IB Maths AI HL 5 Calculus

Practise IB Maths AI HL calculus through shared-core and HL rates of change, differentiation, integration and optimization with contextual interpretation.

Covered syllabus points

5.1 Calculus - SL content 5.2 Calculus - AHL content

[Maximum number: 6]

Conrad is investigating the motion of a particle. The velocity of the particle, in $\mathrm{ms}^{-1}$ , is given by $v(t)=2 \cos t+\sin 2 t-0.2$ where t is the time, in seconds, after the investigation begins.

Figure for Question 5 Calculus question 1 — IB Maths AI HL

(a)

On the axes,

[ 3 ]

(i)

label, with coordinates, the point(s) where the particle has zero acceleration.

[ 3 ]

(b)

Write down an integral expression for the distance the particle travels during the first 6 seconds of the investigation.

[ 2 ]

(c)

Hence, find the distance the particle travels during the first 6 seconds of the investigation.

[ 1 ]

$7.76(7.76487 \ldots)(\mathrm{m})$ \nA1

[Maximum number: 5]

This question is about applying ideas from logarithms, calculus and probability to an unfamiliar mathematical theory called information theory.
Claude Shannon developed a mathematical theory called information theory to measure the information gained when random events occur. He defined the information, I, that is gained when an event with probability p occurs as

I=-\ln p

where $0<p \leq 1$ . For example, no information is gained ( I=0 ) when an event is certain to $\operatorname{occur}(p=1)$ .

(a)

Show, using calculus, that I is a decreasing function of p.

[ 3 ]

$\frac{\mathrm{d} I}{\mathrm{~d} p}=-\frac{1}{p}$

Note: Award A1 for $\frac{\mathrm{d} I}{\mathrm{~d} p}$ or equivalent (e.g. $I^{\prime}$ ) and A1 for $-\frac{1}{p}$ . Do not accept $\frac{\mathrm{d} y}{\mathrm{~d} x}$ for the first A1 unless followed by $-\frac{1}{x}$ , which can earn A1 A0.
(for $0<p \leq 1$ ) we therefore have $\frac{\mathrm{d} I}{\mathrm{~d} p}<0$
hence the function is decreasing

(b)

When a coin is flipped, the outcome is either heads or tails. The coin may be biased. Let p be the probability of the outcome being heads.

[ 2 ]

(i)

Hence, find the value of p when the expected information gained is maximized.

A famous puzzle uses 12 balls which appear identical. 11 have the same weight, but one is either lighter or heavier than the others. A pair of scales can be repeatedly used to compare the weights of different combinations of the balls.

Figure for Question (i) — IB Maths AI HL

The outcome of each weighing can be "balanced", "left-hand side heavier" or "right-hand side heavier". The aim of the puzzle is to identify the ball which is the different weight, and whether it is heavier or lighter than the others, in as few weighings as possible.

[ 2 ]

[Maximum number: 23]

This question uses differential equations to model the maximum velocity of a skydiver in free fall.
In 2012, Felix Baumgartner jumped from a height of 40000 m . He was attempting to travel at the speed of sound, $330 \mathrm{~m} \mathrm{~s}^{-1}$ , whilst free-falling to the Earth.
Before making his attempt, Felix used mathematical models to check how realistic his attempt would be. The simplest model he used suggests that

\frac{\mathrm{d} v}{\mathrm{~d} t}=g

where $v \mathrm{~m} \mathrm{~s}^{-1}$ is Felix's velocity and $g \mathrm{~ms}^{-2}$ is the acceleration due to gravity. The time since he began to free-fall is t seconds and the displacement from his initial position is s metres.

Throughout this question, the direction towards the centre of the Earth is taken to be positive and v is a positive quantity.

When s=0, it is given that Felix jumps with an initial velocity v=10.

(a)

Use the chain rule to show that $\frac{\mathrm{d} v}{\mathrm{~d} t}=v \frac{\mathrm{~d} v}{\mathrm{~d} s}$ .

