figures.

(4)

(b)

The total traffic from t = 0 to t = T is 4 gigabytes. Find T to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du = 6t dt, so 12t dt = 2 du	✓ 1
A1 limits u = 1 to u = 13, giving ∫₁¹³ 2/u du	✓ 1
A1 [2 ln u]₁¹³ = 2 ln 13	✓ 1
A1 5.13 gigabytes	✓ 1
Part (b)
M1 2 ln(3T² + 1) = 4	✓ 1
M1 3T² + 1 = e²	✓ 1
A1 T = 1.46 hours	✓ 1

Exam tip: For an integral of the form f′(x)/f(x), the answer is a natural logarithm; check that the top is a constant multiple of the derivative of the bottom.

+ Add to test✓ In your test237IBMI3H-5-08r28-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

An architect designs an arched footbridge whose centre line has height y = 12 sin(πx/20) metres at a horizontal distance x metres from the left abutment, for 0 ≤ x ≤ 20.

(a)

State the greatest height of the arch and the value of x at which it occurs.

(2)

(b)

Find the gradient of the arch at x = 5. Give your answer to 3 significant figures.

(3)

(c)

The tangent to the arch at x = 5 is extended downwards to meet the horizontal line y = 0. Find the x-coordinate of this point, using unrounded values and giving your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 12 m	✓ 1
A1 x = 10 m	✓ 1
Part (b)
M1 chain rule: y′ = (12π/20) cos(πx/20)	✓ 1
A1 substitute x = 5: (12π/20) cos(π/4)	✓ 1
A1 1.33	✓ 1
Part (c)
A1 y(5) = 12 sin(π/4) = 8.485…	✓ 1
M1 solve 0 − y(5) = y′(5)(x − 5), so x = 5 − y(5)/y′(5)	✓ 1
A1 x = 5 − 20/π = −1.37 m	✓ 1

Exam tip: Check that your calculator is in radians before evaluating sin(πx/20) or its derivative.

+ Add to test✓ In your test238IBMI3H-5-08r28-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A logistics company models the average handling cost per parcel as A(n) = 30e^0.1n/n pounds, where n is the number of hundreds of parcels processed per hour, for 1 ≤ n ≤ 20.

(a)

Use the quotient rule to show that A′(n) = 3e^0.1n(n − 10)/n².

(3)

(b)

Find the value of n that minimises the average cost, justifying that it is a minimum.

(3)

(c)

Find the minimum average cost per parcel, to the nearest penny.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule with u = 30e^0.1n and v = n	✓ 1
A1 A′ = (3e^0.1n·n − 30e^0.1n)/n²	✓ 1
A1 factorise 3e^0.1n to reach the printed form	✓ 1
Part (b)
M1 set A′(n) = 0	✓ 1
A1 n = 10	✓ 1
R1 A′ < 0 for n < 10 and A′ > 0 for n > 10	✓ 1
Part (c)
M1 substitute n = 10: 30e/10	✓ 1
A1 £8.15	✓ 1

Common error: Setting the numerator of a quotient-rule derivative to zero is correct, but check that the factor e⁰.¹ⁿ is never zero so it gives no extra solution.

+ Add to test✓ In your test239IBMI3H-5-08r28-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

Water flows through a tidal channel at a rate of q(t) = 24 sin(πt/6) cos(πt/6) thousand m³ per hour, where t is the time in hours after the tide turns, for 0 ≤ t ≤ 3.

(a)

Use the substitution u = sin(πt/6) to find the total volume of water that flows in the 3 hours. Give your answer to 3 significant figures.

(5)

(b)

Find the time T at which half of this total volume has flowed.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du = (π/6) cos(πt/6) dt	✓ 1
A1 limits u = 0 to u = 1	✓ 1
A1 integral becomes ∫₀¹ (144/π)u du	✓ 1
A1 [72u²/π]₀¹ = 72/π	✓ 1
A1 22.9 thousand m³	✓ 1
Part (b)
M1 (72/π) sin²(πT/6) = 36/π, so sin²(πT/6) = ½	✓ 1
M1 sin(πT/6) = √½, so πT/6 = π/4	✓ 1
A1 T = 1.5 hours	✓ 1

Exam tip: Choose the substitution so that its derivative appears as a factor in the integrand; here the cosine supplies du.

+ Add to test✓ In your test240IBMI3H-5-08r28-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A spire on a building has a circular cross-section of radius y = e^−0.2x metres at height x metres above its base, for 0 ≤ x ≤ 5. Its solid shape is generated by rotating this curve about the x-axis.

(a)

Find the volume of the spire. Give your answer to 3 significant figures.

(4)

(b)

Find the height at which the radius is 0.5 m. Give your answer to 3 significant figures.

(2)

(c)

The spire is cast from concrete of density 2400 kg m⁻³. Using the unrounded volume, find its mass in tonnes (1 tonne = 1000 kg) to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 V = π∫₀⁵ e^−0.4x dx	✓ 1
A1 [−2.5e^−0.4x]₀⁵	✓ 1
A1 V = 2.5π(1 − e⁻²)	✓ 1
A1 6.79 m³	✓ 1
Part (b)
M1 e^−0.2x = 0.5	✓ 1
A1 x = 5 ln 2 = 3.47 m	✓ 1
Part (c)
M1 mass = 2400 × 2.5π(1 − e⁻²) kg	✓ 1
A1 16.3 tonnes	✓ 1

Common error: The volume formula needs the radius squared: (e⁻⁰.²x)² = e⁻⁰.⁴x, not e⁻⁰.²x.

+ Add to test✓ In your test241IBMI3H-5-08r28-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The number of fish P (in thousands) in a fishery after t years satisfies dP/dt = 0.4P(1 − P/50), with P = 10 when t = 0.

(a)

Use one step of Euler’s method with step length 1 year to estimate P when t = 1.

(2)

(b)

Use a second step to estimate P when t = 2. Give your answer to 3 significant figures.

(2)

(c)

Find the greatest possible value of dP/dt for this model.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 P₁ = 10 + 1 × 0.4 × 10 × (1 − 10/50)	✓ 1
A1 P at t = 1 ≈ 13.2	✓ 1
Part (b)
M1 P₂ = 13.2 + 0.4 × 13.2 × (1 − 13.2/50)	✓ 1
A1 P(2) ≈ 17.1 thousand	✓ 1
Part (c)
M1 rewrite as dP/dt = 0.4P − 0.008P² and set its derivative with respect to P to zero: 0.4 − 0.016P = 0	✓ 1
A1 P = 25	✓ 1
A1 greatest rate = 0.4 × 25 × (1 − 25/50) = 5 thousand fish per year	✓ 1

Common error: In Euler’s method the new value is the old value plus step × gradient; the gradient must be evaluated at the old P, not the new one.

+ Add to test✓ In your test242IBMI3H-5-08r28-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Containers are loaded onto a ship at a rate of q(t) = 18t²/(t³ + 8) tonnes per hour, where t is the time in hours after loading begins.

(a)

Use the substitution u = t³ + 8 to find the mass loaded in the first 2 hours. Give your answer to 3 significant figures.

(4)

(b)

The mass loaded from t = 0 to t = T is 5 tonnes. Find T to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du = 3t² dt, so 18t² dt = 6 du	✓ 1
A1 limits u = 8 to u = 16, giving ∫₈¹⁶ 6/u du	✓ 1
A1 [6 ln u]₈¹⁶ = 6 ln 2	✓ 1
A1 4.16 tonnes	✓ 1
Part (b)
M1 6 ln((T³ + 8)/8) = 5	✓ 1
M1 T³ = 8(e^5/6 − 1)	✓ 1
A1 T = 2.18 hours	✓ 1

Exam tip: Because the integrand is of the form f′(t)/f(t) up to a constant, the result is a natural logarithm.

+ Add to test✓ In your test243IBMI3H-5-04r29-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

Calculator

A boutique sells silk scarves. When the price of a scarf is €x, the weekly revenue is modelled by R(x) = 400x e^−0.04x euros, for 0 < x < 100.

(a)

Use the product rule to show that R′(x) = 400e^−0.04x(1 − 0.04x).

(3)

(b)

Find the price that gives the maximum weekly revenue.

(2)

(c)

Find the maximum weekly revenue. Give your answer to the nearest €10.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule with u = 400x and v = e^−0.04x	✓ 1
A1 R′(x) = 400e^−0.04x + 400x(−0.04e^−0.04x)	✓ 1
A1 factorises to 400e^−0.04x(1 − 0.04x) (shown)	✓ 1
Part (b)
M1 sets R′(x) = 0 and notes that e^−0.04x > 0, so 1 − 0.04x = 0	✓ 1
A1 x = €25	✓ 1
Part (c)
A1 €3680 (3678.79...)	✓ 1

Common error: When differentiating e⁻⁰.⁰⁴x, forgetting the factor −0.04 from the chain rule is the usual slip. Check the sign of R′ on both sides of the stationary point to confirm a maximum.

+ Add to test✓ In your test244IBMI3H-5-04r29-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

Calculator

A fishing fleet's harvest rate in a sustainable-fishing trial is modelled by h(t) = 12t e^−0.25t² tonnes per month, where t is the time in months since the start of the trial.

(a)

Use the substitution u = 0.25t² to show that the total harvest in the first T months can be written as ∫₀^0.25T² 24e^−u du.

(3)

(b)

Find the total harvest in the first 4 months. Give your answer to 3 significant figures.

(2)

(c)

Show that the total harvest can never exceed 24 tonnes.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du = 0.5t dt (or du/dt = 0.5t)	✓ 1
A1 12t dt = 24 du, with integrand 24e^−u	✓ 1
A1 limits t = 0 → u = 0 and t = T → u = 0.25T²	✓ 1
Part (b)
M1 24[−e^−u] from 0 to 4 = 24(1 − e⁻⁴)	✓ 1
A1 23.6 tonnes (23.56...)	✓ 1
Part (c)
R1 as T → ∞, e^−0.25T² → 0, so the total tends to 24 and 24(1 − e^−0.25T²) < 24 for all T	✓ 1

Common error: When you substitute, change the limits as well as the integrand, or convert back to t before substituting the original limits. Mixing the two is a frequent source of wrong answers.

+ Add to test✓ In your test245IBMI3H-5-04r29-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

Calculator

A decorative glass bead is formed by rotating the region between the curve y = 2 + cos x, the x-axis, and the lines x = 0 and x = π through 360° about the x-axis. All lengths are in centimetres.

(a)

Write down an integral for the volume of the bead.

(2)

(b)

Find the volume of the bead. Give your answer to 3 significant figures.

(2)

(c)

The glass has density 2.5 g cm⁻³. Find the mass of the bead. Use the unrounded volume and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 V = π∫y² dx	✓ 1
A1 V = π∫₀^π (2 + cos x)² dx	✓ 1
Part (b)
M1 evaluates the integral (by GDC or by expanding the square)	✓ 1
A1 44.4 cm³ (44.41...)	✓ 1
Part (c)
M1 mass = density × volume = 2.5 × 44.41...	✓ 1
A1 111 g (111.03...)	✓ 1

Exam tip: In a volume of revolution, square the whole function before integrating: (2 + cos x)² is not 4 + cos²x.

+ Add to test✓ In your test246IBMI3H-5-04r29-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

Calculator

The number of fish, P, in a managed lake t years after the start of a study satisfies dP/dt = 0.4P(1 − P/3000) − 200. At t = 0 there are 1200 fish. Euler's method with step size h = 0.5 is used to estimate P.

(a)

Show that Euler's method gives P(0.5) = 1244.

(2)

(b)

Hence estimate P(1.5). Use unrounded values throughout and give your answer to the nearest whole number.

(3)

(c)

Explain why using a smaller step size would usually give a more accurate estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 dP/dt at t = 0 is 0.4(1200)(1 − 0.4) − 200 = 88	✓ 1
A1 P(0.5) = 1200 + 0.5 × 88 = 1244 (shown)	✓ 1
Part (b)
M1 gradient at (0.5, 1244) = 91.26...	✓ 1
M1 P(1.0) = 1244 + 0.5 × 91.26... = 1289.63... and gradient 94.10...	✓ 1
A1 P(1.5) = 1337 (1336.68...)	✓ 1
Part (c)
R1 each step assumes the gradient is constant over the step, and a smaller step means the gradient changes less within each step	✓ 1

Common error: At each step, use the new values of both t and P to find the next gradient. Reusing the first gradient for every step is a common mistake.

+ Add to test✓ In your test247IBMI3H-5-04r29-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

Calculator

A fashion designer cuts a decorative panel from fabric. In a coordinate system with units of metres, the panel is the region between the curve y = 2x − x² and the line y = 0.5x.

(a)

Find the x-coordinates of the points where the curve and the line meet.

(2)

(b)

Find the area of the panel.

(3)

(c)

The fabric costs €48 per square metre. Find the cost of the fabric for one panel, assuming no waste.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sets 2x − x² = 0.5x, so x(1.5 − x) = 0	✓ 1
A1 x = 0 and x = 1.5	✓ 1
Part (b)
M1 area = ∫₀^1.5 [(2x − x²) − 0.5x] dx	✓ 1
M1 integrates to 0.75x² − x³/3 and substitutes limits	✓ 1
A1 0.5625 m²	✓ 1
Part (c)
A1 €27.00	✓ 1

Common error: Always subtract the lower function from the upper function. Check which function is higher by testing a value of x between the intersection points.

+ Add to test✓ In your test248IBMI3H-5-04r31-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

The rate at which a wheat crop takes up nitrogen is modelled by u(t) = te^−0.2t, for t ≥ 0, where u is in kg per hectare per day and t is the number of days after application of the fertiliser.

(a)

Find du/dt.

(2)

(b)

Find the value of t at which the rate of uptake is greatest.

(2)

(c)

Find the greatest rate of uptake. Give your answer to 3 significant figures.

(1)

(d)

Justify that this value is a maximum.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of the product rule	✓ 1
A1 du/dt = e^−0.2t − 0.2te^−0.2t (= e^−0.2t(1 − 0.2t))	✓ 1
Part (b)
M1 setting du/dt = 0	✓ 1
A1 t = 5	✓ 1
Part (c)
A1 1.84 kg per hectare per day	✓ 1
Part (d)
R1 du/dt changes from positive to negative at t = 5 (or the second derivative is negative there)	✓ 1

Exam tip: Setting a derivative to zero finds stationary points; add a sentence (sign change or second derivative) to show which type it is.

+ Add to test✓ In your test249IBMI3H-5-04r31-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

After a field is sprayed, the concentration of a pesticide in a drainage stream is modelled by C(t) = 12t/(t² + 9), for t ≥ 0, where C is in mg per litre and t is the time in hours after spraying.

(a)

Show that dC/dt = 12(9 − t²)/(t² + 9)².

(3)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

Find the greatest concentration of pesticide in the stream.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of the quotient rule	✓ 1
A1 numerator 12(t² + 9) − 12t × 2t	✓ 1
A1 simplifying to 12(9 − t²) over (t² + 9)² (AG)	✓ 1
Part (b)
M1 setting dC/dt = 0	✓ 1
A1 t = 3 hours	✓ 1
Part (c)
A1 2 mg per litre	✓ 1

Common error: In the quotient rule the order matters: it is (bottom × derivative of top − top × derivative of bottom) divided by bottom squared.

+ Add to test✓ In your test250IBMI3H-5-04r31-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A crop-spraying drone flies along a straight row of plants. Its velocity, in m s⁻¹, is modelled by v(t) = 6 sin(0.3t) for 0 ≤ t ≤ T, where t is the time in seconds and T is the first time after t = 0 at which the drone is at rest. Angles are in radians.

(a)

Find the value of T. Give your answer to 3 significant figures.

(2)

(b)

Find the acceleration of the drone when t = 2. Give your answer to 3 significant figures.

(2)

(c)

Find the distance travelled by the drone from t = 0 until it comes to rest at t = T. Use the exact value of T = π/0.3.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 setting 6 sin(0.3T) = 0 with 0.3T = π	✓ 1
A1 T = 10.5 s	✓ 1
Part (b)
M1 differentiating, a = 1.8 cos(0.3t)	✓ 1
A1 1.49 m s⁻²	✓ 1
Part (c)
M1 setting up the integral of 6 sin(0.3t) from 0 to T	✓ 1
A1 antiderivative −20 cos(0.3t)	✓ 1
A1 20 + 20 = 40 m	✓ 1

Common error: Integrating sin(kt) gives −cos(kt)/k, so the factor 1/0.3 must appear, together with the minus sign.

+ Add to test✓ In your test251IBMI3H-5-04r31-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A seed hopper moves along a straight rail in a greenhouse. Its velocity, in m s⁻¹, is modelled by v(t) = 0.05t² − 0.4t + 0.6 for 0 ≤ t ≤ 8, where t is the time in seconds. Positive velocity means moving to the right.

(a)

Find the times at which the hopper is instantaneously at rest.

(2)

(b)

Find the acceleration of the hopper when t = 1.

(2)

(c)

Find the total distance travelled by the hopper from t = 0 to t = 6. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 setting 0.05t² − 0.4t + 0.6 = 0 (or t² − 8t + 12 = 0)	✓ 1
A1 t = 2 and t = 6	✓ 1
Part (b)
M1 differentiating, a = 0.1t − 0.4	✓ 1
A1 −0.3 m s⁻²	✓ 1
Part (c)
M1 recognising that the motion reverses at t = 2, so the distances in 0 to 2 and 2 to 6 must be found separately (as positive values)	✓ 1
M1 evaluating the integral of v from 0 to 2 as 0.533 (the integral from 2 to 6 is −0.533)	✓ 1
A1 total distance 1.07 m	✓ 1

Common error: The integral of velocity gives displacement, not distance; if the velocity changes sign you must split the interval and add the magnitudes.

+ Add to test✓ In your test252IBMI3H-5-04r31-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The cross-section of a drainage channel in a field has depth d metres at a horizontal distance x metres from one bank, modelled by d(x) = 0.6x − 0.15x², for 0 ≤ x ≤ 4. The area of the cross-section is the area between the curve d = d(x) and the x-axis.

(a)

Use the trapezoidal rule with 4 intervals of equal width to estimate the area of the cross-section.

(3)

(b)

Find the exact area of the cross-section by integration.

(3)

(c)

Explain why the trapezoidal estimate is smaller than the exact area.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 correct use of the trapezoidal rule with width 1	✓ 1
A1 depths 0, 0.45, 0.6, 0.45, 0 used	✓ 1
A1 1.5 m²	✓ 1
Part (b)
M1 setting up the integral of 0.6x − 0.15x² from 0 to 4	✓ 1
A1 antiderivative 0.3x² − 0.05x³	✓ 1
A1 1.6 m²	✓ 1
Part (c)
R1 the curve is concave down, so each chord lies below the curve	✓ 1

Exam tip: Check a trapezoidal estimate against the shape of the curve: on a concave-down curve the rule underestimates the area.

+ Add to test✓ In your test253IBMI3H-5-04r31-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

Water flows into a storage tank at a rate of r(t) = te^−0.1t² thousand litres per hour, where t is the time in hours, for 0 ≤ t ≤ 4. The volume of water entering the tank in the first 4 hours is V = ∫₀⁴ te^−0.1t² dt.

(a)

Use the substitution u = 0.1t² to show that V = 5(1 − e^−1.6).

(4)

(b)

Calculate V to 3 significant figures.

(1)

(c)

Find the mean rate of inflow over the 4 hours. Use the unrounded value of V and give your answer to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du = 0.2t dt, so t dt = 5 du	✓ 1
A1 new limits u = 0 and u = 1.6	✓ 1
M1 integrating 5e^−u to get [−5e^−u]	✓ 1
A1 5(1 − e^−1.6) (AG)	✓ 1
Part (b)
A1 3.99 thousand litres	✓ 1
Part (c)
A1 0.998 thousand litres per hour	✓ 1

Exam tip: When you change variables in a definite integral, change the limits too, then you never need to convert back to the original variable.

+ Add to test✓ In your test254IBMI3H-5-04r31-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

During a descent, the cabin pressure, p kPa, of an aircraft satisfies dp/dt = −0.16(p − 60), where t is the time in minutes. The pressure is 100 kPa when t = 0.

(a)

Use Euler's method with a step size of 0.5 to find an estimate for p when t = 0.5.

(2)

(b)

Continue to find an estimate for p when t = 1. Give your answer to 3 significant figures.

(2)

(c)

The exact solution is p = 60 + 40e^−0.16t. Calculate the percentage error in the Euler estimate for p at t = 1, using the unrounded Euler value 93.856. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 p₁ = 100 + 0.5 × (−0.16)(100 − 60)	✓ 1
A1 96.8 kPa	✓ 1
Part (b)
M1 p₂ = 96.8 + 0.5 × (−0.16)(96.8 − 60)	✓ 1
A1 93.9 kPa	✓ 1
Part (c)
M1 \|93.856 − (60 + 40e^−0.16)\| / (60 + 40e^−0.16) × 100, using exact value 94.0858	✓ 1
A1 0.244 %	✓ 1

Exam tip: In Euler's method each new value uses the previous estimate: pn+₁ = pn + h × (dp/dt evaluated at pn).

+ Add to test✓ In your test255IBMI3H-5-06r32-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The concentration of a drug in a patient's blood is $C(t)=\dfrac{20t}{t^2+4}$ mg/L, where $t\ge0$ is the time in hours after an injection.

(a)

Use the quotient rule to show that $\dfrac{dC}{dt}=\dfrac{20(4-t^2)}{(t^2+4)^2}$.

(3)

(b)

Hence show that the concentration is greatest when $t=2$.

(2)

(c)

Find the greatest concentration.

(1)

(d)

Find the rate of change of concentration at $t=4$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule $\frac{vu'-uv'}{v^2}$	✓ 1
A1 numerator $20(t^2+4)-20t\times2t$	✓ 1
A1 simplifies to $20(4-t^2)$ over $(t^2+4)^2$	✓ 1
Part (b)
M1 sets $\frac{dC}{dt}=0$	✓ 1
A1 $t=2$ (as $t>0$) with sign change from positive to negative, or $C''<0$	✓ 1
Part (c)
A1 $C(2)=5$ mg/L	✓ 1
Part (d)
A1 $\frac{20(4-16)}{(20)^2}=-0.6$ mg/L per hour	✓ 1

Exam tip: In a "show that" question write every line of the working; do not skip from the quotient rule to the simplified result.

+ Add to test✓ In your test256IBMI3H-5-06r32-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

New players join an online game at a rate $R(t)=300te^{-0.1t^2}$ players per hour, where $t$ is the time in hours since a new level was released.

(a)

Use the substitution $u=0.1t^2$ to show that $\int R(t)\,dt=-1500e^{-0.1t^2}+c$.

(3)

(b)

Find the number of players who join in the first 5 hours, to the nearest whole number.

(2)

(c)

Find the limit of the total number of players who ever join, as $t\to\infty$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{du}{dt}=0.2t$, so $300t\,dt=1500\,du$	✓ 1
A1 $\int1500e^{-u}\,du$	✓ 1
A1 $-1500e^{-u}+c=-1500e^{-0.1t^2}+c$	✓ 1
Part (b)
M1 $1500\left(1-e^{-2.5}\right)$ using limits 0 and 5	✓ 1
A1 $1377$	✓ 1
Part (c)
A1 $1500$ (since $e^{-0.1t^2}\to0$)	✓ 1

Exam tip: With a substitution, either change the limits to the new variable or return to $t$ before applying them; never mix the two.

+ Add to test✓ In your test257IBMI3H-5-06r32-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

In a game, a player's skill rating $y$ changes with the level $x$ according to the differential equation $\dfrac{dy}{dx}=x-y$.

(a)

Find the gradient of the slope field at the points $(0, 2)$ and $(3, 1)$.

(2)

(b)

Describe the set of points where the slope field has zero gradient.

(1)

(c)

Show that $y=x-1$ is a solution of the differential equation.

(2)

(d)

Use Euler's method with step size 0.5, starting at $(0, 2)$, to estimate $y$ when $x=1$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $-2$ at $(0,2)$	✓ 1
A1 $2$ at $(3,1)$	✓ 1
Part (b)
A1 the straight line $y=x$	✓ 1
Part (c)
M1 $\frac{dy}{dx}=1$	✓ 1
R1 $x-(x-1)=1$, so both sides are equal	✓ 1
Part (d)
M1 $y_1=2+0.5(0-2)=1$ and then $y_2=y_1+0.5(0.5-y_1)$	✓ 1
A1 $0.75$	✓ 1

Exam tip: In Euler's method update both coordinates each step: $x_{n+1}=x_n+h$ and $y_{n+1}=y_n+h\,f(x_n,y_n)$.

+ Add to test✓ In your test258IBMI3H-5-06r32-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

The height $x$ metres of the water surface in a surge tank above its rest level satisfies $\dfrac{d^2x}{dt^2}=-4x-0.5\dfrac{dx}{dt}$, where $t$ is the time in seconds. At $t=0$, $x=1$ and $\dfrac{dx}{dt}=0$.

(a)

Let $v=\dfrac{dx}{dt}$. Write the equation as a pair of first-order differential equations.

(2)

(b)

Use Euler's method with step size 0.1 to find $x$ and $v$ when $t=0.1$.

(2)

(c)

Continue to estimate $x$ and $v$ when $t=0.2$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dx}{dt}=v$	✓ 1
A1 $\frac{dv}{dt}=-4x-0.5v$	✓ 1
Part (b)
M1 $x_1=1+0.1\times0$ and $v_1=0+0.1(-4\times1-0.5\times0)$	✓ 1
A1 $x_1=1$, $v_1=-0.4$	✓ 1
Part (c)
M1 $x_2=1+0.1\times(-0.4)$ and $v_2=-0.4+0.1(-4\times1-0.5\times(-0.4))$ (ECF from (b))	✓ 1
A1 $x_2=0.96$ m, $v_2=-0.78$ m/s	✓ 1

Common error: Each Euler step must use the old values of both $x$ and $v$; do not use the new $x$ when updating $v$.

+ Add to test✓ In your test259IBMI3H-5-06r32-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

An open water tank has a square base of side $x$ metres and height $h$ metres, and must hold exactly 12 m$^3$. The base and four vertical sides are made of sheet metal of total area $A$ m$^2$.

(a)

Show that $A=x^2+\dfrac{48}{x}$.

(2)

(b)

Find the value of $x$ that minimises $A$.

(3)

(c)

Justify that this value gives a minimum.

(1)

(d)

Find the minimum area of sheet metal, to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $x^2h=12$ so $h=\frac{12}{x^2}$	✓ 1
A1 $A=x^2+4xh=x^2+\frac{48}{x}$	✓ 1
Part (b)
M1 $\frac{dA}{dx}=2x-\frac{48}{x^2}$	✓ 1
M1 sets $\frac{dA}{dx}=0$	✓ 1
A1 $x^3=24$, $x=2.88$ m	✓ 1
Part (c)
R1 $\frac{d^2A}{dx^2}=2+\frac{96}{x^3}>0$ at $x=2.88$	✓ 1
Part (d)
A1 $25.0$ m$^2$	✓ 1

Exam tip: Reduce the area to a function of one variable using the volume condition before differentiating.

+ Add to test✓ In your test260IBMI3H-5-06r32-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

The depth of a river is measured across its width at 2 m intervals, starting at one bank:

Distance from bank (m)	0	2	4	6	8
Depth (m)	0	1.2	1.9	1.5	0

(a)

Use the trapezoidal rule to estimate the area of the river's cross-section.

(3)

(b)

The average water speed is 0.8 m/s. Estimate the volume flow rate, using your answer to (a).

(1)

(c)

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

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{h}{2}\left[y_0+y_n+2(y_1+\dots)\right]$	✓ 1
M1 $\frac{2}{2}[0+0+2(1.2+1.9+1.5)]$	✓ 1
A1 $9.2$ m$^2$	✓ 1
Part (b)
A1 $9.2\times0.8=7.36$ m$^3$/s (ECF from (a))	✓ 1
Part (c)
A1 take depth measurements at smaller intervals (more strips)	✓ 1

Common error: With 5 measurements there are 4 strips: the width $h$ is the spacing between readings, not the total width.

+ Add to test✓ In your test261IBMI3H-5-07r32-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

A swimming pool is cooled by the night air. Its temperature $T\,^\circ$C at time $t$ hours satisfies $\frac{dT}{dt}=-0.05(T-18)$ for $T>18$. At $t=0$ the temperature is $30\,^\circ$C.

(a)

Solve the differential equation to show that $T=18+12e^{-0.05t}$.

(4)

(b)

Find the time taken for the temperature to fall to $24\,^\circ$C, giving your answer in hours to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 separate variables: $\int\frac{1}{T-18}dT=\int-0.05\,dt$	✓ 1
A1 $\ln\|T-18\|=-0.05t+c$	✓ 1
M1 use $T=30$ at $t=0$ to find $c=\ln 12$	✓ 1
A1 $T-18=12e^{-0.05t}$, hence the result	✓ 1
Part (b)
M1 $24=18+12e^{-0.05t}$, so $e^{-0.05t}=0.5$	✓ 1
A1 $t=13.9$ hours	✓ 1

Exam tip: Keep $\ln|T-18|$ until you have used the initial condition; moving to exponential form early with an unknown constant is where signs often go astray.

+ Add to test✓ In your test262IBMI3H-5-07r32-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A swimmer's speed during the first eight seconds of a sprint is modelled by $v(t)=\frac{6t}{t^2+4}$ m s⁻¹, for $0\le t\le 8$.

(a)

Use the substitution $u=t^2+4$ to show that the distance swum in the first eight seconds is $3\ln 17$ metres.

(4)

(b)

Find this distance to 3 significant figures.

(1)

(c)

Calculate the swimmer's mean speed over the first eight seconds, to 3 significant figures. Use the exact value $3\ln17$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{du}{dt}=2t$, so $6t\,dt=3\,du$	✓ 1
A1 integral becomes $3\int\frac{1}{u}\,du$	✓ 1
A1 limits change to $u=4$ and $u=68$	✓ 1
A1 $3\ln 68-3\ln 4=3\ln 17$	✓ 1
Part (b)
A1 $8.50$ m	✓ 1
Part (c)
M1 distance divided by time: $\frac{3\ln 17}{8}$	✓ 1
A1 $1.06$ m s⁻¹	✓ 1

Common error: When you substitute in a definite integral, either change the limits to $u$-values or return to $t$ before substituting; mixing the two gives wrong answers.

+ Add to test✓ In your test263IBMI3H-5-07r32-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

The power $P$ megawatts produced by a wind turbine at wind speed $v$ m s⁻¹ is modelled, for $v>1.5$, by $P(v)=8\ln(2v-3)$.

(a)

Find $P'(v)$.

(2)

(b)

Find the equation of the tangent to the curve at $v=6$.

(3)

(c)

Use the tangent to estimate the power at $v=6.3$, to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule: $8\times\frac{2}{2v-3}$	✓ 1
A1 $P'(v)=\frac{16}{2v-3}$	✓ 1
Part (b)
A1 $P(6)=8\ln 9$	✓ 1
A1 $P'(6)=\frac{16}{9}$	✓ 1
A1 $P-8\ln9=\frac{16}{9}(v-6)$	✓ 1
Part (c)
A1 $8\ln9+\frac{16}{9}\times0.3=18.1$ MW	✓ 1

Exam tip: Leave the point of tangency in exact form, such as $8\ln9$, until the final step to avoid rounding errors.

+ Add to test✓ In your test264IBMI3H-5-08r32-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

A strand of kelp has length $L$ metres at time $t$ days, and grows according to $\frac{dL}{dt}=0.6\sqrt{L}$. At $t=0$ the length is 4 m.

(a)

Solve the differential equation to show that $L=(0.3t+2)^2$.

(4)

(b)

Find the time taken for the kelp to reach 16 m, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 separate variables: $\int L^{-1/2}\,dL=\int0.6\,dt$	✓ 1
A1 $2\sqrt{L}=0.6t+c$	✓ 1
M1 use $L=4$ at $t=0$: $c=4$	✓ 1
A1 $\sqrt{L}=0.3t+2$, so $L=(0.3t+2)^2$	✓ 1
Part (b)
M1 $(0.3t+2)^2=16$, so $0.3t+2=4$	✓ 1
A1 $t=6.67$ days	✓ 1

Common error: When integrating $L^{-1/2}$, add one to the power and divide by the new power: the result is $2L^{1/2}$, not $L^{1/2}$.

+ Add to test✓ In your test265IBMI3H-5-08r32-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A window is a rectangle of width $2r$ metres and height $h$ metres, topped by a semicircle of radius $r$ metres. The outer frame, which runs along the base, the two vertical sides and the curved top, has total length 12 m.

(a)

Show that the area of the window is $A=12r-\left(2+\frac{\pi}{2}\right)r^2$.

(3)

(b)

Find the value of $r$ that maximises the area, to 3 significant figures.

(2)

(c)

Find the maximum area, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frame length $2r+2h+\pi r=12$, so $h=6-r-\frac{\pi r}{2}$	✓ 1
M1 $A=2rh+\frac{\pi r^2}{2}$ with $h$ substituted	✓ 1
A1 $A=12r-2r^2-\frac{\pi r^2}{2}$, as required	✓ 1
Part (b)
M1 $\frac{dA}{dr}=12-(4+\pi)r=0$	✓ 1
A1 $r=\frac{12}{4+\pi}=1.68$ m	✓ 1
Part (c)
M1 substitute the unrounded $r$ into $A$	✓ 1
A1 $A=\frac{72}{4+\pi}=10.1$ m²	✓ 1

Common error: Use the unrounded optimal value of $r$ when finding the maximum area; substituting 1.68 can alter the third significant figure.

+ Add to test✓ In your test266IBMI3H-5-08r32-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

An architect designs a concrete spire by rotating the curve $y=2e^{-0.4x}$, for $0\le x\le L$, through $360^\circ$ about the $x$-axis. Lengths are in metres.

(a)

Show that the volume of the spire is $5\pi\left(1-e^{-0.8L}\right)$ cubic metres.

(3)

(b)

Find the length $L$ for which the spire has volume 12 m³, to 3 significant figures.

(3)

(c)

State the greatest volume that a spire of this shape could have, to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $V=\pi\int_0^L\left(2e^{-0.4x}\right)^2dx=\pi\int_0^L4e^{-0.8x}dx$	✓ 1
A1 $\pi\left[-5e^{-0.8x}\right]_0^L$	✓ 1
A1 $5\pi\left(1-e^{-0.8L}\right)$	✓ 1
Part (b)
M1 $5\pi\left(1-e^{-0.8L}\right)=12$	✓ 1
M1 $e^{-0.8L}=1-\frac{12}{5\pi}$ and take logarithms	✓ 1
A1 $L=1.80$ m	✓ 1
Part (c)
A1 $5\pi=15.7$ m³ (as $L\to\infty$, $e^{-0.8L}\to0$)	✓ 1

Common error: When squaring $2e^{-0.4x}$ for a volume of revolution, square both the 2 and the exponent: the result is $4e^{-0.8x}$.

+ Add to test✓ In your test267IBMI3H-5-08r32-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A whale shark takes in plankton at a rate $r(t)=6te^{-t^2/4}$ kilograms per hour, $t$ hours after it begins feeding.

(a)

Use the substitution $u=\frac{t^2}{4}$ to show that the plankton taken in during the first four hours is $12\left(1-e^{-4}\right)$ kg.

(4)

(b)

Calculate this amount, to 3 significant figures.

(1)

(c)

Find the time at which the feeding rate is greatest, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{du}{dt}=\frac{t}{2}$, so $6t\,dt=12\,du$	✓ 1
A1 integral becomes $12\int e^{-u}\,du$	✓ 1
A1 limits change to $u=0$ and $u=4$	✓ 1
A1 $12\left(1-e^{-4}\right)$	✓ 1
Part (b)
A1 $11.8$ kg	✓ 1
Part (c)
M1 differentiate with the product and chain rules: $r'(t)=6e^{-t^2/4}\left(1-\frac{t^2}{2}\right)$ and set to zero	✓ 1
A1 $t=\sqrt2=1.41$ hours	✓ 1

Common error: In a definite integral with substitution, change the limits at the same time as the variable; leaving them as $t$-values gives the wrong region.

+ Add to test✓ In your test268IBMI3H-5-06r34-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

The number of visitors per hour arriving at a library t hours after it opens is modelled by V(t) = 120t ÷ (t² + 4), for 0 ≤ t ≤ 8.

(a)

Use the quotient rule to show that V′(t) = 120(4 − t²) ÷ (t² + 4)².

(3)

(b)

Find the time at which the arrival rate is greatest, and the greatest arrival rate.

(3)

(c)

Find the rate of change of the arrival rate when t = 1.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Quotient rule with u = 120t and v = t² + 4	✓ 1
V′ = [120(t² + 4) − 120t·2t] ÷ (t² + 4)²	✓ 1
= 120(4 − t²) ÷ (t² + 4)²	✓ 1
Part (b)
V′ = 0 gives t = 2	✓ 1
V(2) = 240 ÷ 8 = 30 visitors per hour	✓ 1
Sign of V′ changes from positive to negative at t = 2 (maximum)	✓ 1
Part (c)
V′ at t = 1 is 120 × 3 ÷ 25	✓ 1
= 14.4 visitors per hour per hour	✓ 1

Exam tip: When a question says 'show that', write every line of the quotient-rule working before simplifying; the final form is given, so the steps carry the marks.

+ Add to test✓ In your test269IBMI3H-5-06r34-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

The power output of a wind turbine at time t hours is P′(t) = 0.06t√(t² + 9) megawatts, 0 ≤ t ≤ 4, where P(t) is the energy in megawatt-hours produced up to time t.

(a)

Use the substitution u = t² + 9 to find ∫ 0.06t√(t² + 9) dt.

(3)

(b)

Hence find the total energy produced during the first 4 hours.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
du = 2t dt, so the integral becomes 0.03∫u^1/2 du	✓ 1
= 0.02u^3/2 + C	✓ 1
= 0.02(t² + 9)^3/2 + C	✓ 1
Part (b)
Limits become u = 9 and u = 25 (or substitute back with t = 0 and 4)	✓ 1
0.02(25^3/2 − 9^3/2) = 0.02(125 − 27)	✓ 1
= 1.96 MWh	✓ 1

Common error: When you change variable in a definite integral, change the limits too, or substitute back before using the original limits; mixing the two gives the wrong value.

+ Add to test✓ In your test270IBMI3H-5-06r34-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A float for a swimming-lane rope is made by rotating the region bounded by the curve y = 3e^−0.1x, the x-axis, and the lines x = 0 and x = 10 through 360° about the x-axis. Lengths are in cm.

(a)

Write down an integral for the volume V of the float.

(2)

(b)

Find the exact value of V, and give it in cm³ to 3 significant figures.

(3)

(c)

The float material has density 0.2 g cm⁻³. Using the unrounded value of V, find the mass of the float to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
V = π∫₀¹⁰ y² dx	✓ 1
V = π∫₀¹⁰ 9e^−0.2x dx	✓ 1
Part (b)
∫ 9e^−0.2x dx = −45e^−0.2x	✓ 1
V = 45π(1 − e⁻²)	✓ 1
V = 122 cm³	✓ 1
Part (c)
Mass = density × volume = 0.2 × 122.24…	✓ 1
= 24.4 g	✓ 1

Exam tip: Square the whole function before integrating; (3e^(−0.1x))² = 9e^(−0.2x), not 3e^(−0.2x).

+ Add to test✓ In your test271IBMI3H-5-06r34-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

In a model of an additive in a swimming pool, the concentration y mg per litre after x hours satisfies dy/dx = x − y/2.

(a)

Find the gradient of the slope field at the points (0, 2) and (2, 4).

(2)

(b)

Show that y = 2x − 4 is a solution of the differential equation.

(2)

(c)

A solution curve passes through (0, 0). Find the value of d²y/dx² at this point and state what it shows about the curve there.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
At (0, 2): 0 − 1 = −1	✓ 1
At (2, 4): 2 − 2 = 0	✓ 1
Part (b)
dy/dx = 2 for y = 2x − 4	✓ 1
x − (2x − 4)/2 = x − x + 2 = 2, which equals dy/dx	✓ 1
Part (c)
Differentiate implicitly: d²y/dx² = 1 − (1/2)dy/dx	✓ 1
At (0, 0), dy/dx = 0, so d²y/dx² = 1	✓ 1
Positive, so the curve is concave up (a local minimum) at (0, 0)	✓ 1

Exam tip: To verify a proposed solution, differentiate it and substitute both y and y′ into the differential equation.

+ Add to test✓ In your test272IBMI3H-5-06r34-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The concentration C ppm of chlorine in a pool satisfies dC/dt = −0.1C², where t is the time in hours. Initially C = 2.

(a)

Solve the differential equation to show that C = 1 ÷ (0.1t + 0.5).

(4)

(b)

Find the concentration after 10 hours, to 3 significant figures.

(1)

(c)

Find the time taken for the concentration to fall to 0.5 ppm.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Separate variables: ∫C⁻² dC = ∫−0.1 dt	✓ 1
−1/C = −0.1t + k	✓ 1
Use C = 2 at t = 0 to get k = −0.5	✓ 1
C = 1 ÷ (0.1t + 0.5)	✓ 1
Part (b)
C = 1 ÷ 1.5 = 0.667 ppm	✓ 1
Part (c)
1/0.5 = 0.1t + 0.5	✓ 1
t = 15 hours	✓ 1

Common error: Do not forget the constant of integration before using the initial condition; it fixes the particular solution.

+ Add to test✓ In your test273IBMI3H-5-06r34-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

Two library branches lend a popular set of books. Let x and y be the numbers of these books (in hundreds) at branch A and branch B respectively after t days. Books move between branches according to
dx/dt = −0.2x + 0.1y, dy/dt = 0.2x − 0.1y,
with x = 50 and y = 10 when t = 0.

(a)

Show that the total x + y stays constant.

(1)

(b)

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

(4)

(c)

Find the long-term values of x and y. Hence state the long-term number of books, in hundreds, at branch B.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
d(x + y)/dt = (−0.2x + 0.1y) + (0.2x − 0.1y) = 0	✓ 1
Part (b)
Step 1: dx/dt = −9 and dy/dt = 9 at (50, 10)	✓ 1
x = 45.5 and y = 14.5 at t = 0.5	✓ 1
Step 2 gradients: dx/dt = −7.65 and dy/dt = 7.65	✓ 1
x = 41.675 and y = 18.325 at t = 1	✓ 1
Part (c)
Equilibrium: 0.2x = 0.1y, so y = 2x	✓ 1
Total is 60, so 3x = 60	✓ 1
x = 20 and y = 40 (hundreds), i.e. 4000 books at B	✓ 1

Exam tip: In coupled Euler steps, compute both gradients from the old values before updating either variable.

