IB · MATH AI HL

Mathematics: Applications & Interpretation HL

Advanced calculus and modelling — Topics 5 and 1

Name: ____________________Date: October 14, 2026

1.
A tank's volume is modelled by V(t) = 40 + 6t − 0.5t², where V is in litres and t is time in minutes, for 0 ≤ t ≤ 12. (a) Find dV/dt. (b) Find the time at which the volume is a maximum. (c) State the maximum volume.
[4 marks]
Marking points
- Differentiates to obtain dV/dt = 6 − t.
- Sets dV/dt = 0 to obtain t = 6.
- Confirms this gives a maximum, for example using the second derivative or a sign change.
- States the maximum volume V(6) = 58 litres.
Examiner tip: Always confirm a stationary point is a maximum before quoting it as the answer to a 'maximum volume' question.
2.
Marking analysis: A learner attempts the following task: “A tank's volume is modelled by V(t) = 40 + 6t − 0.5t², where V is in litres and t is time in minutes, for 0 ≤ t ≤ 12. (a) Find dV/dt. (b) Find the time at which the volume is a maximum. (c) State the maximum volume.” Their response addresses only this point: “Differentiates to obtain dV/dt = 6 − t.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Differentiates to obtain dV/dt = 6 − t.
- Identifies the missing requirement: Sets dV/dt = 0 to obtain t = 6.
- Identifies the missing requirement: Confirms this gives a maximum, for example using the second derivative or a sign change.
- Identifies the missing requirement: States the maximum volume V(6) = 58 litres.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
3.
The rate of change of the temperature T (°C) of a cooling drink is modelled by dT/dt = −0.12(T − 4), where t is in minutes and initially T = 20. (a) Use Euler's method with a step length of 1 minute to estimate T after 2 minutes, showing both steps. (b) State one way the estimate could be made more accurate without changing the model.
[5 marks]
Marking points
- States Euler's update rule Tₙ₊₁ = Tₙ + h·f(Tₙ).
- Computes step 1: T(1) ≈ 20 + 1(−0.12)(20 − 4) = 18.08.
- Computes step 2: T(2) ≈ 18.08 + 1(−0.12)(18.08 − 4) = 16.39 (accept rounding).
- States the final estimate T(2) ≈ 16.4 °C.
- States a valid improvement, such as using a smaller step length.
Examiner tip: Recompute f at the new value before each Euler step; do not reuse the original rate for both steps.
4.
Marking analysis: A learner attempts the following task: “The rate of change of the temperature T (°C) of a cooling drink is modelled by dT/dt = −0.12(T − 4), where t is in minutes and initially T = 20. (a) Use Euler's method with a step length of 1 minute to estimate T after 2 minutes, showing both steps. (b) State one way the estimate could be made more accurate without changing the model.” Their response addresses only this point: “States Euler's update rule Tₙ₊₁ = Tₙ + h·f(Tₙ).” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: States Euler's update rule Tₙ₊₁ = Tₙ + h·f(Tₙ).
- Identifies the missing requirement: Computes step 1: T(1) ≈ 20 + 1(−0.12)(20 − 4) = 18.08.
- Identifies the missing requirement: Computes step 2: T(2) ≈ 18.08 + 1(−0.12)(18.08 − 4) = 16.39 (accept rounding).
- Identifies the missing requirement: States the final estimate T(2) ≈ 16.4 °C.
- Identifies the missing requirement: States a valid improvement, such as using a smaller step length.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
5.
A city's population of a rare bird species, P(t) hundreds, t years after monitoring begins, is modelled by the logistic differential equation dP/dt = 0.3P(1 − P/8), with P(0) = 1. (a) Show that P = 8 is an equilibrium solution and interpret it in context. (b) By separating variables, show that the general solution can be written P(t) = 8 / (1 + 7e^(−0.3t)). (c) Find the population predicted after 10 years. (d) Find the time at which the population is growing fastest, and state the population at that time. (e) Comment on one limitation of using this model to predict the population after 50 years.
[8 marks]
Marking points
- Sets dP/dt = 0 and solves 0.3P(1 − P/8) = 0 to obtain P = 0 or P = 8, identifying P = 8 as the non-trivial equilibrium.
- Interprets P = 8 (hundreds) as the long-term carrying capacity the population approaches.
- Separates variables to obtain ∫ dP/[P(1 − P/8)] = ∫0.3 dt.
- Uses partial fractions (or GDC-supported integration) with the initial condition P(0) = 1 to justify the given closed form.