[ 1 ]

$\frac{\mathrm{d} v}{\mathrm{~d} t}=\frac{\mathrm{d} s}{\mathrm{~d} t} \times \frac{\mathrm{d} v}{\mathrm{~d} s}$ $\left(v=\frac{\mathrm{d} s}{\mathrm{~d} t}\right)$ $\frac{\mathrm{d} v}{\mathrm{~d} t}=v \frac{\mathrm{~d} v}{\mathrm{~d} s}$

(b)

Assuming that g is a constant, solve the differential equation $v \frac{\mathrm{~d} v}{\mathrm{~d} s}=g$ to find
v as a function of s. v as a function of s.

[ 1 ]

$\quad v \frac{\mathrm{~d} v}{\mathrm{~d} s}=g$
attempt to separate variables
$\int v \mathrm{~d} v=\int g \mathrm{ds}$ $\frac{v^{2}}{2}=g s(+c)$
using initial conditions (can be done at any point)
50=c
so $v=\sqrt{2 g s+100}$

Note: Marks are intentionally unimplied to ensure on-syllabus techniques are used.

(c)

Using g=9.8, determine whether the model predicts that Felix will succeed in travelling at the speed of sound at some point before s=40000. Justify your answer.

[ 3 ]

(iii) EITHER
attempt to use their part (a)(ii) to find a value of s when v=330
$330=\sqrt{2 g s+100}$
therefore $s=5551.02 \ldots$
(5551.02 < 40000)
so (the model does predict) he will reach the speed of sound

OR
attempt to use their part (a)(ii) to find a value of v when s=40000
$v=\sqrt{2 g(40000)+100}$
=885 (885.49...)
(885>330)
so (the model does predict) he will reach the speed of sound (before s=40000 )

Note: For the OR method, accept any large s that leads to v=330.
FT from $\sqrt{2 g s}$ gives $885(885.437 \ldots)$ for v and $5560(5556.12 \ldots)$ for s
FT from their v or their s for the final A1, provided M1 is awarded

(d)

To test the model

\frac{\mathrm{d} v}{\mathrm{~d} t}=g

Felix conducted a trial jump from a lower height, and data for v against t was found.

[ 2 ]

(i)

If the model is correct, describe the shape of the graph of v against t.

Felix's data are plotted on the following graph.

[ 2 ]

(e)

An improved model considers air resistance, using

\frac{\mathrm{d} v}{\mathrm{~d} t}=g-k v^{2}

where k is a positive constant. You are reminded that initially s=0 and v=10.

[ 11 ]

(i)

By using $\frac{\mathrm{d} v}{\mathrm{~d} t}=v \frac{\mathrm{~d} v}{\mathrm{~d} s}$ , solve the differential equation to find v in terms of s, g and k.

You may assume that $g-k v^{2}>0$ .

Felix uses the graph of v against t shown in part (b) to estimate the value of k.

[ 5 ]

$v \frac{\mathrm{~d} v}{\mathrm{~d} s}=g-k v^{2}$
separating variables
$\int \frac{v}{g-k v^{2}} \mathrm{~d} v=\int \mathrm{d} s$

-\frac{1}{2 k} \ln \left(g-k v^{2}\right)=s(+c) \text { OR }-\frac{1}{2 k} \ln \left|g-k v^{2}\right|=s(+c)

rearranging to make v the subject

Note: Award (M1) for making v the subject of their equation and not just an attempt, or an erroneous equation with v also on the RHS.

\begin{gathered} g-k v^{2}=A \mathrm{e}^{-2 k s} \\ v=\sqrt{\frac{g-A \mathrm{e}^{-2 k s}}{k}} \end{gathered}

applying initial conditions (here or elsewhere)

\begin{aligned} & 100=\frac{g-A}{k} \\ & A=g-100 k \end{aligned}

v=\sqrt{\frac{g-(g-100 k) \mathrm{e}^{-2 k s}}{k}}

(ii)

The gradient is estimated to be 9.672 when v=40. Taking g to be 9.8 , use this information to show that Felix found that $k=8 \times 10^{-5}$ .

[ 2 ]

9.672=9.8-1600 k

Note: Award A1 for correct left-hand side and A1 for correct right-hand side.

\begin{aligned} & k=\frac{9.8-9.672}{1600} \\ & k=8 \times 10^{-5} \end{aligned}

Note: Award A1AO for $k=8 \times 10^{-5}$ substituted into the right-hand side of the expression, leading to 9.672 .

(iii)

Hence, find the value of v predicted by this model, as s tends to infinity.