+ Add to test✓ In your test274IBMI3H-5-07r34-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

In a video game, a robot moves along a straight rail. Its velocity at time t seconds after it starts is $v(t) = 8e^{-0.5t} - 3$ m s⁻¹, for $t \ge 0$. A positive velocity means motion to the right. The robot starts at the origin.

(a)

Find the acceleration of the robot when $t=1$, to three significant figures.

(2)

(b)

Find the time at which the robot momentarily stops. Give your answer to four significant figures.

(2)

(c)

Find the greatest displacement of the robot to the right of its starting point, to three significant figures. Use the unrounded stopping time from (b).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $a(t) = \frac{dv}{dt} = -4e^{-0.5t}$	✓ 1
A1 acceleration at $t=1$ is $-2.43$ m s⁻² (−2.426…)	✓ 1
Part (b)
M1 set $8e^{-0.5t} = 3$	✓ 1
A1 $t = 2\ln\frac{8}{3} = 1.962$ s (1.9617…)	✓ 1
Part (c)
M1 $s(t) = \int_0^{t}(8e^{-0.5u}-3)\,du$ with upper limit equal to the stopping time	✓ 1
A1 $s = 16(1 - e^{-0.5t}) - 3t$ (or correct value from GDC integration)	✓ 1
A1 4.11 m (4.1150… before rounding)	✓ 1

Common error: Using the rounded time 1.962 in the integral can change the third significant figure; store the exact value on the calculator.

+ Add to test✓ In your test275IBMI3H-5-07r34-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

A flared storage vessel in a water-supply system is modelled by rotating the region bounded by the curve $y = 0.5e^{0.3x}$, the x-axis, the y-axis and the line $x=4$ through $360°$ about the x-axis. Both x and y are measured in metres.

(a)

Write down an integral that gives the volume of the vessel.

(2)

(b)

Find the volume of the vessel in cubic metres, correct to three significant figures.

(2)

(c)

Find the capacity of the vessel in litres, to three significant figures. Use the unrounded volume from (b). (1 m³ = 1000 litres)

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $V = \pi\int y^2\,dx$	✓ 1
A1 $V = \pi\int_0^4 (0.5e^{0.3x})^2\,dx$ (limits and integrand correct)	✓ 1
Part (b)
M1 attempt to evaluate, e.g. $\frac{0.25\pi}{0.6}\left(e^{2.4}-1\right)$ or by GDC	✓ 1
A1 13.1 m³ (13.120… before rounding)	✓ 1
Part (c)
A1 13100 litres (13120… before rounding)	✓ 1

Exam tip: Remember to square the whole function, including its constant, before integrating: $(0.5e^{0.3x})^2 = 0.25e^{0.6x}$.

+ Add to test✓ In your test276IBMI3H-5-07r34-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A slow-release tablet delivers a drug into the bloodstream at a rate of $r(t) = 12te^{-0.5t^2}$ mg per hour, where t is the time in hours after the tablet is swallowed.

(a)

Use the substitution $u = 0.5t^2$ to show that $\int_0^3 12te^{-0.5t^2}\,dt = 12\int_0^{4.5} e^{-u}\,du$.

(4)

(b)

Find the total mass of drug delivered during the first 3 hours, to three significant figures.

(2)

(c)

The tablet eventually delivers $12$ mg in total as $t \to \infty$. Find the percentage of this total delivered in the first 3 hours, to three significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{du}{dt} = t$ (or $du = t\,dt$)	✓ 1
A1 $12t\,dt = 12\,du$ so the integrand becomes $12e^{-u}$	✓ 1
M1 new limits: $t=0 \Rightarrow u=0$ and $t=3 \Rightarrow u=4.5$	✓ 1
A1 result shown (AG)	✓ 1
Part (b)
M1 $12\left[-e^{-u}\right]_0^{4.5} = 12(1 - e^{-4.5})$	✓ 1
A1 11.9 mg (11.867… before rounding)	✓ 1
Part (c)
M1 $\frac{12(1-e^{-4.5})}{12} \times 100$ (or use of unrounded value from (b))	✓ 1
A1 98.9%	✓ 1

Exam tip: After a substitution, change the limits at the same time; then you never need to return to the original variable.

+ Add to test✓ In your test277IBMI3H-5-07r34-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Water flows into a treatment tank at a rate of $r(t) = 5 + 3\ln(t+1)$ litres per minute, where t is the time in minutes, $0 \le t \le 8$.

(a)

Use the trapezoidal rule with four intervals of equal width to estimate the volume of water that enters the tank in the first 8 minutes. Give your answer to three significant figures.

(3)

(b)

Use your graphic display calculator to find the exact volume entering the tank, $\int_0^8 r(t)\,dt$, to three significant figures.

(2)

(c)

Explain why the trapezoidal estimate is smaller than the exact volume.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 interval width 2 and correct values $r(0)=5$, $r(2)=8.2958$, $r(4)=9.8283$, $r(6)=10.8377$, $r(8)=11.5917$ (4 d.p.)	✓ 1
M1 $\frac{2}{2}\left[r(0) + 2(r(2)+r(4)+r(6)) + r(8)\right]$	✓ 1
A1 74.5 litres (74.515… before rounding)	✓ 1
Part (b)
M1 definite integral of $r(t)$ from 0 to 8 evaluated	✓ 1
A1 75.3 litres (75.325… before rounding)	✓ 1
Part (c)
M1 $r''(t) = -\frac{3}{(t+1)^2} < 0$, so the graph is concave down	✓ 1
R1 the straight-line tops of the trapezoids lie below the curve, so the area is underestimated	✓ 1

Exam tip: Check the sign of the second derivative to decide whether the trapezoidal rule over- or underestimates.

+ Add to test✓ In your test278IBMI3H-5-07r34-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A drug is given as a single dose of 100 mg. Let x mg be the amount in the blood plasma and y mg the amount in body tissue, t hours after the dose. The amounts satisfy the coupled equations
$\frac{dx}{dt} = -0.5x + 0.2y,\qquad \frac{dy}{dt} = 0.5x - 0.2y.$
Initially $x = 100$ and $y = 0$.

(a)

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

(3)

(b)

Show that $\frac{d(x+y)}{dt} = 0$, and state what this means in context.

(2)

(c)

In the long term, both rates of change tend to zero. Find the amount of drug in the plasma in the long term, to three significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 one correct Euler step applied to both equations	✓ 1
A1 after the first step $x=75$, $y=25$	✓ 1
A1 $x = 58.75$, $y = 41.25$ when $t=1$	✓ 1
Part (b)
M1 add the two equations to get $\frac{dx}{dt}+\frac{dy}{dt}=0$	✓ 1
R1 the total amount of drug in the body stays constant at 100 mg (none is removed in this model)	✓ 1
Part (c)
M1 set $\frac{dx}{dt}=0$, giving $0.5x = 0.2y$	✓ 1
M1 substitute $y = 100 - x$ (from (b)) to get $0.5x = 0.2(100-x)$, so $0.7x = 20$	✓ 1
A1 28.6 mg (28.571… before rounding)	✓ 1

Exam tip: In coupled Euler steps, compute both gradients from the old pair before updating either variable.

+ Add to test✓ In your test279IBMI3H-5-07r34-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

A decorative vase is formed by rotating the curve $x = 2e^{y/10}$, for $0 \le y \le 20$, through $360°$ about the y-axis. Lengths are in centimetres.

(a)

Find the radius of the top of the vase, to three significant figures.

(1)

(b)

Write down an integral for the volume of the vase.

(2)

(c)

Find the volume of the vase in litres, to three significant figures. (1 litre = 1000 cm³)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $2e^{2} = 14.8$ cm (14.778…)	✓ 1
Part (b)
M1 $V=\pi\int x^2\,dy$	✓ 1
A1 $V = \pi\int_0^{20} 4e^{y/5}\,dy$	✓ 1
Part (c)
M1 evaluate, e.g. $20\pi(e^4 - 1)$ cm³ (3367.7… cm³)	✓ 1
A1 3.37 litres	✓ 1

Exam tip: When rotating about the y-axis, rewrite the curve as x in terms of y and integrate with respect to y.

+ Add to test✓ In your test280IBMI3H-5-07r35-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A test cart on a straight track has velocity $v(t)=\frac{12t}{t^2+4}$ m s⁻¹, for $0\le t\le 10$, where $t$ is the time in seconds after the cart starts from rest.

(a)

Show that the acceleration of the cart is $\frac{12(4-t^2)}{(t^2+4)^2}$.

(3)

(b)

Find the maximum speed of the cart and the time at which it occurs.

(3)

(c)

Find the distance travelled by the cart in the first 6 seconds. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the quotient rule	✓ 1
A1 $\frac{dv}{dt}=\frac{12(t^2+4)-12t\times 2t}{(t^2+4)^2}$	✓ 1
A1 $=\frac{48-12t^2}{(t^2+4)^2}=\frac{12(4-t^2)}{(t^2+4)^2}$ (AG)	✓ 1
Part (b)
M1 set acceleration equal to zero	✓ 1
A1 $t=2$ s (reject $t=-2$)	✓ 1
A1 $v(2)=6$ m s⁻¹	✓ 1
Part (c)
M1 attempt to evaluate $\int_0^6\frac{12t}{t^2+4}\,dt$ (by substitution $u=t^2+4$ or using a GDC)	✓ 1
A1 $6\ln 10=13.8$ m	✓ 1

Exam tip: When asked to show a given result, write the quotient-rule line in full before simplifying; the final printed answer only earns the last mark if the working leading to it is visible.

+ Add to test✓ In your test281IBMI3H-5-07r35-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A cyclist's power output during an 8-second sprint test is modelled by $P(t)=40te^{-t^2/16}$ watts, for $0\le t\le 8$, where $t$ is in seconds.

(a)

Find the time at which the power output is greatest. Give your answer correct to 3 significant figures.

(4)

(b)

The energy transferred, in joules, is $E=\int_0^8P(t)\,dt$. Use the substitution $u=\frac{t^2}{16}$ to find $E$. Give your answer correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule (or chain rule inside the product)	✓ 1
A1 $P'(t)=40e^{-t^2/16}\left(1-\frac{t^2}{8}\right)$	✓ 1
M1 set $P'(t)=0$	✓ 1
A1 $t=\sqrt{8}=2.83$ s	✓ 1
Part (b)
M1 $\frac{du}{dt}=\frac{t}{8}$, so $t\,dt=8\,du$	✓ 1
A1 new limits $u=0$ and $u=4$ and integral $320\int_0^4e^{-u}\,du$	✓ 1
A1 $320\left(1-e^{-4}\right)$	✓ 1
A1 $E=314$ J	✓ 1

Exam tip: After a substitution, change the limits as well as the variable; otherwise you must return to $t$ before substituting the original limits.

+ Add to test✓ In your test282IBMI3H-5-07r35-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

The outline of a rugby ball is modelled by rotating the curve $y=0.09\cos\left(\frac{\pi x}{0.3}\right)$, for $-0.15\le x\le 0.15$, through $360^\circ$ about the x-axis. Both $x$ and $y$ are measured in metres.

(a)

Write down an integral for the volume $V$ of the ball.

(2)

(b)

Hence find $V$ in m³, correct to 3 significant figures.

(2)

(c)

The ball has mass 0.44 kg. Using the unrounded volume from (b), find its mean density, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of $V=\pi\int y^2\,dx$	✓ 1
A1 $V=\pi\int_{-0.15}^{0.15}\left(0.09\cos\frac{\pi x}{0.3}\right)^2dx$	✓ 1
Part (b)
M1 attempt to evaluate the integral (GDC or $\cos^2$ identity)	✓ 1
A1 $V=3.82\times10^{-3}$ m³	✓ 1
Part (c)
M1 density $=\frac{0.44}{V}$	✓ 1
A1 $115$ kg m⁻³	✓ 1

Common error: Remember to square the whole function $y$ before integrating; squaring only the cosine, or forgetting the factor $\pi$, are the usual slips.

+ Add to test✓ In your test283IBMI3H-5-07r35-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A cyclist freewheels down a long straight hill. Her speed $v$ m s⁻¹ after $t$ seconds satisfies $\frac{dv}{dt}=2.4-0.08v^2$, with $v=0$ when $t=0$.

(a)

Use Euler's method with a step length of 1 second to estimate her speed when $t=3$. Carry unrounded values through your working and give your final answer correct to 3 significant figures.

(4)

(b)

Find the limiting (terminal) speed of the cyclist, correct to 3 significant figures.

(2)

(c)

The exact solution gives $v=4.74$ m s⁻¹ when $t=3$. Explain why Euler's method overestimates the speed here.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of $v_{n+1}=v_n+1\times(2.4-0.08v_n^2)$	✓ 1
A1 $v_1=2.4$	✓ 1
A1 $v_2=4.3392$	✓ 1
A1 $v_3=5.23$ m s⁻¹	✓ 1
Part (b)
M1 set $\frac{dv}{dt}=0$, so $2.4=0.08v^2$	✓ 1
A1 $v=\sqrt{30}=5.48$ m s⁻¹	✓ 1
Part (c)
R1 the acceleration decreases as $v$ increases, so the graph of $v$ against $t$ is concave down	✓ 1
R1 each step uses the gradient at the start of the interval, which is greater than the mean gradient over that interval	✓ 1

Common error: Euler's method uses the gradient at the start of each step; substituting the new speed into the right-hand side too early gives a different (and wrong) sequence.

+ Add to test✓ In your test284IBMI3H-5-04r39-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

After an injection, a drug moves between the blood (amount x mg) and body tissue (amount y mg). At time t hours, dx/dt = −0.6x + 0.1y and dy/dt = 0.4x − 0.1y. Initially x = 100 and y = 0.

(a)

Use Euler's method with a step length of 0.5 hours to find estimates of x and y when t = 0.5.

(3)

(b)

Perform a second step to estimate x and y when t = 1.

(3)

(c)

Show that d(x + y)/dt = −0.2x, and explain what this means for the total amount of drug in the blood and tissue.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 dx/dt = −60 and dy/dt = 40 at t = 0	✓ 1
A1 x₁ = 100 + 0.5(−60) = 70	✓ 1
A1 y₁ = 0 + 0.5(40) = 20	✓ 1
Part (b)
M1 dx/dt = −42 + 2 = −40 and dy/dt = 28 − 2 = 26 at (70, 20)	✓ 1
A1 x₂ = 50	✓ 1
A1 y₂ = 33	✓ 1
Part (c)
A1 adding the equations: −0.6x + 0.4x + 0.1y − 0.1y = −0.2x	✓ 1
R1 for x > 0 the total decreases (the drug is being removed from the body)	✓ 1

Exam tip: In a coupled system, update both variables using the old values of both; do not use the new x to compute the new y within the same step.

+ Add to test✓ In your test285IBMI3H-5-04r39-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

The vertical displacement x metres of a Formula 1 car's front suspension from its rest position, t seconds after the car hits a kerb, satisfies x″ + 0.8x′ + 4x = 0, with x = 0.05 and x′ = 0 when t = 0.

(a)

By letting v = x′, show that the equation can be written as the system x′ = v, v′ = −4x − 0.8v.

(1)

(b)

Use Euler's method with a step length of 0.1 s to estimate x and v when t = 0.2.

(4)

(c)

The system can be written in matrix form with matrix A whose characteristic equation is λ² + 0.8λ + 4 = 0. Find the eigenvalues of A, giving the imaginary part to 3 significant figures.

(2)

(d)

Use the eigenvalues to describe the long-term motion of the suspension.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 v′ = x″ = −0.8x′ − 4x = −4x − 0.8v	✓ 1
Part (b)
A1 x₁ = 0.05 + 0.1 × 0 = 0.05	✓ 1
A1 v₁ = 0 + 0.1(−0.2) = −0.02	✓ 1
A1 x₂ = 0.05 + 0.1(−0.02) = 0.048	✓ 1
A1 v₂ = −0.02 + 0.1(−0.2 + 0.016) = −0.0384	✓ 1
Part (c)
M1 λ = (−0.8 ± √(0.64 − 16))/2 (or equivalent GDC solve)	✓ 1
A1 λ = −0.4 ± 1.96i	✓ 1
Part (d)
R1 the real part is negative, so the oscillation decays to zero; the imaginary part means it oscillates about the rest position (a stable spiral in the phase portrait)	✓ 1

Exam tip: A second-order equation becomes a coupled first-order system by naming the first derivative as a new variable; Euler's method then applies to both equations together.

+ Add to test✓ In your test286IBMI3H-5-04r39-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A hospital vial is modelled by rotating the curve y = 0.5 + 0.3cos(0.5x), for 0 ≤ x ≤ 6, through 360° about the x-axis. Lengths are in centimetres and angles are in radians.

(a)

Find the radius of the vial at x = 0 and at x = 6. Give each answer to 3 significant figures.

(2)

(b)

Find the volume of the vial. Give your answer to 3 significant figures.

(3)

(c)

The vial is filled to 80% of its volume. Find the volume of liquid, in mL, using the unrounded volume and giving your answer to 3 significant figures. (1 cm³ = 1 mL)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 0.800 cm	✓ 1
A1 0.203 cm	✓ 1
Part (b)
M1 V = π∫₀⁶ y² dx	✓ 1
A1 correct integrand (0.5 + 0.3cos(0.5x))² and limits	✓ 1
A1 V = 5.79 cm³	✓ 1
Part (c)
M1 0.8 × V	✓ 1
A1 4.63 mL	✓ 1

Common error: Square the entire expression for the radius, (0.5 + 0.3cos(0.5x))², rather than squaring each term separately.

+ Add to test✓ In your test287IBMI3H-5-05r40-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

During an outbreak, the number of new infections reported per day, I, at time t days after the first report is modelled by $I(t) = 40t\,e^{-0.25t}$, for $t \ge 0$.

(a)

Show that $\frac{dI}{dt} = 40e^{-0.25t}(1 - 0.25t)$.

(3)

(b)

Hence find the time at which the number of new infections per day is greatest.

(2)

(c)

Find the greatest number of new infections per day, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 $\frac{dI}{dt} = 40e^{-0.25t} + 40t(-0.25)e^{-0.25t}$	✓ 1
A1 factorise to $40e^{-0.25t}(1-0.25t)$ (AG)	✓ 1
Part (b)
M1 set $\frac{dI}{dt}=0$ with $e^{-0.25t}>0$, so $1-0.25t=0$	✓ 1
A1 $t = 4$ days	✓ 1
Part (c)
M1 $I(4) = 160e^{-1}$	✓ 1
A1 58.9	✓ 1

Common error: When a derivative has the form (exponential) × (bracket), the exponential is never zero, so only the bracket can give a stationary point.

+ Add to test✓ In your test288IBMI3H-5-05r40-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The water depth, H metres, at the entrance of a tidal-energy channel is modelled by $H(t) = 5 + 2\sin(0.5t)$, where t is the time in hours after midnight and $0 \le t \le 12$. The argument of the sine function is in radians.

(a)

Find $\frac{dH}{dt}$.

(2)

(b)

Find the rate at which the depth is changing at $t = 2$, giving your answer in m h⁻¹ to 3 significant figures.

(2)

(c)

Find the first time at which the depth is increasing at 0.8 m h⁻¹. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the chain rule	✓ 1
A1 $\frac{dH}{dt} = \cos(0.5t)$	✓ 1
Part (b)
M1 substitute $t=2$ to get $\cos 1$ (radians)	✓ 1
A1 0.540 m h⁻¹	✓ 1
Part (c)
M1 set $\cos(0.5t)=0.8$	✓ 1
A1 $0.5t = \cos^{-1}(0.8) = 0.6435$	✓ 1
A1 $t = 1.29$ hours	✓ 1

Common error: Set your calculator to radians whenever the argument of a trigonometric function is not given in degrees.

+ Add to test✓ In your test289IBMI3H-5-05r40-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

A wave-energy device delivers power at a rate $R(t) = 12t\,e^{-0.5t^2}$ kW, where t is the time in hours since the device was switched on.

(a)

Use the substitution $u = 0.5t^2$ to show that the energy delivered during the first three hours is $\int_0^{4.5} 12e^{-u}\,du$ kWh.

(3)

(b)

Hence find the energy delivered in the first three hours, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{du}{dt} = t$, so $t\,dt = du$	✓ 1
A1 replace $12t\,e^{-0.5t^2}\,dt$ by $12e^{-u}\,du$	✓ 1
A1 limits change to $u=0$ and $u=0.5(3)^2=4.5$	✓ 1
Part (b)
M1 $12\left[-e^{-u}\right]_0^{4.5} = 12\left(1-e^{-4.5}\right)$	✓ 1
A1 11.9 kWh	✓ 1

Exam tip: After a substitution, either change the limits to the new variable or return to the original variable before substituting the limits, never mix the two.

+ Add to test✓ In your test290IBMI3H-5-05r40-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

A biofuel storage tank has the shape formed when the region under the curve $y = 3e^{-0.2x}$, for $0 \le x \le 5$, is rotated through $360^\circ$ about the x-axis. Lengths are in metres.

(a)

Write down an integral for the volume of the tank.

(2)

(b)

Find the volume of the tank, to 3 significant figures.

(2)

(c)

The tank is filled at a constant rate of 1.5 m³ per minute. Using the unrounded volume from part (b), find the time to fill the tank completely, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of $V = \pi\int y^2\,dx$	✓ 1
A1 $V = \pi\int_0^5 9e^{-0.4x}\,dx$	✓ 1
Part (b)
M1 $\pi\left[-22.5e^{-0.4x}\right]_0^5 = 22.5\pi\left(1-e^{-2}\right)$	✓ 1
A1 61.1 m³	✓ 1
Part (c)
M1 $61.1195\ldots \div 1.5$	✓ 1
A1 40.7 minutes	✓ 1

Common error: Square the whole function inside the integral: $(3e^{-0.2x})^2 = 9e^{-0.4x}$, not $3e^{-0.4x}$.

+ Add to test✓ In your test291IBMI3H-5-05r40-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

The stock level, y hundred bags, in a retail warehouse satisfies the differential equation $\frac{dy}{dt} = 0.5t - y$, where t is the time in days.

(a)

Find the gradient of the solution curve at the points $(0, 2)$ and $(4, 1)$.

(2)

(b)

Show that every solution curve has a horizontal tangent where it meets the line $y = 0.5t$.

(1)

(c)

A solution curve passes through $(2,1)$. Find $\frac{d^2y}{dt^2}$ at this point, and hence state whether the curve has a local minimum or a local maximum there.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 at $(0,2)$: $0 - 2 = -2$	✓ 1
A1 at $(4,1)$: $2 - 1 = 1$	✓ 1
Part (b)
R1 when $y=0.5t$, $\frac{dy}{dt}=0.5t-0.5t=0$	✓ 1
Part (c)
M1 differentiate implicitly: $\frac{d^2y}{dt^2} = 0.5 - \frac{dy}{dt}$	✓ 1
A1 at $(2,1)$, $\frac{dy}{dt}=0$, so $\frac{d^2y}{dt^2} = 0.5$	✓ 1
R1 positive, so a local minimum	✓ 1

Exam tip: For slope-field questions, differentiate the given equation with respect to the independent variable, treating the dependent variable as a function of it.

+ Add to test✓ In your test292IBMI3H-5-05r40-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A retailer spends x thousand pounds on advertising a new product. The profit, P thousand pounds, is modelled by $P(x) = 120\ln(1+x) - 20x$, for $x \ge 0$.

(a)

Find $\frac{dP}{dx}$.

(2)

(b)

Find the advertising spend that maximises the profit.

(2)

(c)

Show that this is a maximum.

(2)

(d)

Find the maximum profit, in thousands of pounds, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate $\ln(1+x)$	✓ 1
A1 $\frac{dP}{dx} = \frac{120}{1+x} - 20$	✓ 1
Part (b)
M1 set $\frac{120}{1+x} = 20$	✓ 1
A1 $x = 5$, i.e. £5000	✓ 1
Part (c)
M1 $\frac{d^2P}{dx^2} = -\frac{120}{(1+x)^2}$	✓ 1
R1 negative for all $x \ge 0$, so a maximum	✓ 1
Part (d)
M1 $P(5) = 120\ln 6 - 100$	✓ 1
A1 115 thousand pounds	✓ 1

Exam tip: The derivative of $\ln(1+x)$ is $\frac{1}{1+x}$, using the chain rule with an inner derivative of 1.

+ Add to test✓ In your test293IBMI3H-5-04r42-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Calculator

A beekeeper builds an open-topped wooden box for a hive. The box is a cuboid with a square base of side x metres and height h metres. The volume of the box must be 0.05 m³. The total area of wood, S m², includes the base and the four sides.

(a)

Show that S = x² + 0.2x⁻¹.

(2)

(b)

Find dS/dx and hence find the value of x that minimises S, giving your answer to 3 significant figures.

(3)

(c)

Show that this value of x gives a minimum.

(1)

(d)

Find the minimum value of S, giving your answer to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 h = 0.05/x² and S = x² + 4xh	✓ 1
A1 S = x² + 0.2/x (AG)	✓ 1
Part (b)
A1 dS/dx = 2x − 0.2x⁻²	✓ 1
M1 set dS/dx = 0, i.e. x³ = 0.1	✓ 1
A1 x = 0.464 m	✓ 1
Part (c)
R1 d²S/dx² = 2 + 0.4x⁻³ = 6 > 0 (positive), so a minimum	✓ 1
Part (d)
A1 0.646 m²	✓ 1

Exam tip: Rewrite the area as a function of one variable before differentiating, and rewrite 1/x as x⁻¹ so that the power rule applies.

+ Add to test✓ In your test294IBMI3H-5-04r42-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

Calculator

A Formula 1 car brakes for a corner. Its speed, v m s⁻¹, is modelled by v(t) = 72 − 18t + 1.5t² for 0 ≤ t ≤ 6, where t is the time in seconds after braking starts.

(a)

Find the acceleration of the car when t = 2.

(2)

(b)

Show that the minimum speed during this interval is 18 m s⁻¹.

(2)

(c)

Find the distance travelled by the car during the 6 seconds of braking.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a = dv/dt = −18 + 3t	✓ 1
A1 −12 m s⁻²	✓ 1
Part (b)
M1 a = 0 giving t = 6 (or complete the square)	✓ 1
A1 v(6) = 72 − 108 + 54 = 18 m s⁻¹	✓ 1
Part (c)
M1 ∫₀⁶ (72 − 18t + 1.5t²) dt	✓ 1
A1 216 m	✓ 1

Common error: Acceleration is the derivative of velocity and displacement is the integral of velocity; do not use v = u + at, which only holds for constant acceleration.

+ Add to test✓ In your test295IBMI3H-5-04r42-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Calculator

The cross-section of the glass roof of a hospital atrium is modelled by the curve y = 9 − 0.04x², where x and y are in metres, the ground is the line y = 0 and −15 ≤ x ≤ 15.

(a)

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

(3)

(b)

Find the angle that this tangent makes with the horizontal, to 1 decimal place.

(2)

(c)

Find the area of the cross-section of the atrium.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 dy/dx = −0.08x, gradient −0.8 at x = 10 and y = 5	✓ 1
M1 y − 5 = −0.8(x − 10)	✓ 1
A1 y = −0.8x + 13	✓ 1
Part (b)
M1 tan θ = 0.8 (or arctan 0.8)	✓ 1
A1 38.7°	✓ 1
Part (c)
M1 ∫₋₁₅¹⁵ (9 − 0.04x²) dx	✓ 1
A1 180 m²	✓ 1

Exam tip: The gradient of the tangent is the derivative evaluated at the given x; find the y-coordinate from the original curve before writing the equation of the line.

+ Add to test✓ In your test296IBMI3H-5-04r42-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

Calculator

The concentration of a drug in a patient's blood, C mg per litre, is modelled by C(t) = 5t/(t² + 4), where t ≥ 0 is the time in hours after an injection.

(a)

Use the quotient rule to show that C′(t) = 5(4 − t²)/(t² + 4)².

(3)

(b)

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

(2)

(c)

Find the rate of change of concentration when t = 3, giving your answer to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule with 5 and t² + 4 and 2t	✓ 1
A1 [5(t² + 4) − 5t × 2t]/(t² + 4)²	✓ 1
A1 simplify to 5(4 − t²)/(t² + 4)² (AG)	✓ 1
Part (b)
M1 C′(t) = 0 giving t = 2 (t > 0)	✓ 1
A1 C(2) = 1.25 mg per litre at t = 2 hours	✓ 1
Part (c)
A1 -0.148 mg per litre per hour	✓ 1

Exam tip: In the quotient rule take care with the order of the subtraction: (derivative of the top × bottom) minus (top × derivative of the bottom).

+ Add to test✓ In your test297IBMI3H-5-04r42-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

Calculator

The bowl of a honey dipper is modelled by rotating the region between the curve y = 2cos(0.2x), the x-axis, the y-axis and the line x = 7 through 360° about the x-axis. Lengths are in centimetres.

(a)

Write down an integral for the volume of the bowl.

(2)

(b)

Find the volume of the bowl, giving your answer to 3 significant figures.

(2)

(c)

Honey has density 1.42 g cm⁻³. Using the unrounded volume, find the mass of honey that would exactly fill the bowl, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π∫y² dx with limits 0 and 7	✓ 1
A1 V = π∫₀⁷ (2cos(0.2x))² dx	✓ 1
Part (b)
M1 evaluate the integral (using radians)	✓ 1
A1 49.2 cm³	✓ 1
Part (c)
M1 mass = density × volume = 1.42 × 49.244…	✓ 1
A1 69.9 g	✓ 1

Exam tip: For a volume of revolution about the x-axis the integrand is πy²: square the whole function, including any constant in front.

+ Add to test✓ In your test298IBMI3H-5-04r42-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

Calculator

A Formula 1 car accelerates from rest on a straight. Its speed, v m s⁻¹, is modelled by v(t) = 85(1 − e^−0.35t), where t is the time in seconds from the start.

(a)

Find the acceleration of the car when t = 2, to 3 significant figures.

(2)

(b)

Find the distance travelled in the first 8 seconds, to 3 significant figures.

(2)

(c)

Find the time at which the car reaches a speed of 80 m s⁻¹, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a(t) = 29.75e^−0.35t	✓ 1
A1 14.8 m s⁻²	✓ 1
Part (b)
M1 ∫₀⁸ 85(1 − e^−0.35t) dt	✓ 1
A1 452 m	✓ 1
Part (c)
M1 85(1 − e^−0.35t) = 80, so e^−0.35t = 5/85	✓ 1
A1 8.09 s	✓ 1

Common error: Integrating e⁻⁰.³⁵t gives −e⁻⁰.³⁵t/0.35; forgetting to divide by the coefficient of t is a common slip.

+ Add to test✓ In your test299IBMI3H-5-04r42-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

Calculator

The number of bees, P, in a hive t weeks after a new colony is established satisfies the differential equation dP/dt = 0.3P(1 − P/1200), with P = 200 when t = 0.

(a)

A slope field is drawn for this differential equation in the (t, P) plane. Write down the gradient of the line segment at the point (0, 200), and state the value of P greater than zero at which every line segment is horizontal.

(2)

(b)

Use Euler's method with a step size of 1 week to estimate the number of bees at t = 3. Give your answer to the nearest whole number.

(3)

(c)

State the long-term behaviour of P predicted by the model, and explain what P = 1200 represents.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 gradient = 0.3 × 200 × (1 − 200/1200) = 50	✓ 1
A1 P = 1200	✓ 1
Part (b)
M1 P_n+1 = P_n + 1 × 0.3P_n(1 − P_n/1200)	✓ 1
A1 P at t = 1 is 250 and at t = 2 is 309.375	✓ 1
A1 P at t = 3 is approximately 378	✓ 1
Part (c)
R1 P approaches 1200, the maximum (carrying capacity) the hive can support	✓ 1

Common error: In Euler's method use the newest value of P at each step; do not keep using the starting value in the right-hand side.

+ Add to test✓ In your test300IBMI3H-5-06r43-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The cumulative number of cases C in an outbreak in a town of 5000 people satisfies dC/dt = 0.02(5000 − C), where t is in days and C = 200 when t = 0.

(a)

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

(5)

(b)

Find the time at which 3000 people have been infected, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 separate variables: ∫ 1/(5000 − C) dC = ∫ 0.02 dt	✓ 1
A1 −ln\|5000 − C\| = 0.02t + k	✓ 1
M1 use C = 200 at t = 0 to find the constant	✓ 1
A1 5000 − C = 4800e^−0.02t	✓ 1
A1 C = 5000 − 4800e^−0.02t	✓ 1
Part (b)
M1 attempt to solve 5000 − 3000 = 4800e^−0.02t	✓ 1
A1 t = 43.8 days	✓ 1

Exam tip: After integrating, always substitute the initial condition before rearranging for the dependent variable.

+ Add to test✓ In your test301IBMI3H-5-06r43-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium4 marks

A hydrogen storage vessel is made by rotating the region bounded by the curve y = √x e^−x/4, the x-axis and the line x = 6 through 360° about the x-axis. Lengths are in metres.

(a)

Write down an integral for the volume of the vessel.

(2)

(b)

Find the volume of the vessel, in m³ to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use V = π∫y² dx	✓ 1
A1 V = π∫₀⁶ xe^−x/2 dx	✓ 1
Part (b)
M1 evaluate the definite integral using a calculator	✓ 1
A1 10.1 m³	✓ 1

Common error: Square the whole function before integrating: y squared is x times e to the power −x/2, not the square of x.

+ Add to test✓ In your test302IBMI3H-5-06r43-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

A shop sells scarves in the winter season. The rate of sales is modelled by R(t) = 40te^−0.1t² scarves per week, where t is the number of weeks after the start of the season.

(a)

Use the substitution u = −0.1t² to show that ∫ 40te^−0.1t² dt = −200e^−0.1t² + c.

(3)

(b)

Find the total number of scarves sold in the first 5 weeks, to the nearest whole number.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du/dt = −0.2t, so 40t dt = −200 du	✓ 1
A1 ∫ −200e^u du	✓ 1
A1 −200e^−0.1t² + c (AG)	✓ 1
Part (b)
M1 200(1 − e^−2.5) from [−200e^−0.1t²] between 0 and 5	✓ 1
A1 184 scarves	✓ 1

Common error: When substituting, change every part of the integrand, including dt, into the new variable; a leftover t means the substitution is incomplete.

+ Add to test✓ In your test303IBMI3H-5-06r43-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A shop finds that the number of units sold per day at price €p is N(p) = 800e^−0.04p. The daily revenue is R(p) = pN(p) euros.

(a)

Show that R′(p) = 800e^−0.04p(1 − 0.04p).

(3)

(b)

Find the price that maximises daily revenue, and justify that it is a maximum.

(3)

(c)

Find the maximum daily revenue, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 R′ = 800e^−0.04p + p(−32e^−0.04p)	✓ 1
A1 factorise to 800e^−0.04p(1 − 0.04p) (AG)	✓ 1
Part (b)
M1 set R′(p) = 0	✓ 1
A1 p = 25	✓ 1
R1 R′ changes from positive to negative at p = 25 (or R″(25) < 0)	✓ 1
Part (c)
M1 substitute: R(25) = 25 × 800e⁻¹	✓ 1
A1 €7360	✓ 1

Common error: Remember the exponential factor never equals zero, so the stationary point comes only from the bracket.

+ Add to test✓ In your test304IBMI3H-5-06r43-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

The solar irradiance on a panel is modelled by I(t) = 850 sin(π(t − 6)/12) W m⁻² for 6 ≤ t ≤ 18, where t is the time in hours after midnight.

(a)

Find dI/dt at t = 9, to 3 significant figures.

(3)

(b)

Find the energy received per square metre between t = 6 and t = 18, in W h m⁻² to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the chain rule	✓ 1
A1 dI/dt = 850 × (π/12) cos(π(t − 6)/12)	✓ 1
A1 157 W m⁻² h⁻¹	✓ 1
Part (b)
M1 attempt to integrate I from 6 to 18	✓ 1
A1 antiderivative −(10200/π) cos(π(t − 6)/12) (or evaluation with a calculator)	✓ 1
A1 6490 W h m⁻²	✓ 1

Common error: Use radian mode on the calculator whenever the argument contains π.

+ Add to test✓ In your test305IBMI3H-5-05r45-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

In a food science experiment, the concentration of a flavour compound in a simmering sauce is modelled by C(t) = 8te^−0.25t mg per litre, where t is the time in minutes after the sauce reaches simmering point, for t ≥ 0.

(a)

Use the product rule to show that C′(t) = 8e^−0.25t(1 − 0.25t).

(3)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

Find the greatest concentration, and justify that it is a maximum by considering the sign of C′(t) either side of your answer to (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at product rule with 8t and e^−0.25t	✓ 1
A1 C′ = 8e^−0.25t − 2te^−0.25t	✓ 1
A1 factorises to reach 8e^−0.25t(1 − 0.25t) with no errors	✓ 1
Part (b)
M1 sets C′(t) = 0 and notes e^−0.25t ≠ 0	✓ 1
A1 t = 4 minutes	✓ 1
Part (c)
A1 C(4) = 32e⁻¹ = 11.8 mg per litre	✓ 1
R1 C′(3) = 0.945… > 0 and C′(5) = −0.573… < 0, so the gradient changes from positive to negative (accept use of second derivative or a sketch of the graph)	✓ 1

Exam tip: When asked to justify a maximum, give evidence such as the sign of the gradient either side or the sign of the second derivative, not just a statement.

+ Add to test✓ In your test306IBMI3H-5-05r45-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

The wind speed w metres per second at a mountain hut is modelled by w(t) = 16 + 7 sin(0.26t), where t is the time in hours after 06:00, for 0 ≤ t ≤ 12. The argument of the sine function is in radians.

(a)

Find w′(t).

(2)

(b)

Calculate w′(5) to 3 significant figures and state what it tells you about the wind at 11:00.

(2)

(c)

Find the time, in hours after 06:00, at which the wind speed is first increasing at 1.2 m s⁻¹ per hour. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at chain rule with a factor of 0.26	✓ 1
A1 w′(t) = 1.82cos(0.26t)	✓ 1
Part (b)
A1 w′(5) = 0.487	✓ 1
R1 the wind speed is increasing at about 0.487 m s⁻¹ per hour at 11:00	✓ 1
Part (c)
M1 sets 1.82cos(0.26t) = 1.2	✓ 1
M1 attempts inverse cosine, 0.26t = cos⁻¹(1.2/1.82) = 0.851…	✓ 1
A1 t = 3.27 hours	✓ 1

Exam tip: Check that the calculator is in radian mode whenever the argument of a trigonometric function is stated to be in radians.

+ Add to test✓ In your test307IBMI3H-5-05r45-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A 200-litre pickling tank initially contains 10 kg of salt dissolved in water. Brine of concentration 0.5 kg per litre is pumped in at 4 litres per minute. The mixture is kept uniform and leaves the tank at 4 litres per minute. Let x kg be the mass of salt in the tank after t minutes.

(a)

Show that dx/dt = 2 − 0.02x.

(2)

(b)

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

(4)

(c)

Find the time taken for the tank to contain 60 kg of salt. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rate in = 0.5 × 4 = 2 kg per minute	✓ 1
A1 rate out = 4 × x/200 = 0.02x (since the volume stays 200 litres), so dx/dt = 2 − 0.02x	✓ 1
Part (b)
M1 separates variables: ∫ 1/(2 − 0.02x) dx = ∫ dt	✓ 1
A1 −50 ln\|2 − 0.02x\| = t + c	✓ 1
M1 uses x = 10 when t = 0 to find the constant	✓ 1
A1 x = 100 − 90e^−0.02t	✓ 1
Part (c)
M1 sets 100 − 90e^−0.02t = 60 (so e^−0.02t = 4/9)	✓ 1
A1 t = 40.5 minutes	✓ 1

Exam tip: For mixing problems, write the equation as rate in minus rate out, and check that the units of both rates match before integrating.

+ Add to test✓ In your test308IBMI3H-5-05r45-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

A flavour compound passes between a sauce and the pasta it coats. Let x mg be the mass of the compound in the sauce and y mg the mass in the pasta, t minutes after they are combined. The masses satisfy
dx/dt = −0.12x + 0.04y, dy/dt = 0.12x − 0.04y.
Initially x = 100 and y = 0.

(a)

Show that x + y is constant and explain what this means in context.

(2)

(b)

Use Euler’s method with a step length of 0.5 to estimate x and y when t = 1.5. Give each answer to 3 significant figures.

(4)

(c)

Eventually the masses stop changing. Find the long-term values of x and y.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 d(x + y)/dt = dx/dt + dy/dt = 0	✓ 1
R1 the total mass of the compound stays at 100 mg; none is lost or gained, it only moves between sauce and pasta	✓ 1
Part (b)
M1 uses x_n+1 = x_n + 0.5(dx/dt) and y_n+1 = y_n + 0.5(dy/dt)	✓ 1
A1 (x, y) at t = 0.5 is (94, 6)	✓ 1
A1 (x, y) at t = 1 is (88.48, 11.52)	✓ 1
A1 x ≈ 83.4 and y ≈ 16.6 at t = 1.5 (from 83.4016 and 16.5984)	✓ 1
Part (c)
M1 sets dx/dt = 0 to get y = 3x and uses x + y = 100	✓ 1
A1 x = 25 mg and y = 75 mg	✓ 1

Common error: In a coupled system, both variables must be updated using the values from the same step; do not use the new x when calculating the new y.

+ Add to test✓ In your test309IBMI3H-5-05r45-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

A chocolatier makes a solid decorative piece by rotating the region bounded by the curve y = 1.5 + sin(0.3x), the x-axis, and the lines x = 0 and x = 8 through 360° about the x-axis. The units are centimetres and the argument of the sine function is in radians.

(a)

Find the volume of the piece. Give your answer to 3 significant figures.

(4)

(b)

The chocolate has density 1.30 g cm⁻³ and costs €0.012 per gram. Using the unrounded volume from part (a), find the cost of the chocolate in one piece. Give your answer to the nearest cent.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume = π∫₀⁸ y² dx	✓ 1
M1 substitutes y = 1.5 + sin(0.3x) with correct limits	✓ 1
A1 evaluates the integral (e.g. on a calculator) as 40.2…	✓ 1
A1 volume = 126 cm³	✓ 1
Part (b)
M1 mass = 1.30 × 126.30…	✓ 1
A1 mass = 164 g (164.19…)	✓ 1
A1 cost = €1.97	✓ 1

Common error: Forgetting to square the function inside the integral, or leaving π out of the volume-of-revolution formula.

