IB Maths AI HL 5.2 Calculus Ahl Content Questions

Practise IB Mathematics AI HL 5.2 by combining advanced calculus, technology, differential models, numerical methods and optimisation in extended contexts.

Exam points

Apply chain/product/quotient and higher-order derivatives to related rates, concavity, inflexion and advanced extrema.
Use substitution and standard antiderivatives for further integrals, enclosed areas and volumes of revolution.
Use displacement, velocity and acceleration relationships to solve advanced kinematics and total-distance problems.
Set up and solve separable differential equations and interpret general/particular solutions and slope fields.
Apply Euler/numerical methods to first- or second-order and coupled systems, reporting approximation effects.
Use phase portraits and eigenvalue behaviour to classify equilibria, stability, spirals and long-run trajectories.

Covered syllabus points

AHL 5.9 (HL)—Further differentiation AHL 5.10 (HL)—Second derivative and concavity AHL 5.11 (HL)—Further integration AHL 5.12 (HL)—Areas and volumes AHL 5.13 (HL)—Kinematics AHL 5.14 (HL)—Differential equation modelling AHL 5.15 (HL)—Slope fields AHL 5.16 (HL)—Euler's method AHL 5.17 (HL)—Phase portraits AHL 5.18 (HL)—Second-order differential equations

[Maximum number: 3]

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 $\operatorname{occur}(p=1)$ .

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

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

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

[Maximum number: 3]

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

[ 1 ]

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

(b)

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).

[Maximum number: 7]

(a)

Find the indefinite integral $\int x e^{-x^{2}} d x$ .

[ 4 ]

A valid/coherent attempt to use substitution:

u=-x^2,\quad \frac{du}{dx}=-2x

The integral becomes

-\frac12\int e^u\,du=-\frac12e^{-x^2}+c

Accept integration by inspection using $\frac{d}{dx}(e^{-x^2})=-2xe^{-x^2}$ .
Note: Award M1A1M1A0 for $-\frac12e^{-x^2}$ with c omitted.

(b)

Hence find the area bounded by the x-axis, the curve $\mathrm{y}=x e^{-x^{2}}$ and the line x=k. Give your answer in terms of k.

[ 3 ]

Recognition that 0 is the lower limit and substitute both 0 and k into the previous expression.

\int_0^k xe^{-x^2}\,dx=\left[-\frac12e^{-x^2}\right]_0^k=\frac12-\frac12e^{-k^2}

[Maximum number: 4]

Find the indefinite integral $\int x e^{-x^{2}} d x$ .

A valid/coherent attempt to use substitution:

u=-x^2,\quad \frac{du}{dx}=-2x

The integral becomes

-\frac12\int e^u\,du=-\frac12e^{-x^2}+c

Accept integration by inspection using $\frac{d}{dx}(e^{-x^2})=-2xe^{-x^2}$ .
Note: Award M1A1M1A0 for $-\frac12e^{-x^2}$ with c omitted.

[Maximum number: 3]

Hence find the area bounded by the x-axis, the curve $\mathrm{y}=x e^{-x^{2}}$ and the line x=k. Give your answer in terms of k.

Recognition that 0 is the lower limit and substitute both 0 and k into the previous expression.

\int_0^k xe^{-x^2}\,dx=\left[-\frac12e^{-x^2}\right]_0^k=\frac12-\frac12e^{-k^2}

[Maximum number: 11]

A sports stadium has a T-shirt cannon which is used to launch T-shirts into the crowd. The purpose of this question is to determine whether a person sitting in a particular seat will ever receive a T-shirt.
A T-shirt cannon is placed on the horizontal ground of a stadium playing area. A coordinate system is created such that the origin, O , is the point on the ground from where the T-shirts are launched. In this coordinate system, x and y represent the horizontal and vertical displacement from O, and are measured in metres.
Seat $\mathrm{A}_{1}$ is the nearest seat to the T-shirt cannon. The coordinates of the front of the foot space for seat $\mathrm{A}_{1}$ are (30, 2.1).

Each seat behind seat $\mathrm{A}_{1}$ is 1.0 m further from O horizontally and 0.5 m higher than the seat in the row below it, as shown on the diagram.

Seat $\mathrm{A}_{1}$ is in row 1 . Let seat $\mathrm{A}_{n}$ be the seat directly behind $\mathrm{A}_{1}$ in row n.

(a)

Find an expression for the velocity, $\binom{\dot{x}}{\dot{y}}$ , at time t.

[ 3 ]

evidence of integration of the acceleration vector OR use of v=u+a t (M1)

\begin{aligned} & \binom{\dot{x}}{\dot{y}}=\binom{c_{1}}{-9.8 t+c_{2}} \\ & \left(\binom{\dot{x}}{\dot{y}}=\right)\binom{29.4 \cos \theta}{29.4 \sin \theta-9.8 t} \end{aligned}

Note: The first A1 is for $\dot{x}$ and the second is for $\dot{y}$ .

(b)

Hence show that when the T-shirt is launched vertically, the time for it to reach its maximum height is 3 seconds.

