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IB Maths AI HL – Topic 5: Calculus

Differential equations, Euler’s method, slope fields, coupled systems.

12 questions  ·  74 marks  ·  answers included  ·  IB AI HL
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This worksheet covers:

  • HL-only calculus for Applications and Interpretation
  • Separable differential equations
  • Slope fields
  • Euler’s method
  • Coupled linear systems (eigenvalues and phase portraits)
  • Modelling

A GDC is expected throughout; allow about 90 minutes.

Questions

Section A – Skills

1

Solve $\dfrac{dy}{dx} = 2y$ given that $y = 3$ when $x = 0$.

Show answer
$\ln|y| = 2x + c$; $y = 3e^{2x}$
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2

Consider $\dfrac{dy}{dx} = x^2 – y$ with $y(0) = 1$. Use Euler’s method with step length $h = 0.2$ to find an approximation for $y(0.6)$. Give each intermediate value to 4 decimal places.

Show answer
$y(0.2) \approx 0.8$, $y(0.4) \approx 0.648$, $y(0.6) \approx 0.5504$
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3

The slope field for $\dfrac{dy}{dx} = y – x$ is drawn on the region $-3 \le x \le 3$, $-3 \le y \le 3$.(a) Write down the equation of the line along which all the slope segments are horizontal.(b) Describe the gradient of the segments at points above and below this line.(c) Show that $y = x + 1$ is a solution, and describe the shape of the solution curve through $(0, 2)$.

Show answer
(a) $y = x$ (isocline for gradient $0$)(b) Above $y = x$: $y – x > 0$, positive gradient, steeper further from the line; below: negative gradient(c) $\dfrac{dy}{dx} = 1$ and $y – x = 1$, so it satisfies the equation. Through $(0,2)$ the curve lies above $y = x + 1$ and diverges from it, increasing and steepening (it is $y = x + 1 + e^x$)
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4

Find the general solution of $\dfrac{dy}{dx} = \dfrac{\cos x}{y}$, and the particular solution with $y = 2$ when $x = 0$.

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$\dfrac{y^2}{2} = \sin x + c$; particular: $y^2 = 2\sin x + 4$, i.e. $y = \sqrt{2\sin x + 4}$
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5

Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$, $y > 0$, given that $y = 3$ when $x = 0$, giving $y$ explicitly in terms of $x$.

Show answer
$\dfrac{y^2}{2} = \dfrac{x^2}{2} + c$; $y^2 = x^2 + 9$, so $y = \sqrt{x^2 + 9}$
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Section B – Exam-style

6

A cup of coffee cools according to $\dfrac{dT}{dt} = -k(T – 20)$, where $T$ is the temperature in $^\circ$C after $t$ minutes. Initially $T = 90$ and after 10 minutes $T = 60$.(a) Show that $T = 20 + 70e^{-kt}$.(b) Find $k$.(c) Find the temperature after 30 minutes.

Show answer
(a) Separate: $\ln|T – 20| = -kt + c$, $T(0) = 90 \Rightarrow T = 20 + 70e^{-kt}$(b) $40 = 70e^{-10k} \Rightarrow k = \dfrac{1}{10}\ln\dfrac{7}{4} = 0.0560$(c) $T = 20 + 70\left(\dfrac{4}{7}\right)^3 = 33.1^\circ$C
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7

The number of fish $P$ in a lake satisfies $\dfrac{dP}{dt} = 0.4P\left(1 – \dfrac{P}{500}\right)$, with $P = 50$ when $t = 0$ (years).(a) State the carrying capacity.(b) The solution is $P = \dfrac{500}{1 + Ae^{-0.4t}}$. Find $A$.(c) Find $P$ after 5 years.(d) Find the time at which the population is growing fastest.

Show answer
(a) $500$(b) $A = 9$(c) $P(5) = \dfrac{500}{1 + 9e^{-2}} = 225$ (225.4)(d) Fastest when $P = 250$: $9e^{-0.4t} = 1 \Rightarrow t = \dfrac{\ln 9}{0.4} = 5.49$ years
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8

Consider the system $\dfrac{dx}{dt} = x + 4y$, $\dfrac{dy}{dt} = x + y$.(a) Find the eigenvalues and corresponding eigenvectors of the coefficient matrix.(b) Write down the general solution.(c) Find the solution satisfying $x(0) = 4$, $y(0) = 0$.(d) Describe the phase portrait and state the long-term behaviour of $\dfrac{y}{x}$.