- Substitutes t = 10 to obtain P(10) ≈ 5.9 hundred (accept equivalent rounding).
- Recognises that the fastest growth occurs at the inflection point, where P = 4 (half the carrying capacity).
- Solves 8/(1 + 7e^(−0.3t)) = 4 to obtain t ≈ 6.48 years.
- States a valid limitation, such as the model assuming no major environmental change, disease, or human impact over a 50-year horizon.
Examiner tip: For a logistic model, growth is fastest exactly halfway to the carrying capacity — you can locate this point without fully solving for t first.
6.
Marking analysis: A learner attempts the following task: “A city's population of a rare bird species, P(t) hundreds, t years after monitoring begins, is modelled by the logistic differential equation dP/dt = 0.3P(1 − P/8), with P(0) = 1. (a) Show that P = 8 is an equilibrium solution and interpret it in context. (b) By separating variables, show that the general solution can be written P(t) = 8 / (1 + 7e^(−0.3t)). (c) Find the population predicted after 10 years. (d) Find the time at which the population is growing fastest, and state the population at that time. (e) Comment on one limitation of using this model to predict the population after 50 years.” Their response addresses only this point: “Sets dP/dt = 0 and solves 0.3P(1 − P/8) = 0 to obtain P = 0 or P = 8, identifying P = 8 as the non-trivial equilibrium.” Evaluate the response against the complete 8-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[8 marks]
Marking points
- Recognises credit for the stated point: Sets dP/dt = 0 and solves 0.3P(1 − P/8) = 0 to obtain P = 0 or P = 8, identifying P = 8 as the non-trivial equilibrium.
- Identifies the missing requirement: Interprets P = 8 (hundreds) as the long-term carrying capacity the population approaches.
- Identifies the missing requirement: Separates variables to obtain ∫ dP/[P(1 − P/8)] = ∫0.3 dt.
- Identifies the missing requirement: Uses partial fractions (or GDC-supported integration) with the initial condition P(0) = 1 to justify the given closed form.
- Identifies the missing requirement: Substitutes t = 10 to obtain P(10) ≈ 5.9 hundred (accept equivalent rounding).
- Identifies the missing requirement: Recognises that the fastest growth occurs at the inflection point, where P = 4 (half the carrying capacity).
- Identifies the missing requirement: Solves 8/(1 + 7e^(−0.3t)) = 4 to obtain t ≈ 6.48 years.
- Identifies the missing requirement: States a valid limitation, such as the model assuming no major environmental change, disease, or human impact over a 50-year horizon.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
7.
The radius of a circular oil spill is increasing at a rate of 3 cm/s. Find the rate at which the area of the spill is increasing when the radius is 10 cm.
[4 marks]
Marking points
- States A = πr² and differentiates with respect to time using the chain rule: dA/dt = 2πr·(dr/dt).
- Substitutes r = 10 and dr/dt = 3.
- Evaluates dA/dt = 2π(10)(3).
- Obtains dA/dt ≈ 188.5 cm²/s.
Examiner tip: Related rates problems always use the chain rule to connect the rate you want to the rate you are given — identify the equation linking the two quantities first, then differentiate both sides with respect to time.
8.
Marking analysis: A learner attempts the following task: “The radius of a circular oil spill is increasing at a rate of 3 cm/s. Find the rate at which the area of the spill is increasing when the radius is 10 cm.” Their response addresses only this point: “States A = πr² and differentiates with respect to time using the chain rule: dA/dt = 2πr·(dr/dt).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: States A = πr² and differentiates with respect to time using the chain rule: dA/dt = 2πr·(dr/dt).
- Identifies the missing requirement: Substitutes r = 10 and dr/dt = 3.
- Identifies the missing requirement: Evaluates dA/dt = 2π(10)(3).
- Identifies the missing requirement: Obtains dA/dt ≈ 188.5 cm²/s.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
9.
The curve x²y + y³ = 10 passes through the point (1, 2). Use implicit differentiation to find dy/dx at this point.
[4 marks]
Marking points
- Differentiates both sides with respect to x, using the product rule on x²y: 2xy + x²(dy/dx) + 3y²(dy/dx) = 0.
- Collects the dy/dx terms: (dy/dx)(x² + 3y²) = −2xy.
- Rearranges to obtain dy/dx = −2xy/(x² + 3y²).
- Substitutes x = 1, y = 2 to obtain dy/dx = −4/13 ≈ −0.308.
Examiner tip: When a term mixes x and y, such as x²y, apply the product rule and remember that differentiating any y-term introduces a dy/dx factor via the chain rule.