+ Add to test✓ In your test310IBMI3H-5-07r45-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Calculator

A lava flow cools in the surrounding air, which has a constant temperature of $25\,^\circ\mathrm{C}$. The lava temperature, $T\,^\circ\mathrm{C}$, at time $t$ hours satisfies the differential equation $\frac{dT}{dt}=-0.05(T-25)$, with $T=1100$ when $t=0$.

(a)

Use Euler's method with a step length of $h=2$ hours to estimate $T$ when $t=6$. Do not round intermediate values; give your final answer to 1 decimal place.

(3)

(b)

The exact solution of the differential equation is $T=25+1075e^{-0.05t}$. Calculate the exact temperature when $t=6$, correct to 3 significant figures.

(2)

(c)

Find the error in the Euler estimate, using the unrounded exact value and the unrounded Euler value, to 3 significant figures. State whether Euler's method has under- or over-estimated, and give a reason.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of $T_{n+1}=T_n+h\times(-0.05)(T_n-25)$ with $h=2$	✓ 1
A1 $T_1=992.5$ and $T_2=895.75$ seen	✓ 1
A1 $T_3=808.7$ (808.675)	✓ 1
Part (b)
M1 substitute $t=6$ into the exact solution	✓ 1
A1 $821$ ($821.38\ldots$)	✓ 1
Part (c)
A1 error $=821.38-808.675=12.7$ (accept 12.6 to 12.8)	✓ 1
R1 under-estimate, because the exact curve is concave up (decay slows), so each tangent step lies below the curve	✓ 1

Exam tip: When a question gives you the exact solution, use it to check whether Euler's method has under- or over-estimated, and link the reason to the shape of the curve.

+ Add to test✓ In your test311IBMI3H-5-07r45-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Calculator

The power used by a stage spotlight while it is slowly dimmed is $P(t)=\dfrac{4t}{1+t^2}$ kilowatts, where $t$ is the time in hours, $0\le t\le 3$. The energy used in kilowatt-hours (kWh) is $E=\int_0^3 P(t)\,dt$.

(a)

Use the substitution $u=1+t^2$ to show that $E=2\ln 10$.

(4)

(b)

Write down the value of $E$ correct to 3 significant figures.

(1)

(c)

Electricity costs €0.28 per kWh. Using the unrounded value of $E$, find the cost of the energy, to the nearest cent.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $u=1+t^2$ so $du=2t\,dt$	✓ 1
A1 $\int_1^{10}\frac{2}{u}\,du$ (limits changed correctly)	✓ 1
A1 $[2\ln u]_1^{10}$	✓ 1
R1 $2\ln 10-2\ln 1=2\ln 10$ since $\ln 1=0$	✓ 1
Part (b)
A1 $4.61$ kWh ($4.605\ldots$)	✓ 1
Part (c)
M1 $2\ln 10\times 0.28$	✓ 1
A1 €1.29 (1.2894...)	✓ 1

Exam tip: In a 'show that' question, every line must lead logically to the printed result. State the change of limits explicitly when you substitute.

+ Add to test✓ In your test312IBMI3H-5-07r45-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

Calculator

A prop for a theatre set is a solid made by rotating the region bounded by the curve $y=3e^{-0.25x}$, the $x$-axis, the $y$-axis and the line $x=4$ through $360^\circ$ about the $x$-axis. Lengths are in decimetres (dm).

(a)

Find the volume of the prop, in dm$^3$, correct to 3 significant figures.

(4)

(b)

The prop is made from foam costing €2.50 per litre. Given that 1 dm$^3$ = 1 litre, and using the unrounded volume, find the cost of the foam to the nearest cent.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $V=\pi\int_0^4 y^2\,dx$	✓ 1
A1 integrand $9e^{-0.5x}$	✓ 1
A1 $\pi\left[-18e^{-0.5x}\right]_0^4=18\pi(1-e^{-2})$	✓ 1
A1 $V=48.9$ dm$^3$ (48.895...)	✓ 1
Part (b)
M1 $48.8956\ldots\times 2.50$	✓ 1
A1 €122.24	✓ 1

Common error: Square the whole function before integrating a volume of revolution: $(3e^{-0.25x})^2=9e^{-0.5x}$, not $3e^{-0.5x}$.

+ Add to test✓ In your test313IBMI3H-5-07r45-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Calculator

The fuel burn rate of a cargo ship is $r(t)=3te^{-0.1t^2}$ tonnes per hour, where $t$ is the time in hours after leaving port, $0\le t\le 5$.

(a)

Using the substitution $u=0.1t^2$, show that the total fuel used in the five hours is $15\int_0^{2.5}e^{-u}\,du$.

(3)

(b)

Evaluate the total fuel used, correct to 3 significant figures.

(2)

(c)

Find the mean burn rate over the five hours, to 3 significant figures, using the unrounded total.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $du=0.2t\,dt$	✓ 1
A1 $3t\,dt=15\,du$	✓ 1
A1 limits changed to $u=0$ and $u=2.5$	✓ 1
Part (b)
M1 $15\left[-e^{-u}\right]_0^{2.5}=15(1-e^{-2.5})$	✓ 1
A1 $13.8$ tonnes	✓ 1
Part (c)
M1 total divided by $5$	✓ 1
A1 $2.75$ tonnes per hour	✓ 1

Exam tip: The mean value of a rate over an interval is the total change divided by the length of the interval.

+ Add to test✓ In your test314IBMI3H-5-06r46-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

Calculator

The power drawn by a robot's motor is $P(t) = 12te^{0.5t^2}$ watts, where $t$ is the time in seconds from the start of a move.

(a)

Use the substitution $u = 0.5t^2$ to show that $\int P(t)\,dt = 12e^{0.5t^2} + C$.

(3)

(b)

Hence find the energy, in joules, used in the first 2 seconds, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{du}{dt} = t$ so $t\,dt = du$	✓ 1
A1 $\int 12e^{u}\,du$	✓ 1
A1 $12e^{u} + C = 12e^{0.5t^2} + C$	✓ 1
Part (b)
M1 $\left[12e^{0.5t^2}\right]_0^2 = 12e^2 - 12$	✓ 1
A1 76.7 J	✓ 1

Exam tip: Energy is the integral of power with respect to time.

+ Add to test✓ In your test315IBMI3H-5-06r46-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium5 marks

Calculator

A hat mould is made by rotating the region between the curve $y = \sqrt{x}\,e^{-x/4}$, the x-axis and the lines $x = 0$ and $x = 4$ through $360^\circ$ about the x-axis. Lengths are in decimetres (dm).

(a)

Write down an integral for the volume of the mould.

(2)

(b)

Find the volume of the mould in dm³, correct to 3 significant figures.

(1)

(c)

The mould is cast from foam of density 0.03 kg per dm³. Find its mass, using the unrounded volume and giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $V = \pi\int y^2\,dx$	✓ 1
A1 $V = \pi\int_0^4 x\,e^{-x/2}\,dx$ (limits 0 and 4)	✓ 1
Part (b)
A1 7.46 dm³	✓ 1
Part (c)
M1 mass = density × volume = $0.03 \times 7.4\ldots$	✓ 1
A1 0.224 kg	✓ 1

Exam tip: When rotating about the x-axis, square the whole function before integrating: $\left(\sqrt{x}\,e^{-x/4}\right)^2 = xe^{-x/2}$.

+ Add to test✓ In your test316IBMI3H-5-06r46-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Calculator

A mobile robot's path is modelled by the differential equation $\frac{dy}{dx} = x + y$, where $x$ and $y$ are in metres, and it passes through the point $(0, 1)$.

(a)

Find the gradient of the path at $(1, 2)$, and state the value of the gradient at every point where $y = -x$.

(2)

(b)

Use Euler's method with step size 0.2 to estimate $y$ when $x = 0.6$.

(3)

(c)

The exact solution is $y = 2e^{x} - x - 1$. Find the percentage error of the Euler estimate at $x = 0.6$, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 gradient 3	✓ 1
A1 gradient 0 along $y = -x$	✓ 1
Part (b)
M1 $y_1 = 1 + 0.2\times 1 = 1.2$	✓ 1
A1 $y_2 = 1.2 + 0.2(0.2 + 1.2) = 1.48$	✓ 1
A1 $y_3 = 1.48 + 0.2(0.4 + 1.48) = 1.856$	✓ 1
Part (c)
M1 exact value $2e^{0.6} - 1.6 = 2.0442\ldots$ and $\frac{\|1.856 - 2.0442\ldots\|}{2.0442\ldots}\times 100$	✓ 1
A1 9.21% (accept 9.2%)	✓ 1

Exam tip: In Euler's method $y_{n+1} = y_n + h\,f(x_n, y_n)$: update both $x$ and $y$ in every step.

+ Add to test✓ In your test317IBMI3H-5-06r46-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

Calculator

A fashion workshop's dye bath at temperature $T$ °C cools in a room at 20 °C. The rate of cooling is modelled by $\frac{dT}{dt} = -k(T - 20)$, where $t$ is in minutes and $k > 0$. The bath is at 90 °C when $t = 0$.

(a)

By separating variables, show that $T = 20 + Ae^{-kt}$ for a constant $A$.

(3)

(b)

Find the value of $A$.

(1)

(c)

The bath cools to 60 °C after 10 minutes. Find $k$, correct to 3 significant figures.

(2)

(d)

Find the temperature after 25 minutes, using the unrounded value of $k$ and giving your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 separate: $\int \frac{1}{T-20}\,dT = \int -k\,dt$	✓ 1
A1 $\ln(T - 20) = -kt + c$	✓ 1
A1 $T - 20 = e^{c}e^{-kt}$, so $T = 20 + Ae^{-kt}$	✓ 1
Part (b)
A1 $A = 70$	✓ 1
Part (c)
M1 $60 = 20 + 70e^{-10k}$	✓ 1
A1 $k = \frac{1}{10}\ln\frac{7}{4} = 0.0560$	✓ 1
Part (d)
M1 $T = 20 + 70e^{-25k}$	✓ 1
A1 37.3 °C	✓ 1

Exam tip: For Newton cooling, the quantity that decays exponentially is the difference from the surroundings, not the temperature itself.

+ Add to test✓ In your test318IBMI3H-5-06r46-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium6 marks

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A packing robot moves along a straight rail. Its velocity is $v(t) = 1.5\sin(0.8t)$ m s^-1 for $0 \le t \le \frac{\pi}{0.8}$, where $t$ is in seconds. Work in radians.

(a)

Find the acceleration at $t = 1$ s, correct to 3 significant figures.

(2)

(b)

Find the time at which the robot comes to rest again.

(1)

(c)

Find the distance travelled by the robot in this time, in metres, correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $a = \frac{dv}{dt} = 1.2\cos(0.8t)$	✓ 1
A1 0.836 m s^-2	✓ 1
Part (b)
A1 $0.8t = \pi$, so $t = 3.93$ s (accept $\frac{\pi}{0.8}$)	✓ 1
Part (c)
M1 $\int_0^{\pi/0.8} 1.5\sin(0.8t)\,dt$	✓ 1
A1 $\left[-\frac{1.5}{0.8}\cos(0.8t)\right]$ evaluated between the limits	✓ 1
A1 3.75 m	✓ 1

Exam tip: Differentiate velocity to get acceleration and integrate velocity to get displacement; since the velocity here is never negative, displacement equals distance.

+ Add to test✓ In your test319IBMI3H-5-06r46-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium8 marks

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A fish farm builds an open-topped cylindrical tank of volume 2 m³ with base radius r m. The base costs €30 per m² and the curved side costs €20 per m².

(a)

Show that the total cost is $C = 30\pi r^2 + \frac{80}{r}$ euros.

(3)

(b)

Find $\frac{dC}{dr}$ and hence the value of $r$ that minimises the cost, correct to 3 significant figures.

(3)

(c)

Find the minimum cost, using the unrounded value of $r$ and giving your answer to the nearest cent.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height $h = \frac{2}{\pi r^2}$	✓ 1
M1 curved area $2\pi r h = \frac{4}{r}$	✓ 1
A1 $C = 30\pi r^2 + 20 \times \frac{4}{r} = 30\pi r^2 + \frac{80}{r}$	✓ 1
Part (b)
A1 $\frac{dC}{dr} = 60\pi r - \frac{80}{r^2}$	✓ 1
M1 set equal to zero: $r^3 = \frac{80}{60\pi}$	✓ 1
A1 $r = 0.752$ m	✓ 1
Part (c)
M1 substitute $r = 0.7515\ldots$ into $C$	✓ 1
A1 €159.68	✓ 1

Exam tip: Show that the stationary point is a minimum by checking the sign of the second derivative or the sign of the gradient on either side.

+ Add to test✓ In your test320IBMI3H-5-06r46-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Medium7 marks

Calculator

A survey boat moves along a straight channel. Its velocity is $v(t) = 0.6t^2 - 3.6t + 4.8$ km per hour for $0 \le t \le 5$, where $t$ is in hours and positive is upstream. At $t = 0$ the boat is at its mooring.

(a)

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

(2)

(b)

Find the boat's displacement from the mooring at $t = 5$.

(2)

(c)

Find the total distance travelled by the boat during the 5 hours.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 solve $0.6t^2 - 3.6t + 4.8 = 0$	✓ 1
A1 $t = 2$ and $t = 4$	✓ 1
Part (b)
M1 $\int_0^5 v\,dt$ giving $0.2t^3 - 1.8t^2 + 4.8t$	✓ 1
A1 4 km upstream	✓ 1
Part (c)
M1 displacement at $t = 2$ is 4 and at $t = 4$ is 3.2	✓ 1
M1 split the journey at the times found in (a)	✓ 1
A1 5.6 km	✓ 1

Exam tip: Distance needs the journey to be split wherever the velocity changes sign; integrating over the whole interval gives displacement, not distance.

+ Add to test✓ In your test321IBMI3H-5-03r8-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard6 marks

A satellite dish has the shape obtained by rotating the curve y = 0.25x², for 0 ≤ x ≤ 3, about the y-axis. Lengths are in metres. The dish is a bowl whose rim is at y = 2.25.

(a)

Show that the volume of the bowl is 2π(2.25)² cubic metres.

(3)

(b)

The bowl is filled with water to a depth h metres measured from its lowest point, such that the water fills exactly half of the volume of the bowl. Find h to three significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume = π∫x² dy with x² = 4y	✓ 1
M1 π∫₀^2.25 4y dy = π[2y²]₀^2.25	✓ 1
A1 = 2π(2.25)²	✓ 1
Part (b)
M1 volume to depth h is 2πh²	✓ 1
M1 2πh² = π(2.25)²	✓ 1
A1 h = 1.59 m	✓ 1

Exam tip: For rotation about the y-axis, rewrite the curve as x² in terms of y and integrate with respect to y, using y-limits.

+ Add to test✓ In your test322IBMI3H-5-03r8-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard5 marks

On a glacier, new snow accumulates at a rate of s(t) = 12cos(0.2t)/(2 + sin(0.2t)) centimetres per hour, where t is the time in hours and angles are in radians.

(a)

Use the substitution u = 2 + sin(0.2t) to find ∫ s(t) dt.

(3)

(b)

Find the depth of snow that accumulates between t = 0 and t = 10, in centimetres to three significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du = 0.2cos(0.2t) dt	✓ 1
M1 integral becomes ∫ 60/u du	✓ 1
A1 60ln(2 + sin(0.2t)) + c (the bracket is positive, so no modulus needed)	✓ 1
Part (b)
M1 60[ln(2 + sin(0.2t))]₀¹⁰ = 60(ln(2 + sin 2) − ln 2)	✓ 1
A1 22.5 cm	✓ 1

Common error: Dividing by 0.2 when changing from du to dt gives a factor of 5; with the 12 already present this makes 60, not 2.4.

+ Add to test✓ In your test323IBMI3H-5-04r8-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard8 marks

In a coastal ecosystem, x is the number of fish (in hundreds) and y is the number of sharks at time t years, where
dx/dt = 0.5x − 0.01xy,
dy/dt = 0.004xy − 0.2y.
At t = 0, x = 50 and y = 20.

(a)

Use Euler's method with step length 0.5 years to estimate x and y when t = 0.5.

(3)

(b)

Continue the method to estimate x and y when t = 1.

(3)

(c)

Find the number of sharks y for which the number of fish does not change, given that x > 0.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 dx/dt = 15 and dy/dt = 0 at t = 0	✓ 1
A1 x = 57.5	✓ 1
A1 y = 20	✓ 1
Part (b)
M1 dx/dt = 17.25 and dy/dt = 0.6 at (57.5, 20) (ECF from (a))	✓ 1
A1 x = 66.125 (≈ 66.1)	✓ 1
A1 y = 20.3	✓ 1
Part (c)
M1 set dx/dt = 0 and divide by x: 0.5 − 0.01y = 0	✓ 1
A1 y = 50	✓ 1

Common error: In each Euler step use both old values when working out both gradients; updating x first and then reusing the new x for y changes the result.

+ Add to test✓ In your test324IBMI3H-5-04r8-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard8 marks

The horizontal sway x metres of the top of a tall building at time t seconds is modelled by d²x/dt² = −9x − 0.6(dx/dt). At t = 0, x = 0.2 and dx/dt = 0.

(a)

By letting y = dx/dt, write the model as two first-order equations.

(2)

(b)

Use Euler's method with step length 0.1 seconds to estimate x when t = 0.2.

(3)

(c)

Hence write down an estimate of dx/dt when t = 0.2, to three significant figures, and state whether the top of the building is moving towards or away from its original vertical position.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 dx/dt = y	✓ 1
A1 dy/dt = −9x − 0.6y	✓ 1
Part (b)
M1 first step: x = 0.2 and y = −0.18 at t = 0.1	✓ 1
M1 second step: x = 0.2 + 0.1 × (−0.18)	✓ 1
A1 x = 0.182 m	✓ 1
Part (c)
M1 y = −0.18 + 0.1(−9 × 0.2 − 0.6 × (−0.18)) at t = 0.2	✓ 1
A1 dx/dt ≈ −0.349 m s⁻¹	✓ 1
R1 x is positive and the velocity is negative, so it is moving towards the vertical	✓ 1

Exam tip: A second-order differential equation needs two starting values; list x and y = dx/dt together in each row of your Euler table.

+ Add to test✓ In your test325IBMI3H-5-04r8-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard6 marks

A fish tank is made by rotating the curve y = √x·e^−x/4, for 0 ≤ x ≤ a, about the x-axis. Lengths are in decimetres, and 1 dm³ = 1 litre.

(a)

Show that the volume of the tank is V = π∫₀^a xe^−x/2 dx.

(2)

(b)

Find the volume, in litres to three significant figures, when a = 8.

(2)

(c)

The tank is to hold exactly 10 litres. Find the required value of a, to three significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume = π∫y² dx	✓ 1
A1 y² = xe^−x/2	✓ 1
Part (b)
M1 integrate by parts or use a calculator: ∫xe^−x/2 dx = −2xe^−x/2 − 4e^−x/2	✓ 1
A1 π(4 − 20e⁻⁴) = 11.4 litres	✓ 1
Part (c)
M1 π(4 − (2a + 4)e^−a/2) = 10 solved on a calculator	✓ 1
A1 a = 5.93 dm	✓ 1

Common error: Square the radius function before integrating: (√x·e⁻x/⁴)² = xe⁻x/², so the exponent doubles.

+ Add to test✓ In your test326IBMI3H-5-04r11-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard7 marks

A glass vase is made by rotating the region between the $x$-axis and the curve $y = 1 + 0.5\sin x$, for $0 \le x \le \pi$, through $360^\circ$ about the $x$-axis. The units are centimetres.

(a)

Write down an integral for the volume $V$ of the vase.

(2)

(b)

Hence find the volume, to 3 significant figures.

(1)

(c)

Show that the exact value of the volume is $\pi\left(\frac{9\pi}{8} + 2\right)$.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $V = \pi\int y^2\,dx$	✓ 1
A1 $V = \pi\int_0^{\pi}(1 + 0.5\sin x)^2\,dx$	✓ 1
Part (b)
A1 $V = 17.4$ cm³	✓ 1
Part (c)
M1 expand: $(1 + 0.5\sin x)^2 = 1 + \sin x + 0.25\sin^2 x$	✓ 1
M1 use $\sin^2 x = \frac{1 - \cos 2x}{2}$	✓ 1
M1 integrate: $\left[x - \cos x + \frac{x}{8} - \frac{\sin 2x}{16}\right]_0^{\pi}$	✓ 1
A1 $= \left(\pi + 1 + \frac{\pi}{8}\right) - (-1) = \frac{9\pi}{8} + 2$, so $V = \pi\left(\frac{9\pi}{8}+2\right)$	✓ 1

Common error: Squaring only the terms inside the bracket separately. $(1+0.5\sin x)^2$ has a middle term, $2(1)(0.5\sin x) = \sin x$.

+ Add to test✓ In your test327IBMI3H-5-04r11-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard7 marks

Two connected tanks hold $x$ kg and $y$ kg of a pollutant at time $t$ hours. The amounts satisfy $\frac{dx}{dt} = -0.3x + 0.1y$ and $\frac{dy}{dt} = 0.3x - 0.1y$, with $x = 80$ and $y = 20$ when $t = 0$.

(a)

Use Euler's method with step length 0.5 to estimate $x$ and $y$ when $t = 1$.

(4)

(b)

Find the amounts in the tanks after a very long time.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $x_{n+1} = x_n + 0.5(-0.3x_n + 0.1y_n)$ and $y_{n+1} = y_n + 0.5(0.3x_n - 0.1y_n)$	✓ 1
A1 after one step: $x = 69$ and $y = 31$	✓ 1
A1 $x \approx 60.2$ kg at $t = 1$	✓ 1
A1 $y \approx 39.8$ kg at $t = 1$	✓ 1
Part (b)
M1 set both derivatives to zero, giving $y = 3x$	✓ 1
M1 note that $\frac{d}{dt}(x + y) = 0$, so $x + y = 100$ always	✓ 1
A1 $x = 25$ kg and $y = 75$ kg	✓ 1

Common error: Using only one of the two equations in the equilibrium step. You also need the fact that the total amount is constant to find the actual values.

+ Add to test✓ In your test328IBMI3H-5-05r28-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard7 marks

Calculator

A container hangs from a crane and swings. The angle $\theta$ (radians) from the vertical, $t$ seconds after release, satisfies $\dfrac{d^2\theta}{dt^2}=-9\theta-0.6\dfrac{d\theta}{dt}$ with $\theta(0)=0.2$ and $\theta'(0)=0$.

(a)

Let $x=\theta$ and $y=\frac{d\theta}{dt}$. Write the equation as a pair of first-order differential equations.

(2)

(b)

Use Euler's method with step size $h=0.1$ to estimate $\theta$ and $\frac{d\theta}{dt}$ at $t=0.2$. Give the answers to 3 significant figures.

(4)

(c)

State the physical effect of the term $-0.6\frac{d\theta}{dt}$ on the motion.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{dx}{dt}=y$	✓ 1
A1 $\frac{dy}{dt}=-9x-0.6y$	✓ 1
Part (b)
M1 apply $x_{n+1}=x_n+hy_n$, $y_{n+1}=y_n+h(-9x_n-0.6y_n)$ from $(0.2,0)$	✓ 1
A1 after one step $x=0.2$, $y=-0.18$	✓ 1
M1 second step using both values from the first step	✓ 1
A1 $\theta(0.2)\approx0.182$ and $\theta'(0.2)\approx-0.349$	✓ 1
Part (c)
R1 it is a damping (resistance) term that opposes the motion, so the amplitude of the swings decreases	✓ 1

Common error: In Euler's method for a system, both new values must be calculated from the old values; do not use the new $x$ when finding the new $y$.

+ Add to test✓ In your test329IBMI3H-5-05r28-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard8 marks

Calculator

A model for the sideways drift $y$ (km) of a tagged buoy from a ship's track, $x$ hours after release, is given by the differential equation $\dfrac{dy}{dx}=0.5(x-y)$. A slope field is drawn for this equation.

(a)

Find the equation of the isocline on which all the gradient segments are horizontal.

(2)

(b)

Show that $y=x-2$ is a solution of the differential equation.

(2)

(c)

The buoy is released from $(0,0)$. Describe in words the solution curve through $(0,0)$ for $x\ge0$.

(3)

(d)

Find the gradient of the slope field at $(4,1)$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set $\frac{dy}{dx}=0$	✓ 1
A1 $y=x$	✓ 1
Part (b)
M1 LHS: $\frac{d}{dx}(x-2)=1$	✓ 1
A1 RHS: $0.5(x-(x-2))=1$, equal, so it is a solution	✓ 1
Part (c)
A1 starts at the origin with zero gradient (horizontal)	✓ 1
A1 the curve lies below $y=x$, so the gradient is positive and it rises	✓ 1
A1 gradient increases towards 1 and the curve approaches the line $y=x-2$ without crossing it	✓ 1
Part (d)
A1 1.5	✓ 1

Exam tip: For a slope field, zero-gradient lines are found by setting the right-hand side to zero; solutions cannot cross a line that is itself a solution.

+ Add to test✓ In your test330IBMI3H-5-05r28-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard7 marks

Calculator

A decorative glass tank for an aquarium is made by rotating the region bounded by $y=\sqrt{x}\,e^{-x/4}$, the $x$-axis and the line $x=6$ through $360^\circ$ about the $x$-axis. Lengths are in metres.

(a)

Show by differentiation that $\int xe^{-x/2}\,dx=-2(x+2)e^{-x/2}+c$.

(2)

(b)

Find the volume of the tank. Give the exact value, then the value in m³ to 3 significant figures.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate $-2(x+2)e^{-x/2}$ using the product rule	✓ 1
A1 $-2e^{-x/2}+(x+2)e^{-x/2}=xe^{-x/2}$ (AG)	✓ 1
Part (b)
M1 $V=\pi\int_0^6y^2\,dx$	✓ 1
A1 $y^2=xe^{-x/2}$, so $V=\pi\int_0^6xe^{-x/2}\,dx$	✓ 1
M1 apply the result from (a) with limits 0 and 6	✓ 1
A1 exact: $V=\pi(4-16e^{-3})$	✓ 1
A1 10.1 m³	✓ 1

Exam tip: The volume about the x-axis uses $\pi y^2$, so square the whole function including the exponent: $(e^{-x/4})^2=e^{-x/2}$.

+ Add to test✓ In your test331IBMI3H-5-06r28-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard8 marks

Calculator

A decorative basin in a library courtyard has inner surface formed by rotating the curve $y=0.2x^2$ for $0\le x\le5$ about the $y$-axis, where $x$ and $y$ are in metres and $y$ is the depth of water above the lowest point.

(a)

Show that the volume of water at depth $y$ is $V=2.5\pi y^2$.

(3)

(b)

Water enters at 0.5 m³ per minute. Find the rate at which the depth is increasing when the depth is 2 m, to 3 significant figures.

(3)

(c)

Find the time to fill the basin from empty to a depth of 4 m, to the nearest minute.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $x^2=5y$	✓ 1
A1 $V=\pi\int_0^y5u\,du$ (any variable of integration)	✓ 1
A1 $V=2.5\pi y^2$ (AG)	✓ 1
Part (b)
M1 $\frac{dV}{dt}=\frac{dV}{dy}\times\frac{dy}{dt}$	✓ 1
A1 $\frac{dV}{dy}=5\pi y=10\pi$	✓ 1
A1 $\frac{dy}{dt}=\frac{0.5}{10\pi}=0.0159$ m min^-1	✓ 1
Part (c)
M1 $V=2.5\pi(4)^2=40\pi$ divided by 0.5	✓ 1
A1 251 minutes	✓ 1

Exam tip: In related-rates questions, write the chain rule first, then substitute the value only after differentiating.

+ Add to test✓ In your test332IBMI3H-5-07r28-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard8 marks

A parabolic satellite dish has a cross-section given by y = x²/5 for 0 ≤ x ≤ 1.5, where lengths are in metres. The dish is formed by rotating this curve through 360° about the y-axis. Its full depth is 0.45 m, so its rim has diameter 3.00 m.

(a)

Show that the volume of water filling the dish to depth h metres is 2.5πh² m³.

(3)

(b)

Find the volume when the dish is completely full. Give your answer to 3 significant figures.

(2)

(c)

Find the depth of water when the dish is half full by volume. Use the unrounded full volume and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 V = π∫₀^h x² dy (rotation about the y-axis)	✓ 1
A1 write x² = 5y, so V = π∫₀^h 5y dy	✓ 1
A1 V = 2.5πh²	✓ 1
Part (b)
M1 substitute h = 0.45	✓ 1
A1 1.59 m³	✓ 1
Part (c)
M1 2.5πh² = ½ × 2.5π × 0.45²	✓ 1
M1 h² = 0.45²/2	✓ 1
A1 h = 0.318 m	✓ 1

Common error: For rotation about the y-axis you need x² in terms of y and a dy integral; integrating x² with respect to x gives the volume about the wrong axis.

+ Add to test✓ In your test333IBMI3H-5-08r28-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard8 marks

The sway of a building in wind is modelled by d²x/dt² = −9x − 0.6 dx/dt, where x cm is the displacement from rest at time t seconds. Initially x = 0 and dx/dt = 0.3.

(a)

Let v = dx/dt. Write the model as two first-order equations.

(2)

(b)

Use Euler’s method with step length 0.1 to estimate x and v when t = 0.2.

(4)

(c)

If the damping term were removed, the model would be d²x/dt² = −9x, whose solutions oscillate with angular frequency √9. Find the period of these oscillations. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 dx/dt = v	✓ 1
A1 dv/dt = −9x − 0.6v	✓ 1
Part (b)
M1 first step with x = 0, v = 0.3	✓ 1
A1 x(0.1) ≈ 0.03, v(0.1) ≈ 0.282	✓ 1
M1 second step with these values	✓ 1
A1 x(0.2) ≈ 0.0582, v(0.2) ≈ 0.23808	✓ 1
Part (c)
M1 period = 2π/ω with ω = 3	✓ 1
A1 2.09 s	✓ 1

Common error: Remember that the damping term uses the current velocity v, not the current displacement, when you update v.

+ Add to test✓ In your test334IBMI3H-5-06r32-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard8 marks

A reservoir holds $V$ m$^3$ of water at time $t$ days. It is filled at a constant rate of 12 000 m$^3$ per day and loses water at a rate of $0.05V$ m$^3$ per day, so $\dfrac{dV}{dt}=12000-0.05V$. Initially $V=100\,000$.

(a)

Separate the variables and integrate to show that $-20\ln|12000-0.05V|=t+c$.

(3)

(b)

Find $V$ as a function of $t$.

(3)

(c)

Find the time for the reservoir to reach $190\,000$ m$^3$, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 separates to $\int\frac{dV}{12000-0.05V}=\int dt$	✓ 1
A1 $-\frac{1}{0.05}\ln\|12000-0.05V\|=-20\ln\|12000-0.05V\|$	✓ 1
A1 $=t+c$	✓ 1
Part (b)
M1 uses $V=100000$ at $t=0$, so $12000-5000=7000$ and $c=-20\ln7000$	✓ 1
M1 rearranges to $12000-0.05V=7000e^{-0.05t}$	✓ 1
A1 $V=240000-140000e^{-0.05t}$	✓ 1
Part (c)
M1 $140000e^{-0.05t}=50000$ (ECF from (b))	✓ 1
A1 $20.6$ days	✓ 1

Common error: Dividing by $-0.05$ when integrating $\frac{1}{12000-0.05V}$ matters: the coefficient of $\ln$ is $-20$, not $-0.05$.

+ Add to test✓ In your test335IBMI3H-5-06r34-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard8 marks

The sway x metres of the top of a wind-turbine tower at time t seconds satisfies d²x/dt² + 0.5 dx/dt + 4x = 0, with x = 1 and dx/dt = 0 when t = 0.

(a)

Let v = dx/dt. Write the equation as a pair of first-order differential equations.

(2)

(b)

Use Euler's method with step length 0.1 to estimate x and v when t = 0.2.

(4)

(c)

The exact value is x(0.2) = 0.9236 (4 d.p.). Find the error in the Euler estimate for x(0.2), to 3 significant figures, and suggest one way to reduce it.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
dx/dt = v	✓ 1
dv/dt = −4x − 0.5v	✓ 1
Part (b)
Step 1: x = 1 + 0.1(0) = 1 and v = 0 + 0.1(−4) = −0.4	✓ 1
Step 2: dv/dt = −4 × 1 − 0.5 × (−0.4) = −3.8	✓ 1
x = 1 + 0.1(−0.4) = 0.96	✓ 1
v = −0.4 + 0.1(−3.8) = −0.78	✓ 1
Part (c)
Error = 0.96 − 0.9236 = 0.0364	✓ 1
Use a smaller step length	✓ 1

Exam tip: A second-order equation is handled by introducing the first derivative as a new variable; then both variables are updated together at each step.

+ Add to test✓ In your test336IBMI3H-5-07r34-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard7 marks

The vertical displacement x metres of a camera platform on a damped suspension, t seconds after release, satisfies $\frac{d^2x}{dt^2} + 0.4\frac{dx}{dt} + 4x = 0$. At $t=0$, $x = 1$ and the platform is at rest.

(a)

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

(2)

(b)

Use Euler's method with a step length of 0.1 seconds to estimate x and v when $t=0.3$.

(4)

(c)

State the estimated speed of the platform at $t=0.3$, to three significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dx}{dt} = v$	✓ 1
A1 $\frac{dv}{dt} = -4x - 0.4v$	✓ 1
Part (b)
M1 correct Euler updates $x_{n+1}=x_n+0.1v_n$ and $v_{n+1}=v_n+0.1(-4x_n-0.4v_n)$	✓ 1
A1 $t=0.1$: $x=1$, $v=-0.4$	✓ 1
A1 $t=0.2$: $x=0.96$, $v=-0.784$	✓ 1
A1 $t=0.3$: $x=0.8816$, $v=-1.13664$	✓ 1
Part (c)
A1 1.14 m s⁻¹ (1.13664 before rounding)	✓ 1

Common error: Updating x with the new value of v (instead of the old value) turns the method into a different scheme and gives wrong answers.

+ Add to test✓ In your test337IBMI3H-5-06r43-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard8 marks

In a town with 900 susceptible people, 10 become infected. Let S and I be the numbers of susceptible and infected people after t days, modelled by
dS/dt = −0.002SI, dI/dt = 0.002SI − 0.1I, with S = 900 and I = 10 when t = 0.

(a)

Use Euler's method with a step size of 1 day to find S and I when t = 1.

(2)

(b)

Continue the method to find S and I when t = 2, each to 3 significant figures. Use the unrounded values from part (a).

(3)

(c)

For I ≠ 0, find the value of S at which the number of infected people is momentarily constant.

(2)

(d)

State what happens to the number of infected people when S is below this value.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to update both S and I using the starting values	✓ 1
A1 S = 882 and I = 27	✓ 1
Part (b)
M1 use S₁ and I₁ in both equations (decrease in S = 47.628)	✓ 1
A1 S = 834	✓ 1
A1 I = 71.9	✓ 1
Part (c)
M1 set 0.002SI − 0.1I = 0 and divide by I	✓ 1
A1 S = 50	✓ 1
Part (d)
A1 the number of infected people decreases	✓ 1

Common error: Both variables must be updated using the previous values of both; do not use the new S when calculating the new I within the same step.

+ Add to test✓ In your test338IBMI3H-5-06r43-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard7 marks

The vertical displacement x metres of a wave-energy buoy satisfies d²x/dt² = −4x − 0.5 dx/dt, where t is in seconds. At t = 0, x = 1 and dx/dt = 0.

(a)

Let v = dx/dt. Write the equation as a pair of first-order differential equations.

(2)

(b)

Use Euler's method with step size 0.1 to estimate x when t = 0.3.

(4)

(c)

Suggest one way to improve the accuracy of the estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 dx/dt = v	✓ 1
A1 dv/dt = −4x − 0.5v	✓ 1
Part (b)
M1 use x_n+1 = x_n + 0.1v_n and v_n+1 = v_n + 0.1(−4x_n − 0.5v_n)	✓ 1
A1 x = 1 and v = −0.4 at t = 0.1	✓ 1
A1 x = 0.96 and v = −0.78 at t = 0.2	✓ 1
A1 x = 0.882 at t = 0.3	✓ 1
Part (c)
R1 use a smaller step size	✓ 1

Common error: Update both x and v from the same previous step; using the new x to find the new v mixes steps.

+ Add to test✓ In your test339IBMI3H-5-06r43-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard7 marks

The number of people infected in an outbreak is modelled by I(t) = 2000/(1 + 19e^−0.5t), where t is in days.

(a)

Find I(0).

(1)

(b)

Find the rate of increase of the number of infected people at t = 4, to 3 significant figures.

(4)

(c)

Use a calculator to find the time at which the number of infected people is increasing most rapidly, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 100	✓ 1
Part (b)
M1 attempt to use the quotient rule or the chain rule	✓ 1
A1 dI/dt = 19000e^−0.5t/(1 + 19e^−0.5t)²	✓ 1
M1 substitute t = 4	✓ 1
A1 202 people per day	✓ 1
Part (c)
M1 attempt to find the maximum of dI/dt (graph, or I = 1000)	✓ 1
A1 t = 5.89 days	✓ 1

Exam tip: The fastest growth in a logistic model occurs when the population is half of its limiting value.

+ Add to test✓ In your test340IBMI3H-5-06r46-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 1Hard7 marks

Calculator

A robot's joint, of displacement $x$ radians from its rest position at time $t$ seconds, satisfies $\frac{d^2x}{dt^2} = -4x - 0.6\frac{dx}{dt}$. At $t = 0$, $x = 1$ and $\frac{dx}{dt} = 0$.

(a)

Let $v = \frac{dx}{dt}$. Write the second-order equation as a pair of coupled first-order equations.

(2)

(b)

Use Euler's method with step size $h = 0.1$ to estimate $x$ and $v$ at $t = 0.2$. Give $x$ and $v$ to 3 decimal places.

(4)

(c)

State what the sign of $v$ at $t = 0.2$ tells you about the joint.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dx}{dt} = v$	✓ 1
A1 $\frac{dv}{dt} = -4x - 0.6v$	✓ 1
Part (b)
M1 update both variables using the values at the start of each step	✓ 1
A1 $t = 0.1$: $x = 1$, $v = -0.4$	✓ 1
A1 $x(0.2) \approx 0.96$	✓ 1
A1 $v(0.2) \approx -0.776$	✓ 1
Part (c)
R1 $v < 0$, so the joint is moving back towards its rest position (displacement is decreasing)	✓ 1

Exam tip: In coupled Euler steps use the old value of each variable on the right-hand side; do not use a value you have just updated within the same step.

+ Add to test✓ In your test341IBMI3H-5-06r23-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Easy9 marks

Calculator

The height of the tide in a harbour is modelled by H(t) = 3.2 + 1.8 sin(0.5t − 1) metres, where t is the time in hours after midnight and the angle is in radians.

(a)

Find H′(t).

(2)

(b)

Find the rate at which the tide height is changing at 04:00, correct to 3 significant figures.

(2)

(c)

State the greatest rate at which the tide height can rise, with a reason.

(2)

(d)

Find the time of the first high tide after midnight, correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of the chain rule on the sine term	✓ 1
A1 H′(t) = 0.9 cos(0.5t − 1)	✓ 1
Part (b)
M1 substitute t = 4	✓ 1
A1 0.486 m per hour	✓ 1
Part (c)
R1 the greatest value of cos is 1	✓ 1
A1 0.9 m per hour	✓ 1
Part (d)
M1 set H′(t) = 0	✓ 1
M1 0.5t − 1 = π/2	✓ 1
A1 t = 5.14 hours after midnight	✓ 1

Common error: Calculator in radian mode: the angle in this model is in radians, so using degree mode gives a very different value for the rate.

+ Add to test✓ In your test342IBMI3H-5-07r28-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Easy10 marks

A bowl of soup cools in a kitchen at 20 °C. Its temperature T °C after t minutes satisfies dT/dt = −0.1(T − 20), with T = 90 when t = 0.

(a)

Use Euler’s method with a step length of 2 minutes to estimate T when t = 6. Work with unrounded values and give the final answer to 3 significant figures.

(4)

(b)

Solve the differential equation by separating variables to show that T = 20 + 70e^−0.1t.

(3)

(c)

Find the exact-solution value of T when t = 6 to 3 significant figures, and explain why Euler’s method gives a lower value.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 T_n+1 = T_n + 2 × (−0.1)(T_n − 20)	✓ 1
A1 T(2) ≈ 76	✓ 1
A1 T(4) ≈ 64.8	✓ 1
A1 T(6) ≈ 55.8 °C	✓ 1
Part (b)
M1 separate: ∫ 1/(T − 20) dT = ∫ −0.1 dt	✓ 1
A1 ln(T − 20) = −0.1t + c	✓ 1
A1 use T(0) = 90 to get T − 20 = 70e^−0.1t, hence the printed result	✓ 1
Part (c)
M1 substitute t = 6	✓ 1
A1 58.4 °C	✓ 1
R1 the gradient becomes less steep as the soup cools, but each Euler step uses the steeper gradient at the start of the step, so it over-estimates the fall in temperature	✓ 1

Exam tip: Keep full calculator values through each Euler step and round only the final answer.

+ Add to test✓ In your test343IBMI3H-5-08r28-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Easy9 marks

A marine research tank is an open-topped cuboid with a square base of side x metres and height h metres. Its volume is 4 m³. Glass for all five faces costs €85 per m².

(a)

Show that the total area A m² of glass needed is A = x² + 16/x.

(3)

(b)

Find the value of x that minimises A, justifying that it is a minimum.

(4)

(c)

Find the minimum cost of the glass.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x²h = 4, so h = 4/x²	✓ 1
M1 A = x² + 4xh	✓ 1
A1 substitute to get A = x² + 16/x	✓ 1
Part (b)
M1 A′ = 2x − 16/x²	✓ 1
M1 set A′ = 0, so x³ = 8	✓ 1
A1 x = 2	✓ 1
R1 A″ = 2 + 32/x³ > 0, so a minimum	✓ 1
Part (c)
M1 A = 4 + 8 = 12 m², then multiply by 85	✓ 1
A1 €1020	✓ 1

Exam tip: Count only the faces that exist: an open-topped tank has a base and four sides, so the area is x² + 4xh.