[ 2 ]

$s \rightarrow \infty, \mathrm{e}^{-2 k s} \rightarrow 0$ OR $\frac{\mathrm{d} v}{\mathrm{~d} t}=0$ OR graph/table

\left(v_{\max }=\sqrt{\frac{g}{k}}=\right) 350 \quad\left(m s^{-1}\right)

(iv)

Find the upper bound for the velocity according to this model, given that $0<s \leq 40000$ . Give your answer to four significant figures.

The assumption that the value of g is constant is not correct. It can be shown that

g=\frac{3.98 \times 10^{14}}{\left(6.41 \times 10^{6}-s\right)^{2}}

Hence, the new model is given by

v \frac{\mathrm{~d} v}{\mathrm{~d} s}=\frac{3.98 \times 10^{14}}{\left(6.41 \times 10^{6}-s\right)^{2}}-\left(8 \times 10^{-5}\right) v^{2}

When s=0, it is known that v=10.

[ 2 ]

upper limit occurs when s=40000

Note: The M1 can be implied by 40000 substituted into their part (c)(i).

349.7\left(\mathrm{~ms}^{-1}\right)

Note: Answer must be to 4 sf .

(f)

Use Euler's method with a step length of 4000 to estimate the value of v when s=40000.

[ 4 ]

$\quad s_{n+1}=s_{n}+4000$ $v_{n+1}=v_{n}+4000 \times\left(\frac{3.98 \times 10^{14}}{v_{n}\left(6.41 \times 10^{6}-s_{n}\right)^{2}}-\left(8 \times 10^{-5}\right) v_{n}\right)$

Note: Award (M1) for attempt to use Euler method formula AND dividing through by v.
if $v_{0}=10$ , then $v_{10}=361(360.658 \ldots)$

(g)

After Felix completed his record-breaking jump, he found that the answer from part (d) was not supported by data collected during the jump.

[ 1 ]

(i)

Suggest one improvement to the use of Euler's method which might increase the accuracy of the prediction of the model.

[ 1 ]

[Maximum number: 9]

Kailash manufactures drink containers in the shape of a cuboid. The container has a square top and a square base of length, $l\,\mathrm{cm}$ . Its height, $d\,\mathrm{cm}$ , is three times the length of the base.

diagram not to scale

(a)

Find $\frac{\mathrm d A}{\mathrm d r}$ .

[ 3 ]

$\frac{\mathrm d A}{\mathrm d r}=4\pi r-750r^{-2}$
A1 for $4\pi r$ , A1 for -750, A1 for $r^{-2}$ or equivalent.

(b)

Hence or otherwise

[ 3 ]

(i)

find the value of r that will minimize A.

[ 2 ]

Figure for Answer to Question (i) — IB Maths AI HL

METHOD 1: correct sketch with some indication of minimum point.
METHOD 2: attempt to solve $4\pi r-\frac{750}{r^2}=0$ , graphically or otherwise.
$r=3.91\,\mathrm{cm}$ $(3.90796\ldots)$

(ii)

find the minimum value of A needed for the cylinder.

[ 1 ]

$\quad 288\left(\mathrm{~cm}^{2}\right)(287.873 \ldots)$

(c)

Find $\frac{\mathrm{d}^{2} A}{\mathrm{~d} r^{2}}$ .

[ 1 ]

$\left(\frac{\mathrm{d}^{2} A}{\mathrm{~d} r^{2}}=\right) 4 \pi+\frac{1500}{r^{3}}$

(d)

Hence determine whether the graph of A is concave-up or concave-down for r>0. Justify your answer.

[ 2 ]

As r>0, $\frac{\mathrm d^2A}{\mathrm d r^2}>0$ .
Hence the graph is concave-up.
Note: marks in part (ii) require an expression seen in part (i).

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Covered syllabus points

5 Calculus question 1

Question (a)

Question (i)

Question (b)

Question (c)

5 Calculus question 2

Question (a)

Question (b)

Question (i)

5 Calculus question 3

Question (a)

Question (b)

Question (c)

Question (d)

Question (i)

Question (e)

Question (i)

Question (ii)

Question (iii)

Question (iv)

Question (f)

Question (g)

Question (i)

5 Calculus question 4

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

Question (d)