10.
Marking analysis: A learner attempts the following task: “The curve x²y + y³ = 10 passes through the point (1, 2). Use implicit differentiation to find dy/dx at this point.” Their response addresses only this point: “Differentiates both sides with respect to x, using the product rule on x²y: 2xy + x²(dy/dx) + 3y²(dy/dx) = 0.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Differentiates both sides with respect to x, using the product rule on x²y: 2xy + x²(dy/dx) + 3y²(dy/dx) = 0.
- Identifies the missing requirement: Collects the dy/dx terms: (dy/dx)(x² + 3y²) = −2xy.
- Identifies the missing requirement: Rearranges to obtain dy/dx = −2xy/(x² + 3y²).
- Identifies the missing requirement: Substitutes x = 1, y = 2 to obtain dy/dx = −4/13 ≈ −0.308.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
11.
For the curve f(x) = x³ − 3x² − 9x + 5, find the x-coordinate of the point of inflection using the second derivative, and state the corresponding y-coordinate.
[4 marks]
Marking points
- Differentiates twice to obtain f''(x) = 6x − 6.
- Sets f''(x) = 0 and solves to obtain x = 1.
- Confirms this is a genuine inflection point since f'' changes sign at x = 1.
- Substitutes x = 1 into f(x) to obtain y = −6.
Examiner tip: A point where the second derivative equals zero is only a genuine inflection point if concavity actually changes there — check the sign of f'' on either side to confirm.
12.
Marking analysis: A learner attempts the following task: “For the curve f(x) = x³ − 3x² − 9x + 5, find the x-coordinate of the point of inflection using the second derivative, and state the corresponding y-coordinate.” Their response addresses only this point: “Differentiates twice to obtain f''(x) = 6x − 6.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Differentiates twice to obtain f''(x) = 6x − 6.
- Identifies the missing requirement: Sets f''(x) = 0 and solves to obtain x = 1.
- Identifies the missing requirement: Confirms this is a genuine inflection point since f'' changes sign at x = 1.
- Identifies the missing requirement: Substitutes x = 1 into f(x) to obtain y = −6.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
13.
Given the differential equation dy/dx = x + y, with y(0) = 1, use Euler's method with a step length of 0.5 to estimate y(1), showing both steps.
[5 marks]
Marking points
- States Euler's update rule yₙ₊₁ = yₙ + h·f(xₙ, yₙ).
- Computes step 1: y(0.5) ≈ 1 + 0.5(0 + 1) = 1.5.
- Computes step 2: y(1) ≈ 1.5 + 0.5(0.5 + 1.5) = 2.5.
- States the final estimate y(1) ≈ 2.5.
- Correctly uses the updated x-value (0.5, not 0) in the second step's evaluation of f.
Examiner tip: Keep a small table of (x, y) pairs as you work through Euler's method — it prevents the common error of reusing the previous x-value in the next step.
14.
Marking analysis: A learner attempts the following task: “Given the differential equation dy/dx = x + y, with y(0) = 1, use Euler's method with a step length of 0.5 to estimate y(1), showing both steps.” Their response addresses only this point: “States Euler's update rule yₙ₊₁ = yₙ + h·f(xₙ, yₙ).” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: States Euler's update rule yₙ₊₁ = yₙ + h·f(xₙ, yₙ).
- Identifies the missing requirement: Computes step 1: y(0.5) ≈ 1 + 0.5(0 + 1) = 1.5.
- Identifies the missing requirement: Computes step 2: y(1) ≈ 1.5 + 0.5(0.5 + 1.5) = 2.5.
- Identifies the missing requirement: States the final estimate y(1) ≈ 2.5.
- Identifies the missing requirement: Correctly uses the updated x-value (0.5, not 0) in the second step's evaluation of f.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
15.
Solve the differential equation dy/dx = 2xy, given that y = 3 when x = 0, by separating variables. Hence find y when x = 1.
[5 marks]
Marking points
- Separates variables: ∫(1/y) dy = ∫2x dx.
- Integrates to obtain ln|y| = x² + c.
- Uses y(0) = 3 to find c = ln(3).
- Rearranges to obtain y = 3e^(x²).
- Substitutes x = 1 to obtain y ≈ 8.15.
Examiner tip: After separating variables and integrating, always use the initial condition immediately to find the constant — this avoids carrying an unnecessary unknown through the final substitution.
16.