+ Add to test✓ In your test344IBMI3H-5-07r45-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Easy9 marks

Calculator

The cross-section of a lava channel wall is modelled by $h(x)=0.01x^3-0.3x^2+2.4x$, where $h$ is the height in metres and $x$ is the horizontal distance in metres, $0\le x\le 8$.

(a)

Find $h'(x)$.

(2)

(b)

Find the height and the gradient of the wall when $x=5$.

(2)

(c)

Find the equation of the tangent at $x=5$, in the form $y=mx+c$.

(2)

(d)

The normal to the wall at $x=5$ crosses the vertical axis at the point $(0,c)$. Find $c$, correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate at least one term correctly	✓ 1
A1 $h'(x)=0.03x^2-0.6x+2.4$	✓ 1
Part (b)
A1 $h(5)=5.75$	✓ 1
A1 $h'(5)=0.15$	✓ 1
Part (c)
M1 $y-5.75=0.15(x-5)$	✓ 1
A1 $y=0.15x+5$	✓ 1
Part (d)
M1 normal gradient $=-\frac{1}{0.15}$	✓ 1
M1 $c=5.75+5\times\frac{1}{0.15}$	✓ 1
A1 $c=39.1$	✓ 1

Exam tip: The normal gradient is the negative reciprocal of the tangent gradient. Use the unrounded gradient, $-1/0.15$, when you find the intercept.

+ Add to test✓ In your test345IBMI-034-03-r1-05Differential equations and slope fields Paper 2Medium13 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 test346IBMI-034-04-r2-05Euler's method and coupled systems Paper 2Medium11 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 test347IBMI-034-03-r2-03Differential equations and slope fields Paper 2Medium12 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 test348IBMI-034-04-r1-03Euler's method and coupled systems Paper 2Medium10 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 test349IBMI-034-04-r1-06Euler's method and coupled systems Paper 2Medium9 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 test350IBMI-034-06-r1-05Optimisation in context Paper 2Medium11 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 test351IBMI-034-06-r2-05Optimisation in context Paper 2Medium10 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 test352IBMI-034-08-r1-03Differential equations and slope fields Paper 2Medium14 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 test353IBMI-034-03-r2-05Differential equations and slope fields Paper 2Medium12 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 test354IBMI-034-04-r2-03Euler's method and coupled systems Paper 2Medium10 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 test355IBMI-034-08-r2-05Differential equations and slope fields Paper 2Medium12 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 test356IBMI-034-08-r2-03Differential equations and slope fields Paper 2Medium9 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 test357IBMI-034-01-r2-05Optimisation in context Paper 2Medium10 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 test358IBMI3H-5-10-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium13 marks

The inside surface of a loudspeaker horn is formed by rotating the curve y = 4e^−0.1x, 0 ≤ x ≤ 15, through 360° about the x-axis. Lengths are in centimetres.

(a)

Find the radius of the horn at x = 0 and at x = 15.

(2)

(b)

Show that the volume of the horn can be written as $V=\pi\int_0^{15}16e^{-0.2x}\,dx$.

(2)

(c)

Find the volume of the horn.

(4)

(d)

Find the value of x at which the radius is half of its value at x = 0.

(3)

(e)

Find the percentage of the total volume that lies between x = 0 and the value of x found in (d).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 4 cm	✓ 1
A1 0.893 cm (0.8925)	✓ 1
Part (b)
M1 use of $V=\pi\int y^2\,dx$	✓ 1
A1 $y^2=16e^{-0.2x}$	✓ 1
Part (c)
M1 attempt to integrate, antiderivative $-80e^{-0.2x}$	✓ 1
A1 $-80\pi e^{-0.2x}$ correct with π	✓ 1
M1 substitute limits 15 and 0	✓ 1
A1 $80\pi(1-e^{-3})=239$ cm³	✓ 1
Part (d)
M1 $4e^{-0.1x}=2$	✓ 1
M1 take natural logarithms	✓ 1
A1 x = 10 ln 2 = 6.93 cm	✓ 1
Part (e)
M1 $\frac{1-e^{-0.2\times 6.93}}{1-e^{-3}}\times 100$	✓ 1
A1 78.9%	✓ 1

Common error: Square the whole function when rotating about the x-axis: (4e⁻⁰.¹x)² = 16e⁻⁰.²x, not 4²e⁻⁰.¹x.

+ Add to test✓ In your test359IBMI3H-5-10-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium12 marks

A decorative bowl is made by rotating the curve y = 0.1x², 0 ≤ y ≤ 8, through 360° about the y-axis. Lengths are in centimetres, and the bowl is open at the top.

(a)

Find the radius of the rim of the bowl.

(2)

(b)

Show that when the bowl contains water to a depth of h cm, the volume of water is V = 5πh².

(4)

(c)

Find the capacity of the bowl, in millilitres, to 3 significant figures.

(2)

(d)

Water is poured into the empty bowl at a constant rate of 60 cm³ s⁻¹. Find the rate at which the depth is increasing when the depth is 4 cm.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute y = 8 to get x² = 80	✓ 1
A1 8.94 cm	✓ 1
Part (b)
M1 use of $V=\pi\int x^2\,dy$	✓ 1
A1 $x^2=10y$	✓ 1
M1 $\pi\int_0^h 10y\,dy$	✓ 1
A1 $5\pi h^2$	✓ 1
Part (c)
M1 substitute h = 8	✓ 1
A1 1010 ml (1005 cm³)	✓ 1
Part (d)
M1 $\frac{dV}{dh}=10\pi h$	✓ 1
M1 $\frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}$	✓ 1
A1 $60=10\pi(4)\frac{dh}{dt}$	✓ 1
A1 0.477 cm s⁻¹	✓ 1

Common error: When rotating about the y-axis, write x² in terms of y and integrate with respect to y; do not integrate y with respect to x.

+ Add to test✓ In your test360IBMI3H-5-10b-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium12 marks

The profile of an artificial hill on a mountain-bike track is modelled by f(x) = 0.1x³ − 1.2x² + 3.6x, for 0 ≤ x ≤ 6, where x is the horizontal distance and f(x) the height, both in metres.

(a)

Find f′(x).

(2)

(b)

Find the coordinates of the stationary points of the profile.

(3)

(c)

Use the second derivative to determine the nature of each stationary point.

(2)

(d)

Find the x-coordinate of the steepest part of the downhill section, and the gradient there.

(3)

(e)

Find the area between the profile and the x-axis for 0 ≤ x ≤ 6.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 0.3x² − 2.4x + 3.6	✓ 1
Part (b)
M1 set f′(x) = 0	✓ 1
A1 x = 2 and x = 6	✓ 1
A1 (2, 3.2) and (6, 0)	✓ 1
Part (c)
M1 f″(x) = 0.6x − 2.4	✓ 1
A1 f″(2) = −1.2 < 0 so a maximum at (2, 3.2); f″(6) = 1.2 > 0 so a minimum at (6, 0)	✓ 1
Part (d)
M1 set f″(x) = 0	✓ 1
A1 x = 4	✓ 1
A1 gradient = −1.2	✓ 1
Part (e)
M1 ∫₀⁶ f(x) dx	✓ 1
A1 10.8 m²	✓ 1

Exam tip: The steepest point of a slope is where the gradient has a stationary value, so set the second derivative equal to zero.

+ Add to test✓ In your test361IBMI3H-5-04r11-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

The vertical displacement $x$ metres of a game vehicle's suspension satisfies $\frac{d^2x}{dt^2} + 0.6\frac{dx}{dt} + 4x = 0$. At $t = 0$, $x = 1$ and $\frac{dx}{dt} = 0$.

(a)

Let $y = \frac{dx}{dt}$. Write the equation as a pair of first-order differential equations.

(2)

(b)

Use Euler's method with step length 0.1 to estimate $x$ and $y$ when $t = 0.2$.

(4)

(c)

The system matrix is $\begin{pmatrix}0&1\\-4&-0.6\end{pmatrix}$. Find its eigenvalues and hence describe the motion. Give the angular frequency in rad/s to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dx}{dt} = y$	✓ 1
A1 $\frac{dy}{dt} = -4x - 0.6y$	✓ 1
Part (b)
M1 $x_{n+1} = x_n + 0.1y_n$ and $y_{n+1} = y_n + 0.1(-4x_n - 0.6y_n)$	✓ 1
A1 after one step: $x = 1$ and $y = -0.4$	✓ 1
A1 $x(0.2) \approx 0.96$	✓ 1
A1 $y(0.2) \approx -0.776$	✓ 1
Part (c)
M1 $\lambda^2 + 0.6\lambda + 4 = 0$	✓ 1
A1 $\lambda = -0.3 \pm 1.98i$ (imaginary part $\sqrt{4 - 0.3^2} = 1.98$)	✓ 1
R1 complex eigenvalues with negative real part: oscillation with angular frequency 1.98 rad/s and decreasing amplitude	✓ 1

Exam tip: A second-order equation becomes a first-order system by letting $y = \frac{dx}{dt}$, so that $\frac{d^2x}{dt^2} = \frac{dy}{dt}$.

+ Add to test✓ In your test362IBMI3H-5-03r14-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

A food company designs a closed cylindrical tin of volume 500 cm³, with base radius $r$ cm and height $h$ cm. The company wishes to minimise the surface area $A$ cm² of metal used.

(a)

Show that $A=2\pi r^2+\dfrac{1000}{r}$.

(2)

(b)

Find the value of $r$ that minimises $A$, giving your answer to 3 significant figures.

(4)

(c)

Justify that this value of $r$ gives a minimum.

(1)

(d)

Using the unrounded value of $r$, find the minimum surface area to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\pi r^2h=500$, so $h=\frac{500}{\pi r^2}$	✓ 1
A1 $A=2\pi r^2+2\pi rh=2\pi r^2+\frac{1000}{r}$	✓ 1
Part (b)
M1 attempts to differentiate	✓ 1
A1 $\frac{dA}{dr}=4\pi r-\frac{1000}{r^2}$	✓ 1
M1 sets $\frac{dA}{dr}=0$ giving $r^3=\frac{250}{\pi}$	✓ 1
A1 $r=4.30$ cm	✓ 1
Part (c)
R1 $\frac{d^2A}{dr^2}=4\pi+\frac{2000}{r^3}>0$ for $r>0$ (accept sign change of $\frac{dA}{dr}$ from negative to positive)	✓ 1
Part (d)
M1 substitutes $r=4.3012\ldots$ into $A$	✓ 1
A1 349 cm²	✓ 1

Exam tip: For a 'show that' question, write each algebraic step; the given result must follow clearly from your working.

+ Add to test✓ In your test363IBMI3H-5-03r14-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

The number of subscribers $N$ (in thousands) to a village mobile network, $t$ years after launch, is modelled by $\frac{dN}{dt}=0.3N\left(1-\frac{N}{50}\right)$. A slope field is drawn for this equation in the $t$–$N$ plane.

(a)

Calculate the gradient of the slope field at the point where $N=10$.

(2)

(b)

Find the values of $N$ for which the slope segments are horizontal.

(2)

(c)

Calculate the gradient where $N=60$, and describe the direction of the slope segments there.

(2)

(d)

Find the value of $N$ at which the gradient is greatest, and the greatest gradient. Give units.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $0.3\times10\times\left(1-\frac{10}{50}\right)$	✓ 1
A1 2.4	✓ 1
Part (b)
M1 sets $0.3N\left(1-\frac{N}{50}\right)=0$	✓ 1
A1 $N=0$ and $N=50$	✓ 1
Part (c)
M1 $0.3\times60\times\left(1-\frac{60}{50}\right)$	✓ 1
A1 −3.6, so the segments slope downward (towards $N=50$)	✓ 1
Part (d)
M1 writes $f(N)=0.3N-0.006N^2$ and differentiates with respect to $N$: $0.3-0.012N$	✓ 1
A1 $N=25$ (thousand)	✓ 1
A1 greatest gradient 3.75 thousand subscribers per year	✓ 1

Exam tip: For a rate that depends only on $N$, find its maximum by differentiating the right-hand side with respect to $N$, not with respect to $t$.

+ Add to test✓ In your test364IBMI3H-5-05r17-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium11 marks

Two container terminals, A and B, exchange containers. At time t days, terminal A holds x thousand containers and terminal B holds y thousand containers. The system is modelled by
$\dfrac{dx}{dt}=-0.2x+0.1y,\qquad \dfrac{dy}{dt}=0.2x-0.1y.$
Initially $x=6.0$ and $y=2.0$.

(a)

Show that $x+y$ is constant, and state what this means in context.

(2)

(b)

Use Euler's method with step length 0.5 to estimate x and y when $t=1$.

(4)

(c)

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

(2)

(d)

Find the value that x approaches in the long term. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{d(x+y)}{dt}=\frac{dx}{dt}+\frac{dy}{dt}=0$	✓ 1
R1 so $x+y=8$ always: the total number of containers in the two terminals does not change	✓ 1
Part (b)
M1 apply Euler's method to both equations: $x_1=6+0.5(-1.0)$, $y_1=2+0.5(1.0)$	✓ 1
A1 $t=0.5$: $x=5.5$, $y=2.5$	✓ 1
M1 second step using the new values: $\frac{dx}{dt}=-0.85$, $\frac{dy}{dt}=0.85$	✓ 1
A1 $t=1$: $x=5.075$, $y=2.925$	✓ 1
Part (c)
M1 attempt to solve $\lambda^2+0.3\lambda=0$ (or use GDC)	✓ 1
A1 $\lambda=0$ and $\lambda=-0.3$	✓ 1
Part (d)
M1 equilibrium: $\frac{dx}{dt}=0$, so $-0.2x+0.1y=0$	✓ 1
A1 $y=2x$, with $x+y=8$ from (a)	✓ 1
A1 $x=2.67$ thousand containers	✓ 1

Exam tip: A zero eigenvalue signals a conserved quantity, so the long-term state is found by combining the equilibrium condition with the constant total.

+ Add to test✓ In your test365IBMI3H-5-06r21-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium10 marks

A server of temperature T °C is cooled in a room at 22 °C. Its temperature satisfies dT/dt = −0.1(T − 22), where t is in minutes and T = 70 when t = 0.

(a)

Use Euler's method with a step length of 2 minutes to estimate T at t = 6. Show each step.

(4)

(b)

Solve the differential equation by separating variables to show that T = 22 + 48e^−0.1t.

(3)

(c)

Explain why Euler's method gives a lower temperature than the exact solution at t = 6.

(1)

(d)

Find the exact value of T at t = 6, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 T_n+1 = T_n + 2 × (−0.1)(T_n − 22)	✓ 1
A1 T at t = 2 is 60.4	✓ 1
A1 T at t = 4 is 52.72	✓ 1
A1 T at t = 6 is 46.576, so 46.6 °C	✓ 1
Part (b)
M1 ∫ 1/(T − 22) dT = ∫ −0.1 dt	✓ 1
A1 ln(T − 22) = −0.1t + c	✓ 1
A1 uses T(0) = 70: T − 22 = 48e^−0.1t (AG)	✓ 1
Part (c)
R1 each step uses the gradient at the start of the step, which is steeper than the average gradient over the step because the cooling rate is decreasing	✓ 1
Part (d)
M1 22 + 48e^−0.6	✓ 1
A1 48.3 °C	✓ 1

Exam tip: Show a table of t, T and the gradient at each step; it earns method marks even if an arithmetic slip occurs.

+ Add to test✓ In your test366IBMI3H-5-06r21-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

A cyclist rides along a straight training road. Her velocity is v(t) = 3cos(0.4t) m s⁻¹ for 0 ≤ t ≤ 6, where positive means away from the start and the angle is in radians.

(a)

Find her acceleration at t = 2, to 3 significant figures.

(2)

(b)

Find the time at which she changes direction, to 3 significant figures.

(2)

(c)

Find her displacement from the start at t = 6, to 3 significant figures.

(2)

(d)

Find the total distance she travels in the six seconds, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a = −1.2 sin(0.4t)	✓ 1
A1 −0.861 m s⁻²	✓ 1
Part (b)
M1 3cos(0.4t) = 0, so 0.4t = π/2	✓ 1
A1 t = 3.93 s	✓ 1
Part (c)
M1 ∫₀⁶ 3cos(0.4t) dt = [7.5 sin(0.4t)]₀⁶	✓ 1
A1 5.07 m	✓ 1
Part (d)
M1 splits the integral at t = 5π/4 (3.93…)	✓ 1
A1 distance out = 7.5 m and distance back = 7.5(1 − sin 2.4) = 2.43 m	✓ 1
A1 total = 9.93 m	✓ 1

Common error: Displacement and distance are different when the direction changes: integrate |v| (or split at the turning point) to find the distance.

+ Add to test✓ In your test367IBMI3H-5-06r22-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium10 marks

A cup of coffee at 96 °C is placed in a room at 22 °C. Its temperature T °C after t minutes satisfies $\frac{dT}{dt}=-0.1(T-22)$.

(a)

Find the gradient of the solution curve at t = 0.

(1)

(b)

Describe two features of the slope field of this differential equation in the t–T plane for T > 22.

(2)

(c)

Use Euler's method with a step of 2 minutes to estimate T when t = 4.

(3)

(d)

The exact solution is $T=22+74e^{-0.1t}$. Find the percentage error of the Euler estimate for t = 4, to 3 significant figures. Use 69.36 as the estimate.

(3)

(e)

State whether Euler's method under- or overestimates here, and give a reason based on the shape of the exact curve.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 −7.4 °C min⁻¹	✓ 1
Part (b)
R1 all gradients are negative, so temperature decreases	✓ 1
R1 gradients are the same along any horizontal line and become less steep as T approaches 22	✓ 1
Part (c)
M1 $T_{n+1}=T_n+2\times\frac{dT}{dt}$	✓ 1
A1 T(2) ≈ 81.2	✓ 1
A1 T(4) ≈ 69.36 °C	✓ 1
Part (d)
A1 exact value 71.6 (accept 71.60)	✓ 1
M1 $\frac{\|71.60-69.36\|}{71.60}\times100$	✓ 1
A1 3.13 %	✓ 1
Part (e)
R1 underestimates: the curve is concave up (decreasing rate of fall), so each tangent step falls faster than the true curve	✓ 1

Exam tip: In Euler's method, recalculate the gradient from the latest (t, T) values at every step; do not reuse the first gradient.

+ Add to test✓ In your test368IBMI3H-5-06r23-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium10 marks

Calculator

After an eruption, the concentration of ash in the air at a monitoring site is modelled by C(t) = 12te^−0.4t mg m⁻³, where t is the time in hours after the eruption, t ≥ 0.

(a)

Find C′(t), giving your answer in the form 12e^−0.4t(a + bt).

(3)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

Find the greatest concentration, correct to 3 significant figures, and justify that it is a maximum.

(2)

(d)

Show that the concentration is decreasing most rapidly when t = 5.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of the product rule	✓ 1
A1 C′ = 12e^−0.4t − 4.8te^−0.4t	✓ 1
A1 C′ = 12e^−0.4t(1 − 0.4t)	✓ 1
Part (b)
M1 set C′(t) = 0	✓ 1
A1 t = 2.5 hours	✓ 1
Part (c)
R1 C′ changes from positive to negative at t = 2.5 (or C″(2.5) < 0)	✓ 1
A1 C(2.5) = 30e⁻¹ = 11.0 mg m⁻³	✓ 1
Part (d)
M1 differentiate again, C″ = 12e^−0.4t(0.16t − 0.8)	✓ 1
M1 C″ = 0	✓ 1
A1 t = 5, where C″ changes from negative to positive, so C′ is least	✓ 1

Exam tip: For a maximum value, give a reason such as a sign change of the first derivative or the sign of the second derivative; the value alone does not earn the reasoning mark.

+ Add to test✓ In your test369IBMI3H-5-06r23-04Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium10 marks

Calculator

The cross-section of a volcanic crater is modelled by the curve y = x²/50 for 0 ≤ x ≤ 20, where x and y are in metres. The crater is formed by rotating this curve through 360° about the y-axis, so that the crater floor is at the origin and the rim is 20 m from the axis.

(a)

Show that x² = 50y and find the depth of the crater.

(2)

(b)

Find the volume of the crater, correct to 3 significant figures.

(3)

(c)

Show that when the lava in the crater has depth h metres above the floor, the volume of lava is 25πh² m³.

(2)

(d)

Lava enters the crater at a constant rate of 80 m³ per minute. Find the rate at which the depth of the lava is increasing when h = 5, correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearrange y = x²/50	✓ 1
A1 depth = 20²/50 = 8 m	✓ 1
Part (b)
M1 V = π∫x² dy = π∫₀⁸ 50y dy	✓ 1
A1 π[25y²] from 0 to 8	✓ 1
A1 1600π = 5030 m³	✓ 1
Part (c)
M1 V = π∫₀^h 50y dy	✓ 1
A1 25πh²	✓ 1
Part (d)
M1 dV/dt = 50πh × dh/dt	✓ 1
M1 substitute dV/dt = 80 and h = 5	✓ 1
A1 0.102 m per minute	✓ 1

Common error: When rotating about the y-axis, integrate with respect to y and write x² in terms of y; using the integral of y² dx gives a different solid.

+ Add to test✓ In your test370IBMI3H-5-06r23-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

Calculator

Lava leaving a vent at 1150 °C cools in air at a constant temperature of 25 °C. Its temperature T °C after t hours satisfies dT/dt = −k(T − 25), where k is a positive constant.

(a)

By separating variables, show that T = 25 + 1125e^−kt.

(4)

(b)

After 6 hours the temperature of the lava is 400 °C. Find the value of k, correct to 3 significant figures.

(3)

(c)

Using the unrounded value of k, find the time for the lava to cool to 100 °C, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ∫ 1/(T − 25) dT = ∫ −k dt	✓ 1
A1 ln\|T − 25\| = −kt + c	✓ 1
M1 T − 25 = Ae^−kt	✓ 1
A1 T = 1150 when t = 0 gives A = 1125	✓ 1
Part (b)
M1 375 = 1125e^−6k	✓ 1
M1 e^−6k = 1/3 so −6k = ln(1/3)	✓ 1
A1 k = 0.183	✓ 1
Part (c)
M1 75 = 1125e^−kt	✓ 1
A1 t = 14.8 hours	✓ 1

Exam tip: Use the unrounded value of k in later parts; for this model the time can also be written exactly as 6 ln 15 / ln 3.

+ Add to test✓ In your test371IBMI3H-5-06r23-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium11 marks

Calculator

Ground tilt y (in mm) measured at a volcano x days after an alert satisfies the differential equation dy/dx = x − y. The slope field of this equation is to be studied.

(a)

Find the gradient of the slope field at each of the points (0, 2), (1, 0) and (2, 2).

(3)

(b)

Find the equation of the line along which every slope segment has gradient 1.

(2)

(c)

Show that y = x − 1 is itself a solution of the differential equation.

(2)

(d)

A solution curve passes through (0, 3). Describe the shape of this curve for x ≥ 0.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 at (0, 2) the gradient is −2	✓ 1
A1 at (1, 0) the gradient is 1	✓ 1
A1 at (2, 2) the gradient is 0	✓ 1
Part (b)
M1 set x − y = 1	✓ 1
A1 y = x − 1	✓ 1
Part (c)
M1 dy/dx = 1 for y = x − 1	✓ 1
R1 x − (x − 1) = 1, so both sides are equal	✓ 1
Part (d)
A1 the curve starts with gradient −3, so it is decreasing initially	✓ 1
A1 it has a minimum where the curve meets the line y = x, since the gradient is zero there	✓ 1
A1 after this the gradient is positive and the curve is increasing	✓ 1
R1 solution curves cannot cross the solution y = x − 1, so the curve stays above it and approaches it as x becomes large	✓ 1

Exam tip: Zero-gradient points lie on the line where the right-hand side of the differential equation equals zero; use this line to place turning points of solution curves.

+ Add to test✓ In your test372IBMI3H-5-06r23-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

Calculator

A stage lift moves vertically so that its velocity upward at time t seconds is v(t) = 0.2t² − t + 0.8 m s⁻¹, for 0 ≤ t ≤ 5.

(a)

Find the acceleration of the lift when t = 2.

(2)

(b)

Find the times at which the lift is instantaneously at rest.

(2)

(c)

Find the displacement of the lift from its starting position at t = 5, correct to 3 significant figures.

(2)

(d)

Find the total distance travelled by the lift during the 5 seconds, correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a = dv/dt = 0.4t − 1	✓ 1
A1 −0.2 m s⁻²	✓ 1
Part (b)
M1 set v = 0	✓ 1
A1 t = 1 and t = 4 seconds	✓ 1
Part (c)
M1 ∫₀⁵ v dt	✓ 1
A1 −0.167 m (the lift finishes 0.167 m below its start)	✓ 1
Part (d)
M1 split the integral at t = 1 and t = 4	✓ 1
M1 sum of absolute values of the three integrals	✓ 1
A1 0.367 + 0.9 + 0.367 = 1.63 m	✓ 1

Common error: Total distance needs the integral of |v|, found by splitting at the times when v = 0; integrating v over the whole interval gives displacement instead.

+ Add to test✓ In your test373IBMI3H-5-05r24-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

A swimmer pushes off from the wall of a pool and swims along a straight lane. Her velocity, v m s⁻¹, at time t seconds is modelled by v(t) = 1.6(1 − e^−0.8t), for t ≥ 0.

(a)

Find her acceleration at t = 0.

(2)

(b)

State the speed that the model predicts she approaches as t becomes very large.

(1)

(c)

Find the distance she swims in the first 5 seconds. Give your answer to 3 significant figures.

(3)

(d)

Find the time at which her speed first reaches 1.2 m s⁻¹. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a = dv/dt = 1.28e^−0.8t	✓ 1
A1 1.28 m s⁻²	✓ 1
Part (b)
A1 1.6 m s⁻¹ (as t → ∞, e^−0.8t → 0)	✓ 1
Part (c)
M1 attempt to integrate v from 0 to 5	✓ 1
A1 1.6[t + e^−0.8t/0.8]₀⁵	✓ 1
A1 6.04 m	✓ 1
Part (d)
M1 1.6(1 − e^−0.8t) = 1.2	✓ 1
A1 e^−0.8t = 0.25	✓ 1
A1 t = ln4/0.8 = 1.73 s	✓ 1

Exam tip: For a velocity that levels off, the limiting value comes from letting the exponential term tend to zero, with no calculation needed.

+ Add to test✓ In your test374IBMI3H-5-05r24-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium11 marks

The temperature T °C of a swimming pool being heated satisfies dT/dt = 0.15(32 − T), where t is the time in hours since heating began. At t = 0 the temperature is 18 °C.

(a)

Use Euler's method with a step length of 1 hour to estimate the temperature at t = 3. Give your answer to 3 significant figures.

(4)

(b)

Solve the differential equation by separating variables to show that T = 32 − 14e^−0.15t.

(4)

(c)

Using the unrounded estimate 23.40225 from part (a), find the percentage error of Euler's estimate at t = 3 compared with the exact value. Give your answer to 3 significant figures.

(2)

(d)

Explain why Euler's method gives an overestimate here.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 T_n+1 = T_n + 1 × 0.15(32 − T_n)	✓ 1
A1 T₁ = 20.1	✓ 1
A1 T₂ = 21.885	✓ 1
A1 T₃ = 23.40225, so 23.4 °C	✓ 1
Part (b)
M1 ∫ 1/(32 − T) dT = ∫ 0.15 dt	✓ 1
A1 −ln\|32 − T\| = 0.15t + c	✓ 1
A1 use T(0) = 18 to get c = −ln 14	✓ 1
A1 32 − T = 14e^−0.15t, hence the given result	✓ 1
Part (c)
M1 exact T(3) = 23.0732… and \|23.40225 − 23.0732…\| / 23.0732… × 100	✓ 1
A1 1.43%	✓ 1
Part (d)
R1 the solution curve is concave down (gradient decreasing as T rises), so each tangent step lies above the curve	✓ 1

Exam tip: In a percentage error, divide by the exact value, not by the estimate, and keep unrounded values until the final line.

+ Add to test✓ In your test375IBMI3H-5-05r24-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

A pool is being drained. The depth of water h metres at time t minutes satisfies dh/dt = −0.05√h. Initially the depth is 4 m.

(a)

By separating variables, show that √h = 2 − 0.025t.

(4)

(b)

Find the time at which the depth is 1 m.

(2)

(c)

Find the time at which the depth is falling at 0.025 m per minute.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ∫ h^−1/2 dh = ∫ −0.05 dt	✓ 1
A1 2√h = −0.05t + c	✓ 1
M1 use h = 4 at t = 0 to find c = 4	✓ 1
A1 √h = 2 − 0.025t	✓ 1
Part (b)
M1 2 − 0.025t = 1	✓ 1
A1 t = 40 minutes	✓ 1
Part (c)
M1 −0.05√h = −0.025	✓ 1
A1 h = 0.25 m	✓ 1
A1 t = (2 − 0.5)/0.025 = 60 minutes	✓ 1

Exam tip: Give an integration constant its value as soon as an initial condition is available, before manipulating the equation further.

+ Add to test✓ In your test376IBMI3H-5-05r24-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium10 marks

A library needs an open-topped storage box with a square base of side x metres and height h metres. The volume must be 0.5 m³. The base material costs €18 per m² and the side material costs €10 per m².

(a)

Show that the total material cost is C = 18x² + 20/x euros.

(3)

(b)

Find dC/dx.

(2)

(c)

Find the value of x that minimises the cost. Give your answer to 3 significant figures.

(2)

(d)

Using the unrounded value of x, find the minimum cost, to the nearest cent, and show that it is a minimum.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 h = 0.5/x²	✓ 1
M1 cost = 18x² + 10(4xh)	✓ 1
A1 substitute for h: 18x² + 20/x	✓ 1
Part (b)
M1 attempt to differentiate, including 20x⁻¹	✓ 1
A1 36x − 20/x²	✓ 1
Part (c)
M1 set dC/dx = 0, so x³ = 5/9	✓ 1
A1 x = 0.822 m	✓ 1
Part (d)
M1 substitute x = (5/9)^1/3 into C	✓ 1
A1 €36.49	✓ 1
R1 C″ = 36 + 40/x³ = 108 > 0, so a minimum	✓ 1

Common error: Eliminate one variable using the volume constraint before differentiating; differentiating with both x and h present gives an incorrect result.

+ Add to test✓ In your test377IBMI3H-5-05r24-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium10 marks

A trolley carrying books is pushed along a straight library corridor. Its displacement from the door, s metres, after t seconds is s(t) = t³ − 9t² + 24t, for 0 ≤ t ≤ 6.

(a)

Find an expression for the velocity of the trolley.

(2)

(b)

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

(2)

(c)

Find the acceleration of the trolley when it first comes to rest.

(2)

(d)

Find the total distance travelled by the trolley in the first 6 seconds.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiate s	✓ 1
A1 v = 3t² − 18t + 24	✓ 1
Part (b)
M1 3t² − 18t + 24 = 0	✓ 1
A1 t = 2 and t = 4	✓ 1
Part (c)
M1 a = 6t − 18 evaluated at t = 2	✓ 1
A1 −6 m s⁻²	✓ 1
Part (d)
M1 evaluate s at the turning points and end points	✓ 1
A1 s(2) = 20, s(4) = 16, s(6) = 36	✓ 1
M1 sum of the magnitudes of the changes: 20 + 4 + 20	✓ 1
A1 44 m	✓ 1

Common error: Displacement and distance differ when the object turns round: split the journey at each time the velocity is zero.

+ Add to test✓ In your test378IBMI3H-5-04r26-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium11 marks

A delivery robot moves along a straight rail. Its velocity at time t seconds is v(t) = 5e^−0.3t − 2 m s⁻¹, for 0 ≤ t ≤ 10. Positive velocity is towards the loading bay.

(a)

Find the acceleration of the robot when t = 2, to 3 significant figures.

(2)

(b)

Find the time at which the robot is first momentarily at rest.

(2)

(c)

Find the displacement of the robot from its position at t = 0 to its position at t = 10, to 3 significant figures.

(3)

(d)

Find the total distance travelled by the robot during the 10 seconds, to 3 significant figures. Use unrounded values from your working.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a = dv/dt = −1.5e^−0.3t	✓ 1
A1 a(2) = −1.5e^−0.6 = −0.823 m s⁻²	✓ 1
Part (b)
M1 5e^−0.3t = 2	✓ 1
A1 t = ln(2.5)/0.3 = 3.05 s	✓ 1
Part (c)
M1 integrate: s = −(50/3)e^−0.3t − 2t	✓ 1
M1 evaluate between 0 and 10: (50/3)(1 − e⁻³) − 20	✓ 1
A1 −4.16 m	✓ 1
Part (d)
M1 recognise the direction changes at t = 3.054… so the two stages are found separately	✓ 1
A1 distance for 0 ≤ t ≤ 3.054…: (50/3)(0.6) − 2(3.054…) = 3.89 m	✓ 1
A1 displacement after this time is −4.163 − 3.891 = −8.05 m, so distance 8.05 m	✓ 1
A1 total distance = 11.9 m	✓ 1

Common error: Displacement and distance differ when the velocity changes sign; integrate |v| by splitting the interval at the time when v = 0.

+ Add to test✓ In your test379IBMI3H-5-05r28-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium11 marks

Calculator

Krill $x$ and penguins $y$ in a bay are modelled, in thousands of individuals, by
$\frac{dx}{dt}=0.5x-0.01xy,\qquad \frac{dy}{dt}=0.004xy-0.3y,$
where $t$ is in years. At $t=0$, $x=100$ and $y=30$.

(a)

Find the non-zero equilibrium point of the system.

(3)

(b)

Use Euler's method with step size $h=0.5$ to estimate $x$ and $y$ at $t=1$. Give both to 1 decimal place.

(6)

(c)

Find the initial rate of change of the penguin population, and state whether it is increasing or decreasing at $t=0$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set both derivatives to zero and factorise, e.g. $x(0.5-0.01y)=0$ and $y(0.004x-0.3)=0$	✓ 1
A1 $y=50$	✓ 1
A1 $x=75$	✓ 1
Part (b)
M1 apply Euler's method to both equations simultaneously	✓ 1
A1 $x_1=110$	✓ 1
A1 $y_1=31.5$	✓ 1
M1 second step using $x_1$ and $y_1$	✓ 1
A1 $x_2\approx120.2$	✓ 1
A1 $y_2\approx33.7$	✓ 1
Part (c)
M1 substitute $x=100$, $y=30$: $0.004(100)(30)-0.3(30)$	✓ 1
A1 $+3$ thousand per year, so the penguin population is increasing	✓ 1

Common error: Euler's method for a system must update both variables using the values from the start of the step.

+ Add to test✓ In your test380IBMI3H-5-06r28-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

Calculator

A rectangular swimming area of 300 m² is marked out against a straight pool wall using rope on the other three sides. The sides perpendicular to the wall each have length $x$ metres.

(a)

Show that the total length of rope is $L=2x+\dfrac{300}{x}$.

(2)

(b)

Find $\frac{dL}{dx}$.

(2)

(c)

Find the minimum length of rope. Use the unrounded value of $x$ and give the answer to 3 significant figures.

(4)

(d)

Verify that this length is a minimum.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 length parallel to the wall $=\frac{300}{x}$ from area $=300$	✓ 1
A1 $L=2x+\frac{300}{x}$ (AG)	✓ 1
Part (b)
M1 differentiate, $\frac{300}{x}=300x^{-1}$	✓ 1
A1 $\frac{dL}{dx}=2-\frac{300}{x^2}$	✓ 1
Part (c)
M1 set $\frac{dL}{dx}=0$	✓ 1
A1 $x=\sqrt{150}=12.2\ldots$	✓ 1
M1 substitute into $L$	✓ 1
A1 49.0 m	✓ 1
Part (d)
R1 $\frac{d^2L}{dx^2}=\frac{600}{x^3}>0$ for $x>0$	✓ 1

Exam tip: Always justify that a stationary point is a minimum or maximum using the second derivative or a sign change.

+ Add to test✓ In your test381IBMI3H-5-06r28-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium11 marks

Calculator

The energy $y$ (MWh) stored in a battery at a wind farm, $x$ hours after a reading, satisfies $\dfrac{dy}{dx}=2x-y$ with $y(0)=1$.

(a)

Use Euler's method with step size $0.25$ to estimate $y(0.5)$.

(3)

(b)

Show that $y=2x-2+3e^{-x}$ satisfies the differential equation and the initial condition.

(4)

(c)

Find the percentage error in the Euler estimate of $y(0.5)$, using the unrounded value 0.6875 and the exact value from (b).

(3)

(d)

Suggest one way to reduce the error.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use $y_{n+1}=y_n+0.25(2x_n-y_n)$	✓ 1
A1 $y(0.25)\approx0.75$	✓ 1
A1 $y(0.5)\approx0.6875$	✓ 1
Part (b)
A1 $\frac{dy}{dx}=2-3e^{-x}$	✓ 1
M1 $2x-y=2x-(2x-2+3e^{-x})$	✓ 1
A1 $=2-3e^{-x}$, equal to $\frac{dy}{dx}$	✓ 1
A1 $y(0)=-2+3=1$	✓ 1
Part (c)
A1 exact $y(0.5)=0.8196$	✓ 1
M1 $\frac{\text{exact}-\text{estimate}}{\text{exact}}\times100$	✓ 1
A1 16.1%	✓ 1
Part (d)
R1 use a smaller step size	✓ 1

Common error: Use the exact solution as the reference value for error; dividing by the estimate instead gives a different percentage.

+ Add to test✓ In your test382IBMI3H-5-07r28-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

In a food science experiment, a flavour compound moves between two layers of a cake. The amounts (in mg) in the upper and lower layers at time t hours are x and y, where
dx/dt = −0.2x + 0.1y, dy/dt = 0.2x − 0.1y,
with x = 10 and y = 2 when t = 0.

(a)

Show that x + y is constant and state its value.

(2)

(b)

Use Euler’s method with step length 0.5 hours to estimate x and y when t = 1.

(4)

(c)

Find the amounts in each layer as t becomes very large, assuming the system settles to a steady state. Give the amount in the upper layer first.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 add the two equations to get d(x + y)/dt = 0	✓ 1
R1 so x + y is constant, equal to 12 mg	✓ 1
Part (b)
M1 first step: dx/dt = −1.8, dy/dt = 1.8	✓ 1
A1 x(0.5) ≈ 9.1, y(0.5) ≈ 2.9	✓ 1
M1 second step using these values: dx/dt = −1.53	✓ 1
A1 x at t = 1 ≈ 8.335, y at t = 1 ≈ 3.665	✓ 1
Part (c)
M1 steady state: dx/dt = 0, so −0.2x + 0.1y = 0 and y = 2x	✓ 1
M1 use x + y = 12, so 3x = 12	✓ 1
A1 x = 4 mg and y = 8 mg	✓ 1

Exam tip: In a coupled system, update both variables from the same old values at each step; do not use a new x to compute the new y within one step.

+ Add to test✓ In your test383IBMI3H-5-07r28-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium11 marks

A battery in a mountain hut stores C kWh of energy at time t hours. Solar panels and the hut’s use satisfy dC/dt = 6 − 0.4C, with C = 2 when t = 0.

(a)

A slope field is drawn for this equation. Find the gradient at C = 2, and the value of C at which every gradient is zero. Describe how the gradients change for C above this value.

(3)

(b)

Solve the differential equation to show that C = 15 − 13e^−0.4t.

(5)

(c)

Find the time at which the battery first holds 10 kWh. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 gradient at C = 2 is 6 − 0.8 = 5.2	✓ 1
A1 gradients are zero at C = 15	✓ 1
R1 for C > 15 the gradients are negative, so solution curves move down towards C = 15	✓ 1
Part (b)
M1 separate: ∫ 1/(6 − 0.4C) dC = ∫ dt	✓ 1
A1 −2.5 ln\|6 − 0.4C\| = t + k	✓ 1
M1 use C(0) = 2 to find the constant	✓ 1
A1 6 − 0.4C = 5.2e^−0.4t	✓ 1
A1 C = 15 − 13e^−0.4t	✓ 1
Part (c)
M1 set 15 − 13e^−0.4t = 10, so e^−0.4t = 5/13	✓ 1
M1 take natural logarithms: t = ln(13/5)/0.4	✓ 1
A1 t = 2.39 hours	✓ 1

Exam tip: Check a solved differential equation by substituting t = 0 back in; here 15 − 13 = 2 confirms the constant.

+ Add to test✓ In your test384IBMI3H-5-08r28-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

A warehouse robot starts from rest and moves along a straight aisle. Its acceleration is a = 0.9e^−0.5t m s⁻², where t is the time in seconds.

(a)

Find an expression for the velocity v of the robot at time t.

(3)

(b)

Write down the speed that the robot approaches for large t.

(1)

(c)

Find the time taken to reach 90% of this speed. Give your answer to 3 significant figures.

(2)

(d)

Find the distance travelled in the first 6 seconds. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 integrate a with respect to t	✓ 1
A1 v = −1.8e^−0.5t + c	✓ 1
A1 use v(0) = 0: v = 1.8(1 − e^−0.5t)	✓ 1
Part (b)
A1 1.8 m s⁻¹	✓ 1
Part (c)
M1 1 − e^−0.5t = 0.9, so e^−0.5t = 0.1	✓ 1
A1 t = 2 ln 10 = 4.61 s	✓ 1
Part (d)
M1 distance = ∫₀⁶ 1.8(1 − e^−0.5t) dt	✓ 1
A1 1.8[t + 2e^−0.5t]₀⁶	✓ 1
A1 7.38 m	✓ 1

Exam tip: Use the initial condition immediately after integrating to find the constant, before using the expression in later parts.

+ Add to test✓ In your test385IBMI3H-5-04r29-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

Calculator

In a fish-farm tank, the concentration C mg per litre of a dissolved nutrient at time t hours satisfies the differential equation dC/dt = 6 − 0.5C. A slope field is drawn for this equation in the (t, C) plane.

(a)

Find the gradient of the slope field at the points (0, 2), (0, 10) and (3, 12).

(2)

(b)

Explain why the slope segments along any horizontal line in the slope field are all parallel, and state the value of C at which the segments are horizontal.

