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Date	May 2021	Marks available	6	Reference code	21M.2.AHL.TZ1.6
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Find	Question number	6	Adapted from	N/A

Question

An ice-skater is skating such that her position vector when viewed from above at time $t$ seconds can be modelled by

$(\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} a e^{b t} \cos t \\ a e^{b t} sin t \end{matrix})$

with respect to a rectangular coordinate system from a point $O$ , where the non-zero constants $a$ and $b$ can be determined. All distances are in metres.

At time $t = 0$ , the displacement of the ice-skater is given by $(\begin{matrix} 5 \\ 0 \end{matrix})$ and the velocity of the ice‑skater is given by $(\begin{matrix} - 3.5 \\ 5 \end{matrix})$ .

Find the velocity vector at time $t$ .

[3]

a.

Show that the magnitude of the velocity of the ice-skater at time $t$ is given by

$a e^{b t} \sqrt{(1 + b^{2})}$ .

[4]

b.

Find the value of $a$ and the value of $b$ .

[3]

c.

Find the magnitude of the velocity of the ice-skater when $t = 2$ .

[2]

d.

At a point $P$ , the ice-skater is skating parallel to the $y$ -axis for the first time.

Find $OP$ .

[6]

e.

Markscheme

use of product rule (M1)

$(\begin{matrix} \dot{x} \\ \dot{y} \end{matrix}) = (\begin{matrix} a b e^{b t} \cos t - a e^{b t} sin t \\ a b e^{b t} sin t + a e^{b t} \cos t \end{matrix})$ A1A1

[3 marks]

a.

${|v|}^{2} = {\dot{x}}^{2} + {\dot{y}}^{2} = {[a b e^{b t} \cos t - a e^{b t} sin t]}^{2} + {[a b e^{b t} sin t + a e^{b t} \cos t]}^{2}$ M1

Note: It is more likely that an expression for $|v|$ is seen.
$\sqrt{{\dot{x}}^{2} + {\dot{y}}^{2}}$ is not sufficient to award the M1, their part (a) must be substituted.

$= [a^{2} \sin^{2} t - 2 a^{2} b \sin t \cos t + a^{2} b^{2} \cos^{2} t + a^{2} \cos^{2} t + 2 a^{2} b \sin t \cos t + a^{2} b^{2} \sin^{2} t] e^{2 b t}$ A1

use of $\sin^{2} t + \cos^{2} t = 1$ within a factorized expression that leads to the final answer M1

$= a^{2} (b^{2} + 1) e^{2 b t}$ A1

magnitude of velocity is $a e^{b t} \sqrt{(1 + b^{2})}$ AG

[4 marks]

b.

when $t = 0, a e^{b t} \cos t = 5$

$a = 5$ A1

$a b e^{b t} \cos t - a e^{b t} \sin t = - 3.5$ (M1)

$b = - 0.7$ A1

Note: Use of $a e^{b t} \sqrt{(1 + b^{2})}$ result from part (b) is an alternative approach.

[3 marks]

c.

$5 e^{- 0.7 \times 2} \sqrt{(1 + {(- 0.7)}^{2})}$ (M1)

$1.51 (1.50504 \dots)$ A1

[2 marks]

d.

$\dot{x} = 0$ (M1)

$a e^{b t} (b \cos t - \sin t) = 0$

$\tan t = b$

$t = 2.53 (2.53086 \dots)$ (A1)

correct substitution of their $t$ to find $x$ or $y$ (M1)
$x = - 0.697 (- 0.696591 \dots)$ and $y = 0.488 (0.487614 \dots)$ (A1)

use of Pythagoras / distance formula (M1)
$OP = 0.850 m (0.850297 \dots)$ A1

[6 marks]

e.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.

Syllabus sections

Topic 3—Geometry and trigonometry » AHL 3.12—Vector applications to kinematics

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