Marking analysis: A learner attempts the following task: “Solve the differential equation dy/dx = 2xy, given that y = 3 when x = 0, by separating variables. Hence find y when x = 1.” Their response addresses only this point: “Separates variables: ∫(1/y) dy = ∫2x dx.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Separates variables: ∫(1/y) dy = ∫2x dx.
- Identifies the missing requirement: Integrates to obtain ln|y| = x² + c.
- Identifies the missing requirement: Uses y(0) = 3 to find c = ln(3).
- Identifies the missing requirement: Rearranges to obtain y = 3e^(x²).
- Identifies the missing requirement: Substitutes x = 1 to obtain y ≈ 8.15.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
17.
A particle's position vector is given by r(t) = (t², 3t) metres, where t is time in seconds. (a) Find the velocity vector v(t). (b) Find the speed of the particle at t = 2 seconds.
[4 marks]
Marking points
- Differentiates each component of r(t) to obtain v(t) = (2t, 3).
- Substitutes t = 2 to obtain v(2) = (4, 3).
- Uses speed = |v(2)| = √(4² + 3²).
- Obtains a speed of 5 m/s.
Examiner tip: Differentiate a position vector component by component to get velocity, then take the magnitude of the velocity vector to get speed — speed is always a positive scalar, not a vector.
18.
Marking analysis: A learner attempts the following task: “A particle's position vector is given by r(t) = (t², 3t) metres, where t is time in seconds. (a) Find the velocity vector v(t). (b) Find the speed of the particle at t = 2 seconds.” Their response addresses only this point: “Differentiates each component of r(t) to obtain v(t) = (2t, 3).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Differentiates each component of r(t) to obtain v(t) = (2t, 3).
- Identifies the missing requirement: Substitutes t = 2 to obtain v(2) = (4, 3).
- Identifies the missing requirement: Uses speed = |v(2)| = √(4² + 3²).
- Identifies the missing requirement: Obtains a speed of 5 m/s.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
19.
A cylindrical can must have a volume of 500 cm³. (a) Express the surface area S of the can in terms of its radius r only. (b) Find the radius and height that minimise the surface area, and find the minimum surface area.
[6 marks]
Marking points
- Uses V = πr²h = 500 to express h = 500/(πr²).
- Writes S = 2πr² + 2πrh and substitutes h to obtain S(r) = 2πr² + 1000/r.
- Differentiates to obtain S'(r) = 4πr − 1000/r².
- Sets S'(r) = 0 and solves to obtain r ≈ 4.30 cm.
- Confirms a minimum using S''(r) > 0, and finds h ≈ 8.60 cm.
- Substitutes back to obtain a minimum surface area of approximately 348.7 cm².
Examiner tip: Optimisation problems with two variables always need a constraint equation (here, fixed volume) to eliminate one variable before differentiating — never differentiate a function of two variables directly in this context.
20.
Marking analysis: A learner attempts the following task: “A cylindrical can must have a volume of 500 cm³. (a) Express the surface area S of the can in terms of its radius r only. (b) Find the radius and height that minimise the surface area, and find the minimum surface area.” Their response addresses only this point: “Uses V = πr²h = 500 to express h = 500/(πr²).” Evaluate the response against the complete 6-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[6 marks]
Marking points
- Recognises credit for the stated point: Uses V = πr²h = 500 to express h = 500/(πr²).
- Identifies the missing requirement: Writes S = 2πr² + 2πrh and substitutes h to obtain S(r) = 2πr² + 1000/r.
- Identifies the missing requirement: Differentiates to obtain S'(r) = 4πr − 1000/r².
- Identifies the missing requirement: Sets S'(r) = 0 and solves to obtain r ≈ 4.30 cm.
- Identifies the missing requirement: Confirms a minimum using S''(r) > 0, and finds h ≈ 8.60 cm.
- Identifies the missing requirement: Substitutes back to obtain a minimum surface area of approximately 348.7 cm².
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
21.
A spherical balloon is being inflated so that its volume increases at a constant rate of 100 cm³/s. Find the rate at which the radius is increasing when the radius is 5 cm.
[4 marks]
Marking points
- States V = (4/3)πr³ and differentiates with respect to time: dV/dt = 4πr²(dr/dt).
- Substitutes dV/dt = 100 and r = 5.
- Rearranges to obtain dr/dt = 100/(4π(5)²).
- Obtains dr/dt ≈ 0.318 cm/s.
Examiner tip: Notice that as the balloon gets bigger, the same volume flow rate produces a smaller rate of change in radius — this is because surface area (and hence 4πr²) grows with r.