The displacement of the T-shirt, t seconds after it is launched, is given by the vector equation

\binom{x}{y}=\binom{29.4(\cos \theta) t}{29.4(\sin \theta) t-4.9 t^{2}}

[ 3 ]

$\theta=90^{\circ}$

29.4-9.8 t=0

Note: Award M1 for setting their $\dot{y}$ to zero (may still include $\theta$ ), A1 for correct equation, leading to given result.
If they substitute t=3 award at most (A1)MOAO.

\begin{aligned} & \text { maximum point when } t=\frac{29.4}{9.8} \\ & =3 \text { (seconds) } \end{aligned}

(c)

Using the given answer to part (b)(ii) or otherwise, find the maximum height reached by a T-shirt when it is launched vertically.

[ 2 ]

correct substitution OR use of correct graph maximum height is $29.4 \times 3-4.9 \times 3^{2}$

\text { = } 44.1 \text { (m) }

(d)

If there was no seating, and the T-shirt was launched at an angle $\theta$ , show that the value of x when it would hit the ground is given by the expression

x=176.4 \sin \theta \cos \theta

[ 3 ]

$\quad 29.4 \sin \theta t-4.9 t^{2}=0$ $t=6 \sin \theta \quad($ or t=0)
$x=29.4 \cos \theta \times 6 \sin \theta$ $=176.4 \cos \theta \sin \theta$

[Maximum number: 21]

This question explores models for the height of water in a cylindrical container as water drains out.
The diagram shows a cylindrical water container of height 3.2 metres and base radius 1 metre. At the base of the container is a small circular valve, which enables water to drain out.

Eva closes the valve and fills the container with water.
At time t=0, Eva opens the valve. She records the height, h metres, of water remaining in the container every 5 minutes.

Eva first tries to model the height using a linear function, h(t)=a t+b, where $a, b \in \mathbb{R}$ .

(a)

Find the value of k.

[ 2 ]

$\quad 0.002 t^{2}-0.174 t+3.2=0$
26.4 (26.4046...)

(b)

Show that $\frac{\mathrm{d} h}{\mathrm{~d} t}=-R^{2} \sqrt{70560 h}$ .

[ 3 ]

$\quad V=\pi(1)^{2} h$

EITHER
$\frac{\mathrm{d} V}{\mathrm{dt}}=\pi \frac{\mathrm{d} h}{\mathrm{dt}}$

OR
attempt to use chain rule
$\frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{\mathrm{d} h}{\mathrm{~d} V} \times \frac{\mathrm{d} V}{\mathrm{~d} t}$
THEN

\begin{aligned} & \frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{1}{\pi} \times-\pi R^{2} \sqrt{70560 h} \\ & \frac{\mathrm{~d} h}{\mathrm{~d} t}=-R^{2} \sqrt{70560 h} \end{aligned}

(c)

By solving the differential equation $\frac{\mathrm{d} h}{\mathrm{~d} t}=-R^{2} \sqrt{70560 h}$ , show that the general solution is given by $h=17640\left(c-R^{2} t\right)^{2}$ , where $c \in \mathbb{R}$ .

Eva measures the radius of the valve to be 0.023 metres. Let T be the time, in minutes, it takes for all the water to drain out of the container.

[ 5 ]

attempt to separate variables M1

\begin{aligned} & \int \frac{1}{\sqrt{70560 h}} \mathrm{~d} h=\int-R^{2} \mathrm{~d} t \\ & \frac{2 \sqrt{h}}{\sqrt{70560}}=-R^{2} t+c \end{aligned}

Note: Award A1 for each correct side of the equation.

\sqrt{h}=\frac{\sqrt{70560}}{2}\left(c-R^{2} t\right)

Note: Award the final A1 for any correct intermediate step that clearly leads to the given equation.

h=17640\left(c-R^{2} t\right)^{2}

(d)

Use the general solution from part (d) and the initial condition h(0)=3.2 to predict the value of T.

Eva wants to use the container as a timer. She adjusts the initial height of water in the container so that all the water will drain out of the container in 15 minutes.

[ 4 ]

$t=0 \Rightarrow 3.2=17640 c^{2}$

c=0.0134687 \ldots

substituting h=0 and their non-zero value of c

\begin{aligned} & T=\frac{c}{R^{2}}=\frac{0.0134687 \ldots}{0.023^{2}} \\ & =25.5(\text { minutes })(25.4606 \ldots) \end{aligned}

(e)

Find this new height.

Eva has another water container that is identical to the first one. She places one water container above the other one, so that all the water from the highest container will drain into the lowest container. Eva completely fills the highest container, but only fills the lowest container to a height of 1 metre, as shown in the diagram.

diagram not to scale

At time t=0 Eva opens both valves. Let H be the height of water, in metres, in the lowest container at time t.

[ 3 ]

$\quad h=0 \Rightarrow c=R^{2} t$ $c=0.023^{2} \times 15(=0.007935)$ $t=0 \Rightarrow h=17640\left(0.023^{2} \times 15\right)^{2}$
h=1.11 (metres) (1.11068...)

