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IB Maths AI SL – Topic 5: Calculus

Differentiation, integration, trapezoidal rule, optimisation.

12 questions  ·  50 marks  ·  answers included  ·  IB Mathematics: Applications and Interpretation SL
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This worksheet covers:

  • Topic 5 (differentiation, integration, the trapezoidal rule and optimisation) in the style of Paper 1
  • Paper 2

A GDC is allowed throughout; give answers to 3 significant figures unless exact. Allow about 75 minutes.

Questions

Section A – Skills

Short questions on differentiation and integration.
1

Find $\dfrac{dy}{dx}$ when $y = 3x^4 – 2x^2 + 5x – 7$.

Show answer
$\dfrac{dy}{dx} = 12x^3 – 4x + 5$
[2]
2

The function $f$ is defined by $f(x) = 2x^3 + \dfrac{1}{x}$, $x eq 0$.(a) Find $f'(x)$.(b) Find the gradient of the curve at $x = 2$.

Show answer
(a) $f'(x) = 6x^2 – \dfrac{1}{x^2}$(b) $f'(2) = 24 – 0.25 = 23.75$
[3]
3

The curve $y = x^2 – 4x$ passes through the point $P(3, -3)$.(a) Find the gradient of the curve at $P$.(b) Hence find the equation of the tangent to the curve at $P$ in the form $y = mx + c$.

Show answer
(a) $\dfrac{dy}{dx} = 2x – 4$, gradient $= 2$(b) $y = 2x – 9$
[3]
4

Find $\displaystyle\int (6x^2 – 4x + 3)\,dx$.

Show answer
$2x^3 – 2x^2 + 3x + C$
[2]
5

Evaluate $\displaystyle\int_1^3 (2x + 1)\,dx$.

Show answer
$\left[x^2 + x\right]_1^3 = 12 – 2 = 10$
[3]
6

A function has derivative $f'(x) = 4x – 3$ and passes through the point $(2, 5)$.Find $f(x)$.

Show answer
$f(x) = 2x^2 – 3x + 3$ (since $8 – 6 + c = 5$ gives $c = 3$)
[3]

Section B – Exam-style

7

Consider the curve $y = x^3 – 6x^2 + 9x + 1$.(a) Find $\dfrac{dy}{dx}$.(b) Find the coordinates of the two stationary points of the curve.(c) Determine the nature of each stationary point, justifying your answer.

Show answer
(a) $\dfrac{dy}{dx} = 3x^2 – 12x + 9 = 3(x-1)(x-3)$(b) $(1, 5)$ and $(3, 1)$(c) $(1, 5)$ is a local maximum (gradient changes $+$ to $-$, or $\dfrac{d^2y}{dx^2} = -6 < 0$); $(3, 1)$ is a local minimum ($\dfrac{d^2y}{dx^2} = 6 > 0$)
[6]
8

Let $f(x) = \sqrt{x^2 + 1}$.(a) Complete a table of values of $f(x)$ for $x = 0, 0.5, 1, 1.5, 2$, giving values to 4 decimal places.(b) Use the trapezoidal rule with four intervals to estimate $\displaystyle\int_0^2 \sqrt{x^2+1}\,dx$.(c) Use your GDC to find the exact integral to 4 significant figures, and hence find the percentage error in your estimate.

Show answer
(a) $1,\ 1.1180,\ 1.4142,\ 1.8028,\ 2.2361$(b) $\dfrac{0.5}{2}\left[1 + 2.2361 + 2(1.1180 + 1.4142 + 1.8028)\right] = 2.98$ (2.9765)(c) Exact $= 2.958$; percentage error $\approx 0.63\%$
[6]
9

A company sells $x$ units of a product per week. The weekly revenue is $R(x) = 80x – 0.5x^2$ dollars and the weekly cost is $C(x) = 20x + 500$ dollars.(a) Write down an expression for the weekly profit $P(x)$.(b) Find $\dfrac{dP}{dx}$.(c) Find the number of units that maximises the weekly profit, and the maximum profit.

Show answer
(a) $P(x) = 60x – 0.5x^2 – 500$(b) $\dfrac{dP}{dx} = 60 – x$(c) $x = 60$ units; maximum profit $= \$1300$
[5]
10

The curve $y = 4x – x^2$ meets the $x$-axis at the origin and at the point $A$.(a) Find the $x$-coordinate of $A$.(b) Find the area of the region enclosed by the curve and the $x$-axis.

Show answer
(a) $x = 4$(b) $\displaystyle\int_0^4 (4x – x^2)\,dx = \left[2x^2 – \dfrac{x^3}{3}\right]_0^4 = \dfrac{32}{3} = 10.7$
[4]

Section C – Challenge

11

A cylindrical can with a closed top and bottom is to hold $500\,\text{cm}^3$ of drink. The radius of the can is $r$ cm and its height is $h$ cm.(a) Show that $h = \dfrac{500}{\pi r^2}$.(b) Show that the total surface area of the can is $S = 2\pi r^2 + \dfrac{1000}{r}$.(c) Find $\dfrac{dS}{dr}$.(d) Find the value of $r$ that minimises the surface area, and the corresponding minimum surface area.

Show answer
(a) $V = \pi r^2 h = 500 \Rightarrow h = \dfrac{500}{\pi r^2}$(b) $S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + \dfrac{1000}{r}$(c) $\dfrac{dS}{dr} = 4\pi r – \dfrac{1000}{r^2}$(d) $r^3 = \dfrac{250}{\pi}$, $r = 4.30\,\text{cm}$; $S_{\min} = 349\,\text{cm}^2$ (348.7); ($h = 8.60\,\text{cm}$)
[7]
12

A surveyor measures the depth of a river at 2 m intervals across its width. The depths (in metres) are $0,\ 1.2,\ 2.5,\ 3.1,\ 2.8,\ 1.6,\ 0$.(a) Use the trapezoidal rule to estimate the cross-sectional area of the river.(b) The water flows at an average speed of $0.5\,\text{m s}^{-1}$. Estimate the volume of water passing a point on the river each second.(c) Explain how the surveyor could obtain a more accurate estimate of the area.

Show answer
(a) $\dfrac{2}{2}\left[0 + 0 + 2(1.2 + 2.5 + 3.1 + 2.8 + 1.6)\right] = 22.4\,\text{m}^2$(b) $22.4 \times 0.5 = 11.2\,\text{m}^3\,\text{s}^{-1}$(c) Take depth readings at smaller intervals (more strips)
[6]

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