22.
Marking analysis: A learner attempts the following task: “A spherical balloon is being inflated so that its volume increases at a constant rate of 100 cm³/s. Find the rate at which the radius is increasing when the radius is 5 cm.” Their response addresses only this point: “States V = (4/3)πr³ and differentiates with respect to time: dV/dt = 4πr²(dr/dt).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: States V = (4/3)πr³ and differentiates with respect to time: dV/dt = 4πr²(dr/dt).
- Identifies the missing requirement: Substitutes dV/dt = 100 and r = 5.
- Identifies the missing requirement: Rearranges to obtain dr/dt = 100/(4π(5)²).
- Identifies the missing requirement: Obtains dr/dt ≈ 0.318 cm/s.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
23.
A population of bacteria grows according to dP/dt = 0.05P, with P(0) = 200. Find the general solution for P(t), and use it to find the population after 20 hours.
[4 marks]
Marking points
- Separates variables and integrates to obtain ln(P) = 0.05t + c.
- Uses P(0) = 200 to obtain the general solution P(t) = 200e^(0.05t).
- Substitutes t = 20.
- Obtains P(20) ≈ 544 (bacteria, to the nearest integer).
Examiner tip: A differential equation of the form dP/dt = kP always has the general solution P = P₀e^(kt) — recognising this pattern immediately avoids reworking the separation of variables each time.
24.
Marking analysis: A learner attempts the following task: “A population of bacteria grows according to dP/dt = 0.05P, with P(0) = 200. Find the general solution for P(t), and use it to find the population after 20 hours.” Their response addresses only this point: “Separates variables and integrates to obtain ln(P) = 0.05t + c.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Separates variables and integrates to obtain ln(P) = 0.05t + c.
- Identifies the missing requirement: Uses P(0) = 200 to obtain the general solution P(t) = 200e^(0.05t).
- Identifies the missing requirement: Substitutes t = 20.
- Identifies the missing requirement: Obtains P(20) ≈ 544 (bacteria, to the nearest integer).
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
25.
For the curve f(x) = x⁴ − 4x³, find all values of x where the curve has a point of inflection, using the second derivative.
[5 marks]
Marking points
- Differentiates twice to obtain f''(x) = 12x² − 24x.
- Sets f''(x) = 0 and factorises to obtain 12x(x − 2) = 0.
- Obtains candidate values x = 0 and x = 2.
- Confirms f'' changes sign at both values, so both are genuine points of inflection.
- States the corresponding y-coordinates as f(0) = 0 and f(2) = −16.
Examiner tip: A quartic can have more than one point of inflection — always solve the second derivative equation fully rather than stopping at the first root found.
26.
Marking analysis: A learner attempts the following task: “For the curve f(x) = x⁴ − 4x³, find all values of x where the curve has a point of inflection, using the second derivative.” Their response addresses only this point: “Differentiates twice to obtain f''(x) = 12x² − 24x.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Differentiates twice to obtain f''(x) = 12x² − 24x.
- Identifies the missing requirement: Sets f''(x) = 0 and factorises to obtain 12x(x − 2) = 0.
- Identifies the missing requirement: Obtains candidate values x = 0 and x = 2.
- Identifies the missing requirement: Confirms f'' changes sign at both values, so both are genuine points of inflection.
- Identifies the missing requirement: States the corresponding y-coordinates as f(0) = 0 and f(2) = −16.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
27.
Use the trapezoidal rule with 4 strips to estimate ∫₀⁴ x² dx, and compare your estimate with the exact value found by integration.
[5 marks]
Marking points
- Uses strip width h = 1 and evaluates f(x) = x² at x = 0, 1, 2, 3, 4 to obtain 0, 1, 4, 9, 16.
- Applies the trapezoidal rule: (h/2)[f(x0) + 2(f(x1)+f(x2)+f(x3)) + f(x4)].
- Obtains a trapezoidal estimate of 22.
- Finds the exact value by integration: ∫₀⁴ x² dx = [x³/3]₀⁴ = 64/3 ≈ 21.33.
- Comments that the trapezoidal rule overestimates the integral here because the curve is concave up, so the trapezoids lie above the curve.
Examiner tip: For a concave-up curve, the trapezoidal rule always overestimates the true area; for a concave-down curve, it always underestimates — use this to sanity-check your numerical answer.
28.