(2)

(c)

Show that C = 12 − 10e^−0.5t is a solution of the differential equation and passes through (0, 2).

(3)

(d)

Find the time at which the concentration first reaches 10 mg per litre. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes values of C into 6 − 0.5C	✓ 1
A1 gradients 5, 1 and 0 respectively	✓ 1
Part (b)
R1 dC/dt depends only on C and not on t, so the gradient is the same at every t for a given C	✓ 1
A1 C = 12	✓ 1
Part (c)
M1 dC/dt = 5e^−0.5t	✓ 1
M1 6 − 0.5(12 − 10e^−0.5t) = 5e^−0.5t	✓ 1
A1 the two sides are equal and C(0) = 12 − 10 = 2 (shown)	✓ 1
Part (d)
M1 12 − 10e^−0.5t = 10, so e^−0.5t = 0.2	✓ 1
A1 t = 3.22 hours (2 ln 5 = 3.2188...)	✓ 1

Exam tip: To check that a given function satisfies a differential equation, differentiate it and substitute both the function and its derivative into the equation. Show that both sides simplify to the same expression.

+ Add to test✓ In your test386IBMI3H-5-07r32-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium11 marks

Swimmers move between a main pool and a training pool. At time $t$ minutes, $x$ swimmers are in the main pool and $y$ are in the training pool, where $\frac{dx}{dt}=-0.2x+0.1y$ and $\frac{dy}{dt}=0.2x-0.1y$. Initially $x=60$ and $y=20$.

(a)

Show that $x+y$ is constant, and state its value.

(2)

(b)

Use Euler's method with step length $h=1$ to estimate the values of $x$ and $y$ when $t=2$.

(4)

(c)

The system settles to an equilibrium where both rates of change are zero. Find the equilibrium values of $x$ and $y$ to 3 significant figures.

(3)

(d)

Compare the Euler estimate of $x$ at $t=2$ with the equilibrium value and comment on what the model suggests about the main pool.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{dx}{dt}+\frac{dy}{dt}=0$, so $x+y$ is constant	✓ 1
A1 $x+y=80$	✓ 1
Part (b)
M1 $x_{n+1}=x_n+h\frac{dx}{dt}$ and $y_{n+1}=y_n+h\frac{dy}{dt}$ applied from $(60,20)$	✓ 1
A1 $x_1=50$, $y_1=30$	✓ 1
M1 second step applied from $(50,30)$	✓ 1
A1 $x_2=43$, $y_2=37$	✓ 1
Part (c)
M1 set $-0.2x+0.1y=0$, so $y=2x$	✓ 1
A1 using $x+y=80$: $x=26.7$	✓ 1
A1 $y=53.3$	✓ 1
Part (d)
R1 $x$ is decreasing (60, 50, 43) towards the equilibrium value 26.7	✓ 1
R1 the main pool keeps losing swimmers until it holds about 26.7, half as many as the training pool	✓ 1

Common error: When a pair of rates sums to zero, the total is conserved; use that fact to reduce the equilibrium problem to one equation rather than solving two unrelated ones.

+ Add to test✓ In your test387IBMI3H-5-07r32-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium10 marks

A turbine blade vibrates so that its tip displacement $x$ metres at time $t$ seconds satisfies $\frac{d^2x}{dt^2}+0.5\frac{dx}{dt}+4x=0$. At $t=0$, $x=1$ and $\frac{dx}{dt}=0$.

(a)

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

(2)

(b)

Use Euler's method with step length $h=0.1$ to complete three steps, giving $x$ and $y$ at $t=0.1$, $0.2$ and $0.3$.

(5)

(c)

For undamped motion the quantity $Q=x^2+\left(\frac{y}{2}\right)^2$ is constant. Calculate $Q$ at $t=0.3$ using your Euler values and, given that the damping should make $Q$ decrease, comment on the result. Give $Q$ to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dx}{dt}=y$	✓ 1
A1 $\frac{dy}{dt}=-0.5y-4x$	✓ 1
Part (b)
M1 use $x_{n+1}=x_n+0.1y_n$ and $y_{n+1}=y_n+0.1(-0.5y_n-4x_n)$ from $(1,0)$	✓ 1
A1 $t=0.1$: $x=1$, $y=-0.4$	✓ 1
A1 $t=0.2$: $x=0.96$, $y=-0.78$	✓ 1
A1 $t=0.3$: $x=0.882$	✓ 1
A1 $t=0.3$: $y=-1.125$	✓ 1
Part (c)
M1 $Q=0.882^2+\left(\frac{-1.125}{2}\right)^2$ (ECF from (b))	✓ 1
A1 $Q=1.09$ (initial value is 1)	✓ 1
R1 $Q$ has increased, so the Euler method has introduced error (it overestimates the vibration energy)	✓ 1

Common error: In a second-order Euler scheme update both variables using values from the same previous step; using a freshly updated $x$ to find $y$ mixes up the time levels.

+ Add to test✓ In your test388IBMI3H-5-07r32-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium12 marks

The number of members, $y$ thousand, of a library's online service $t$ years after launch is modelled by $\frac{dy}{dt}=0.3y\left(1-\frac{y}{6}\right)$, with $y=1$ when $t=0$. A slope field is drawn for this equation.

(a)

Calculate the gradient of the slope field at the points where $y=1$, $y=3$ and $y=6$.

(3)

(b)

Describe how the gradients vary along any horizontal line, and describe the slope field for $y>6$.

(3)

(c)

Use Euler's method with step length $h=1$ to estimate $y$ when $t=2$. Carry full calculator accuracy and give your final answer to 3 significant figures.

(4)

(d)

Explain why the solution through $(0,1)$ can never exceed $y=6$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $y=1$: gradient $0.25$	✓ 1
A1 $y=3$: gradient $0.45$	✓ 1
A1 $y=6$: gradient $0$	✓ 1
Part (b)
A1 the gradient depends only on $y$, so segments along a horizontal line are parallel	✓ 1
A1 for $y>6$ the factor $\left(1-\frac{y}{6}\right)$ is negative, so segments slope downwards	✓ 1
A1 segments are horizontal along $y=6$ (and $y=0$)	✓ 1
Part (c)
M1 $y_{n+1}=y_n+h\times0.3y_n\left(1-\frac{y_n}{6}\right)$	✓ 1
A1 $y_1=1.25$	✓ 1
M1 second step from $y=1.25$: $1.25+0.3\times1.25\times\left(1-\frac{1.25}{6}\right)$	✓ 1
A1 $y_2=1.55$ (thousand members)	✓ 1
Part (d)
R1 $y=6$ is a horizontal solution curve (gradient zero)	✓ 1
R1 solution curves cannot cross, so the curve through $(0,1)$ stays below $y=6$	✓ 1

Exam tip: When asked to describe a slope field, use the zeros of the gradient function first; they identify the horizontal segments that act as barriers.

+ Add to test✓ In your test389IBMI3H-5-08r32-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium10 marks

A window in a building has an arched top whose height above a horizontal base is $y=6\sin\left(\frac{\pi x}{10}\right)$ metres, for $0\le x\le10$, where $x$ is the horizontal distance in metres. Use radians throughout.

(a)

Find the exact area of the window.

(3)

(b)

The glass costs €85 per square metre. Using the unrounded area from part (a), find the cost of the glass, to the nearest cent.

(2)

(c)

Estimate the area using the trapezoidal rule with five strips of equal width.

(3)

(d)

Find the percentage error of the trapezoidal estimate in part (c) compared with the exact area, to 3 significant figures. Use unrounded values.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\int_0^{10}6\sin\left(\frac{\pi x}{10}\right)dx$	✓ 1
A1 $\left[-\frac{60}{\pi}\cos\left(\frac{\pi x}{10}\right)\right]_0^{10}$	✓ 1
A1 $\frac{120}{\pi}$ m² ($=38.2$ m²)	✓ 1
Part (b)
M1 $\frac{120}{\pi}\times85$	✓ 1
A1 €3246.76	✓ 1
Part (c)
M1 strip width 2 and ordinates at $x=0,2,4,6,8,10$	✓ 1
A1 ordinates $0,\,3.527,\,5.706,\,5.706,\,3.527,\,0$	✓ 1
A1 area $\approx36.9$ m²	✓ 1
Part (d)
M1 $\frac{38.1972-36.9322}{38.1972}\times100$	✓ 1
A1 $3.31\%$	✓ 1

Exam tip: For percentage error, divide the difference by the exact value, not by the estimate, and keep full calculator accuracy until the end.

+ Add to test✓ In your test390IBMI3H-5-08r32-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium12 marks

The stock $y$ (hundreds of units) in a distribution centre at day $x$ satisfies $\frac{dy}{dx}=0.5(x-y)$, with $y=1$ when $x=0$. A slope field is drawn for this equation.

(a)

Calculate the gradient of the slope field at the points $(0,0)$, $(2,0)$ and $(2,4)$.

(3)

(b)

Show that the line $y=x-2$ is a solution curve of the differential equation.

(2)

(c)

Use Euler's method with step length $h=0.5$ to estimate $y$ when $x=1.5$, to 3 significant figures. Use unrounded intermediate values.

(4)

(d)

Describe a sketch of the solution curve through $(0,1)$ for $x\ge0$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $(0,0)$: $0$	✓ 1
A1 $(2,0)$: $1$	✓ 1
A1 $(2,4)$: $-1$	✓ 1
Part (b)
M1 $\frac{dy}{dx}=1$ for $y=x-2$	✓ 1
A1 $0.5(x-(x-2))=1$, so both sides agree	✓ 1
Part (c)
M1 $y_{n+1}=y_n+0.5\times0.5(x_n-y_n)$	✓ 1
A1 $y_1=0.75$	✓ 1
A1 $y_2=0.6875$	✓ 1
A1 $y_3=0.766$	✓ 1
Part (d)
A1 the curve starts with negative gradient ($-0.5$), so initially decreases	✓ 1
A1 it reaches a minimum where it meets the line $y=x$ (gradient zero)	✓ 1
A1 it then increases, approaching the line $y=x-2$ from above	✓ 1

Exam tip: To describe a slope field quickly, find where the gradient is zero and where it is constant; those isoclines show the shape of the solution curves.

+ Add to test✓ In your test391IBMI3H-5-08r32-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium11 marks

A fuel storage tank holds $V$ litres at time $t$ minutes. Fuel is delivered at 15 litres per minute and drawn off at a rate equal to 5% of the current volume per minute. Initially the tank holds 100 litres.

(a)

Show that $\frac{dV}{dt}=15-0.05V$.

(2)

(b)

Solve the differential equation to show that $V=300-200e^{-0.05t}$.

(4)

(c)

State the volume the tank approaches in the long term, giving a reason.

(2)

(d)

Find the time taken for the tank to contain 250 litres, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rate of change = inflow $-$ outflow	✓ 1
A1 $15-0.05V$, with outflow $0.05V$	✓ 1
Part (b)
M1 separate variables: $\int\frac{1}{15-0.05V}\,dV=\int dt$	✓ 1
A1 $-20\ln\|15-0.05V\|=t+c$	✓ 1
M1 use $V=100$ at $t=0$ to find $c=-20\ln10$	✓ 1
A1 $15-0.05V=10e^{-0.05t}$, hence the result	✓ 1
Part (c)
A1 $300$ litres	✓ 1
R1 as $t\to\infty$, $e^{-0.05t}\to0$	✓ 1
Part (d)
M1 $300-200e^{-0.05t}=250$	✓ 1
M1 $e^{-0.05t}=0.25$, take natural logarithms	✓ 1
A1 $t=20\ln4=27.7$ minutes	✓ 1

Exam tip: Check a solved differential equation against the initial condition: here $V(0)=300-200=100$, which confirms the constant.

+ Add to test✓ In your test392IBMI3H-5-06r34-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

After pushing off from the wall, a swimmer's speed is modelled by v(t) = 3te^−0.8t m s⁻¹, where t is the time in seconds, t ≥ 0.

(a)

Use the product rule to find v′(t), in the form ke^−0.8t(1 − 0.8t).

(3)

(b)

Find the time at which the swimmer's speed is greatest.

(2)

(c)

Find the greatest speed, to 3 significant figures.

(2)

(d)

Find the acceleration at t = 2 and interpret its sign. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Product rule on 3t and e^−0.8t	✓ 1
v′ = 3e^−0.8t − 2.4te^−0.8t	✓ 1
v′ = 3e^−0.8t(1 − 0.8t)	✓ 1
Part (b)
v′ = 0 and e^−0.8t > 0, so 1 − 0.8t = 0	✓ 1
t = 1.25 s	✓ 1
Part (c)
v(1.25) = 3 × 1.25 × e⁻¹	✓ 1
= 1.38 m s⁻¹	✓ 1
Part (d)
v′(2) = 3e^−1.6(1 − 1.6) = −0.363 m s⁻²	✓ 1
Negative, so the swimmer is slowing down	✓ 1

Exam tip: To locate a maximum, set the derivative to zero and then say why the exponential factor cannot be zero; it justifies dividing it out.

+ Add to test✓ In your test393IBMI3H-5-07r35-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium10 marks

A club of 80 athletes is adopting a training app. The number of athletes $N$ using the app after $t$ months is modelled by $\frac{dN}{dt}=0.3N\left(1-\frac{N}{80}\right)$, with $N=10$ when $t=0$.

(a)

Find the gradient of the solution curve at the point where $N=10$.

(2)

(b)

State the two values of $N$ for which the slope field has horizontal line segments, and explain what each means for the club.

(2)

(c)

Find the value of $N$ at which the app is being adopted fastest, and the rate of adoption at that value.

(3)

(d)

Use Euler's method with a step length of 1 month to estimate $N$ when $t=2$. Carry unrounded values through your working and give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute $N=10$ into the differential equation	✓ 1
A1 $2.625$ (athletes per month)	✓ 1
Part (b)
A1 $N=0$ and $N=80$	✓ 1
A1 $N=0$: nobody uses the app so no one starts; $N=80$: all athletes already use it so there is no further growth	✓ 1
Part (c)
M1 recognise the rate is a quadratic in $N$ with maximum midway between its roots (or differentiate with respect to $N$)	✓ 1
A1 $N=40$	✓ 1
A1 $6$ athletes per month	✓ 1
Part (d)
M1 use of $N_{n+1}=N_n+1\times0.3N_n\left(1-\frac{N_n}{80}\right)$	✓ 1
A1 $N_1=12.625$	✓ 1
A1 $N_2=15.8$	✓ 1

Common error: Euler's method gives a value for the number of athletes, which need not be a whole number; do not round between steps.

+ Add to test✓ In your test394IBMI3H-5-07r35-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium10 marks

A rail company designs a closed storage box with a square base of side $x$ m and height $h$ m. The volume of the box is to be 1.5 m³. The base and lid cost €15 per m² and the four sides cost €8 per m².

(a)

Show that the total cost, in euros, is $C(x)=30x^2+\frac{48}{x}$.

(3)

(b)

Find the value of $x$ that minimises the cost, correct to 3 significant figures.

(3)

(c)

Show that this value of $x$ gives a minimum cost.

(2)

(d)

Find the minimum cost, giving your answer to the nearest cent.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $h=\frac{1.5}{x^2}$	✓ 1
M1 $C=2\times15x^2+8\times4xh$	✓ 1
A1 substitute $h$ to obtain $C=30x^2+\frac{48}{x}$ (AG)	✓ 1
Part (b)
M1 attempt to differentiate $C$	✓ 1
A1 $C'(x)=60x-\frac{48}{x^2}$	✓ 1
A1 setting $C'(x)=0$ gives $x^3=0.8$, $x=0.928$	✓ 1
Part (c)
M1 $C''(x)=60+\frac{96}{x^3}$	✓ 1
R1 $C''>0$ for $x>0$ (e.g. $C''=180$ at $x^3=0.8$), so the cost is a minimum	✓ 1
Part (d)
M1 substitute the optimal $x$ into $C(x)$	✓ 1
A1 €77.56	✓ 1

Exam tip: Keep the unrounded optimal value in your calculator when finding the minimum cost; using the rounded 0.928 changes the final cents.

+ Add to test✓ In your test395IBMI3H-5-04r39-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium11 marks

A hospital pump delivers a drug at a rate of r(t) = 15te^−t²/8 mL per hour, where t is the time in hours after the pump starts, 0 ≤ t ≤ 4.

(a)

Use the substitution u = t²/8 to show that the total volume delivered in the first 4 hours is 60(1 − e⁻²) mL.

(4)

(b)

Evaluate this volume to 3 significant figures.

(1)

(c)

Find dr/dt and hence find the time at which the rate of delivery is greatest.

(4)

(d)

Find the greatest rate of delivery. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 du = (t/4) dt, so t dt = 4 du	✓ 1
A1 integral becomes 60∫e^−u du	✓ 1
A1 new limits u = 0 to u = 2	✓ 1
A1 60[−e^−u]₀² = 60(1 − e⁻²)	✓ 1
Part (b)
A1 51.9 mL	✓ 1
Part (c)
M1 product rule and chain rule	✓ 1
A1 dr/dt = 15e^−t²/8(1 − t²/4)	✓ 1
M1 dr/dt = 0	✓ 1
A1 t = 2 hours	✓ 1
Part (d)
M1 r(2) = 30e^−0.5	✓ 1
A1 18.2 mL per hour	✓ 1

Exam tip: When using substitution on a definite integral, change the limits to the new variable so that you never need to return to the original one.

+ Add to test✓ In your test396IBMI3H-5-04r39-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium10 marks

A Formula 1 car accelerates from rest along a straight. Its speed is modelled by v(t) = 80(1 − e^−0.25t) m s⁻¹, where t is the time in seconds after the start.

(a)

Find an expression for the acceleration of the car, and state its initial acceleration.

(3)

(b)

Find the time taken to reach a speed of 60 m s⁻¹. Give your answer to 3 significant figures.

(3)

(c)

Write down the speed that the model predicts the car approaches for large t.

(1)

(d)

Find the distance travelled in the first 10 seconds. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiate using the chain rule	✓ 1
A1 a = 20e^−0.25t	✓ 1
A1 initial acceleration 20 m s⁻²	✓ 1
Part (b)
M1 80(1 − e^−0.25t) = 60, so e^−0.25t = 0.25	✓ 1
M1 use of logarithms: t = −4 ln 0.25	✓ 1
A1 t = 5.55 s	✓ 1
Part (c)
A1 80 m s⁻¹	✓ 1
Part (d)
M1 distance = ∫₀¹⁰ v dt	✓ 1
A1 antiderivative 80t + 320e^−0.25t	✓ 1
A1 506 m	✓ 1

Common error: Distance is the integral of speed with respect to time; do not multiply the final speed by the time.

+ Add to test✓ In your test397IBMI3H-5-04r39-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium10 marks

A beekeeper designs a closed cylindrical jar that must hold 500 cm³ of honey. The jar has radius r cm and height h cm. The beekeeper wants to use as little material as possible, so the total surface area A cm² must be as small as possible.

(a)

Write down an expression for h in terms of r.

(1)

(b)

Show that A = 2πr² + 1000/r.

(2)

(c)

Find dA/dr.

(2)

(d)

Find the value of r that minimises A. Give your answer to 3 significant figures.

(2)

(e)

Show that this value of r gives a minimum, and find the minimum surface area. Use the unrounded value of r and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 h = 500/(πr²)	✓ 1
Part (b)
A1 A = 2πr² + 2πrh	✓ 1
A1 substitute h: 2πr × 500/(πr²) = 1000/r	✓ 1
Part (c)
M1 differentiate (1000/r = 1000r⁻¹)	✓ 1
A1 dA/dr = 4πr − 1000/r²	✓ 1
Part (d)
M1 4πr − 1000/r² = 0, so r³ = 250/π	✓ 1
A1 r = 4.30 cm	✓ 1
Part (e)
R1 d²A/dr² = 4π + 2000/r³ > 0 for r > 0 (or the gradient changes from negative to positive)	✓ 1
M1 substitute r into A = 2πr² + 1000/r	✓ 1
A1 A = 349 cm²	✓ 1

Exam tip: After setting the derivative to zero, always justify that the stationary point is a minimum, using the second derivative or a sign change in the gradient.

+ Add to test✓ In your test398IBMI3H-5-04r39-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium12 marks

In a Formula 1 test, a car's speed on a straight is modelled by v(t) = 20 + 6t ln(t + 1) m s⁻¹ for 0 ≤ t ≤ 5, where t is the time in seconds.

(a)

Find an expression for the acceleration a(t) of the car.

(3)

(b)

Find the acceleration at t = 3. Give your answer to 3 significant figures.

(2)

(c)

Show that the acceleration is increasing for 0 ≤ t ≤ 5.

(3)

(d)

Find the time at which the car reaches a speed of 60 m s⁻¹. Give your answer to 3 significant figures.

(2)

(e)

Find the distance travelled in the first 5 seconds. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule	✓ 1
A1 6 ln(t + 1)	✓ 1
A1 + 6t/(t + 1)	✓ 1
Part (b)
M1 substitute t = 3: 6 ln 4 + 18/4	✓ 1
A1 12.8 m s⁻²	✓ 1
Part (c)
M1 differentiate: a′(t) = 6/(t + 1) + 6/(t + 1)²	✓ 1
A1 correct derivative of 6t/(t + 1) as 6/(t + 1)²	✓ 1
R1 both terms are positive for t ≥ 0, so a′(t) > 0	✓ 1
Part (d)
M1 20 + 6t ln(t + 1) = 60 (or intersection on GDC)	✓ 1
A1 t = 4.09 s	✓ 1
Part (e)
M1 distance = ∫₀⁵ v dt (GDC or integration by parts)	✓ 1
A1 207 m	✓ 1

Exam tip: For a product such as t ln(t + 1), label the two factors u and v before differentiating so that the chain rule inside the logarithm is not missed.

+ Add to test✓ In your test399IBMI3H-5-05r40-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

The speed of a warehouse delivery robot, in m s⁻¹, is modelled by $v(t) = 3t\,e^{-t^2/4}$, for $0 \le t \le 4$, where t is the time in seconds after it starts moving.

(a)

Show that the acceleration is $a(t) = 3e^{-t^2/4}\left(1 - \frac{t^2}{2}\right)$.

(3)

(b)

Find the time at which the robot reaches its greatest speed, and state that speed to 3 significant figures.

(3)

(c)

Use the substitution $u = \frac{t^2}{4}$ to find the distance travelled in the first two seconds, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at product rule with chain rule for $e^{-t^2/4}$	✓ 1
A1 $a = 3e^{-t^2/4} + 3t\left(-\frac{t}{2}\right)e^{-t^2/4}$	✓ 1
A1 factorise to the given form (AG)	✓ 1
Part (b)
M1 set $a=0$, so $1-\frac{t^2}{2}=0$	✓ 1
A1 $t=\sqrt{2}\approx 1.41$ s	✓ 1
A1 $v(\sqrt2) = 2.57$ m s⁻¹	✓ 1
Part (c)
M1 $du = \frac{t}{2}dt$, so $3t\,dt = 6\,du$	✓ 1
A1 $\int_0^1 6e^{-u}\,du = 6(1-e^{-1})$	✓ 1
A1 3.79 m	✓ 1

Exam tip: Check that the substituted limits match the new variable: here $t=0$ and $t=2$ become $u=0$ and $u=1$.

+ Add to test✓ In your test400IBMI3H-5-06r43-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium10 marks

The weekly sales of a product, in thousands of euros, during a promotion are modelled by S(w) = 0.1w³ − 1.2w² + 6w + 20, where w is the number of weeks after the start.

(a)

Find S′(w).

(2)

(b)

Find the gradient of the graph of S and the value of S at w = 2.

(2)

(c)

Find the equation of the tangent to the graph at w = 2.

(2)

(d)

Use the tangent to estimate S(2.5). Find the percentage error in this estimate, to 3 significant figures.

(3)

(e)

Explain why the tangent overestimates S(2.5).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 0.3w² − 2.4w + 6	✓ 1
Part (b)
A1 gradient = 2.4	✓ 1
A1 S(2) = 28	✓ 1
Part (c)
M1 attempt to use y − 28 = 2.4(w − 2)	✓ 1
A1 y = 2.4w + 23.2	✓ 1
Part (d)
A1 estimate = 29.2 (actual value 29.0625)	✓ 1
M1 attempt at \|estimate − actual\|/actual × 100	✓ 1
A1 0.473%	✓ 1
Part (e)
R1 S″(2) = −1.2 < 0, so the graph is concave down and lies below its tangent	✓ 1

Common error: Percentage error is measured relative to the actual value, not the estimate.

+ Add to test✓ In your test401IBMI3H-5-05r45-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium12 marks

A food company makes closed cylindrical tins of volume 400 cm³. The tin has radius r cm and height h cm. The metal for the circular ends costs €0.0009 per cm² and the metal for the curved side costs €0.0004 per cm². The cost of the metal for one tin is €C.

(a)

Write down an expression for h in terms of r.

(2)

(b)

Show that C = 0.0018πr² + 0.32/r.

(3)

(c)

Find dC/dr.

(2)

(d)

Find the value of r for which the cost is a minimum, giving your answer to 3 significant figures.

(2)

(e)

Find the minimum cost of the metal for one tin, to 3 significant figures, and show that this value is a minimum.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume = πr²h = 400	✓ 1
A1 h = 400/(πr²)	✓ 1
Part (b)
M1 area of the two ends = 2πr² and area of the side = 2πrh	✓ 1
M1 C = 0.0009(2πr²) + 0.0004(2πrh)	✓ 1
A1 substitutes h to get 0.0004 × 800/r = 0.32/r, and so reaches the given expression	✓ 1
Part (c)
A1 0.0036πr	✓ 1
A1 −0.32/r²	✓ 1
Part (d)
M1 sets dC/dr = 0, e.g. 0.0036πr³ = 0.32	✓ 1
A1 r = 3.05 cm	✓ 1
Part (e)
A1 C = €0.158 (using unrounded r or r = 3.05)	✓ 1
M1 attempts d²C/dr² = 0.0036π + 0.64/r³	✓ 1
R1 d²C/dr² = 0.0339… > 0 at the stationary point, so it is a minimum (accept a sign change of dC/dr from negative to positive)	✓ 1

Exam tip: In a ‘show that’ question, write every step of the substitution so that the final line matches the given result exactly.

+ Add to test✓ In your test402IBMI3H-5-05r45-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium12 marks

A loaf of bread comes out of an oven with its core at 85 °C. The kitchen is at a constant 20 °C. The core temperature T °C after t minutes satisfies dT/dt = −0.06(T − 20).

(a)

Use Euler’s method with a step length of 2 minutes to estimate the core temperature after 8 minutes. Give your answer to 2 decimal places.

(4)

(b)

Solve the differential equation by separation of variables to show that T = 20 + 65e^−0.06t.

(4)

(c)

Find the exact-solution core temperature after 8 minutes to 1 decimal place, and calculate the percentage error in your answer to part (a) using unrounded values. Give the percentage to 3 significant figures.

(3)

(d)

Explain why Euler’s method gives an underestimate of the temperature in this case.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses T_n+1 = T_n + 2 × (−0.06(T_n − 20))	✓ 1
A1 T(2) ≈ 77.2	✓ 1
A1 T(4) ≈ 70.336 and T(6) ≈ 64.29568	✓ 1
A1 T(8) ≈ 58.98 °C	✓ 1
Part (b)
M1 separates variables: ∫ 1/(T − 20) dT = ∫ −0.06 dt	✓ 1
A1 ln\|T − 20\| = −0.06t + c	✓ 1
M1 uses T = 85 when t = 0 to find the constant (c = ln 65, or A = 65 in T − 20 = Ae^−0.06t)	✓ 1
A1 reaches T = 20 + 65e^−0.06t	✓ 1
Part (c)
A1 T(8) = 60.2 °C (60.22…)	✓ 1
M1 percentage error = \|60.22… − 58.9801984\|/60.22… × 100	✓ 1
A1 2.06 %	✓ 1
Part (d)
R1 the graph of T against t is concave up (decreasing at a decreasing rate), so each tangent-line step falls faster than the true curve and ends below it	✓ 1

Exam tip: For the percentage error, keep the unrounded values from your calculator until the final step.

+ Add to test✓ In your test403IBMI3H-5-07r45-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

Calculator

Ash settles on a platform roof. The depth of ash, $y$ centimetres, after $x$ hours is modelled by the differential equation $\dfrac{dy}{dx}=0.2y(5-y)$, $y\ge 0$. A slope field for this equation is drawn on the $(x,y)$ plane.

(a)

Calculate the gradient of the slope field at the points $(0,1)$, $(0,5)$ and $(0,6)$.

(3)

(b)

State the values of $y$ for which every line segment of the slope field is horizontal.

(1)

(c)

A solution curve passes through $(0,1)$. Describe its behaviour as $x$ increases.

(2)

(d)

Use Euler's method with step length $h=1$ to estimate $y$ when $x=2$, given $y=1$ when $x=0$. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 at $(0,1)$: $0.2\times1\times4=0.8$	✓ 1
A1 at $(0,5)$: $0$	✓ 1
A1 at $(0,6)$: $0.2\times6\times(-1)=-1.2$	✓ 1
Part (b)
A1 $y=0$ and $y=5$	✓ 1
Part (c)
A1 the depth increases (positive gradient for \(0	✓ 1
A1 approaches $y=5$ from below and never crosses it (levels off)	✓ 1
Part (d)
M1 use of $y_{n+1}=y_n+h\times0.2y_n(5-y_n)$	✓ 1
A1 $y_1=1.8$	✓ 1
A1 $y_2=2.95$ (2.952)	✓ 1

Exam tip: Where the gradient is zero the solution is constant, so a solution curve cannot cross such a line.

+ Add to test✓ In your test404IBMI3H-5-07r45-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium11 marks

Calculator

An open-top steel container for ship cargo has a square base of side $x$ metres and height $h$ metres. Its volume is $50\ \mathrm{m^3}$. The steel area $A$ (base and four sides) is to be minimised.

(a)

Write an expression for $h$ in terms of $x$.

(2)

(b)

Show that $A=x^2+\dfrac{200}{x}$.

(2)

(c)

Find the value of $x$ that minimises $A$, to 3 significant figures.

(3)

(d)

Find the minimum area, to 3 significant figures, using the unrounded value of $x$.

(2)

(e)

Show that this value of $x$ gives a minimum.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $x^2h=50$	✓ 1
A1 $h=\frac{50}{x^2}$	✓ 1
Part (b)
M1 $A=x^2+4xh$ and substitute $h=\frac{50}{x^2}$	✓ 1
A1 $A=x^2+\frac{200}{x}$	✓ 1
Part (c)
A1 $\frac{dA}{dx}=2x-\frac{200}{x^2}$	✓ 1
M1 set $\frac{dA}{dx}=0$ so $x^3=100$	✓ 1
A1 $x=4.64$ (4.6416...)	✓ 1
Part (d)
M1 substitute $x=4.6416\ldots$ into $A$	✓ 1
A1 $64.6$ m$^2$	✓ 1
Part (e)
M1 $\frac{d^2A}{dx^2}=2+\frac{400}{x^3}$	✓ 1
R1 positive for $x>0$, so a minimum	✓ 1

Exam tip: Use the second derivative (or a sign change of the first) to justify that a stationary point is a minimum, not just assume it.

+ Add to test✓ In your test405IBMI3H-5-07r45-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Medium9 marks

Calculator

The amplitude $A$ of a seismic wave recorded at a volcano is modelled by $A(t)=\dfrac{50\sin(0.5t)}{1+0.1t}$ millimetres, for $0\le t\le 3$, where $t$ is in seconds and the angle is in radians.

(a)

Find $A'(t)$.

(3)

(b)

Find the rate of change of the amplitude when $t=3$, to 3 significant figures.

(2)

(c)

Find the time at which the amplitude is greatest.

(2)

(d)

Find the greatest amplitude, to 3 significant figures, using the unrounded time from (c).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the quotient rule	✓ 1
A1 numerator $25\cos(0.5t)(1+0.1t)-5\sin(0.5t)$	✓ 1
A1 over $(1+0.1t)^2$	✓ 1
Part (b)
M1 substitute $t=3$ into $A'(t)$	✓ 1
A1 $-1.59$ mm s$^{-1}$	✓ 1
Part (c)
M1 solve $A'(t)=0$ (e.g. with a graphing calculator)	✓ 1
A1 $t=2.83$ s	✓ 1
Part (d)
M1 substitute $t=2.8324\ldots$ into $A(t)$	✓ 1
A1 $38.5$ mm	✓ 1

Exam tip: Switch your calculator to radian mode for any trigonometric function in a calculus question.

+ Add to test✓ In your test406IBMI-034-01-r1-06Optimisation in context Paper 2Hard14 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 test407IBMI-034-00-r1-06Gradients, tangents and rates of change Paper 2Hard11 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 test408IBMI-034-04-r2-06Euler's method and coupled systems Paper 2Hard14 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 test409IBMI-034-08-r2-06Differential equations and slope fields Paper 2Hard14 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 test410IBMI-034-03-r2-06Differential equations and slope fields Paper 2Hard14 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 test411IBMI-034-02-r2-06Integration and the trapezoidal rule Paper 2Hard14 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 test412IBMI-034-08-r1-05Differential equations and slope fields Paper 2Hard13 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 test413IBMI-034-05-r1-06Gradients, tangents and rates of change Paper 2Hard14 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 test414IBMI-034-00-r2-06Gradients, tangents and rates of change Paper 2Hard14 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 test415IBMI-034-03-r1-06Differential equations and slope fields Paper 2Hard14 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 test416IBMI-034-04-r1-04Euler's method and coupled systems Paper 2Hard16 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 test417IBMI-034-07-r2-06Integration and the trapezoidal rule Paper 2Hard13 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 test418IBMI-034-07-r1-05Integration and the trapezoidal rule Paper 2Hard10 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 test419IBMI-034-01-r2-06Optimisation in context Paper 2Hard13 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 test420IBMI-034-06-r2-06Optimisation in context Paper 2Hard14 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.

+ Add to test✓ In your test421IBMI3H-5-10-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard14 marks

The population P of fish in a lake is modelled by the differential equation
$\frac{dP}{dt}=0.2P\left(1-\frac{P}{500}\right)$,
where P is the number of fish and t is the time in years. A slope field is drawn for this equation.

(a)

Find the gradient of the slope field at the point (t, P) = (0, 100).

(2)

(b)

Find the equilibrium values of P.

(2)

(c)

Find the value of P at which the gradient of the slope field is greatest, and state that greatest gradient.

(3)

(d)

Explain why a solution that starts at P = 600 decreases towards 500.

(2)

(e)

Describe the shape of the solution curve that passes through (0, 100).

(2)

(f)

Use Euler's method with a step length of 1 year to estimate P when t = 2, starting from P(0) = 100.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute P = 100	✓ 1
A1 16	✓ 1
Part (b)
M1 set $\frac{dP}{dt}=0$	✓ 1
A1 P = 0 and P = 500	✓ 1
Part (c)
M1 differentiate the right-hand side with respect to P and set to zero (0.2 − 0.0008P = 0)	✓ 1
A1 P = 250	✓ 1
A1 greatest gradient 25	✓ 1
Part (d)
M1 substitute P = 600 to get gradient −24	✓ 1
R1 gradient negative above 500, and zero at 500, so the solution falls towards the equilibrium	✓ 1
Part (e)
R1 increasing, steepest where P = 250, then flattening	✓ 1
R1 approaches the horizontal asymptote P = 500 and never crosses it	✓ 1
Part (f)
M1 P₁ = 100 + 1 × 16	✓ 1
A1 P ≈ 116 when t = 1	✓ 1
A1 P ≈ 134 when t = 2	✓ 1

Common error: The slope field for this equation depends only on P, so gradients are identical along every horizontal line. Do not read them as depending on t.

+ Add to test✓ In your test422IBMI3H-5-10-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard14 marks

A mass on a spring with damping has displacement x metres from equilibrium at time t seconds, where
$\frac{d^2x}{dt^2}+0.6\frac{dx}{dt}+4x=0,$
with x = 0.5 and $\frac{dx}{dt}=0$ when t = 0.

(a)

Let v = $\frac{dx}{dt}$. Write the second-order equation as a pair of first-order equations.

(2)

(b)

Use Euler's method with a step length of 0.1 s to estimate x and v when t = 0.2.

(4)

(c)

Find the roots of the characteristic equation λ² + 0.6λ + 4 = 0.

(3)

(d)

Describe the motion, and find the time between successive maximum displacements.

(3)

(e)

The exact solution is x = e^−0.3t(0.5 cos 1.977t + 0.0759 sin 1.977t). Find x when t = 0.2.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dx}{dt}=v$	✓ 1
A1 $\frac{dv}{dt}=-4x-0.6v$	✓ 1
Part (b)
M1 correct use of the Euler step for both variables	✓ 1
A1 x(0.1) = 0.5, v(0.1) = −0.2	✓ 1
A1 x(0.2) = 0.48	✓ 1
A1 v(0.2) = −0.388	✓ 1
Part (c)
M1 attempt to solve (quadratic formula or calculator)	✓ 1
A1 λ = −0.3 ± ki	✓ 1
A1 k = 1.98 (1.977)	✓ 1
Part (d)
R1 oscillation with decreasing amplitude (damped)	✓ 1
M1 period = 2π ÷ 1.977	✓ 1
A1 3.18 s	✓ 1
Part (e)
M1 substitute t = 0.2 with the calculator in radians	✓ 1
A1 0.462 m	✓ 1

Common error: For an Euler step on a coupled system, update x and v from the old values of both. Using the new x inside the v step gives different numbers.

+ Add to test✓ In your test423IBMI3H-5-10-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard15 marks

Two robots P and Q move along the same straight rail, measured from a fixed point O. At time t seconds, 0 ≤ t ≤ 10, their displacements in metres are
s_P = 0.1t³ − 1.5t² + 7t and s_Q = 2t.

(a)

Find the velocity of P at time t.

(2)

(b)

Find the times at which P is instantaneously at rest.

(3)

(c)

Verify that P and Q are at the same position when t = 5.

(2)

(d)

For 0 < t < 5, P is ahead of Q. Find the greatest distance by which P is ahead of Q.

(4)

(e)

Find the total distance travelled by P in the 10 seconds.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 0.3t² − 3t + 7	✓ 1
Part (b)
M1 set 0.3t² − 3t + 7 = 0 and solve	✓ 1
A1 t = 3.71	✓ 1
A1 t = 6.29	✓ 1
Part (c)
M1 evaluate s_P(5) = 12.5 − 37.5 + 35	✓ 1
A1 both equal 10 m	✓ 1
Part (d)
M1 greatest gap when the velocities are equal: 0.3t² − 3t + 7 = 2	✓ 1
A1 t = 2.11	✓ 1
M1 gap = s_P − s_Q at this time	✓ 1
A1 4.81 m	✓ 1
Part (e)
M1 split the motion at the times found in (b)	✓ 1
A1 s_P(3.71) = 10.4 and s_P(6.29) = 9.57	✓ 1
A1 distances 10.4, 0.861 and 10.4	✓ 1
A1 21.7 m	✓ 1

Common error: A gap or distance is largest when the two velocities are equal, not when the displacements are equal. Distance travelled needs the turning points.

+ Add to test✓ In your test424IBMI3H-5-10b-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard11 marks

An arched hall entrance has a cross-section bounded by the curve y = 9 − x² and the x-axis (the floor), with both axes in metres. A rectangular glass panel with its base on the floor is placed symmetrically under the arch, with its top corners on the curve. The panel has half-width x metres, where 0 < x < 3.

(a)

Show that the area of the panel is A = 18x − 2x³.

(2)

(b)

Find the exact value of x that maximises the area, justify that it is a maximum, and find the maximum area.

(4)

(c)

The panel has a light strip around its whole perimeter. Find the value of x for which the perimeter is greatest, and the greatest perimeter.

(3)

(d)

For the panel of greatest area, find by what percentage its perimeter is below the greatest possible perimeter.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 width = 2x and height = 9 − x²	✓ 1
A1 A = 2x(9 − x²) = 18x − 2x³	✓ 1
Part (b)
M1 dA/dx = 18 − 6x²	✓ 1
A1 x = √3	✓ 1
M1 d²A/dx² = −12x < 0, so a maximum	✓ 1
A1 A = 12√3 ≈ 20.8 m²	✓ 1
Part (c)
A1 P = 2(2x) + 2(9 − x²) = 4x + 18 − 2x²	✓ 1
M1 dP/dx = 4 − 4x = 0	✓ 1
A1 x = 1 and P = 20 m	✓ 1
Part (d)
M1 P(√3) = 4√3 + 12 ≈ 18.9 m and (20 − 18.9)/20 × 100	✓ 1
A1 5.36%	✓ 1

Common error: The value of x that maximises the area is not the value that maximises the perimeter; treat each quantity as its own function.

+ Add to test✓ In your test425IBMI3H-5-10b-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard12 marks

Two linked ponds, A and B, are polluted. The amounts of pollutant (in kg) in the ponds t days after the pollution is detected are x and y, and they satisfy
dx/dt = −3x + 2y,
dy/dt = x − 2y.
Initially x = 60 and y = 30.

(a)

Find the eigenvalues of the matrix M = [[−3, 2], [1, −2]] (rows separated by commas).

(3)

(b)

Find an eigenvector for each eigenvalue.

(3)

(c)

Find the particular solution for x and y in terms of t.

(3)

(d)

Find the amount of pollutant in pond A after half a day.

(2)

(e)

Describe, in context, the long-term behaviour of the pollution.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 det(M − λI) = 0	✓ 1
A1 λ² + 5λ + 4 = 0	✓ 1
A1 λ = −1 and λ = −4	✓ 1
Part (b)
M1 solve (M − λI)v = 0	✓ 1
A1 λ = −1: (1, 1)	✓ 1
A1 λ = −4: (2, −1)	✓ 1
Part (c)
M1 general solution (x, y) = Ae^−t(1, 1) + Be^−4t(2, −1)	✓ 1
M1 use the initial values: A + 2B = 60 and A − B = 30	✓ 1
A1 x = 40e^−t + 20e^−4t and y = 40e^−t − 10e^−4t	✓ 1
Part (d)
M1 substitute t = 0.5 into x	✓ 1
A1 27.0 kg	✓ 1
Part (e)
R1 both eigenvalues are negative, so x and y both tend to 0: the pollutant eventually leaves both ponds	✓ 1

Exam tip: Check your particular solution by substituting t = 0 and by differentiating one component back into the original equations.