(f)

Show that $\frac{\mathrm{d} H}{\mathrm{~d} t} \approx 0.2514-0.009873 t-0.1405 \sqrt{H}$ , where $0 \leq t \leq T$ .

[ 4 ]

let h be the height of water in the highest container from parts (d) and (e) we get

\begin{aligned} & \frac{\mathrm{d} h}{\mathrm{~d} t}=-35280 R^{2}\left(0.0134687 \ldots-R^{2} t\right) \\ & \text { so } \frac{\mathrm{d} H}{\mathrm{~d} t}=35280 R^{2}\left(0.0135-R^{2} t\right)-R^{2} \sqrt{70560 H} \\ & \left(\frac{\mathrm{~d} H}{\mathrm{~d} t}=18.6631 \ldots(0.0134687 \ldots-0.000529 t)-0.000529 \sqrt{70560 H}\right) \\ & \left(\frac{\mathrm{d} H}{\mathrm{~d} t}=0.251367 \ldots-0.0987279 \ldots-0.140518 \ldots \sqrt{H}\right) \\ & \frac{\mathrm{d} H}{\mathrm{~d} t} \approx 0.2514-0.009873 t-0.1405 \sqrt{H} \end{aligned}

[Maximum number: 2]

Consider the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x}{\mathrm{e}^{2 y}}$ .

Identify which of the following diagrams, A, B or C, represents the slope field for the differential equation. Give a reason for your answer.

[Maximum number: 3]

Eva closes the valve and fills the container with water.
At time t=0, Eva opens the valve. She records the height, h metres, of water remaining in the container every 5 minutes.

Eva first tries to model the height using a linear function, h(t)=a t+b, where $a, b \in \mathbb{R}$ .

Use Euler's method with a step length of 0.5 minutes to estimate the maximum value of H.

[Maximum number: 1]

Consider the system given by the coupled differential equations

\begin{aligned} & \frac{\mathrm{d} x}{\mathrm{~d} t}=3 x-2 y \\ & \frac{\mathrm{~d} y}{\mathrm{~d} t}=4 x-y \end{aligned}

The following phase portrait shows the trajectory that passes through the point (1,0) at time t=5.

Write down which one of the following could be an eigenvalue for the system.

2 i

1+2 i

-1+2 i

[Maximum number: 10]

An environmental scientist is asked by a river authority to model the effect of a leak from a power plant on the mercury levels in a local river. The variable x measures the concentration of mercury in micrograms per litre.
The situation is modelled using the second order differential equation

\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+3 \frac{\mathrm{~d} x}{\mathrm{~d} t}+2 x=0

where $t \geq 0$ is the time measured in days since the leak started. It is known that when t=0, x=0 and $\frac{\mathrm{d} x}{\mathrm{~d} t}=1$ .

(a)

Show that the system of coupled first order equations:

$\frac{\mathrm{d}x}{\mathrm{d}t}=y$ $\frac{\mathrm{d}y}{\mathrm{d}t}=-2x-3y$

can be written as the given second order differential equation.

[ 2 ]

differentiating first equation.
$\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=\frac{\mathrm{d} y}{\mathrm{~d} t}$
substituting in for $\frac{\mathrm{d} y}{\mathrm{~d} t}$ $=-2 x-3 y=-2 x-3 \frac{\mathrm{~d} x}{\mathrm{~d} t}$
therefore $\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+3 \frac{\mathrm{~d} x}{\mathrm{~d} t}+2 x=0$

Note: The AG line must be seen to award the final M1 mark.

(b)

Hence find the exact solution of the second order differential equation.

[ 5 ]

EITHER
The general solution is $x=A\mathrm{e}^{-t}+B\mathrm{e}^{-2t}$ . M1
Note: Must have constants, but condone sign error for the M1.
Then $\frac{\mathrm{d}x}{\mathrm{d}t}=-A\mathrm{e}^{-t}-2B\mathrm{e}^{-2t}$ .
OR
Attempt to find eigenvectors: respective eigenvectors are $\binom1{-1}$ and $\binom1{-2}$ .
$\binom{x}{y}=A\mathrm{e}^{-t}\binom1{-1}+B\mathrm{e}^{-2t}\binom1{-2}$
THEN
Initial conditions: 0=A+B, 1=-A-2B.
This gives A=1, B=-1, so $x=\mathrm{e}^{-t}-\mathrm{e}^{-2t}$ .

(c)

If the mercury levels are greater than 0.1 micrograms per litre, fishing in the river is considered unsafe and is stopped.
Use the model to calculate the total amount of time when fishing should be stopped.

[ 3 ]

intersecting graph with y=0.1

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

so the time fishing is stopped between $2.1830 \ldots$ and $0.11957 \ldots$
=2.06 ( $343 \ldots$ ) days

All question bank results loaded

Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)

Question 4

Question 5

Question 6

Question (a)

Question (b)

Question (c)

Question (d)

Question 7

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

Question 8

Question 9

Question 10

Question 11

Question (a)

Question (b)

Question (c)