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Date	May 2022	Marks available	1	Reference code	22M.1.AHL.TZ2.17
Level	Additional Higher Level	Paper	Paper 1	Time zone	Time zone 2
Command term	Show that and Hence	Question number	17	Adapted from	N/A

Question

The cross-section of a beach is modelled by the equation $y = 0.02 x^{2}$ for $0 \leq x \leq 10$ where $y$ is the height of the beach (in metres) at a horizontal distance $x$ metres from an origin. $t$ is the time in hours after low tide.

At $t = 0$ the water is at the point $(0, 0)$ . The height of the water rises at a rate of $0.2$ metres per hour. The point $W (x (t), y (t))$ indicates where the water level meets the beach at time $t$ .

A snail is modelled as a single point. At $t = 0$ it is positioned at $(1, 0.02)$ . The snail travels away from the incoming water at a speed of $1$ metre per hour in the direction along the curve of the cross-section of the beach. The following diagram shows this for a value of $t$ , such that $t > 0$ .

When $W$ has an $x$ -coordinate equal to $1$ , find the horizontal component of the velocity of $W$ .

[3]

Find the time taken for the snail to reach the point $(10, 2)$ .

[4]

b.i.

Hence show that the snail reaches the point $(10, 2)$ before the water does.

[1]

b.ii.

Markscheme

use of chain rule (M1)

$\frac{d y}{d t} = \frac{d y}{d x} \frac{d x}{d t}$

attempt to find $\frac{d y}{d x}$ at $x = 1$ (M1)

$0.2 = 0.04 \times \frac{d x}{d t}$

$(\frac{d x}{d t} =) 5 {m h}^{- 1}$ A1

[3 marks]

if the position of the snail is $(X, Y)$

from part (a) $\frac{d X}{d t} = \frac{1}{0.04 X} \frac{d Y}{d t}$

since speed is $1$ :

finding modulus of velocity vector and equating to $1$ (M1)

$1 = \sqrt{{(\frac{\dot{Y}}{0.04 X})}^{2} + {\dot{Y}}^{2}}$ OR $1 = \sqrt{{\dot{X}}^{2} + 0.0016 X^{2} {\dot{X}}^{2}}$

$1 = {\dot{Y}}^{2} (\frac{1}{0.0016 X^{2}} + 1)$ OR $1 = {\dot{X}}^{2} (1 + 0.0016 X^{2})$

$\dot{Y} = \sqrt{\frac{1}{\frac{1}{0.08 Y} + 1}}$ OR $\dot{X} = \sqrt{\frac{1}{1 + 0.0016 X^{2}}}$ (A1)

$\int_{0.02}^{2} \sqrt{\frac{1}{0.08 Y} + 1} d Y = \int_{0}^{T} d t$ OR $\int_{1}^{10} \sqrt{1 + 0.0016 X^{2}} d X = \int_{0}^{T} d t$ (M1)

$T = 9.26$ hours A1

[4 marks]

b.i.

EITHER

time for water to reach top is $\frac{2}{0.2} = 10$ hours (seen anywhere) A1

or at time $t = 9.26$ , height of water is $0.2 \times 9.26 = 1.852$ A1

THEN

so the water will not reach the snail AG

[1 mark]

b.ii.

Examiners report

In part (a), a small minority of candidates found the horizontal component of velocity correctly. Few candidates made any significant progress in part (b).

[N/A]

b.i.

[N/A]

b.ii.

Syllabus sections

Topic 5—Calculus » AHL 5.9—Differentiating standard functions and derivative rules

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