+ Add to test✓ In your test426IBMI3H-5-10b-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

The vertical displacement x metres of a mountain-bike saddle from its rest position, t seconds after the bike lands from a small jump, satisfies d²x/dt² + 0.6 dx/dt + 4x = 0, with x = 0.1 and dx/dt = 0 when t = 0.

(a)

By letting y = dx/dt, write the equation as a pair of coupled first-order equations.

(2)

(b)

Find the value of d²x/dt² when t = 0, and describe what it means for the saddle.

(2)

(c)

Use Euler's method with step length 0.1 to complete three steps, giving x and y at t = 0.1, 0.2 and 0.3.

(4)

(d)

State the estimated velocity of the saddle at t = 0.3, giving its direction.

(1)

(e)

Suggest one way to improve the accuracy of the estimates.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 dx/dt = y	✓ 1
A1 dy/dt = −4x − 0.6y	✓ 1
Part (b)
M1 substitute x = 0.1 and y = 0 into −4x − 0.6y	✓ 1
A1 −0.4 m s⁻²: the saddle is accelerating back towards its rest position	✓ 1
Part (c)
M1 x_n+1 = x_n + 0.1y_n and y_n+1 = y_n + 0.1(−4x_n − 0.6y_n)	✓ 1
A1 t = 0.1: x = 0.1 and y = -0.04	✓ 1
A1 t = 0.2: x = 0.096 and y = -0.0776	✓ 1
A1 t = 0.3: x = 0.08824 and y = -0.111344	✓ 1
Part (d)
A1 -0.111 m s⁻¹, i.e. moving back towards the rest position at about 0.111 m s⁻¹	✓ 1
Part (e)
R1 use a smaller step length, since Euler's method assumes the gradient is constant over each step	✓ 1

Common error: Both variables must be updated using the values from the previous step; do not use the new x when calculating the new y in the same step.

+ Add to test✓ In your test427IBMI3H-5-03r8-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard9 marks

In a brined cheese, the salt concentrations (g per kg) in the rind, x, and the core, y, at time t hours satisfy
dx/dt = 2 − 0.3x + 0.1y,
dy/dt = 0.3x − 0.4y.
At t = 0, x = 0 and y = 10.

(a)

Use Euler's method with step length 0.5 hours to find estimates of x and y when t = 0.5.

(3)

(b)

Continue the method to estimate x and y when t = 1.

(3)

(c)

The concentrations approach constant values as t becomes large. Find the long-term value of y, to three significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradients at t = 0: dx/dt = 3 and dy/dt = −4	✓ 1
A1 x(0.5) = 1.5	✓ 1
A1 y(0.5) = 8	✓ 1
Part (b)
M1 gradients at (1.5, 8): dx/dt = 2.35 and dy/dt = −2.75 (ECF from (a))	✓ 1
A1 x = 2.675 at t = 1	✓ 1
A1 y = 6.625 at t = 1	✓ 1
Part (c)
M1 set both derivatives equal to zero	✓ 1
M1 from dy/dt = 0, x = (4/3)y; substitute into 2 − 0.3x + 0.1y = 0	✓ 1
A1 y = 20/3 = 6.67 g per kg	✓ 1

Common error: Both Euler updates in a step must use the values from the start of that step; do not use the new x when computing the new y.

+ Add to test✓ In your test428IBMI3H-5-03r14-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard11 marks

A custard at 85 °C is left to cool in a kitchen at 20 °C. Its temperature $T$ °C, $t$ minutes after it is left, is modelled by $\frac{dT}{dt}=-0.08(T-20)$.

(a)

Use Euler's method with a step length of 5 minutes to estimate the temperature at $t=10$.

(3)

(b)

Solve the differential equation, given that $T=85$ when $t=0$, to find $T$ in terms of $t$.

(4)

(c)

Explain why Euler's method underestimates the temperature here.

(1)

(d)

Calculate the percentage error in the Euler estimate from part (a), relative to the exact value at $t=10$. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses $T_{n+1}=T_n+5\times\left(-0.08(T_n-20)\right)$	✓ 1
A1 $T(5)=59$	✓ 1
A1 $T(10)=43.4$ °C	✓ 1
Part (b)
M1 separates variables: $\int\frac{1}{T-20}dT=\int-0.08\,dt$	✓ 1
A1 $\ln\|T-20\|=-0.08t+c$	✓ 1
M1 uses $T(0)=85$ to find $c=\ln65$	✓ 1
A1 $T=20+65e^{-0.08t}$	✓ 1
Part (c)
R1 the solution curve is concave up (decreasing at a decreasing rate), so each tangent step lies below the curve	✓ 1
Part (d)
A1 exact value $T(10)=49.2$ (accept 49.206)	✓ 1
M1 $\frac{49.206-43.4}{49.206}\times100$	✓ 1
A1 11.8 %	✓ 1

Exam tip: Percentage error is measured against the exact value: divide by the exact value, not by the estimate.

+ Add to test✓ In your test429IBMI3H-5-03r14-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard11 marks

Two servers in a telecoms network hold queued data. Let $x$ and $y$ gigabytes (GB) be the data queued in server A and server B, $t$ hours after $t=0$. Data are transferred and cleared according to
$\frac{dx}{dt}=-0.6x+0.2y,\qquad \frac{dy}{dt}=0.4x-0.4y.$
At $t=0$, $x=30$ and $y=0$.

(a)

Show that the eigenvalues of the matrix $\begin{pmatrix}-0.6&0.2\\0.4&-0.4\end{pmatrix}$ are $-0.2$ and $-0.8$.

(3)

(b)

Find an eigenvector for each eigenvalue.

(2)

(c)

Find $x$ and $y$ in terms of $t$.

(3)

(d)

Find the time at which server B holds the most data, and the maximum amount held. Give both to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sets $\det(M-\lambda I)=0$	✓ 1
A1 $(-0.6-\lambda)(-0.4-\lambda)-0.08=0$, i.e. $\lambda^2+\lambda+0.16=0$	✓ 1
A1 $(\lambda+0.2)(\lambda+0.8)=0$, so $\lambda=-0.2$ or $-0.8$	✓ 1
Part (b)
A1 $\lambda=-0.2$: eigenvector $\begin{pmatrix}1\\2\end{pmatrix}$ (any multiple)	✓ 1
A1 $\lambda=-0.8$: eigenvector $\begin{pmatrix}1\\-1\end{pmatrix}$ (any multiple)	✓ 1
Part (c)
M1 general solution $\begin{pmatrix}x\\y\end{pmatrix}=Ae^{-0.2t}\begin{pmatrix}1\\2\end{pmatrix}+Be^{-0.8t}\begin{pmatrix}1\\-1\end{pmatrix}$ (ECF from (b))	✓ 1
M1 uses initial conditions: $A+B=30$ and $2A-B=0$	✓ 1
A1 $x=10e^{-0.2t}+20e^{-0.8t}$ and $y=20e^{-0.2t}-20e^{-0.8t}$	✓ 1
Part (d)
M1 sets $\frac{dy}{dt}=-4e^{-0.2t}+16e^{-0.8t}=0$ so $e^{0.6t}=4$	✓ 1
A1 $t=\frac{\ln4}{0.6}=2.31$ hours	✓ 1
A1 maximum $y=9.45$ GB	✓ 1

Exam tip: Check an eigenvector by substituting it back into $M\mathbf{v}=\lambda\mathbf{v}$ before building the general solution.

+ Add to test✓ In your test430IBMI3H-5-03r14-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

A climber falls and is held by a dynamic rope. The climber's displacement $x$ metres from the equilibrium position, $t$ seconds after the rope first becomes taut, is modelled by $\frac{d^2x}{dt^2}+0.6\frac{dx}{dt}+9x=0$. At $t=0$, $x=0.4$ and $\frac{dx}{dt}=0$.

(a)

Let $v=\frac{dx}{dt}$. Write the second-order equation as a pair of coupled first-order equations.

(2)

(b)

Use Euler's method with step length 0.1 to estimate $x$ at $t=0.3$. Give your answer to 3 significant figures.

(4)

(c)

Show that the characteristic equation $\lambda^2+0.6\lambda+9=0$ gives oscillating motion.

(2)

(d)

The oscillations have angular frequency equal to the imaginary part of the roots. Find the period of oscillation, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dx}{dt}=v$	✓ 1
A1 $\frac{dv}{dt}=-9x-0.6v$	✓ 1
Part (b)
M1 uses $x_{n+1}=x_n+0.1v_n$ and $v_{n+1}=v_n+0.1(-9x_n-0.6v_n)$	✓ 1
A1 $x_1=0.4$, $v_1=-0.36$	✓ 1
A1 $x_2=0.364$, $v_2=-0.6984$	✓ 1
A1 $x_3=0.29416\ldots=0.294$ m	✓ 1
Part (c)
M1 discriminant $0.6^2-4\times9=-35.64$	✓ 1
R1 negative discriminant, so the roots are complex and the motion oscillates	✓ 1
Part (d)
M1 $\omega=\sqrt{9-0.3^2}=2.985\ldots$ and period $\frac{2\pi}{\omega}$	✓ 1
A1 2.10 s	✓ 1

Common error: In Euler's method for a coupled system, update both variables using the old values from the same step, not a mixture of old and new.

+ Add to test✓ In your test431IBMI3H-5-05r17-03Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard12 marks

A float for a plankton net is made by rotating the region bounded by the curve $y=\sqrt{x}\,e^{-x^2/4}$, the x-axis and the line $x=3$ through 360° about the x-axis. One unit on each axis represents 10 cm.

(a)

Use the substitution $u=\frac{x^2}{2}$ to show that $\int x e^{-x^2/2}\,dx=-e^{-x^2/2}+c$.

(2)

(b)

Find the volume of the float in cubic units, giving your answer to 3 significant figures.

(3)

(c)

Find the volume of the float in cm³. Use the unrounded value from (b) and give your answer to the nearest cm³.

(2)

(d)

The float is cut from foam of density 0.30 g cm⁻³. Find its mass, to the nearest gram, using the unrounded volume from (b).

(2)

(e)

Find the greatest radius of the float, in cm, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{du}{dx}=x$ (so $x\,dx=du$) and integrand becomes $\int e^{-u}\,du$	✓ 1
A1 $-e^{-u}+c=-e^{-x^2/2}+c$	✓ 1
Part (b)
M1 $V=\pi\int_0^3 y^2\,dx=\pi\int_0^3 xe^{-x^2/2}\,dx$	✓ 1
A1 $V=\pi\left[-e^{-x^2/2}\right]_0^3=\pi\left(1-e^{-4.5}\right)$	✓ 1
A1 $V=3.11$ cubic units	✓ 1
Part (c)
M1 one cubic unit is $10^3=1000$ cm³	✓ 1
A1 3107 cm³	✓ 1
Part (d)
M1 mass = density × volume = $0.30\times 3106.69$	✓ 1
A1 932 g	✓ 1
Part (e)
M1 differentiate $y^2=xe^{-x^2/2}$ (or $y$) and set equal to zero	✓ 1
A1 $e^{-x^2/2}(1-x^2)=0$ so $x=1$	✓ 1
A1 $y=e^{-0.25}=0.779$ units, so greatest radius = 7.79 cm	✓ 1

Exam tip: Since $y^2$ is the integrand, maximising $y^2$ gives the same x-value as maximising y and avoids differentiating a square root.

+ Add to test✓ In your test432IBMI3H-5-05r17-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard14 marks

An aquarium tank holds 500 litres of fresh water. Salt water of concentration 35 g per litre enters the tank at 10 litres per minute. The mixture is kept well stirred and leaves at 10 litres per minute, so the volume stays constant. Let S g per litre be the salt concentration in the tank after t minutes, so $S=0$ when $t=0$.

(a)

Show that $\dfrac{dS}{dt}=0.02(35-S)$.

(2)

(b)

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

(5)

(c)

Find the time taken for the concentration to reach 30 g per litre. Give your answer in minutes to 3 significant figures.

(2)

(d)

Use Euler's method with step length 10 minutes to estimate S when $t=20$.

(3)

(e)

Find the error in your Euler estimate from (d), by comparing with the exact value of $S(20)$. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rate of change of salt = rate in − rate out $=10\times 35-10S$ grams per minute	✓ 1
A1 divide by 500 litres: $\frac{dS}{dt}=\frac{350-10S}{500}=0.02(35-S)$	✓ 1
Part (b)
M1 separate variables: $\int\frac{dS}{35-S}=\int 0.02\,dt$	✓ 1
A1 $-\ln\|35-S\|=0.02t+c$	✓ 1
A1 $35-S=Ae^{-0.02t}$	✓ 1
M1 use $S=0$ when $t=0$ to find $A=35$	✓ 1
A1 $S=35\left(1-e^{-0.02t}\right)$	✓ 1
Part (c)
M1 $e^{-0.02t}=\frac{1}{7}$ (ECF from (b))	✓ 1
A1 $t=97.3$ minutes	✓ 1
Part (d)
M1 $S_{n+1}=S_n+10\times 0.02(35-S_n)$ with $S_0=0$	✓ 1
A1 $S(10)=7$	✓ 1
A1 $S(20)=12.6$	✓ 1
Part (e)
M1 exact $S(20)=35(1-e^{-0.4})=11.54$ (ECF from (b))	✓ 1
A1 error $=12.6-11.54=1.06$ g per litre (overestimate)	✓ 1

Exam tip: When asked to show a differential equation from a context, state the 'rate in' and 'rate out' explicitly, with units, before simplifying.

+ Add to test✓ In your test433IBMI3H-5-05r17-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard12 marks

The opening of an aircraft hangar has the shape of the region between the curve $y=8-\frac{x^2}{2}$ and the x-axis, where x and y are in metres. A rectangular door is to be fitted inside the opening, with its base on the x-axis, symmetric about the y-axis, and its upper corners on the curve. The door has width 2x m and area A m².

(a)

Show that $A=16x-x^3$.

(2)

(b)

Find the value of x for which A is a maximum. Give your answer to 3 significant figures.

(3)

(c)

Use the second derivative to show that this value of x gives a maximum.

(2)

(d)

Find the maximum area of the door, using the unrounded value of x. Give your answer to 3 significant figures.

(1)

(e)

Find the area of the whole opening of the hangar.

(3)

(f)

Find the maximum door area as a percentage of the area of the opening, using unrounded values. Give your answer to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height of door $=8-\frac{x^2}{2}$ and $A=2x\left(8-\frac{x^2}{2}\right)$	✓ 1
A1 $A=16x-x^3$ (shown)	✓ 1
Part (b)
M1 $\frac{dA}{dx}=16-3x^2$	✓ 1
A1 set equal to 0: $x^2=\frac{16}{3}$	✓ 1
A1 $x=2.31$ (taking the positive root)	✓ 1
Part (c)
M1 $\frac{d^2A}{dx^2}=-6x$	✓ 1
R1 negative for $x=2.31$, so a maximum	✓ 1
Part (d)
A1 $A=24.6$ m²	✓ 1
Part (e)
M1 $\int_{-4}^{4}\left(8-\frac{x^2}{2}\right)dx$ with limits found from $y=0$	✓ 1
A1 $\left[8x-\frac{x^3}{6}\right]_{-4}^{4}$	✓ 1
A1 $42.7$ m²	✓ 1
Part (f)
A1 57.7%	✓ 1

Exam tip: Use symmetry to halve the work in an integral over $[-4,4]$ of an even function: integrate from 0 to 4 and double.

+ Add to test✓ In your test434IBMI3H-5-04r20-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard13 marks

A rail company is fencing a rectangular compound of area 2400 m² beside its track. The two sides of length x metres run at right angles to the track, and the side of length y metres opposite the track is also fenced. There is no fence along the track. Fencing costs €60 per metre for the sides of length x and €85 per metre for the side of length y.

(a)

Show that the total cost, in euros, is $C(x)=\frac{204000}{x}+120x$.

(3)

(b)

Find $\frac{dC}{dx}$ and hence find the value of x that gives a stationary value of C, to 3 significant figures.

(4)

(c)

Use the second derivative to show that this value of x gives a minimum cost.

(1)

(d)

Find the minimum cost, to the nearest euro.

(2)

(e)

Because of a boundary, the width x cannot exceed 35 m. Explain why the minimum cost is now found at x = 35, and find this cost to the nearest cent.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $y=\frac{2400}{x}$	✓ 1
M1 $C=2(60x)+85y$	✓ 1
A1 $C=120x+\frac{204000}{x}$ (AG)	✓ 1
Part (b)
M1 differentiate $204000x^{-1}$	✓ 1
A1 $C'(x)=-\frac{204000}{x^2}+120$	✓ 1
M1 set $C'(x)=0$, so $x^2=1700$	✓ 1
A1 x = 41.2 (41.2 m)	✓ 1
Part (c)
R1 $C''(x)=\frac{408000}{x^3}>0$ for $x>0$, so the stationary point is a minimum	✓ 1
Part (d)
M1 substitute the unrounded x into $C(x)$	✓ 1
A1 €9895 (€9895.45)	✓ 1
Part (e)
R1 $C'(x)<0$ for \(0C is decreasing on the allowed interval and is least at the largest allowed x	✓ 1
M1 substitute x = 35 into $C(x)$	✓ 1
A1 €10028.57	✓ 1

Exam tip: When a restriction excludes the stationary point, check the sign of the derivative on the allowed interval; the minimum then sits at an end of that interval.

+ Add to test✓ In your test435IBMI3H-5-06r21-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

A cyclist stops pedalling and freewheels along a level road. Air resistance gives a deceleration modelled by dv/dt = −0.004v², where v m s⁻¹ is the speed at time t seconds. When t = 0, v = 12.

(a)

Solve the differential equation to show that 1/v = 0.004t + 1/12.

(3)

(b)

Find the speed after 20 seconds, to 3 significant figures.

(2)

(c)

Find the time taken for the speed to fall to 4 m s⁻¹, to 3 significant figures.

(2)

(d)

Find the distance travelled in the first 20 seconds, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 separates variables: ∫v⁻² dv = ∫ −0.004 dt	✓ 1
A1 −1/v = −0.004t + c	✓ 1
A1 uses v(0) = 12 to find c = −1/12, giving 1/v = 0.004t + 1/12 (AG)	✓ 1
Part (b)
M1 1/v = 0.08 + 1/12	✓ 1
A1 v = 6.12 m s⁻¹	✓ 1
Part (c)
M1 0.25 = 0.004t + 1/12	✓ 1
A1 t = 41.7 s	✓ 1
Part (d)
M1 writes v = 250/(t + 20.83…) or v = 1/(0.004t + 1/12)	✓ 1
M1 integrates to 250 ln(t + 20.83…) and applies limits 0 and 20	✓ 1
A1 250 ln(1.96) = 168 m	✓ 1

Common error: Do not write v = 12 − 0.004t; the deceleration depends on v², so the equation must be solved by separating the variables.

+ Add to test✓ In your test436IBMI3H-5-06r21-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard12 marks

In a data centre the temperature deviations x and y (in °C) of two cooling zones from their target temperature satisfy the coupled system
dx/dt = −3x + y, dy/dt = 2x − 2y,
where t is in hours.

(a)

Write the system in the form dX/dt = AX and find the eigenvalues of A.

(3)

(b)

Find an eigenvector for each eigenvalue.

(2)

(c)

Initially x = 5 and y = 4. Find x and y as functions of t.

(4)

(d)

Find the value of y when t = 0.5, to 3 significant figures.

(2)

(e)

Describe the long-term behaviour of the two zones.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 A = [[−3, 1],[2, −2]] and det(A − λI) = 0	✓ 1
A1 (−3 − λ)(−2 − λ) − 2 = λ² + 5λ + 4 = 0	✓ 1
A1 λ = −1 and λ = −4	✓ 1
Part (b)
A1 λ = −1: (1, 2) (any multiple)	✓ 1
A1 λ = −4: (1, −1) (any multiple)	✓ 1
Part (c)
M1 general solution X = a(1, 2)e^−t + b(1, −1)e^−4t	✓ 1
M1 a + b = 5 and 2a − b = 4	✓ 1
A1 a = 3, b = 2	✓ 1
A1 x = 3e^−t + 2e^−4t and y = 6e^−t − 2e^−4t	✓ 1
Part (d)
M1 substitutes t = 0.5 into y	✓ 1
A1 y = 3.37 °C	✓ 1
Part (e)
R1 both eigenvalues are negative, so x and y tend to 0: both zones return to the target temperature	✓ 1

Exam tip: Check an eigenvector by substituting it back into AX = λX; it takes seconds and catches sign errors before they propagate.

+ Add to test✓ In your test437IBMI3H-5-06r21-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard9 marks

At an excavation site the number of rodents x and owls y in a valley are modelled by
dx/dt = 0.5x − 0.01xy, dy/dt = 0.004xy − 0.3y,
where t is in years.

(a)

Identify which term in the first equation represents the effect of owls on rodents, and explain its sign.

(1)

(b)

Find the non-zero equilibrium values of x and y.

(3)

(c)

Initially x = 40 and y = 30. Use Euler's method with step length 0.5 years to estimate x and y when t = 1. Give both to 3 significant figures.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R1 −0.01xy, which is negative because owls eat rodents and so reduce the rodent population	✓ 1
Part (b)
M1 sets both derivatives to zero	✓ 1
A1 0.5 − 0.01y = 0, so y = 50	✓ 1
A1 0.004x − 0.3 = 0, so x = 75	✓ 1
Part (c)
M1 gradients at t = 0: dx/dt = 8 and dy/dt = −4.2	✓ 1
A1 at t = 0.5: x = 44.0, y = 27.9	✓ 1
M1 gradients at t = 0.5: dx/dt = 9.724 and dy/dt = −3.4596	✓ 1
A1 x = 48.9	✓ 1
A1 y = 26.2	✓ 1

Exam tip: In a coupled Euler step, both new values must be computed from the old pair; do not use the updated x when finding the new y.

+ Add to test✓ In your test438IBMI3H-5-06r22-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard13 marks

A space probe has two fuel tanks. Let a and b be the masses of fuel, in kg, in tank A and tank B at time t hours. Fuel is pumped between the tanks, and the thrusters draw fuel only from tank B:
$\frac{da}{dt}=-0.2a+0.1b$
$\frac{db}{dt}=0.2a-0.3b$
Initially a = 50 and b = 20.

(a)

Explain the meaning of the terms 0.2a and 0.1b in the equation for $\frac{da}{dt}$.

(2)

(b)

Use Euler's method with step h = 0.5 to estimate a and b at t = 0.5, 1 and 1.5.

(5)

(c)

Find the equilibrium points of the system.

(2)

(d)

Find the eigenvalues of the matrix $\begin{pmatrix}-0.2&0.1\\0.2&-0.3\end{pmatrix}$ and state what they imply about the long-term fuel in the tanks.

(3)

(e)

Use your answer to (b) to state the total fuel in both tanks at t = 1.5, to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R1 0.2a: fuel leaves tank A at 20 % of its mass per hour, pumped to B	✓ 1
R1 0.1b: fuel returns from B to A at 10 % of B's mass per hour	✓ 1
Part (b)
M1 $a_{n+1}=a_n+0.5\left(-0.2a_n+0.1b_n\right)$ and $b_{n+1}=b_n+0.5\left(0.2a_n-0.3b_n\right)$	✓ 1
A1 t = 0.5: (46, 22)	✓ 1
A1 t = 1: (42.5, 23.3)	✓ 1
A1 a(1.5) ≈ 39.415	✓ 1
A1 b(1.5) ≈ 24.055	✓ 1
Part (c)
M1 set both derivatives to zero: b = 2a from the first, then the second gives $0.2a-0.6a=0$	✓ 1
A1 only (0, 0)	✓ 1
Part (d)
M1 $\lambda^2+0.5\lambda+0.04=0$	✓ 1
A1 λ = −0.1 and −0.4	✓ 1
R1 both negative, so both tanks tend to zero (the equilibrium is stable)	✓ 1
Part (e)
A1 63.5 kg (ECF from (b))	✓ 1

Exam tip: In Euler's method for a system, update both variables from the same old values: compute both gradients before changing a or b.

+ Add to test✓ In your test439IBMI3H-5-06r22-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard13 marks

A footbridge deck oscillates vertically with displacement x metres at time t seconds, where
$\frac{d^2x}{dt^2}+0.4\frac{dx}{dt}+4x=0.$
At t = 0, x = 0.06 and $\frac{dx}{dt}=0$.

(a)

Let $y=\frac{dx}{dt}$. Write the equation as a pair of first-order differential equations.

(2)

(b)

Find the eigenvalues of the matrix $\begin{pmatrix}0&1\\-4&-0.4\end{pmatrix}$.

(3)

(c)

Explain what the eigenvalues in (b) tell you about the motion.

(2)

(d)

Use the imaginary part of the eigenvalues to find the period of the oscillation, to 3 significant figures.

(2)

(e)

Use Euler's method with step h = 0.1 to estimate x at t = 0.3, to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dx}{dt}=y$	✓ 1
A1 $\frac{dy}{dt}=-4x-0.4y$	✓ 1
Part (b)
M1 $\lambda^2+0.4\lambda+4=0$	✓ 1
A1 $\lambda=-0.2\pm\sqrt{3.96}\,i$	✓ 1
A1 −0.2 ± 1.99i	✓ 1
Part (c)
R1 complex eigenvalues, so the motion oscillates	✓ 1
R1 negative real part, so the amplitude decays (damped oscillation)	✓ 1
Part (d)
M1 $\frac{2\pi}{\sqrt{3.96}}$	✓ 1
A1 3.16 s	✓ 1
Part (e)
M1 $x_{n+1}=x_n+0.1y_n$ and $y_{n+1}=y_n+0.1(-4x_n-0.4y_n)$	✓ 1
A1 t = 0.1: (0.06, −0.024)	✓ 1
A1 t = 0.2: (0.0576, −0.04704)	✓ 1
A1 x(0.3) ≈ 0.0529 m	✓ 1

Common error: Both variables in an Euler step must use the values from the start of that step; using the updated x to find the new y gives a different (wrong) result.

+ Add to test✓ In your test440IBMI3H-5-06r22-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard11 marks

After a gust of wind, the displacement of a bridge cable is modelled by x(t) = 0.08e^−0.1t sin(2t) metres, where t is the time in seconds and the angle is in radians.

(a)

Find an expression for the velocity $\frac{dx}{dt}$.

(3)

(b)

Find the first time t > 0 at which the cable is momentarily at rest, to 3 significant figures.

(3)

(c)

Find the displacement at this time, to 3 significant figures.

(2)

(d)

The displacement has a local maximum at the time found in (b), and successive local maxima occur every $\pi$ seconds. Find the ratio of the displacement at one local maximum to the displacement at the previous one, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule with the chain rule for each factor	✓ 1
A1 $-0.008e^{-0.1t}\sin 2t+0.16e^{-0.1t}\cos 2t$	✓ 1
A1 $0.08e^{-0.1t}(2\cos 2t-0.1\sin 2t)$ or equivalent	✓ 1
Part (b)
M1 $\frac{dx}{dt}=0$	✓ 1
A1 $\tan 2t=20$	✓ 1
A1 t = 0.760 s	✓ 1
Part (c)
M1 x(0.7604…) (using unrounded t)	✓ 1
A1 0.0740 m	✓ 1
Part (d)
M1 consider $\frac{x(t+\pi)}{x(t)}$	✓ 1
A1 $e^{-0.1\pi}$ since $\sin(2t+2\pi)=\sin 2t$	✓ 1
A1 0.730	✓ 1

Exam tip: When the same trigonometric factor appears at two times a whole period apart, it cancels in the ratio, leaving only the exponential decay.

+ Add to test✓ In your test441IBMI3H-5-06r23-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard13 marks

Calculator

Two linked fuel tanks on a ship hold masses x kg and y kg of a dissolved contaminant at time t hours. The masses satisfy
dx/dt = −3x + 2y, dy/dt = x − 2y,
with x = 10 and y = 4 when t = 0.

(a)

Write the system in matrix form and find the eigenvalues of the matrix.

(3)

(b)

Find an eigenvector for each eigenvalue.

(3)

(c)

Find the particular solution for x and y in terms of t.

(3)

(d)

Show that x > y for all t ≥ 0.

(2)

(e)

Find the mass of contaminant in the first tank after 1 hour, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 dx/dt = Ax with A having rows (−3, 2) and (1, −2)	✓ 1
M1 det(A − λI) = 0, giving (−3 − λ)(−2 − λ) − 2 = 0	✓ 1
A1 λ² + 5λ + 4 = 0, so λ = −1 and λ = −4	✓ 1
Part (b)
M1 solve (A − λI)v = 0 for one eigenvalue	✓ 1
A1 λ = −1: (1, 1) (or any multiple)	✓ 1
A1 λ = −4: (2, −1) (or any multiple)	✓ 1
Part (c)
M1 general solution c₁(1, 1)e^−t + c₂(2, −1)e^−4t	✓ 1
M1 initial conditions c₁ + 2c₂ = 10 and c₁ − c₂ = 4	✓ 1
A1 x = 6e^−t + 4e^−4t and y = 6e^−t − 2e^−4t	✓ 1
Part (d)
M1 x − y = 6e^−4t	✓ 1
R1 an exponential is always positive, so x − y > 0	✓ 1
Part (e)
M1 substitute t = 1: 6e⁻¹ + 4e⁻⁴	✓ 1
A1 2.28 kg	✓ 1

Common error: An eigenvector can be scaled by any non-zero constant, but the constants c₁ and c₂ in the general solution then depend on that choice; fix your eigenvectors before using the initial conditions.

+ Add to test✓ In your test442IBMI3H-5-06r23-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard11 marks

Calculator

The roll angle x radians of a ship, t seconds after it is disturbed, satisfies the second-order differential equation d²x/dt² + 0.5 dx/dt + 4x = 0. Initially x = 0.2 and dx/dt = 0.

(a)

Let y = dx/dt. Write the equation as a pair of coupled first-order differential equations.

(2)

(b)

Use Euler's method with a step length of 0.1 seconds to estimate the roll angle when t = 0.2.

(4)

(c)

Find the eigenvalues of the matrix of the system in (a).

(3)

(d)

Hence find the period of the oscillation, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 dx/dt = y	✓ 1
A1 dy/dt = −4x − 0.5y	✓ 1
Part (b)
M1 x_n+1 = x_n + 0.1y_n and y_n+1 = y_n + 0.1(−4x_n − 0.5y_n)	✓ 1
A1 first step: x = 0.2, y = −0.08	✓ 1
M1 second step using these values	✓ 1
A1 x = 0.192 radians	✓ 1
Part (c)
M1 det(A − λI) = 0 for the matrix with rows (0, 1) and (−4, −0.5)	✓ 1
A1 λ² + 0.5λ + 4 = 0	✓ 1
A1 λ = −0.25 ± 1.98i	✓ 1
Part (d)
R1 the imaginary part of the eigenvalues, √3.9375 = 1.98, gives the angular frequency	✓ 1
A1 period 2π/√3.9375 = 3.17 seconds	✓ 1

Exam tip: Complex eigenvalues p ± qi describe a decaying oscillation when p is negative; the period depends only on the imaginary part q.

+ Add to test✓ In your test443IBMI3H-5-05r24-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard9 marks

A solid glass ornament is made by rotating the region bounded by the curve y = xe^−x/3, the x-axis and the line x = 6 through 360° about the x-axis. Lengths are in centimetres.

(a)

Write down an integral that gives the volume of the ornament.

(2)

(b)

Find the volume of the ornament. Give your answer to 3 significant figures.

(2)

(c)

Find the greatest radius of the ornament, in cm, to 3 significant figures.

(3)

(d)

Glass costs £0.35 per cm³. Using the unrounded volume from (b), find the cost of the glass for one ornament, to the nearest penny.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume = π∫ y² dx	✓ 1
A1 π∫₀⁶ x²e^−2x/3 dx	✓ 1
Part (b)
M1 evaluate the integral using technology	✓ 1
A1 16.2 cm³	✓ 1
Part (c)
M1 dy/dx = e^−x/3(1 − x/3) = 0	✓ 1
A1 x = 3	✓ 1
A1 y = 3e⁻¹ = 1.10 cm	✓ 1
Part (d)
M1 0.35 × 16.1566… (unrounded volume)	✓ 1
A1 £5.65	✓ 1

Common error: Square the whole function when finding a volume of revolution, including the exponential factor: y² changes e⁻x/³ into e⁻²x/³.

+ Add to test✓ In your test444IBMI3H-5-05r24-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard12 marks

Water flows between a main pool and a paddling pool, and a chemical is gradually lost from both. Let x and y be the amounts of the chemical, in grams, in the main pool and the paddling pool after t hours. The amounts satisfy
dx/dt = −0.5x + 0.2y
dy/dt = 0.3x − 0.4y
Initially x = 10 and y = 0.

(a)

Write this system in the form X′ = MX, stating the matrix M.

(1)

(b)

Show that the eigenvalues of M are −0.2 and −0.7.

(2)

(c)

Find an eigenvector for each eigenvalue.

(2)

(d)

Find the solution for x and y in terms of t that satisfies the initial conditions.

(4)

(e)

Find the time at which the amount in the paddling pool is greatest, and that greatest amount. Give both answers to 3 significant figures, using the unrounded time to find the amount.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 M = [[−0.5, 0.2], [0.3, −0.4]]	✓ 1
Part (b)
M1 det(M − λI) = 0, giving λ² + 0.9λ + 0.14 = 0	✓ 1
A1 λ = −0.2 and λ = −0.7	✓ 1
Part (c)
A1 λ = −0.2: (2, 3)^T, or any multiple	✓ 1
A1 λ = −0.7: (1, −1)^T, or any multiple	✓ 1
Part (d)
M1 general solution X = c₁(2, 3)^Te^−0.2t + c₂(1, −1)^Te^−0.7t	✓ 1
M1 apply (10, 0) at t = 0 to get 2c₁ + c₂ = 10 and 3c₁ − c₂ = 0	✓ 1
A1 c₁ = 2, c₂ = 6	✓ 1
A1 x = 4e^−0.2t + 6e^−0.7t, y = 6e^−0.2t − 6e^−0.7t	✓ 1
Part (e)
M1 set dy/dt = 0, so −1.2e^−0.2t + 4.2e^−0.7t = 0	✓ 1
A1 t = 2ln3.5 = 2.51 hours	✓ 1
A1 y = 2.60 g	✓ 1

Common error: Each eigenvector only fixes a direction, so any non-zero multiple is acceptable; the constants c₁ and c₂ absorb the scaling.

+ Add to test✓ In your test445IBMI3H-5-05r24-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard9 marks

A buoy on a lane rope in a swimming pool bobs vertically. Its displacement x metres from the rest position at time t seconds satisfies d²x/dt² = −4x − 0.8dx/dt. At t = 0, x = 0.5 and dx/dt = 0.

(a)

By letting v = dx/dt, write the equation as a pair of first-order differential equations.

(2)

(b)

Use Euler's method with a step length of 0.1 to find estimates of x and v at t = 0.1, 0.2 and 0.3. Give the value of x at t = 0.3 to 3 significant figures.

(4)

(c)

State whether the buoy is moving towards or away from its rest position at t = 0.3, giving a reason.

(1)

(d)

Explain the effect of the term −0.8dx/dt on the motion of the buoy.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 introduce v = dx/dt	✓ 1
A1 dv/dt = −4x − 0.8v	✓ 1
Part (b)
M1 x_n+1 = x_n + 0.1v_n and v_n+1 = v_n + 0.1(−4x_n − 0.8v_n)	✓ 1
A1 t = 0.1: x = 0.5, v = −0.2	✓ 1
A1 t = 0.2: x = 0.48, v = −0.384	✓ 1
A1 t = 0.3: x = 0.4416 so 0.442 m, and v = −0.54528	✓ 1
Part (c)
R1 towards the rest position, since v < 0 while x > 0	✓ 1
Part (d)
A1 it is a resistive (damping) force that opposes the velocity	✓ 1
R1 so the amplitude of the oscillations decreases with time	✓ 1

Exam tip: A second-order equation becomes two first-order equations by naming the first derivative; update both variables at each step using the previous row only.

+ Add to test✓ In your test446IBMI3H-5-05r25-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard12 marks

Two linked storage tanks aboard a ship, A and B, contain a cleaning chemical. The masses of chemical, x kg in tank A and y kg in tank B, after t hours satisfy $\frac{dx}{dt}=-3x+y,\quad \frac{dy}{dt}=2x-2y.$ Initially x = 30 and y = 0.

(a)

Write the system in the form $\frac{d}{dt}\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{M}\begin{pmatrix}x\\y\end{pmatrix}$.

(1)

(b)

Find the eigenvalues of M.

(3)

(c)

Find an eigenvector for each eigenvalue.

(2)

(d)

Find the solution for x and y in terms of t.

(3)

(e)

Find the time at which the mass in tank B is greatest, and this greatest mass. Give each answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\mathbf{M}=\begin{pmatrix}-3&1\\2&-2\end{pmatrix}$	✓ 1
Part (b)
M1 $\det(\mathbf{M}-\lambda\mathbf{I})=0$	✓ 1
A1 $\lambda^2+5\lambda+4=0$	✓ 1
A1 $\lambda=-1$ and $\lambda=-4$	✓ 1
Part (c)
A1 $\begin{pmatrix}1\\2\end{pmatrix}$ for $\lambda=-1$ (any multiple)	✓ 1
A1 $\begin{pmatrix}1\\-1\end{pmatrix}$ for $\lambda=-4$ (any multiple)	✓ 1
Part (d)
M1 general solution as a combination $Ae^{-t}\begin{pmatrix}1\\2\end{pmatrix}+Be^{-4t}\begin{pmatrix}1\\-1\end{pmatrix}$ (ECF)	✓ 1
M1 applies $x(0)=30$, $y(0)=0$ to get $A+B=30$ and $2A-B=0$	✓ 1
A1 $x=10e^{-t}+20e^{-4t}$, $y=20e^{-t}-20e^{-4t}$	✓ 1
Part (e)
M1 sets $\frac{dy}{dt}=0$, so $-20e^{-t}+80e^{-4t}=0$	✓ 1
A1 t = 0.462 hours ($\frac{\ln4}{3}$)	✓ 1
A1 greatest mass 9.45 kg	✓ 1

Exam tip: Check your eigenvectors by substituting them back into $\mathbf{M}\mathbf{v}=\lambda\mathbf{v}$; a slip here carries into every later part.

+ Add to test✓ In your test447IBMI3H-5-04r26-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard9 marks

The vertical displacement x metres of a wave-energy buoy from its rest position at time t seconds satisfies the second-order equation
d²x/dt² = −4x − 0.5 dx/dt.
At t = 0, x = 1 and the buoy is momentarily at rest.

(a)

By letting y = dx/dt, write the equation as a pair of coupled first-order differential equations and state the initial conditions.

(2)

(b)

Use Euler's method with step length h = 0.1 to find estimates of x and y at t = 0.1 and at t = 0.2.

(4)

(c)

Hence estimate the displacement and the velocity of the buoy at t = 0.3.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 dx/dt = y and dy/dt = −4x − 0.5y	✓ 1
A1 x(0) = 1, y(0) = 0	✓ 1
Part (b)
M1 x_n+1 = x_n + 0.1y_n and y_n+1 = y_n + 0.1(−4x_n − 0.5y_n)	✓ 1
A1 t = 0.1: x = 1, y = −0.4	✓ 1
A1 t = 0.2: x = 0.96	✓ 1
A1 t = 0.2: y = −0.78	✓ 1
Part (c)
M1 apply the iteration once more using x = 0.96, y = −0.78	✓ 1
A1 velocity y = −0.78 + 0.1(−3.84 + 0.39) = −1.125 m s⁻¹	✓ 1
A1 displacement x = 0.96 + 0.1(−0.78) = 0.882 m	✓ 1

Exam tip: Write a small table with columns t, x, y and fill it row by row; each row uses only the previous row.

+ Add to test✓ In your test448IBMI3H-5-04r26-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

A closed cylindrical hot-water tank for a solar heating system must hold 0.8 m³. The tank has radius r metres. The top and base cost £30 per m² and the curved side costs £12 per m² to make.

(a)

Show that the cost £C of making the tank is C = 60πr² + 19.2/r.

(4)

(b)

Find the value of r that minimises the cost, to 3 significant figures, and justify that it is a minimum.

(4)

(c)

Find the minimum cost, to the nearest pound. Use the unrounded value of r.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height h = 0.8/(πr²)	✓ 1
A1 curved area 2πrh = 1.6/r	✓ 1
M1 cost = 30 × 2πr² + 12 × 1.6/r	✓ 1
A1 C = 60πr² + 19.2/r (AG)	✓ 1
Part (b)
A1 dC/dr = 120πr − 19.2/r²	✓ 1
M1 set dC/dr = 0, so r³ = 19.2/(120π)	✓ 1
A1 r = 0.371 m	✓ 1
R1 d²C/dr² = 120π + 38.4/r³ > 0, so a minimum	✓ 1
Part (c)
M1 substitute r = 0.37067… into the cost formula	✓ 1
A1 £78 (77.70…)	✓ 1

Exam tip: Keep the unrounded optimal value in your calculator memory when you substitute back to find the minimum cost.

+ Add to test✓ In your test449IBMI3H-5-04r26-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard12 marks

A drug is given to a patient. Let x mg be the amount in the blood and y mg the amount in body tissue at t hours. The amounts satisfy
dx/dt = −0.6x + 0.2y, dy/dt = 0.4x − 0.4y.
Initially x = 6 and y = 3.

(a)

Find the eigenvalues of the matrix of coefficients.

(3)

(b)

Find an eigenvector for each eigenvalue.

(2)

(c)

Find the particular solution for x and y in terms of t.

(3)

(d)

Describe the long-term behaviour of the amounts of drug and classify the equilibrium point at the origin.

