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Date	May 2022	Marks available	2	Reference code	22M.1.AHL.TZ2.16
Level	Additional Higher Level	Paper	Paper 1	Time zone	Time zone 2
Command term	Find	Question number	16	Adapted from	N/A

Question

The position vector of a particle, $P$ , relative to a fixed origin $O$ at time $t$ is given by

$\vec{OP} = (\begin{matrix} \sin (t^{2}) \\ \cos (t^{2}) \end{matrix})$ .

Find the velocity vector of $P$ .

[2]

Show that the acceleration vector of $P$ is never parallel to the position vector of $P$ .

[5]

Markscheme

attempt at chain rule (M1)

$(v = \frac{d O P}{d t} =) (\begin{matrix} 2 t \cos t^{2} \\ - 2 t \sin t^{2} \end{matrix})$ A1

[2 marks]

attempt at product rule (M1)

$a = (\begin{matrix} 2 \cos t^{2} - 4 t^{2} \sin t^{2} \\ - 2 \sin t^{2} - 4 t^{2} \cos t^{2} \end{matrix})$ A1

METHOD 1

let $S = \sin t^{2}$ and $C = \cos t^{2}$

finding $\cos θ$ using

$a \cdot \vec{OP} = 2 S C - 4 t^{2} S^{2} - 2 S C - 4 t^{2} C^{2} = - 4 t^{2}$ M1

$|\vec{OP}| = 1$

$|a| = \sqrt{{(2 C - 4 t^{2} S)}^{2} + {(- 2 S - 4 t^{2} C)}^{2}}$

$= \sqrt{4 + 16 t^{4}} > 4 t^{2}$

if $θ$ is the angle between them, then

$\cos θ = - \frac{4 t^{2}}{\sqrt{4 + 16 t^{4}}}$ A1

so $- 1 < \cos θ < 0$ therefore the vectors are never parallel R1

METHOD 2

solve

$(\begin{matrix} 2 \cos t^{2} - 4 t^{2} \sin t^{2} \\ - 2 \sin t^{2} - 4 t^{2} \cos t^{2} \end{matrix}) = k (\begin{matrix} \sin t^{2} \\ \cos t^{2} \end{matrix})$ M1

then

$(k =) \frac{2 \cos t^{2} - 4 t^{2} \sin t^{2}}{\sin t^{2}} = \frac{- 2 \sin t^{2} - 4 t^{2} \cos t^{2}}{\cos t^{2}}$

Note: Condone candidates not excluding the division by zero case here. Some might go straight to the next line.

$2 \cos^{2} t^{2} - 4 t^{2} \cos t^{2} \sin t^{2} = - 2 \sin^{2} t^{2} - 4 t^{2} \cos t^{2} \sin t^{2}$

$2 \cos^{2} t^{2} + 2 \sin^{2} t^{2} = 0$

$2 = 0$ A1

this is never true so the two vectors are never parallel R1

METHOD 3

embedding vectors in a 3d space and taking the cross product: M1

$(\begin{matrix} \sin t^{2} \\ \cos t^{2} \\ 0 \end{matrix}) \times (\begin{matrix} 2 \cos t^{2} - 4 t^{2} \sin t^{2} \\ - 2 \sin t^{2} - 4 t^{2} \cos t^{2} \\ 0 \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ - 2 \sin^{2} t^{2} - 4 t^{2} \cos t^{2} \sin t^{2} - 2 \cos^{2} t^{2} + 4 t^{2} \cos t^{2} \sin t^{2} \end{matrix})$

$= (\begin{matrix} 0 \\ 0 \\ - 2 \end{matrix})$ A1

since the cross product is never zero, the two vectors are never parallel R1

[5 marks]

Examiners report

In part (a), many candidates found the velocity vector correctly. In part (b), however, many candidates failed to use the product rule correctly to find the acceleration vector. To show that the acceleration vector is never parallel to the position vector, a few candidates put $\overset{. .}{r} = k r$ presumably hoping to show that no value of the constant k existed for any t but this usually went nowhere.

[N/A]

Syllabus sections

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

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