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IB Maths AA HL – Topic 5: Calculus

Implicit differentiation, related rates, integration techniques, Maclaurin series, differential equations.

12 questions  ·  64 marks  ·  answers included  ·  IB AA HL
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This worksheet covers:

  • HL-only calculus
  • Implicit differentiation
  • Related rates
  • Integration by parts
  • Substitution and partial fractions
  • Maclaurin series
  • L’Hôpital’s rule
  • Euler’s method
  • First-order differential equations

Section A is non-calculator; allow about 90 minutes.

Questions

Section A – Skills

Non-calculator
1

The curve $C$ has equation $x^2 + xy + y^2 = 7$. Find the gradient of $C$ at the point $(1, 2)$.

Show answer
$2x + y + x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0$; at $(1,2)$: $4 + 5\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{4}{5}$
[4]
2

Find $\displaystyle\int x\,e^{2x}\,dx$.

Show answer
$\dfrac{x}{2}e^{2x} – \dfrac{1}{4}e^{2x} + C$
[4]
3

Using the substitution $u = 1 + x^2$, or otherwise, find the exact value of $\displaystyle\int_0^1 x\sqrt{1 + x^2}\,dx$.

Show answer
$\left[\tfrac{1}{3}(1 + x^2)^{3/2}\right]_0^1 = \dfrac{2\sqrt{2} – 1}{3}$
[5]
4

Find $\displaystyle\int \frac{3x + 1}{(x – 1)(x + 2)}\,dx$.

Show answer
$\dfrac{3x+1}{(x-1)(x+2)} = \dfrac{4/3}{x-1} + \dfrac{5/3}{x+2}$; integral $= \dfrac{4}{3}\ln|x – 1| + \dfrac{5}{3}\ln|x + 2| + C$
[5]
5

Find the Maclaurin series of $e^x \sin x$ up to and including the term in $x^3$.

Show answer
$\left(1 + x + \tfrac{x^2}{2} + \tfrac{x^3}{6}\right)\left(x – \tfrac{x^3}{6}\right) = x + x^2 + \dfrac{x^3}{3} + \dots$
[4]
6

Solve the differential equation $\dfrac{dy}{dx} = xy$, given that $y = 2$ when $x = 0$. Give your answer in the form $y = f(x)$.

Show answer
$\ln|y| = \dfrac{x^2}{2} + c$; $y = 2e^{x^2/2}$
[4]

Section B – Exam-style

Calculator allowed
7

A spherical balloon is inflated so that its radius increases at a constant rate of $0.5\,\text{cm s}^{-1}$.(a) Find the rate of increase of the volume when the radius is $4$ cm.(b) Find the rate of increase of the surface area at the same instant.

Show answer
(a) $\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt} = 4\pi(16)(0.5) = 32\pi \approx 101\,\text{cm}^3\text{s}^{-1}$(b) $\dfrac{dA}{dt} = 8\pi r\dfrac{dr}{dt} = 16\pi \approx 50.3\,\text{cm}^2\text{s}^{-1}$
[5]
8

A ladder of length $5$ m rests against a vertical wall. The foot of the ladder slides away from the wall at $0.3\,\text{m s}^{-1}$. Find the rate at which the top of the ladder is sliding down the wall when the foot is $3$ m from the wall.

Show answer
$x^2 + y^2 = 25 \Rightarrow 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$; with $x = 3$, $y = 4$: $\dfrac{dy}{dt} = -\dfrac{3(0.3)}{4} = -0.225\,\text{m s}^{-1}$ (sliding down at $0.225\,\text{m s}^{-1}$)
[5]
9

Consider the differential equation $\dfrac{dy}{dx} + \dfrac{2y}{x} = x$, $x > 0$, with $y = 1$ when $x = 1$.(a) Find the integrating factor.(b) Hence find $y$ in terms of $x$.

Show answer
(a) $e^{\int 2/x\,dx} = x^2$(b) $\dfrac{d}{dx}(x^2 y) = x^3 \Rightarrow x^2 y = \dfrac{x^4}{4} + c$; $c = \dfrac{3}{4}$, so $y = \dfrac{x^2}{4} + \dfrac{3}{4x^2}$
[6]
10

Consider $\dfrac{dy}{dx} = x + y$ with $y(0) = 1$.(a) Use Euler’s method with step length $h = 0.1$ to estimate $y(0.2)$.(b) The exact solution is $y = 2e^x – x – 1$. Find the percentage error in your estimate.

Show answer
(a) $y_1 = 1 + 0.1(0 + 1) = 1.1$; $y_2 = 1.1 + 0.1(0.1 + 1.1) = 1.22$(b) Exact $y(0.2) = 1.2428$; error $= \dfrac{1.2428 – 1.22}{1.2428} \times 100 \approx 1.84\%$
[6]

Section C – Challenge

11

(a) Find the Maclaurin series of $\ln(1 + 2x)$ up to and including the term in $x^3$.(b) Use it to find an approximation for $\ln 1.2$, and compare with the calculator value.(c) Use l’Hôpital’s rule to find $\displaystyle\lim_{x \to 0}\frac{e^x – 1 – x}{x^2}$.

Show answer
(a) $2x – 2x^2 + \dfrac{8}{3}x^3 + \dots$(b) $x = 0.1$: $0.2 – 0.02 + 0.00267 = 0.1827$; calculator gives $0.1823$ (agree to 3 sf)(c) $\lim \dfrac{e^x – 1}{2x} = \lim \dfrac{e^x}{2} = \dfrac{1}{2}$
[8]
12

Consider $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}$, $x > 0$, $y > 0$, with $y = 2$ when $x = 1$.(a) Use the substitution $y = vx$ to show that $v\dfrac{dv}{dx} = \dfrac{1}{x}$.(b) Hence find $y^2$ in terms of $x$.

Show answer
(a) $v + x\dfrac{dv}{dx} = \dfrac{1 + v^2}{v} \Rightarrow x\dfrac{dv}{dx} = \dfrac{1}{v}$, so $v\dfrac{dv}{dx} = \dfrac{1}{x}$(b) $\dfrac{v^2}{2} = \ln x + c$; $v = 2$ at $x = 1 \Rightarrow c = 2$; $y^2 = 2x^2(\ln x + 2)$
[8]

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