(2)

(e)

Find the amount of drug in body tissue after 5 hours, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 form the characteristic equation det(A − λI) = 0 with A = (−0.6, 0.2; 0.4, −0.4)	✓ 1
A1 λ² + λ + 0.16 = 0	✓ 1
A1 λ = −0.2 and λ = −0.8	✓ 1
Part (b)
A1 λ = −0.2: eigenvector (1, 2) or a multiple	✓ 1
A1 λ = −0.8: eigenvector (1, −1) or a multiple	✓ 1
Part (c)
M1 general solution (x, y) = A(1, 2)e^−0.2t + B(1, −1)e^−0.8t	✓ 1
M1 use A + B = 6 and 2A − B = 3	✓ 1
A1 x = 3e^−0.2t + 3e^−0.8t, y = 6e^−0.2t − 3e^−0.8t	✓ 1
Part (d)
R1 both eigenvalues are negative, so both exponentials decay to zero	✓ 1
A1 stable node (both amounts tend to 0)	✓ 1
Part (e)
M1 y(5) = 6e⁻¹ − 3e⁻⁴	✓ 1
A1 2.15 mg	✓ 1

Exam tip: Check an eigenvector by substituting it back into (A − λI)v = 0; the initial conditions then give the constants.

+ Add to test✓ In your test450IBMI3H-5-06r28-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard11 marks

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The masses (g) of chlorine in two linked pools are $x(t)$ and $y(t)$, with $t$ in hours:
$\frac{dx}{dt}=-0.3x+0.1y,\qquad\frac{dy}{dt}=0.2x-0.2y.$
Initially $x=60$ and $y=0$.

(a)

Show that the eigenvalues of the coefficient matrix are $-0.1$ and $-0.4$.

(3)

(b)

Find an eigenvector for each eigenvalue.

(3)

(c)

Find the particular solutions for $x(t)$ and $y(t)$.

(3)

(d)

Find the mass of chlorine in the first pool at $t=5$ hours, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\det\begin{pmatrix}-0.3-\lambda&0.1\\0.2&-0.2-\lambda\end{pmatrix}=0$	✓ 1
A1 $\lambda^2+0.5\lambda+0.04=0$	✓ 1
A1 $\lambda=-0.1,\ -0.4$ (AG)	✓ 1
Part (b)
M1 solve $(\mathbf{A}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}$ for one eigenvalue	✓ 1
A1 $\lambda=-0.1$: $\begin{pmatrix}1\\2\end{pmatrix}$	✓ 1
A1 $\lambda=-0.4$: $\begin{pmatrix}1\\-1\end{pmatrix}$ (any non-zero multiples)	✓ 1
Part (c)
M1 general solution $a\binom{1}{2}e^{-0.1t}+b\binom{1}{-1}e^{-0.4t}$	✓ 1
M1 use $x(0)=60,\ y(0)=0$ to get $a+b=60$ and $2a-b=0$	✓ 1
A1 $x=20e^{-0.1t}+40e^{-0.4t}$ and $y=40e^{-0.1t}-40e^{-0.4t}$	✓ 1
Part (d)
M1 substitute $t=5$ into $x(t)$	✓ 1
A1 17.5 g	✓ 1

Exam tip: Check the particular solution by substituting $t=0$ and by checking one of the original equations at $t=0$.

+ Add to test✓ In your test451IBMI3H-5-06r28-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

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An infection spreads among 910 members of a swimming club. The numbers susceptible $S$, infected $I$ and recovered $R$ satisfy
$\frac{dS}{dt}=-0.0005SI,\quad\frac{dI}{dt}=0.0005SI-0.2I,\quad\frac{dR}{dt}=0.2I,$
with $t$ in days. Initially $S=900$, $I=10$ and $R=0$.

(a)

Show that $S+I+R$ is constant.

(2)

(b)

Use Euler's method with $h=1$ to estimate $S$, $I$ and $R$ at $t=1$.

(3)

(c)

Continue to estimate $S$, $I$ and $R$ at $t=2$. Give $S$ and $I$ to 1 decimal place.

(3)

(d)

Show that the number infected increases initially only if $S$ is greater than a threshold value, and find that threshold.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 add the three derivatives	✓ 1
R1 the sum is 0, so $S+I+R=910$ for all $t$	✓ 1
Part (b)
M1 apply Euler's method to all three equations	✓ 1
A1 $S=895.5$ and $I=12.5$	✓ 1
A1 $R=2$	✓ 1
Part (c)
M1 second step using the values at $t=1$	✓ 1
A1 $S\approx889.9$ and $I\approx15.6$	✓ 1
A1 $R=4.5$	✓ 1
Part (d)
M1 $\frac{dI}{dt}>0$ requires $0.0005S>0.2$	✓ 1
A1 $S>400$	✓ 1

Common error: Use values from the previous step for every variable in the same step; mixing updated and old values changes the answers.

+ Add to test✓ In your test452IBMI3H-5-07r28-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

A hiking path rises along a ridge. At a horizontal distance x km from the trailhead (0 ≤ x ≤ 10) its height is h(x) = 800 + 120 ln(1 + x²) metres.

(a)

Show that h′(x) = 240x / (1 + x²).

(2)

(b)

The path is steepest where h′(x) is greatest. Find the value of x at which this happens, justifying that it is a maximum of h′(x).

(5)

(c)

Find the steepest gradient in metres per kilometre, and the angle the path makes with the horizontal there. Give the angle to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule: derivative of ln(1 + x²) is 2x/(1 + x²)	✓ 1
A1 multiply by 120 to obtain the printed result	✓ 1
Part (b)
M1 quotient rule on 240x/(1 + x²)	✓ 1
A1 h″(x) = 240(1 − x²) / (1 + x²)²	✓ 1
M1 set h″(x) = 0	✓ 1
A1 x = 1 (as x ≥ 0)	✓ 1
R1 h″ > 0 for x < 1 and h″ < 0 for x > 1, so h′ is greatest at x = 1	✓ 1
Part (c)
A1 gradient at x = 1 is 120 m per km	✓ 1
M1 angle = tan⁻¹(120/1000) (same units for height and distance)	✓ 1
A1 6.84°	✓ 1

Common error: The gradient 120 m per km is 0.12, not 120; convert both lengths to the same unit before finding an angle.

+ Add to test✓ In your test453IBMI3H-5-08r28-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard9 marks

In a marine reserve, the populations (in hundreds) of prey fish x and predator fish y after t years satisfy
dx/dt = 0.6x − 0.02xy, dy/dt = 0.01xy − 0.3y,
with x = 40 and y = 20 when t = 0.

(a)

Find the equilibrium point with both populations non-zero.

(3)

(b)

Use Euler’s method with step length 0.5 years to estimate x and y when t = 1. Give x as the final answer to 2 decimal places.

(4)

(c)

Using your estimates, state with a reason whether the predator population is increasing or decreasing at t = 1.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set both derivatives to zero: x(0.6 − 0.02y) = 0 and y(0.01x − 0.3) = 0	✓ 1
A1 y = 30	✓ 1
A1 x = 30	✓ 1
Part (b)
M1 first step: dx/dt = 8 and dy/dt = 2	✓ 1
A1 x(0.5) ≈ 44, y(0.5) ≈ 21	✓ 1
M1 second step: dx/dt = 7.92 and dy/dt = 2.94	✓ 1
A1 x at t = 1 ≈ 47.96, y at t = 1 ≈ 22.47	✓ 1
Part (c)
M1 dy/dt = y(0.01x − 0.3) with x ≈ 47.96 gives a positive factor 0.1796	✓ 1
A1 increasing, since y > 0 and the factor is positive (ECF from (b))	✓ 1

Exam tip: At each Euler step compute both derivatives from the same pair of old values before updating either variable.

+ Add to test✓ In your test454IBMI3H-5-08r28-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard11 marks

The cross-section of a curved canopy roof is the curve y = 6xe^−0.5x metres for x ≥ 0, where x is the horizontal distance from the building wall.

(a)

Find y′.

(2)

(b)

Find the coordinates of the highest point of the roof. Give the y-coordinate to 3 significant figures.

(3)

(c)

The roof has a point of inflection for x > 0, where it descends most steeply. Find the gradient there. Give your answer to 3 significant figures.

(6)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule with u = 6x and v = e^−0.5x	✓ 1
A1 y′ = 6e^−0.5x(1 − 0.5x)	✓ 1
Part (b)
M1 set y′ = 0	✓ 1
A1 x = 2	✓ 1
A1 y = 12e⁻¹ = 4.41	✓ 1
Part (c)
M1 differentiate y′ using the product rule	✓ 1
A1 y″ = 6e^−0.5x(0.25x − 1)	✓ 1
M1 set y″ = 0	✓ 1
A1 x = 4	✓ 1
M1 substitute x = 4 into y′	✓ 1
A1 gradient = −6e⁻² = −0.812	✓ 1

Common error: A point of inflection needs y″ = 0 (with a sign change), not y′ = 0, which gives a maximum or minimum.

+ Add to test✓ In your test455IBMI3H-5-04r29-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard11 marks

Calculator

A rectangular fish pen is built against a straight quay, so no netting is needed along the quay. The pen has two end nets and two internal dividing nets, all perpendicular to the quay and each of length x metres, and one outer net parallel to the quay of length y metres. The two end nets cost €12 per metre and all other netting costs €8 per metre. The total netting budget is €4000, which is all spent.

(a)

Show that y = 500 − 5x.

(3)

(b)

Hence show that the area A m² of the pen is A = 500x − 5x².

(1)

(c)

Find the value of x that gives the greatest area of the pen, justifying that it is a maximum.

(3)

(d)

Find the greatest area of the pen.

(1)

(e)

The farm needs the pen to have an area of at least 12 000 m². Find the range of values of x that satisfy this.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cost of end nets = 2 × 12x = 24x	✓ 1
M1 cost of dividers and outer net = 2 × 8x + 8y = 16x + 8y	✓ 1
A1 40x + 8y = 4000, so y = 500 − 5x (shown)	✓ 1
Part (b)
A1 A = xy = x(500 − 5x) = 500x − 5x² (shown)	✓ 1
Part (c)
M1 dA/dx = 500 − 10x	✓ 1
A1 sets equal to zero, so x = 50	✓ 1
R1 d²A/dx² = −10 < 0, so this is a maximum	✓ 1
Part (d)
A1 12 500 m²	✓ 1
Part (e)
M1 500x − 5x² ≥ 12 000, so x² − 100x + 2400 ≤ 0	✓ 1
A1 critical values x = 40 and x = 60	✓ 1
A1 40 ≤ x ≤ 60	✓ 1

Exam tip: In optimisation problems, use the constraint to write the quantity you want to maximise in terms of one variable. Then differentiate, set to zero, and justify the type of stationary point.

+ Add to test✓ In your test456IBMI3H-5-04r31-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard12 marks

The bell of a brass horn is modelled by rotating the region under the curve r = 0.03e^0.8x, for 0 ≤ x ≤ 2, through 360° about the x-axis. Here x is the distance along the horn in metres and r is the inner radius in metres.

(a)

Find the inner radius of the horn at each end. Give the larger radius to 3 significant figures.

(2)

(b)

Show that the volume of air enclosed by the horn is (0.0009π/1.6)(e^3.2 − 1) m³, and evaluate this to 3 significant figures.

(5)

(c)

The brass wall has constant thickness 0.0005 m, so the outer radius is r + 0.0005. Find the volume of brass in the wall. Give your answer in the form a × 10^k with 3 significant figures.

(3)

(d)

The density of brass is 8500 kg m⁻³. Find the mass of brass in the bell. Use the unrounded volume from (c) and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 r(0) = 0.03 m	✓ 1
A1 r(2) = 0.149 m	✓ 1
Part (b)
M1 V = π∫₀² (0.03e^0.8x)² dx	✓ 1
A1 integrand 0.0009e^1.6x	✓ 1
M1 integrating to (0.0009/1.6)e^1.6x	✓ 1
A1 evaluating between 0 and 2 to give (0.0009π/1.6)(e^3.2 − 1) (AG)	✓ 1
A1 0.0416 m³	✓ 1
Part (c)
M1 volume of brass = π∫₀² [(r + 0.0005)² − r²] dx	✓ 1
A1 correct evaluation of the integral (e.g. using a calculator)	✓ 1
A1 4.67 × 10⁻⁴ m³	✓ 1
Part (d)
M1 mass = density × volume = 8500 × 4.67 × 10⁻⁴ (unrounded volume)	✓ 1
A1 3.97 kg	✓ 1

Exam tip: For a hollow solid, subtract the volume of revolution of the inner curve from that of the outer curve; do not subtract the radii before squaring.

+ Add to test✓ In your test457IBMI3H-5-04r31-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard14 marks

The concentrations of a nutrient, in mg per litre, in the topsoil (x) and in a deeper soil layer (y) of a field are modelled, for time t in days, by the coupled system
dx/dt = −3x + y
dy/dt = 2x − 2y
with x = 4 and y = 2 when t = 0.

(a)

Use Euler's method with step size 0.1 to estimate x and y when t = 0.1.

(3)

(b)

Show that the eigenvalues of the matrix A = ((−3, 1), (2, −2)) are −1 and −4.

(3)

(c)

Find the equilibrium point of the system.

(1)

(d)

Hence state, with a reason, the long-term behaviour of the concentrations.

(1)

(e)

Find an eigenvector for each eigenvalue.

(2)

(f)

Hence find the particular solution for x and y in terms of t.

(3)

(g)

Find the concentration in the topsoil after 1 day. Give your answer to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x₁ = 4 + 0.1(−3 × 4 + 2), y₁ = 2 + 0.1(2 × 4 − 2 × 2)	✓ 1
A1 x = 3	✓ 1
A1 y = 2.4	✓ 1
Part (b)
M1 det(A − λI) = 0, i.e. (−3 − λ)(−2 − λ) − 2 = 0	✓ 1
A1 λ² + 5λ + 4 = 0	✓ 1
A1 (λ + 1)(λ + 4) = 0, so λ = −1 and λ = −4 (AG)	✓ 1
Part (c)
A1 (0, 0), since the matrix is non-singular (det A = 4 ≠ 0)	✓ 1
Part (d)
R1 both eigenvalues are negative, so both concentrations tend to 0 (a stable equilibrium)	✓ 1
Part (e)
A1 λ = −1: (1, 2) (or any multiple)	✓ 1
A1 λ = −4: (1, −1) (or any multiple)	✓ 1
Part (f)
M1 general solution (x, y) = Ae^−t(1, 2) + Be^−4t(1, −1)	✓ 1
M1 using x(0) = 4, y(0) = 2 to get A + B = 4 and 2A − B = 2	✓ 1
A1 x = 2e^−t + 2e^−4t, y = 4e^−t − 2e^−4t	✓ 1
Part (g)
A1 0.772 mg per litre	✓ 1

Exam tip: A quick check on an eigenvector: multiply it by the matrix and confirm that you get the eigenvalue times the same vector.

+ Add to test✓ In your test458IBMI3H-5-06r32-09Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard11 marks

A drug is taken by mouth. Let $x$ mg be the amount in the gut and $y$ mg the amount in the blood at time $t$ hours. They satisfy $\dfrac{dx}{dt}=-0.5x$ and $\dfrac{dy}{dt}=0.5x-0.2y$, which can be written $\dfrac{d}{dt}\begin{pmatrix}x\\y\end{pmatrix}=A\begin{pmatrix}x\\y\end{pmatrix}$ with $A=\begin{pmatrix}-0.5&0\\0.5&-0.2\end{pmatrix}$. Initially $x=100$ and $y=0$.

(a)

Find the eigenvalues of $A$.

(2)

(b)

Find an eigenvector for each eigenvalue.

(3)

(c)

Hence find $y$ as a function of $t$.

(3)

(d)

Find the greatest amount of drug in the blood, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\det(A-\lambda I)=(-0.5-\lambda)(-0.2-\lambda)=0$	✓ 1
A1 $\lambda=-0.5$ and $\lambda=-0.2$	✓ 1
Part (b)
M1 solves $(A+0.5I)\mathbf{v}=\mathbf{0}$, e.g. $0.5a+0.3b=0$	✓ 1
A1 $\begin{pmatrix}3\\-5\end{pmatrix}$ for $\lambda=-0.5$	✓ 1
A1 $\begin{pmatrix}0\\1\end{pmatrix}$ for $\lambda=-0.2$ (any non-zero multiples)	✓ 1
Part (c)
M1 general solution $c_1\begin{pmatrix}3\\-5\end{pmatrix}e^{-0.5t}+c_2\begin{pmatrix}0\\1\end{pmatrix}e^{-0.2t}$ (ECF from (b))	✓ 1
A1 $c_1=\frac{100}{3}$ and $c_2=\frac{500}{3}$	✓ 1
A1 $y=\frac{500}{3}\left(e^{-0.2t}-e^{-0.5t}\right)$	✓ 1
Part (d)
M1 sets $\frac{dy}{dt}=0$, giving $e^{0.3t}=2.5$	✓ 1
A1 $t=\frac{\ln2.5}{0.3}=3.05$ hours	✓ 1
A1 $y=54.3$ mg	✓ 1

Exam tip: Check the particular solution against both initial conditions: here $x(0)=100$ and $y(0)=-5c_1+c_2=0$.

+ Add to test✓ In your test459IBMI3H-5-06r32-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

The difficulty rating of level $x$ of a game is modelled by $D(x)=\dfrac{\ln x}{x}$, for $x>0$.

(a)

Show that $D'(x)=\dfrac{1-\ln x}{x^2}$.

(3)

(b)

Find the coordinates of the stationary point and determine its nature.

(3)

(c)

Find the equation of the tangent to the curve at $x=1$.

(2)

(d)

Use the substitution $u=\ln x$ to find $\displaystyle\int_1^{e}\frac{\ln x}{x}\,dx$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule (or product rule on $x^{-1}\ln x$)	✓ 1
A1 numerator $\frac1x\times x-\ln x\times1$	✓ 1
A1 simplifies to $\frac{1-\ln x}{x^2}$	✓ 1
Part (b)
M1 sets $D'(x)=0$ so $\ln x=1$	✓ 1
A1 $x=e$, $D=\frac1e$	✓ 1
R1 $D'$ changes from positive to negative (or $D''(e)<0$), so a maximum	✓ 1
Part (c)
M1 $D$ equals $0$ and $D'$ equals $1$ at $x=1$	✓ 1
A1 $y=x-1$	✓ 1
Part (d)
M1 $du=\frac{dx}{x}$, limits $0$ to $1$, giving $\int_0^1u\,du$	✓ 1
A1 $\frac12$	✓ 1

Common error: In $\int \frac{\ln x}{x}dx$ the factor $\frac{1}{x}$ is exactly $du$; do not integrate $\ln x$ and $\frac1x$ separately.

+ Add to test✓ In your test460IBMI3H-5-07r32-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

A swimming club makes a foam float by rotating the region bounded by the curve $y=xe^{-x/2}$, the $x$-axis and the line $x=4$ through $360^\circ$ about the $x$-axis. Lengths $x$ and $y$ are in decimetres.

(a)

Show that the volume of the float is $\pi\int_0^4 x^2e^{-x}\,dx$.

(2)

(b)

Calculate the volume of the float in litres (1 dm³ = 1 litre), to 3 significant figures.

(2)

(c)

A second float uses the same curve from $x=0$ to $x=L$ and has volume exactly 3 litres. Find $L$ to 3 significant figures.

(2)

(d)

Use the product rule to find $\frac{dy}{dx}$, and hence find the greatest radius of the float to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use $V=\pi\int y^2\,dx$	✓ 1
A1 $y^2=x^2e^{-x}$ with limits 0 and 4	✓ 1
Part (b)
M1 evaluate $\pi\int_0^4 x^2e^{-x}\,dx$ (calculator)	✓ 1
A1 $V=4.79$ litres	✓ 1
Part (c)
M1 equate $\pi\int_0^L x^2e^{-x}\,dx=3$ and solve using a calculator	✓ 1
A1 $L=2.58$	✓ 1
Part (d)
M1 product rule on $xe^{-x/2}$	✓ 1
A1 $\frac{dy}{dx}=e^{-x/2}\left(1-\frac{x}{2}\right)$	✓ 1
M1 set $\frac{dy}{dx}=0$ giving $x=2$	✓ 1
A1 greatest radius $2e^{-1}=0.736$ dm	✓ 1

Common error: The volume integral needs $y^2$, not $y$: square the whole function, including the exponential, before integrating.

+ Add to test✓ In your test461IBMI3H-5-07r32-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard13 marks

Two library branches hold book stocks that differ from target levels by $x$ and $y$ thousand books respectively at time $t$ months, where $\frac{dx}{dt}=-3x+2y$ and $\frac{dy}{dt}=x-2y$. At $t=0$, $x=5$ and $y=2$.

(a)

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

(3)

(b)

Find an eigenvector for each eigenvalue.

(3)

(c)

Find the solution for $x$ and $y$ in terms of $t$.

(4)

(d)

Classify the equilibrium point at $(0,0)$ and describe the long-term behaviour of the stock differences.

(2)

(e)

Find $x$ when $t=1$, to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\det(A-\lambda I)=0$	✓ 1
A1 $\lambda^2+5\lambda+4=0$	✓ 1
A1 $\lambda=-1$ and $\lambda=-4$	✓ 1
Part (b)
M1 solve $(A-\lambda I)\mathbf v=\mathbf 0$	✓ 1
A1 $\lambda=-1$: $\begin{pmatrix}1\\1\end{pmatrix}$	✓ 1
A1 $\lambda=-4$: $\begin{pmatrix}2\\-1\end{pmatrix}$ (any non-zero multiples)	✓ 1
Part (c)
M1 general solution $A e^{-t}\begin{pmatrix}1\\1\end{pmatrix}+Be^{-4t}\begin{pmatrix}2\\-1\end{pmatrix}$ (ECF from (b))	✓ 1
M1 use $x(0)=5$, $y(0)=2$: $A+2B=5$, $A-B=2$	✓ 1
A1 $A=3$, $B=1$	✓ 1
A1 $x=3e^{-t}+2e^{-4t}$, $y=3e^{-t}-e^{-4t}$	✓ 1
Part (d)
A1 both eigenvalues are real and negative, so a stable node	✓ 1
R1 both differences tend to zero, approaching along the direction $(1,1)$ because $e^{-t}$ decays more slowly than $e^{-4t}$	✓ 1
Part (e)
A1 $x=3e^{-1}+2e^{-4}=1.14$ thousand books	✓ 1

Common error: Eigenvectors are only defined up to a multiple; check yours by multiplying the matrix by the vector and confirming you recover $\lambda$ times the vector.

+ Add to test✓ In your test462IBMI3H-5-07r32-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard13 marks

The number of e-books $y$ (hundreds) in a library's catalogue $x$ years after a review satisfies $\frac{dy}{dx}=ay-2$, where $a>0$ is a constant (growth of the catalogue, less 200 withdrawals a year). Initially $y=3$.

(a)

Use Euler's method with step length $h=0.5$ to show that the estimate of $y$ at $x=1$ is $0.75a^2+2.5a+1$.

(4)

(b)

The Euler estimate of $y$ at $x=1$ is $3.48$. Find the value of $a$.

(3)

(c)

Solve the differential equation with $a=0.8$ to find $y$ in terms of $x$.

(4)

(d)

Find the exact-solution value of $y$ when $x=1$, to 3 significant figures.

(1)

(e)

Compare with the Euler estimate and explain the difference.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $y_1=3+0.5(3a-2)$	✓ 1
A1 $y_1=2+1.5a$	✓ 1
M1 $y_2=y_1+0.5(ay_1-2)=y_1(1+0.5a)-1$	✓ 1
A1 $(2+1.5a)(1+0.5a)-1=0.75a^2+2.5a+1$	✓ 1
Part (b)
M1 $0.75a^2+2.5a+1=3.48$	✓ 1
A1 solutions $a=0.8$ and $a\approx-4.13$	✓ 1
R1 reject the negative root since $a>0$	✓ 1
Part (c)
M1 separate variables: $\int\frac{1}{0.8y-2}\,dy=\int dx$	✓ 1
A1 $1.25\ln\|0.8y-2\|=x+c$	✓ 1
M1 use $y=3$ when $x=0$: $0.8y-2=0.4e^{0.8x}$	✓ 1
A1 $y=2.5+0.5e^{0.8x}$	✓ 1
Part (d)
A1 $y=2.5+0.5e^{0.8}=3.61$	✓ 1
Part (e)
R1 Euler underestimates (3.48 < 3.61) because the solution curve is concave up, so each tangent step lies below the curve	✓ 1

Common error: With a quadratic from a modelling context, always check each root against stated conditions such as $a>0$ before giving a final value.

+ Add to test✓ In your test463IBMI3H-5-08r32-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

The populations of fish $x$ and sharks $y$ (both in hundreds) in a bay at time $t$ years are modelled by $\frac{dx}{dt}=0.4x-0.01xy$ and $\frac{dy}{dt}=0.005xy-0.3y$. At $t=0$, $x=100$ and $y=20$.

(a)

Find the non-zero equilibrium populations.

(3)

(b)

Use Euler's method with step length $h=0.5$ to estimate $x$ and $y$ when $t=1$.

(4)

(c)

Calculate $\frac{dy}{dt}$ at the Euler estimate for $t=1$, to 3 significant figures, and state whether the shark population is increasing or decreasing.

(2)

(d)

Suggest one reason why Euler's method with $h=0.5$ may become unreliable for predicting these populations over many years.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set both rates of change to zero and factorise: $x(0.4-0.01y)=0$ and $y(0.005x-0.3)=0$	✓ 1
A1 $y=40$	✓ 1
A1 $x=60$	✓ 1
Part (b)
M1 $x_{n+1}=x_n+h\frac{dx}{dt}$ and $y_{n+1}=y_n+h\frac{dy}{dt}$ from $(100,20)$	✓ 1
A1 $t=0.5$: $x=110$, $y=22$	✓ 1
M1 second step from $(110,22)$	✓ 1
A1 $t=1$: $x=119.9$, $y=24.75$	✓ 1
Part (c)
M1 $0.005\times119.9\times24.75-0.3\times24.75$ (ECF from (b))	✓ 1
A1 $7.41$ per year, so the shark population is increasing	✓ 1
Part (d)
R1 any valid reason, e.g. errors accumulate at each step because the true populations oscillate and the step is large relative to the cycle	✓ 1

Exam tip: An equilibrium of a coupled system needs both rates of change to be zero at the same time; solve each equation for one variable, then pair the results.

+ Add to test✓ In your test464IBMI3H-5-08r32-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard13 marks

The deviations $x$ and $y$ (in hundreds) of two fish populations from their equilibrium levels satisfy $\frac{dx}{dt}=x+2y$ and $\frac{dy}{dt}=3x$, where $t$ is in years.

(a)

Find the eigenvalues of the matrix $\begin{pmatrix}1&2\\3&0\end{pmatrix}$.

(3)

(b)

Find an eigenvector for each eigenvalue.

(3)

(c)

Write down the general solution and classify the equilibrium point at $(0,0)$.

(3)

(d)

Initially $x=4$. Find the value of $y$ at $t=0$ for which both deviations tend to zero.

(3)

(e)

For these initial values, find $x$ when $t=1$, to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\det(A-\lambda I)=0$	✓ 1
A1 $\lambda^2-\lambda-6=0$	✓ 1
A1 $\lambda=3$ and $\lambda=-2$	✓ 1
Part (b)
M1 solve $(A-\lambda I)\mathbf v=\mathbf 0$	✓ 1
A1 $\lambda=3$: $\begin{pmatrix}1\\1\end{pmatrix}$	✓ 1
A1 $\lambda=-2$: $\begin{pmatrix}2\\-3\end{pmatrix}$ (any non-zero multiples)	✓ 1
Part (c)
A1 $\begin{pmatrix}x\\y\end{pmatrix}=Ae^{3t}\begin{pmatrix}1\\1\end{pmatrix}+Be^{-2t}\begin{pmatrix}2\\-3\end{pmatrix}$ (ECF from (b))	✓ 1
A1 eigenvalues are real with opposite signs, so a saddle point	✓ 1
R1 the equilibrium is unstable	✓ 1
Part (d)
M1 the $e^{3t}$ term must vanish, so $A=0$	✓ 1
A1 $\begin{pmatrix}4\\y\end{pmatrix}=B\begin{pmatrix}2\\-3\end{pmatrix}$ gives $B=2$	✓ 1
A1 $y=-6$	✓ 1
Part (e)
A1 $x=4e^{-2}=0.541$	✓ 1

Common error: Both eigenvalues must have the same sign for a node; opposite signs give a saddle, where only initial conditions along one eigenvector return to equilibrium.

+ Add to test✓ In your test465IBMI3H-5-06r34-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

A lifeguard is at point A on a straight beach. A swimmer in difficulty is at point S, 30 m from the shore. The point P on the shore nearest to S is 80 m from A. The lifeguard runs along the shore at 5 m s⁻¹ to a point R, then swims in a straight line to S at 1.5 m s⁻¹. Let x metres be the distance PR, where 0 ≤ x ≤ 80.

(a)

Show that the total time T seconds is T = (80 − x) ÷ 5 + √(900 + x²) ÷ 1.5.

(2)

(b)

Find dT/dx and hence find the value of x that gives a stationary point. Give your answer to 3 significant figures.

(4)

(c)

Show that this value of x gives a minimum time.

(2)

(d)

Find the minimum time, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Running distance AR = 80 − x, so running time = (80 − x) ÷ 5	✓ 1
Swimming distance RS = √(30² + x²) by Pythagoras, so swimming time = √(900 + x²) ÷ 1.5	✓ 1
Part (b)
dT/dx = −1/5 + x ÷ (1.5√(900 + x²))	✓ 1
Set to zero: x ÷ √(900 + x²) = 0.3	✓ 1
x² = 0.09(900 + x²), so 0.91x² = 81	✓ 1
x = 9.43 m	✓ 1
Part (c)
d²T/dx² = 900 ÷ [1.5(900 + x²)^3/2]	✓ 1
This is positive for all x, so the stationary point is a minimum (or compare with T(0) = 36.0 and T(80) = 57.0)	✓ 1
Part (d)
T = 70.57 ÷ 5 + √(900 + 89.0) ÷ 1.5 = 14.11 + 20.97	✓ 1
T = 35.1 s	✓ 1

Exam tip: Keep the unrounded value of x on your calculator when you substitute back to find the minimum value.

+ Add to test✓ In your test466IBMI3H-5-06r34-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard9 marks

Members of a swimming club train either in a morning group or an evening group. Let x and y be the numbers in the morning and evening groups after t weeks, modelled by
dx/dt = −0.4x + 0.2y, dy/dt = 0.4x − 0.2y,
with x = 30 and y = 15 when t = 0.

(a)

Show that the eigenvalues of the matrix $\begin{pmatrix}-0.4 & 0.2\\ 0.4 & -0.2\end{pmatrix}$ are 0 and −0.6.

(2)

(b)

Find an eigenvector for each eigenvalue.

(2)

(c)

Hence find x and y in terms of t.

(3)

(d)

Find the time after which x is within 1 member of its long-term value. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Characteristic equation det(A − λI) = 0: (−0.4 − λ)(−0.2 − λ) − 0.08 = 0	✓ 1
λ² + 0.6λ = 0, so λ = 0 or λ = −0.6	✓ 1
Part (b)
For λ = 0: (1, 2)^T (or any multiple)	✓ 1
For λ = −0.6: (1, −1)^T (or any multiple)	✓ 1
Part (c)
General solution: (x, y) = a(1, 2) + b(1, −1)e^−0.6t	✓ 1
Initial conditions give a + b = 30 and 2a − b = 15, so a = 15 and b = 15	✓ 1
x = 15 + 15e^−0.6t and y = 30 − 15e^−0.6t	✓ 1
Part (d)
15e^−0.6t < 1	✓ 1
t > ln 15 ÷ 0.6 = 4.51 weeks	✓ 1

Exam tip: Check each eigenvector by multiplying it by the matrix; the result must be the eigenvalue times the vector.

+ Add to test✓ In your test467IBMI3H-5-07r34-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

The concentration of a drug in a patient's blood is modelled by $C(t) = \frac{20t}{t^2+4}$ mg L⁻¹, where t is the time in hours since the dose, $t \ge 0$.

(a)

Use the quotient rule to show that $C'(t) = \frac{80 - 20t^2}{(t^2+4)^2}$.

(3)

(b)

Hence find the time of the greatest concentration and the value of that concentration.

(3)

(c)

The total exposure to the drug during the first 6 hours is measured by $\int_0^6 C(t)\,dt$. Use the substitution $u = t^2+4$ to find its exact value, and give it to three significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule with $u=20t$, $v=t^2+4$	✓ 1
A1 $C'(t) = \frac{20(t^2+4) - 20t\times 2t}{(t^2+4)^2}$	✓ 1
A1 numerator simplified to $80 - 20t^2$ (AG)	✓ 1
Part (b)
M1 set $80 - 20t^2 = 0$	✓ 1
A1 $t = 2$ hours	✓ 1
A1 $C(2) = 5$ mg L⁻¹	✓ 1
Part (c)
M1 $du = 2t\,dt$, so $20t\,dt = 10\,du$	✓ 1
A1 integral $10\int_{4}^{40}\frac{1}{u}\,du$ with correct limits	✓ 1
A1 $10\ln 40 - 10\ln 4 = 10\ln 10$	✓ 1
A1 23.0 mg h L⁻¹ (23.026… before rounding)	✓ 1

Exam tip: Spot when the numerator is a multiple of the derivative of the denominator: that signals an integral of the form k ln|u|.

+ Add to test✓ In your test468IBMI3H-5-07r34-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard9 marks

In a strategy game, the number of herds of deer, x, and of wolf packs, y, in a region at time t months are modelled by the coupled equations
$\frac{dx}{dt} = 0.4x - 0.02xy,\qquad \frac{dy}{dt} = 0.01xy - 0.3y.$

(a)

Find the values of x and y, both non-zero, for which both populations are constant.

(3)

(b)

When $x = 40$, find the range of values of y for which the deer population is decreasing.

(2)

(c)

Initially $x = 40$ and $y = 10$. Use Euler's method with a step length of 0.5 months to estimate the number of deer herds after 1 month, to three significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set both derivatives to zero and factorise, e.g. $x(0.4 - 0.02y)=0$ and $y(0.01x - 0.3)=0$	✓ 1
A1 $y = 20$	✓ 1
A1 $x = 30$	✓ 1
Part (b)
M1 $0.4\times 40 - 0.02\times 40\,y < 0$	✓ 1
A1 $y > 20$	✓ 1
Part (c)
M1 first step using both gradients at (40, 10)	✓ 1
A1 gradients $\frac{dx}{dt}=8$ and $\frac{dy}{dt}=1$, giving $x=44$, $y=10.5$ after 0.5 months	✓ 1
M1 second step using gradients at $(44, 10.5)$: $\frac{dx}{dt}=8.36$	✓ 1
A1 48.2 herds (48.18 before rounding)	✓ 1

Exam tip: For equilibrium of a coupled system, set each derivative to zero separately and discard the trivial solution x = y = 0 if asked for non-zero values.

+ Add to test✓ In your test469IBMI3H-5-07r35-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard13 marks

A metro train leaves a station. Its velocity, in m s⁻¹, is modelled by $v(t)=1.2te^{-0.03t}$ for $0\le t\le 120$, where $t$ is the time in seconds after departure.

(a)

Find an expression for the acceleration $a(t)$ and hence state the initial acceleration of the train.

(3)

(b)

Find the maximum speed of the train, correct to 3 significant figures, and the time at which it is reached.

(3)

(c)

Find the distance travelled in the first 60 seconds, correct to 3 significant figures.

(3)

(d)

Find the time at which the acceleration is at its minimum, and the value of this minimum acceleration, correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 $a(t)=1.2e^{-0.03t}-0.036te^{-0.03t}=1.2e^{-0.03t}(1-0.03t)$	✓ 1
A1 $a(0)=1.2$ m s⁻²	✓ 1
Part (b)
M1 set $a(t)=0$	✓ 1
A1 $t=33.3$ s	✓ 1
A1 $v=14.7$ m s⁻¹	✓ 1
Part (c)
M1 attempt to integrate $v$ over the first 60 s	✓ 1
A1 correct integral $\int_0^{60}1.2te^{-0.03t}\,dt$	✓ 1
A1 $716$ m	✓ 1
Part (d)
M1 attempt to differentiate $a(t)$	✓ 1
A1 $a'(t)=1.2e^{-0.03t}(0.0009t-0.06)$	✓ 1
M1 set $a'(t)=0$ giving $t=66.7$ s	✓ 1
A1 minimum acceleration $=-0.162$ m s⁻²	✓ 1

Exam tip: A minimum of the acceleration is found from where the derivative of the acceleration is zero, not where the acceleration itself is zero.

+ Add to test✓ In your test470IBMI3H-5-07r35-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard15 marks

A supplement is injected into the blood of an athlete. Let $x$ and $y$ be the amounts of the supplement, in mg, in the blood and in the muscle tissue respectively, $t$ hours after injection. They satisfy
$\frac{dx}{dt}=-0.5x+0.2y$
$\frac{dy}{dt}=0.3x-0.4y$
with $x=10$ and $y=0$ when $t=0$.

(a)

Use Euler's method with a step length of 0.5 hours to estimate $x$ and $y$ when $t=0.5$.

(2)

(b)

Find the eigenvalues of the matrix $\begin{pmatrix}-0.5&0.2\\0.3&-0.4\end{pmatrix}$.

(3)

(c)

An eigenvector for $\lambda=-0.2$ is $\begin{pmatrix}2\\3\end{pmatrix}$ and an eigenvector for $\lambda=-0.7$ is $\begin{pmatrix}1\\-1\end{pmatrix}$. Write down the general solution for $x$ and $y$.

(2)

(d)

Use the initial conditions to find the particular solution.

(4)

(e)

Find the time at which the amount in the muscle tissue is greatest, and the greatest amount, giving both answers correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of $x_1=x_0+0.5\frac{dx}{dt}$ and $y_1=y_0+0.5\frac{dy}{dt}$ with $x_0=10$, $y_0=0$	✓ 1
A1 $x_1=7.5$, $y_1=1.5$	✓ 1
Part (b)
M1 attempt to solve $\det(A-\lambda I)=0$	✓ 1
A1 $\lambda^2+0.9\lambda+0.14=0$	✓ 1
A1 $\lambda=-0.2$ and $\lambda=-0.7$	✓ 1
Part (c)
A1 $\begin{pmatrix}x\\y\end{pmatrix}=A\begin{pmatrix}2\\3\end{pmatrix}e^{-0.2t}+B\begin{pmatrix}1\\-1\end{pmatrix}e^{-0.7t}$	✓ 1
A1 equivalent separate expressions $x=2Ae^{-0.2t}+Be^{-0.7t}$, $y=3Ae^{-0.2t}-Be^{-0.7t}$	✓ 1
Part (d)
M1 substitute $t=0$ to form $2A+B=10$ and $3A-B=0$	✓ 1
A1 $A=2$, $B=6$	✓ 1
A1 $x=4e^{-0.2t}+6e^{-0.7t}$	✓ 1
A1 $y=6e^{-0.2t}-6e^{-0.7t}$	✓ 1
Part (e)
M1 set $\frac{dy}{dt}=-1.2e^{-0.2t}+4.2e^{-0.7t}=0$	✓ 1
A1 $e^{0.5t}=3.5$	✓ 1
A1 $t=2\ln 3.5=2.51$ hours	✓ 1
A1 $y=2.60$ mg	✓ 1

Exam tip: To confirm a general solution, check that it satisfies the initial conditions before you use it to answer later parts.

+ Add to test✓ In your test471IBMI3H-5-07r35-10Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard12 marks

A trampoline athlete's vertical displacement $x$ metres from the rest position, $t$ seconds after release, satisfies the second-order differential equation $\frac{d^2x}{dt^2}+0.8\frac{dx}{dt}+16x=0$, with $x=0.4$ and $\frac{dx}{dt}=0$ when $t=0$.

(a)

By letting $y=\frac{dx}{dt}$, write the equation as a pair of coupled first-order differential equations.

(2)

(b)

Use Euler's method with a step length of 0.1 s to estimate $x$ when $t=0.3$. Carry unrounded values through your working and give the final answer correct to 3 significant figures.

(5)

(c)

The exact solution is $x=e^{-0.4t}\left(0.4\cos\omega t+b\sin\omega t\right)$, where $\omega=\sqrt{15.84}$ and $b=\frac{0.16}{\omega}$. Use unrounded values of $\omega$ and $b$ to find $x$ when $t=0.3$, correct to 3 significant figures.

(2)

(d)

Using the unrounded values of $x(0.3)$ from (b) and (c), find the percentage error in the Euler estimate, relative to the exact value. Give your answer correct to 3 significant figures.

(2)

(e)

Suggest one change to the method that would reduce this error.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 introduce $y=\frac{dx}{dt}$ and substitute into the original equation	✓ 1
A1 $\frac{dx}{dt}=y$ and $\frac{dy}{dt}=-16x-0.8y$	✓ 1
Part (b)
M1 use of $x_{n+1}=x_n+0.1y_n$ and $y_{n+1}=y_n+0.1(-16x_n-0.8y_n)$	✓ 1
A1 $x_1=0.4$, $y_1=-0.64$	✓ 1
A1 $x_2=0.336$, $y_2=-1.2288$	✓ 1
A1 $x_3=0.21312$	✓ 1
A1 $x(0.3)\approx0.213$ m	✓ 1
Part (c)
M1 substitute $t=0.3$ into the exact solution	✓ 1
A1 $x=0.164$ m	✓ 1
Part (d)
M1 $\frac{0.21312-0.163694}{0.163694}\times100$	✓ 1
A1 $30.2\%$	✓ 1
Part (e)
R1 use a smaller step length (or a more accurate numerical method)	✓ 1

Exam tip: Any second-order equation can be treated like a coupled pair by naming the first derivative as a new variable; then both quantities are updated together at every Euler step using the old values.

+ Add to test✓ In your test472IBMI3H-5-04r39-02Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard15 marks

A drug is infused into a patient's blood. The concentration C mg/L at time t hours satisfies the differential equation dC/dt = 6 − 0.4C, with C = 2 when t = 0.

(a)

Find the value of C for which dC/dt = 0.

(2)

(b)

The slope field for this equation is drawn in the (t, C) plane. State the gradient of the slope field along the line C = 0. State the sign of the gradient for C > 15 and explain what this shows about solution curves that start above C = 15.

(3)

(c)

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

(3)

(d)

Solve the differential equation by separation of variables to show that C = 15 − 13e^−0.4t.

(4)

(e)

Find the exact model's value of C when t = 1. Give your answer to 3 significant figures.