Show answer
(a) $(1-\lambda)^2 – 4 = 0 \Rightarrow \lambda = 3$ with eigenvector $\begin{pmatrix}2\\1\end{pmatrix}$; $\lambda = -1$ with eigenvector $\begin{pmatrix}2\\-1\end{pmatrix}$(b) $\begin{pmatrix}x\\y\end{pmatrix} = A e^{3t}\begin{pmatrix}2\\1\end{pmatrix} + B e^{-t}\begin{pmatrix}2\\-1\end{pmatrix}$(c) $2A + 2B = 4$, $A – B = 0 \Rightarrow A = B = 1$: $x = 2e^{3t} + 2e^{-t}$, $y = e^{3t} – e^{-t}$(d) Saddle point (unstable); trajectories approach the line $y = \tfrac{1}{2}x$, so $\dfrac{y}{x} \to \dfrac{1}{2}$
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9

Consider $\dfrac{dx}{dt} = y$, $\dfrac{dy}{dt} = -x$, with $x(0) = 1$, $y(0) = 0$.(a) Use Euler’s method with $h = 0.1$ to estimate $x(0.2)$ and $y(0.2)$.(b) The exact solution is $x = \cos t$, $y = -\sin t$. Comment on the accuracy of the estimates.(c) Find the eigenvalues of the system and hence describe the phase portrait.

Show answer
(a) $x_1 = 1$, $y_1 = -0.1$; $x_2 = 0.99$, $y_2 = -0.2$(b) Exact: $x = 0.980$, $y = -0.199$; estimates accurate to about $0.01$, Euler slightly overestimates because trajectories curve (it spirals outward)(c) $\lambda^2 + 1 = 0 \Rightarrow \lambda = \pm i$; purely imaginary, so closed circular orbits about the origin (a centre)
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Section C – Challenge

10

A predator–prey model is $\dfrac{dx}{dt} = 0.5x – 0.02xy$, $\dfrac{dy}{dt} = -0.3y + 0.01xy$, where $x$ is the prey and $y$ the predator population (hundreds).(a) Find the equilibrium points.(b) At the non-zero equilibrium, describe the behaviour of the populations for a starting point close to it.(c) If initially $x = 60$ and $y = 25$, state whether $y$ is initially increasing or decreasing, and justify.

Show answer
(a) $(0, 0)$ and $(30, 25)$(b) Populations oscillate in closed cycles around $(30, 25)$, with predator numbers lagging behind prey numbers(c) $\dfrac{dy}{dt} = -0.3(25) + 0.01(60)(25) = 7.5 > 0$, so $y$ is increasing
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11

The displacement $x$ of a damped oscillator satisfies $\dfrac{d^2x}{dt^2} + 2\dfrac{dx}{dt} + 5x = 0$.(a) By letting $y = \dfrac{dx}{dt}$, write the equation as a system of two first-order equations.(b) Find the eigenvalues of the system and describe the phase portrait.(c) With $x(0) = 1$, $y(0) = 0$, use Euler’s method with $h = 0.1$ to estimate $x(0.2)$.

Show answer
(a) $\dfrac{dx}{dt} = y$, $\dfrac{dy}{dt} = -5x – 2y$(b) $\lambda^2 + 2\lambda + 5 = 0 \Rightarrow \lambda = -1 \pm 2i$; negative real part, so a stable spiral (trajectories spiral in to the origin)(c) $x_1 = 1$, $y_1 = -0.5$; $x_2 = 0.95$, $y_2 = -0.5 + 0.1(-5 – (-1)) = -0.9$; so $x(0.2) \approx 0.95$
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12

Salt water is pumped into a tank so that the mass $m$ kg of salt after $t$ minutes satisfies $\dfrac{dm}{dt} = 10 – 0.05m$, with $m = 0$ initially.(a) Solve the equation to find $m$ in terms of $t$.(b) Find the limiting mass of salt in the tank.(c) Find the time taken for the mass to reach 150 kg.

Show answer
(a) $\dfrac{dm}{10 – 0.05m} = dt \Rightarrow -20\ln|10 – 0.05m| = t + c$; $m = 200\left(1 – e^{-0.05t}\right)$(b) $200$ kg(c) $e^{-0.05t} = 0.25 \Rightarrow t = 20\ln 4 = 27.7$ minutes
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