Marking analysis: A learner attempts the following task: “Use the trapezoidal rule with 4 strips to estimate ∫₀⁴ x² dx, and compare your estimate with the exact value found by integration.” Their response addresses only this point: “Uses strip width h = 1 and evaluates f(x) = x² at x = 0, 1, 2, 3, 4 to obtain 0, 1, 4, 9, 16.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Uses strip width h = 1 and evaluates f(x) = x² at x = 0, 1, 2, 3, 4 to obtain 0, 1, 4, 9, 16.
- Identifies the missing requirement: Applies the trapezoidal rule: (h/2)[f(x0) + 2(f(x1)+f(x2)+f(x3)) + f(x4)].
- Identifies the missing requirement: Obtains a trapezoidal estimate of 22.
- Identifies the missing requirement: Finds the exact value by integration: ∫₀⁴ x² dx = [x³/3]₀⁴ = 64/3 ≈ 21.33.
- Identifies the missing requirement: Comments that the trapezoidal rule overestimates the integral here because the curve is concave up, so the trapezoids lie above the curve.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
29.
A particle moves along a line with acceleration a(t) = 6t − 4 m/s², where t is in seconds. Given that the initial velocity is v(0) = 2 m/s and the initial displacement is s(0) = 0, find (a) the velocity function v(t), (b) the displacement function s(t), and (c) the displacement at t = 3 seconds.
[6 marks]
Marking points
- Integrates acceleration to obtain v(t) = 3t² − 4t + c.
- Uses v(0) = 2 to obtain v(t) = 3t² − 4t + 2.
- Integrates velocity to obtain s(t) = t³ − 2t² + 2t + k.
- Uses s(0) = 0 to obtain s(t) = t³ − 2t² + 2t.
- Substitutes t = 3 into s(t).
- Obtains a displacement of 15 m at t = 3 seconds.
Examiner tip: Each integration introduces a new constant of integration that must be found from the given initial condition at that stage — do not carry both unknown constants to the end and try to solve them together.
30.
Marking analysis: A learner attempts the following task: “A particle moves along a line with acceleration a(t) = 6t − 4 m/s², where t is in seconds. Given that the initial velocity is v(0) = 2 m/s and the initial displacement is s(0) = 0, find (a) the velocity function v(t), (b) the displacement function s(t), and (c) the displacement at t = 3 seconds.” Their response addresses only this point: “Integrates acceleration to obtain v(t) = 3t² − 4t + c.” Evaluate the response against the complete 6-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[6 marks]
Marking points
- Recognises credit for the stated point: Integrates acceleration to obtain v(t) = 3t² − 4t + c.
- Identifies the missing requirement: Uses v(0) = 2 to obtain v(t) = 3t² − 4t + 2.
- Identifies the missing requirement: Integrates velocity to obtain s(t) = t³ − 2t² + 2t + k.
- Identifies the missing requirement: Uses s(0) = 0 to obtain s(t) = t³ − 2t² + 2t.
- Identifies the missing requirement: Substitutes t = 3 into s(t).
- Identifies the missing requirement: Obtains a displacement of 15 m at t = 3 seconds.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
31.
A slope field for a differential equation shows that wherever a solution curve crosses the line y = x, its gradient is zero, and that the gradient is positive below this line and negative above it. (a) Suggest a differential equation dy/dx = f(x, y) consistent with this description. (b) Describe the long-term behaviour of a solution curve that starts below the line y = x.
[2 marks] · no calculator
Marking points
- Suggests a valid equation such as dy/dx = x − y, which is zero when y = x, positive when y < x, and negative when y > x.
- Describes that a solution starting below the line y = x will have a positive gradient, causing it to rise and approach the line y = x over time, settling near it.
Examiner tip: Reading a slope field is about matching the sign and zero pattern of the gradient to a simple algebraic expression — you do not need to solve the differential equation to describe the qualitative behaviour.
32.
Marking analysis: A learner attempts the following task: “A slope field for a differential equation shows that wherever a solution curve crosses the line y = x, its gradient is zero, and that the gradient is positive below this line and negative above it. (a) Suggest a differential equation dy/dx = f(x, y) consistent with this description. (b) Describe the long-term behaviour of a solution curve that starts below the line y = x.” Their response addresses only this point: “Suggests a valid equation such as dy/dx = x − y, which is zero when y = x, positive when y < x, and negative when y > x.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Marking points
- Recognises credit for the stated point: Suggests a valid equation such as dy/dx = x − y, which is zero when y = x, positive when y < x, and negative when y > x.
- Identifies the missing requirement: Describes that a solution starting below the line y = x will have a positive gradient, causing it to rise and approach the line y = x over time, settling near it.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.