(2)

(f)

Compare your answer to (c) with your answer to (e) and explain, using the shape of the solution curve, why they differ in this way.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6 − 0.4C = 0	✓ 1
A1 C = 15	✓ 1
Part (b)
A1 gradient 6	✓ 1
A1 the gradient is negative for C > 15	✓ 1
R1 solution curves starting above 15 decrease towards C = 15 (they approach the horizontal line C = 15 without crossing it)	✓ 1
Part (c)
M1 C₁ = 2 + 0.5(6 − 0.4 × 2)	✓ 1
A1 C₁ = 4.6	✓ 1
A1 C₂ = 4.6 + 0.5(6 − 0.4 × 4.6) = 6.68	✓ 1
Part (d)
M1 ∫ 1/(6 − 0.4C) dC = ∫ 1 dt	✓ 1
A1 −2.5 ln\|6 − 0.4C\| = t + k	✓ 1
M1 use C = 2 when t = 0 to find the constant	✓ 1
A1 rearrangement to C = 15 − 13e^−0.4t	✓ 1
Part (e)
M1 substitute t = 1: 15 − 13e^−0.4	✓ 1
A1 C = 6.29 mg/L	✓ 1
Part (f)
R1 Euler's estimate (6.68) is larger than the exact value (6.29); the solution curve is concave down (the gradient decreases as C increases), so each tangent segment lies above the curve	✓ 1

Common error: When separating variables, integrate both sides before using the initial condition; substituting the initial values too early loses the constant.

+ Add to test✓ In your test473IBMI3H-5-04r39-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard13 marks

A Formula 1 car decelerates along a straight after a driver lifts off the throttle at t = 0. Air resistance alone slows it, so its speed v m s⁻¹ satisfies dv/dt = −0.002v², with v = 80 when t = 0.

(a)

Solve the differential equation by separation of variables to show that v = 500/(t + 6.25).

(4)

(b)

Find the time taken for the speed to fall to 40 m s⁻¹.

(2)

(c)

Find the distance travelled during this time. Give your answer to 3 significant figures.

(3)

(d)

Use Euler's method with one step of length 2 s to estimate the speed when t = 2.

(2)

(e)

Find the speed at t = 2 from the solution in (a), to 3 significant figures, and explain why Euler's estimate is lower.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ∫ v⁻² dv = ∫ −0.002 dt	✓ 1
A1 −1/v = −0.002t + c	✓ 1
M1 use v = 80 when t = 0: c = −1/80	✓ 1
A1 1/v = 0.002t + 0.0125, so v = 500/(t + 6.25)	✓ 1
Part (b)
M1 500/(t + 6.25) = 40	✓ 1
A1 t = 6.25 s	✓ 1
Part (c)
M1 distance = ∫₀^6.25 500/(t + 6.25) dt	✓ 1
A1 500 ln(t + 6.25) evaluated between the limits	✓ 1
A1 500 ln 2 = 347 m	✓ 1
Part (d)
M1 v = 80 + 2 × (−0.002 × 80²)	✓ 1
A1 54.4 m s⁻¹	✓ 1
Part (e)
A1 500/8.25 = 60.6 m s⁻¹	✓ 1
R1 the deceleration becomes smaller as the speed falls, so the tangent at the start of the step is steeper than the curve and gives an underestimate	✓ 1

Common error: When integrating 1/v², the result is −1/v, not a logarithm; the logarithm appears only when integrating 1/v.

+ Add to test✓ In your test474IBMI3H-5-04r39-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard12 marks

During an infection in a hospital patient, x is the number of bacteria and y the number of immune cells, both in thousands per mL, at time t hours. They are modelled by dx/dt = 0.5x − 0.02xy and dy/dt = 0.01xy − 0.4y.

(a)

Find the two equilibrium points of the system.

(3)

(b)

At t = 0, x = 60 and y = 20. Use Euler's method with a step length of 0.1 hours to estimate x and y when t = 0.2. Give your answers to 2 decimal places.

(5)

(c)

State, with a reason, whether the number of bacteria and the number of immune cells are increasing or decreasing at t = 0.

(1)

(d)

Interpret the equilibrium point (40, 25) in this context.

(1)

(e)

The immune response is strengthened so that dy/dt = 0.01xy − 0.5y. Find the x-coordinate of the new non-zero equilibrium point, and state its y-coordinate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set dx/dt = 0 and dy/dt = 0, with factorisation x(0.5 − 0.02y) = 0 and y(0.01x − 0.4) = 0	✓ 1
A1 (0, 0)	✓ 1
A1 (40, 25)	✓ 1
Part (b)
M1 dx/dt = 6 and dy/dt = 4 at (60, 20)	✓ 1
A1 x₁ = 60.6 and y₁ = 20.4	✓ 1
M1 gradients at (60.6, 20.4): dx/dt = 5.5752 and dy/dt = 4.2024	✓ 1
A1 x₂ = 61.16	✓ 1
A1 y₂ = 20.82	✓ 1
Part (c)
A1 both increasing, because dx/dt = 6 > 0 and dy/dt = 4 > 0	✓ 1
Part (d)
R1 the numbers of bacteria (40 thousand per mL) and immune cells (25 thousand per mL) would both stay constant	✓ 1
Part (e)
A1 x = 0.5/0.01 = 50	✓ 1
A1 y = 25 (unchanged, since dx/dt = 0 requires 0.5 = 0.02y)	✓ 1

Common error: Both gradients must be evaluated at the same old point before either variable is updated in an Euler step.

+ Add to test✓ In your test475IBMI3H-5-05r40-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard9 marks

In a town, S hundred people are susceptible and I hundred people are infected, where t is in days. The model is
$\frac{dS}{dt} = -0.002SI, \qquad \frac{dI}{dt} = 0.002SI - 0.4I.$
Initially $S = 400$ and $I = 10$. Use Euler's method with a step size of 0.5 days.

(a)

Find $\frac{dS}{dt}$ and $\frac{dI}{dt}$ at $t=0$.

(2)

(b)

Hence estimate S and I at $t=0.5$.

(2)

(c)

Estimate S and I at $t=1$, giving each to 3 significant figures.

(3)

(d)

Find the value of S, in hundreds of people, at which the number of infected people stops increasing.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dS}{dt} = -0.002(400)(10) = -8$	✓ 1
A1 $\frac{dI}{dt} = 8 - 4 = 4$	✓ 1
Part (b)
M1 $S = 400 + 0.5(-8)$ and $I = 10 + 0.5(4)$	✓ 1
A1 $S = 396$, $I = 12$	✓ 1
Part (c)
M1 rates at $(396, 12)$: $\frac{dS}{dt} = -9.504$, $\frac{dI}{dt} = 9.504 - 4.8 = 4.704$	✓ 1
A1 $S = 396 - 4.752 = 391$	✓ 1
A1 $I = 12 + 2.352 = 14.4$	✓ 1
Part (d)
M1 set $\frac{dI}{dt}=0$: $I(0.002S - 0.4)=0$ with $I>0$	✓ 1
A1 $S = 200$	✓ 1

Common error: In coupled Euler steps, compute both rates from the same old values before updating either variable.

+ Add to test✓ In your test476IBMI3H-5-05r40-12Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

The sideways displacement, x metres, of the top of a wind-turbine tower satisfies $\frac{d^2x}{dt^2} + 0.5\frac{dx}{dt} + 4x = 0$. Let $v = \frac{dx}{dt}$, so that $\frac{dx}{dt} = v$ and $\frac{dv}{dt} = -4x - 0.5v$. At $t=0$, $x=1$ and $v=0$.

(a)

Show that the characteristic equation of the second-order equation is $\lambda^2 + 0.5\lambda + 4 = 0$, and use its roots to describe the long-term motion of the tower.

(3)

(b)

Use Euler's method with step size 0.1 to find x and v at $t=0.1$.

(2)

(c)

Continue to find x and v at $t=0.2$.

(3)

(d)

Continue to find x at $t=0.3$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute $x=e^{\lambda t}$ to get $\lambda^2+0.5\lambda+4=0$	✓ 1
A1 $\lambda = -0.25 \pm 1.98i$ (complex roots)	✓ 1
R1 oscillation whose amplitude decays, since the real part $-0.25$ is negative	✓ 1
Part (b)
M1 $x = 1+0.1(0)$ and $v = 0+0.1(-4)$	✓ 1
A1 $x=1$, $v=-0.4$	✓ 1
Part (c)
M1 $x = 1 + 0.1(-0.4)$	✓ 1
A1 $x = 0.96$	✓ 1
A1 $v = -0.4 + 0.1\left(-4 \times 1 - 0.5 \times (-0.4)\right) = -0.78$	✓ 1
Part (d)
M1 $x = 0.96 + 0.1(-0.78)$	✓ 1
A1 $x = 0.882$ m	✓ 1

Exam tip: For a second-order equation solved numerically, rewrite it as two first-order equations and update both quantities at each step.

+ Add to test✓ In your test477IBMI3H-5-05r40-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

A hydro-power draft tube has circular cross-sections of radius $y = \sqrt{\dfrac{x}{1+x^2}}$ metres at distance x metres along its axis, for $0 \le x \le 3$. Its volume is formed by rotating the curve about the x-axis.

(a)

Write down an integral for the volume of the tube.

(2)

(b)

Use the substitution $u = 1+x^2$ to show that $V = \frac{\pi}{2}\int_1^{10} \frac{1}{u}\,du$.

(3)

(c)

Hence find the volume, to 3 significant figures.

(2)

(d)

A second tube has the same shape but runs from $x=0$ to $x=k$. Find k so that its volume is 5 m³, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of $V = \pi\int y^2\,dx$	✓ 1
A1 $V = \pi\int_0^3 \frac{x}{1+x^2}\,dx$	✓ 1
Part (b)
M1 $\frac{du}{dx} = 2x$, so $x\,dx = \frac{1}{2}du$	✓ 1
A1 integrand becomes $\frac{1}{2u}$	✓ 1
A1 limits $u=1$ and $u=10$	✓ 1
Part (c)
M1 $\frac{\pi}{2}\left[\ln u\right]_1^{10} = \frac{\pi}{2}\ln 10$	✓ 1
A1 3.62 m³	✓ 1
Part (d)
M1 $\frac{\pi}{2}\ln(1+k^2) = 5$	✓ 1
A1 $1+k^2 = e^{10/\pi} = 24.12$	✓ 1
A1 $k = 4.81$	✓ 1

Exam tip: Write the squared radius first: $y^2 = \frac{x}{1+x^2}$ removes the square root and makes the substitution visible.

+ Add to test✓ In your test478IBMI3H-5-04r42-08Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard14 marks

Calculator

A beekeeper models the foraging efficiency of a hive by E(x) = xe^−0.4x, x ≥ 0, where x is the distance in kilometres from the hive to a field of flowers.

(a)

Use the product rule to show that E′(x) = e^−0.4x(1 − 0.4x).

(3)

(b)

Find the exact value of x at the stationary point, and the value of E there to 3 significant figures.

(3)

(c)

Find E″(x) and hence justify that the stationary point is a maximum.

(3)

(d)

Show that the graph of E has a point of inflection at x = 5, and find the value of E there.

(3)

(e)

Find ∫₀⁵ E(x) dx, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule applied to x and e^−0.4x	✓ 1
A1 E′(x) = e^−0.4x − 0.4xe^−0.4x	✓ 1
A1 factorise to e^−0.4x(1 − 0.4x) (AG)	✓ 1
Part (b)
M1 set E′(x) = 0 (e^−0.4x is never zero)	✓ 1
A1 x = 2.5	✓ 1
A1 E = 0.920	✓ 1
Part (c)
M1 differentiate E′(x) using the product rule	✓ 1
A1 E″(x) = e^−0.4x(0.16x − 0.8)	✓ 1
R1 E″(2.5) = -0.147 < 0, so a maximum	✓ 1
Part (d)
M1 set E″(x) = 0, giving x = 5	✓ 1
R1 E″ changes sign (negative for x < 5, positive for x > 5)	✓ 1
A1 E(5) = 0.677	✓ 1
Part (e)
M1 integration (by parts or by calculator) of xe^−0.4x over [0, 5]	✓ 1
A1 3.71	✓ 1

Common error: A zero of the second derivative is a point of inflection only if the second derivative changes sign; always check the sign on both sides.

+ Add to test✓ In your test479IBMI3H-5-04r42-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard13 marks

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After an injection, the mass of a drug in a patient's blood, x mg, and in the patient's tissue, y mg, t hours after the injection, satisfy the system
dx/dt = −2x + y, dy/dt = 2x − 3y.
Initially, x = 60 and y = 0.

(a)

Find the eigenvalues of the matrix [[−2, 1], [2, −3]].

(3)

(b)

Find the eigenvectors, and hence write down the general solution of the system.

(4)

(c)

Use the initial conditions to find the particular solution.

(2)

(d)

Find the maximum mass of the drug in the tissue and the time at which it occurs. Give both answers to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 det(A − λI) = 0, i.e. (−2 − λ)(−3 − λ) − 2 = 0	✓ 1
A1 λ² + 5λ + 4 = 0	✓ 1
A1 λ = −1 and λ = −4	✓ 1
Part (b)
A1 eigenvector (1, 1) for λ = −1	✓ 1
A1 eigenvector (1, −2) for λ = −4	✓ 1
M1 general solution as a combination of eigenvector × e^λt	✓ 1
A1 (x, y) = Ae^−t(1, 1) + Be^−4t(1, −2)	✓ 1
Part (c)
M1 A + B = 60 and A − 2B = 0	✓ 1
A1 x = 40e^−t + 20e^−4t, y = 40e^−t − 40e^−4t	✓ 1
Part (d)
M1 dy/dt = 0, i.e. −40e^−t + 160e^−4t = 0	✓ 1
A1 e^3t = 4	✓ 1
A1 t = 0.462 hours	✓ 1
A1 y = 18.9 mg	✓ 1

Common error: Each eigenvector must satisfy (A − λI)v = 0 for its own eigenvalue; check by substituting back into the system before writing the general solution.

+ Add to test✓ In your test480IBMI3H-5-04r42-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard13 marks

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The vertical displacement, x metres, of a Formula 1 car's front suspension from its equilibrium position satisfies d²x/dt² + 2 dx/dt + 5x = 0, where t is the time in seconds. At t = 0, x = 0.04 and dx/dt = 0.

(a)

Let y = dx/dt. Write the equation as a system of two first-order differential equations.

(2)

(b)

Use Euler's method with a step size of 0.1 to estimate x and y at t = 0.1, 0.2 and 0.3.

(4)

(c)

Find the eigenvalues of the matrix of the system in (a).

(3)

(d)

Describe the long-term motion of the suspension and the type of equilibrium point at the origin of the phase portrait.

(2)

(e)

The exact solution is x = e^−t(0.04cos 2t + 0.02sin 2t). Find the percentage error in the Euler estimate of x at t = 0.3, to 2 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 dx/dt = y	✓ 1
A1 dy/dt = −5x − 2y	✓ 1
Part (b)
M1 x_n+1 = x_n + 0.1y_n and y_n+1 = y_n + 0.1(−5x_n − 2y_n) used	✓ 1
A1 t = 0.1: x = 0.04, y = -0.02	✓ 1
A1 t = 0.2: x = 0.038, y = -0.036	✓ 1
A1 t = 0.3: x = 0.0344, y = -0.0478	✓ 1
Part (c)
A1 matrix [[0, 1], [−5, −2]]	✓ 1
M1 λ² + 2λ + 5 = 0	✓ 1
A1 λ = −1 ± 2i	✓ 1
Part (d)
A1 oscillations whose amplitude decays to zero, so the suspension settles at equilibrium	✓ 1
A1 stable spiral (spiral sink), because the eigenvalues are complex with negative real part	✓ 1
Part (e)
M1 exact x(0.3) = 0.03282, then (estimate − exact)/exact × 100	✓ 1
A1 4.8%	✓ 1

Exam tip: To use Euler's method on a second-order equation, introduce a new variable for the first derivative and update both variables from the old values at each step.

+ Add to test✓ In your test481IBMI3H-5-05r45-11Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard11 marks

The top of a radio mast sways horizontally. Its displacement x metres from the rest position, t seconds after a gust, satisfies the second-order differential equation
d²x/dt² = −9x − 0.6 dx/dt,
with x = 0.2 and dx/dt = 0 when t = 0.

(a)

Let y = dx/dt. Write the second-order equation as a pair of first-order differential equations in x and y.

(2)

(b)

Use Euler’s method with a step length of 0.1 to estimate x when t = 0.3. Show the values of x and y after each step.

(5)

(c)

The exact displacement at t = 0.3 is x = 0.12861 (5 significant figures). Using the unrounded value from part (b), calculate the percentage error in the Euler estimate, to 3 significant figures.

(3)

(d)

Suggest one change that would improve the accuracy of the estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 dx/dt = y	✓ 1
A1 dy/dt = −9x − 0.6y	✓ 1
Part (b)
M1 uses x_n+1 = x_n + 0.1y_n and y_n+1 = y_n + 0.1(−9x_n − 0.6y_n)	✓ 1
A1 t = 0.1: x = 0.2, y = −0.18	✓ 1
A1 t = 0.2: x = 0.182, y = −0.3492	✓ 1
A1 t = 0.3: x = 0.14708	✓ 1
A1 y = −0.492048 at t = 0.3 (accept −0.492)	✓ 1
Part (c)
M1 percentage error = \|estimate − exact\|/exact × 100	✓ 1
M1 substitutes 0.14708 and 0.12861	✓ 1
A1 14.4 %	✓ 1
Part (d)
R1 use a smaller step length (e.g. 0.05 or 0.01), at the cost of more steps	✓ 1

Exam tip: To use Euler’s method on a second-order equation, first introduce a new variable for the first derivative so that you have two first-order equations.

+ Add to test✓ In your test482IBMI3H-5-05r45-13Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard12 marks

A relay drone moves along a straight cable between two masts. Its velocity, in metres per second, is v(t) = 3 sin(0.8t) − 1.5 for 0 ≤ t ≤ 6, where t is in seconds and positive velocity is towards the east mast. The argument of the sine function is in radians.

(a)

Find the acceleration of the drone when t = 1, to 3 significant figures.

(3)

(b)

Find the times at which the drone is instantaneously at rest.

(3)

(c)

Find the displacement of the drone from its starting position when t = 6. Give your answer to 3 significant figures and state its direction.

(3)

(d)

Find the total distance travelled by the drone during the first 6 seconds. Use the unrounded times from part (b) and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts to differentiate v	✓ 1
A1 a(t) = 2.4cos(0.8t)	✓ 1
A1 acceleration = 1.67 m s⁻² when t = 1	✓ 1
Part (b)
M1 sets 3 sin(0.8t) − 1.5 = 0 so sin(0.8t) = 0.5	✓ 1
A1 t = 0.654 s	✓ 1
A1 t = 3.27 s (the second solution of the equation within the domain)	✓ 1
Part (c)
M1 displacement = ∫₀⁶ v dt	✓ 1
A1 −5.58 m	✓ 1
A1 5.58 m towards the west mast (negative direction)	✓ 1
Part (d)
M1 splits the interval at the times where v = 0 (0.654…, 3.27…)	✓ 1
M1 adds the magnitudes of the three integrals: 0.479… + 2.57… + 7.67… (negative, positive, negative velocity in turn)	✓ 1
A1 10.7 m	✓ 1

Common error: Integrating the velocity over the whole interval gives displacement, not distance; the distance needs the integrals of the separate intervals taken as positive values.

+ Add to test✓ In your test483IBMI3H-5-05r45-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard12 marks

In a sourdough starter, the population of yeast x and of a bacterium y (both in millions of cells per gram) after t hours are modelled by
dx/dt = 0.5x − 0.02xy, dy/dt = 0.004xy − 0.3y.

(a)

Find the values of x and y, both non-zero, for which both populations are constant.

(3)

(b)

Initially x = 100 and y = 20. Use Euler’s method with a step length of 0.5 to estimate x and y when t = 1.5. Give each answer to 3 significant figures.

(5)

(c)

Using your answer to part (a), explain why both populations initially increase when x = 100 and y = 20.

(2)

(d)

Show that dx/dt = 0 for every value of x whenever y = 25, and state what this means for the slope field in the (x, y) plane along the line y = 25.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sets both dx/dt = 0 and dy/dt = 0 and factorises, x(0.5 − 0.02y) = 0 and y(0.004x − 0.3) = 0	✓ 1
A1 y = 25	✓ 1
A1 x = 75	✓ 1
Part (b)
M1 uses x_n+1 = x_n + 0.5(dx/dt) and y_n+1 = y_n + 0.5(dy/dt) with both gradients evaluated at the same step	✓ 1
A1 (x, y) at t = 0.5 is (105, 21)	✓ 1
A1 (x, y) at t = 1 is (109.2, 22.26)	✓ 1
A1 x ≈ 112 at t = 1.5 (112.19…)	✓ 1
A1 y ≈ 23.8 at t = 1.5 (23.78…)	✓ 1
Part (c)
R1 dx/dt = 10 > 0 because y = 20 is below 25, so the bacteria are too few to hold the yeast in check	✓ 1
R1 dy/dt = 2 > 0 because x = 100 is above 75, so food is plentiful for the bacteria	✓ 1
Part (d)
M1 substitutes y = 25: dx/dt = 0.5x − 0.5x = 0	✓ 1
R1 every line segment on y = 25 is vertical (no change in x)	✓ 1

Exam tip: Evaluate both derivatives from the values at the start of each step before updating either variable; a table with columns for t, x, y, dx/dt and dy/dt keeps this organised.

+ Add to test✓ In your test484IBMI3H-5-07r45-05Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard11 marks

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A ship has two ballast tanks. Pumps move water between them, and the masses of water, $x$ and $y$ tonnes, in tank 1 and tank 2 at time $t$ hours satisfy
$\dfrac{dx}{dt}=-0.3x+0.1y,\qquad \dfrac{dy}{dt}=0.3x-0.1y,$
with $x=80$ and $y=20$ when $t=0$.

(a)

Use Euler's method with step length $h=0.5$ to estimate $x$ and $y$ when $t=1$.

(4)

(b)

The system can be written as $\dfrac{d}{dt}\binom{x}{y}=\mathbf{A}\binom{x}{y}$. Find the eigenvalues of $\mathbf{A}=\begin{pmatrix}-0.3&0.1\\0.3&-0.1\end{pmatrix}$.

(2)

(c)

Find the mass of water in tank 1 as $t\to\infty$. Justify your answer.

(2)

(d)

The exact solution is $x=25+55e^{-0.4t}$. Find the time at which tank 1 first holds 40 tonnes. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of $x_{n+1}=x_n+0.5(-0.3x_n+0.1y_n)$ and $y_{n+1}=y_n+0.5(0.3x_n-0.1y_n)$	✓ 1
A1 $x_1=69$, $y_1=31$	✓ 1
A1 $x_2=60.2$	✓ 1
A1 $y_2=39.8$	✓ 1
Part (b)
M1 $\det(\mathbf{A}-\lambda\mathbf{I})=0$ giving $\lambda^2+0.4\lambda=0$	✓ 1
A1 $\lambda=0$ and $\lambda=-0.4$	✓ 1
Part (c)
M1 equilibrium requires $-0.3x+0.1y=0$ so $y=3x$, and $x+y=100$ is constant (total mass $80+20$)	✓ 1
A1 $x\to 25$ tonnes	✓ 1
Part (d)
M1 $25+55e^{-0.4t}=40$ so $55e^{-0.4t}=15$	✓ 1
M1 $-0.4t=\ln\left(\frac{15}{55}\right)$	✓ 1
A1 $t=3.25$ hours (3.248...)	✓ 1

Exam tip: Because the two equations add to zero, the total mass is constant. Spotting a conserved quantity gives the long-term state without solving the system.

+ Add to test✓ In your test485IBMI3H-5-07r45-07Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

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The roll angle $\theta$ radians of a ship at time $t$ seconds satisfies the second-order differential equation $\dfrac{d^2\theta}{dt^2}=-2\theta-0.6\dfrac{d\theta}{dt}$. At $t=0$, $\theta=0.2$ and $\dfrac{d\theta}{dt}=0$.

(a)

Let $\omega=\dfrac{d\theta}{dt}$. Write the equation as a pair of first-order equations, then use Euler's method with $h=0.1$ to find $\theta$ and $\omega$ when $t=0.2$. Do not round intermediate values.

(5)

(b)

Show that the ship's motion is a decaying oscillation.

(2)

(c)

Find the period of the oscillation, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{d\theta}{dt}=\omega$ and $\frac{d\omega}{dt}=-2\theta-0.6\omega$	✓ 1
M1 use of $\theta_{n+1}=\theta_n+0.1\omega_n$ and $\omega_{n+1}=\omega_n+0.1(-2\theta_n-0.6\omega_n)$	✓ 1
A1 $\theta_1=0.2$, $\omega_1=-0.04$	✓ 1
A1 $\theta_2=0.196$	✓ 1
A1 $\omega_2=-0.0776$	✓ 1
Part (b)
M1 auxiliary equation $r^2+0.6r+2=0$ with discriminant $0.36-8<0$	✓ 1
R1 complex roots, so oscillation, with real part $-0.3<0$ giving decaying amplitude	✓ 1
Part (c)
M1 roots $r=-0.3\pm i\sqrt{1.91}$	✓ 1
A1 angular frequency $\sqrt{1.91}=1.38$	✓ 1
A1 period $\frac{2\pi}{\sqrt{1.91}}=4.55$ s	✓ 1

Exam tip: A negative discriminant in the auxiliary equation means oscillation. The imaginary part of the root, not the whole root, sets the angular frequency.

+ Add to test✓ In your test486IBMI3H-5-07r45-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

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A lava dome grows so that its height $h$ metres at time $t$ days satisfies $\dfrac{dh}{dt}=k(12-h)$, where $k$ is a positive constant and $h<12$. When $t=0$, $h=2$, and when $t=2$, $h=7$.

(a)

Solve the differential equation by separating the variables, to show that $h=12-Ae^{-kt}$, where $A$ is a constant.

(4)

(b)

Find the value of $A$.

(1)

(c)

Show that $k=\dfrac{\ln 2}{2}$, and give $k$ to 3 significant figures.

(2)

(d)

State the value that $h$ approaches as $t\to\infty$.

(1)

(e)

Find the time at which the height first reaches 10 metres. Use the exact value of $k$ and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\int\frac{dh}{12-h}=\int k\,dt$	✓ 1
A1 $-\ln(12-h)=kt+c$	✓ 1
A1 $12-h=Ae^{-kt}$ (with $A=e^{-c}>0$)	✓ 1
R1 $h<12$ so no modulus needed, giving $h=12-Ae^{-kt}$	✓ 1
Part (b)
A1 $A=10$	✓ 1
Part (c)
M1 $7=12-10e^{-2k}$ so $e^{-2k}=0.5$	✓ 1
A1 $k=\frac{\ln 2}{2}=0.347$	✓ 1
Part (d)
A1 $12$ metres	✓ 1
Part (e)
M1 $10=12-10e^{-kt}$ so $e^{-kt}=0.2$	✓ 1
A1 $t=4.64$ days (4.6438...)	✓ 1

Exam tip: When two data points are given, use one to find the arbitrary constant and the other to find the rate constant.

+ Add to test✓ In your test487IBMI3H-5-06r46-06Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard10 marks

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The biomass P tonnes of fish in a managed fishery t years after monitoring begins is modelled by $\frac{dP}{dt} = 0.2P\left(1 - \frac{P}{900}\right) - 30$.

(a)

State what the term $-30$ represents in the context.

(1)

(b)

Show that the model has two equilibrium biomasses and find them, correct to 3 significant figures.

(3)

(c)

Describe what happens to the biomass in the long run when $P = 500$ initially, and justify your answer.

(2)

(d)

Use Euler's method with a step size of $h = 0.5$ to estimate $P(1)$ given $P(0) = 400$. Give your answer to 1 decimal place.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 the fish harvested (caught) each year, 30 tonnes per year	✓ 1
Part (b)
M1 set $\frac{dP}{dt} = 0$	✓ 1
A1 $P^2 - 900P + 135000 = 0$	✓ 1
A1 $P = 190$ tonnes and $P = 710$ tonnes	✓ 1
Part (c)
M1 evaluate $\frac{dP}{dt}$ at $P = 500$, which is $+14.4$ (positive)	✓ 1
R1 biomass rises, and since $\frac{dP}{dt} = 0$ at 710 it approaches the stable equilibrium of 710 tonnes	✓ 1
Part (d)
M1 $P_1 = 400 + 0.5\times\frac{dP}{dt}\Big\|_{P=400}$ with the rate evaluated as 14.44	✓ 1
A1 $P(0.5) \approx 407.22$	✓ 1
M1 $P_2 = 407.22 + 0.5\times$ (rate at 407.22)	✓ 1
A1 414.5 tonnes (using unrounded intermediate values)	✓ 1

Exam tip: In Euler's method, evaluate the derivative at the current point each time and keep unrounded values until the final step.

+ Add to test✓ In your test488IBMI3H-5-06r46-14Derivatives of trig, exp and ln with chain, product and quotient rules; integration by substitution and volumes of revolution; kinematics with calculus; slope fields, Euler's method, coupled and second-order differential equations Paper 2Hard9 marks

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The number of fish in a stocked lake is modelled by $N(t) = \frac{800}{1 + 7e^{-0.5t}}$, where $t$ is in years since stocking.

(a)

Find the number of fish at stocking and the limiting number as $t \to \infty$.

(2)

(b)

Show that $N'(t) = \frac{2800e^{-0.5t}}{(1 + 7e^{-0.5t})^2}$.

(3)

(c)

Find the initial rate of growth of the population, in fish per year.

(1)

(d)

The population grows fastest when $7e^{-0.5t} = 1$. Find this time, correct to 3 significant figures, and then the growth rate at that time in fish per year.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $N(0) = 100$	✓ 1
A1 limit 800	✓ 1
Part (b)
M1 chain rule (or quotient rule) with $\frac{d}{dt}(1 + 7e^{-0.5t}) = -3.5e^{-0.5t}$	✓ 1
A1 $N' = -800(1 + 7e^{-0.5t})^{-2}\times(-3.5e^{-0.5t})$	✓ 1
A1 $\frac{2800e^{-0.5t}}{(1+7e^{-0.5t})^2}$	✓ 1
Part (c)
A1 $N'(0) = \frac{2800}{64} = 43.75$ fish per year (accept 43.8)	✓ 1
Part (d)
M1 solve $e^{-0.5t} = \frac{1}{7}$	✓ 1
A1 $t = 2\ln 7 = 3.89$ years	✓ 1
A1 growth rate $\frac{2800 \times \frac{1}{7}}{4} = 100$ fish per year	✓ 1

Exam tip: Write $N(t)$ as $800(1 + 7e^{-0.5t})^{-1}$ before differentiating so the chain rule is easy to apply.

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

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.	—

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

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

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

Marking point	Mark	Notes
Part (a)
M1 a = \(\frac{dv}{dt}\) = 1.2 − 0.12t	✓ 1
A1 set a = 0 to get t = 10	✓ 1
A1 substitute into v	✓ 1
A1 6 m s⁻¹	✓ 1
Part (b)
M1 \(\int_0^{20}v\,dt\)	✓ 1
A1 80 m	✓ 1

Marking point	Mark	Notes
Part (a)
M1 attempt to use the chain rule with \(\frac{d}{dt}(t^2+4) = 2t\)	✓ 1
A1 \(\frac{dE}{dt} = \frac{80t}{t^2 + 4}\)	✓ 1
Part (b)
M1 substitute \(t = 3\): \(\frac{80(3)}{9 + 4}\)	✓ 1
A1 18.5 points per hour	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{dh}{dt} = 3(0.5)\cos(0.5t)\)	✓ 1
A1 \(1.5\cos 1 = 0.81\) m per day	✓ 1
Part (b)
M1 set \(1.5\cos(0.5t) = 0\) so \(0.5t = \frac{\pi}{2}\)	✓ 1
A1 \(t = \pi = 3.14\) days and depth \(= 15\) m	✓ 1

Marking point	Mark	Notes
Part (a)
A1 \(\frac{dy}{dx} = 3x^2 - 4\)	✓ 1
Part (b)
A1 gradient \(= 3(2)^2 - 4 = 8\) and \(y = 8 - 8 + 1 = 1\)	✓ 1
M1 \(y - 1 = 8(x - 2)\)	✓ 1
A1 \(y = 8x - 15\)	✓ 1
Part (c)
M1 set \(8x - 15 = 0\)	✓ 1
A1 \(x = \frac{15}{8}\) (\(= 1.875\))	✓ 1

Marking point	Mark	Notes
Part (a)
M1 attempts the product rule with \(u=12t\) and \(v=e^{-0.4t}\)	✓ 1
A1 \(u'=12\) and \(v'=-0.4e^{-0.4t}\)	✓ 1
A1 \(\frac{dV}{dt}=12e^{-0.4t}-4.8te^{-0.4t}=12e^{-0.4t}(1-0.4t)\)	✓ 1
Part (b)
M1 sets \(\frac{dV}{dt}=0\)	✓ 1
A1 \(t=2.5\) s	✓ 1
R1 since \(e^{-0.4t}>0\), the derivative is positive for \(t<2.5\) and negative for \(t>2.5\), so a maximum (accept second derivative negative)	✓ 1
Part (c)
M1 \(V(2.5)=12\times2.5\times e^{-1}\)	✓ 1
A1 11.0 V	✓ 1

Marking point	Mark	Notes
Part (a)
M1 interval width \(h=2\)	✓ 1
M1 \(\frac{2}{2}\left[1.2+1.4+2(2.0+2.6+2.2)\right]\)	✓ 1
A1 16.2 cm	✓ 1
Part (b)
R1 on a concave-down curve each chord lies below the curve	✓ 1
A1 so the trapezoidal rule gives an underestimate	✓ 1
Part (c)
M1 \(40+16.2\) (ECF from (a))	✓ 1
A1 56.2 cm	✓ 1

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule on \(40t\) and \(e^{-0.2t}\)	✓ 1
A1 \(\frac{dP}{dt}=40e^{-0.2t}(1-0.2t)\) (or equivalent, e.g. \(40e^{-0.2t}-8te^{-0.2t}\))	✓ 1
Part (b)
M1 set \(\frac{dP}{dt}=0\) (ECF from (a)), noting \(e^{-0.2t}\ne 0\)	✓ 1
A1 \(t=5\) hours	✓ 1
Part (c)
A1 \(P=40(5)e^{-1}=73.6\) thousand cells per m³ (ECF from (b))	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(h=\frac{2}{x^2}\) and \(S=x^2+4xh\) (base plus four sides, no top)	✓ 1
A1 \(S=x^2+\frac{8}{x}\) (shown)	✓ 1
Part (b)
M1 \(\frac{dS}{dx}=2x-\frac{8}{x^2}=0\)	✓ 1
A1 \(x^3=4\)	✓ 1
A1 \(x=1.59\) m	✓ 1
Part (c)
A1 €136.07	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{5}{2}\left[\text{first}+\text{last}+2(\text{sum of others})\right]\)	✓ 1
A1 \(\frac{5}{2}\left[0+0+2(1.8+3.1+3.6+3.2+2.0)\right]\)	✓ 1
A1 68.5 m²	✓ 1
Part (b)
A1 \(0.6\times 68.5=41.1\) m³ per second (ECF from (a))	✓ 1
Part (c)
R1 take depth measurements at smaller intervals (more strips), since this makes each trapezium follow the bed more closely	✓ 1

Marking point	Mark	Notes
Part (a)
M1 chain rule applied to \(\sin(12t)\)	✓ 1
A1 \(\frac{dh}{dt}=4.08\cos(12t)\)	✓ 1
Part (b)
A1 \(4.08\cos(1.8)=-0.927\) m s⁻¹ (the valve is moving downwards)	✓ 1
Part (c)
M1 set \(\frac{dh}{dt}=0\), so \(\cos(12t)=0\) and \(12t=\frac{\pi}{2}\)	✓ 1
A1 t = 0.131 s	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{du}{dt}=2t\), so \(4t\,dt=2\,du\)	✓ 1
M1 integral becomes \(\int2\cos u\,du\)	✓ 1
A1 \(2\sin u+c=2\sin(t^2)+c\) (AG)	✓ 1
Part (b)
M1 \(\left[2\sin(t^2)\right]_0^{1.2}=2\sin(1.44)\)	✓ 1
A1 1.9829 hundred, so 198 meteors	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{5}{2}\{\ldots\}\) with first and last values and twice the middle values	✓ 1
A1 \(\frac{5}{2}\left[0+0+2(2.1+3.4+2.6)\right]\)	✓ 1
A1 40.5 m²	✓ 1
Part (b)
M1 area × speed, \(40.5\times1.2\)	✓ 1
A1 48.6 m³ per second (ECF from (a))	✓ 1

Marking point	Mark	Notes
Part (a)
A1 \(a=6t-18\)	✓ 1
A1 −12 m s⁻²	✓ 1
Part (b)
M1 \(3t^2-18t+24=0\)	✓ 1
A1 t = 2 and t = 4	✓ 1
Part (c)
M1 displacement \(s=t^3-9t^2+24t\)	✓ 1
A1 s(2) = 20 and s(4) = 16	✓ 1
A1 distance = 20 + 4 = 24 m	✓ 1

Marking point	Mark	Notes
Part (a)
A1 \(\frac{dy}{dx}=0.008x-0.16\)	✓ 1
A1 gradient 0.08	✓ 1
Part (b)
A1 y = 10.8 at x = 30	✓ 1
A1 y = 0.08x + 8.4	✓ 1
Part (c)
M1 \(\frac{dy}{dx}=0\)	✓ 1
A1 x = 20	✓ 1
A1 10.4 m	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(V=\pi\int y^2\,dx\) with \(y=2-0.1x^2\)	✓ 1
A1 \(\pi\int_0^3(2-0.1x^2)^2\,dx\)	✓ 1
Part (b)
M1 evaluates the integral (GDC or expansion)	✓ 1
A1 27.9 m³ (27.916...)	✓ 1
Part (c)
M1 multiplies the volume by 18.50	✓ 1
A1 €516	✓ 1

Marking point	Mark	Notes
Part (a)
M1 differentiates: \(a=1.2-0.6t\)	✓ 1
A1 0.6 m s⁻²	✓ 1
Part (b)
M1 sets \(a=0\), giving t = 2	✓ 1
A1 1.2 m s⁻¹	✓ 1
Part (c)
M1 \(\int_0^4(1.2t-0.3t^2)\,dt\)	✓ 1
A1 3.2 m	✓ 1

Marking point	Mark	Notes
Part (a)
M1 chain rule: derivative of \(\sin(0.5t)\) is \(0.5\cos(0.5t)\)	✓ 1
A1 \(\frac{dh}{dt}=1.4\cos(0.5t)\)	✓ 1
Part (b)
M1 substitutes t = 2 (radians)	✓ 1
A1 0.756 m per hour	✓ 1
Part (c)
M1 \(1.4\cos(0.5t)=0.5\)	✓ 1
A1 t = 2.41 hours	✓ 1

Marking point	Mark	Notes
Part (a)
M1 use \(P_{n+1}=P_n+h\frac{dP}{dt}\) with \(P_0=100\), i.e. \(100+0.3\times100\times(1-\frac{100}{800})\)	✓ 1
A1 126.25 (AG)	✓ 1
Part (b)
M1 attempt to apply the method again with \(P_1=126.25\)	✓ 1
A1 \(P_2=158.14785\ldots\) (accept 158.1)	✓ 1
A1 \(P(3)\approx196.2\)	✓ 1
Part (c)
R1 when \(P=800\), \(\frac{dP}{dt}=0\) so the population stops changing (the population cannot cross this value)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 chain rule with \(\ln\) and inner function \(x^2+1\)	✓ 1
A1 \(\frac{dy}{dx}=\frac{2x}{x^2+1}\)	✓ 1
Part (b)
A1 gradient \(=\frac{2}{2}=1\) and point \((1,\ln2)\)	✓ 1
M1 use \(y-\ln2=1(x-1)\)	✓ 1
A1 \(y=x-0.307\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 chain rule with the exponent \(-0.1(t-3)^2\) differentiated	✓ 1
A1 \(N'(t)=-24(t-3)e^{-0.1(t-3)^2}\)	✓ 1
Part (b)
A1 \(N'(t)=0\) when \(t=3\) hours	✓ 1
Part (c)
M1 substitute \(t=5\): \(-24(2)e^{-0.4}\)	✓ 1
A1 \(-32.2\) people per hour per hour	✓ 1
R1 negative: the arrival rate is falling at that time	✓ 1

Marking point	Mark	Notes
Part (a)
M1 use \(C_{n+1}=C_n+h(1.2-0.4C_n)\)	✓ 1
A1 \(C=1.0\) at \(t=0.5\)	✓ 1
A1 \(C=1.4\) at \(t=1\)	✓ 1
Part (b)
A1 \(1.72\) (ECF from (a))	✓ 1
Part (c)
M1 set \(\frac{dC}{dt}=0\)	✓ 1
A1 \(C=3\) mg L^-1	✓ 1

Marking point	Mark	Notes
Part (a)
M1 write \(f=5(1+e^{-0.8x})^{-1}\) and use the chain rule	✓ 1
A1 \(-5(1+e^{-0.8x})^{-2}\times(-0.8e^{-0.8x})\)	✓ 1
A1 \(f'(x)=\frac{4e^{-0.8x}}{(1+e^{-0.8x})^2}\)	✓ 1
Part (b)
A1 \(f(0)=2.5\) and \(f'(0)=1\)	✓ 1
M1 normal gradient is the negative reciprocal, \(-1\)	✓ 1
A1 \(y=2.5-x\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 product rule applied to \(12t\) and \(e^{-0.5t}\)	✓ 1
A1 derivative of \(e^{-0.5t}\) is \(-0.5e^{-0.5t}\)	✓ 1
A1 \(\frac{dC}{dt}=12e^{-0.5t}(1-0.5t)\)	✓ 1
Part (b)
M1 sets \(\frac{dC}{dt}=0\) (ECF from (a))	✓ 1
A1 \(t=2\) hours	✓ 1
Part (c)
A1 \(24e^{-1}=8.83\) mg/L	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(C_{n+1}=C_n+1\times(2-0.3C_n)\)	✓ 1
A1 \(C_1=2\)	✓ 1
A1 \(C_2=3.4\)	✓ 1
A1 \(C_3=4.38\) mg/L	✓ 1
Part (b)
M1 sets \(\frac{dC}{dt}=0\)	✓ 1
A1 \(6.67\) mg/L	✓ 1