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Date	May 2019	Marks available	6	Reference code	19M.1.AHL.TZ1.H_5
Level	Additional Higher Level	Paper	Paper 1	Time zone	Time zone 1
Command term	Find	Question number	H_5	Adapted from	N/A

Question

A camera at point C is 3 m from the edge of a straight section of road as shown in the following diagram. The camera detects a car travelling along the road at $t$ $t$ = 0. It then rotates, always pointing at the car, until the car passes O, the point on the edge of the road closest to the camera.

A car travels along the road at a speed of 24 ms⁻¹. Let the position of the car be X and let OĈX = θ.

Find $\frac{d θ}{d t}$ $\frac{{{\text{d}}\theta }}{{{\text{d}}t}}$ , the rate of rotation of the camera, in radians per second, at the instant the car passes the point O .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

let OX = $x$ $x$

METHOD 1

$\frac{d x}{d t} = 24$ $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 24$ (or −24) (A1)

$\frac{d θ}{d t} = \frac{d x}{d t} \times \frac{d θ}{d x}$ $\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{{\text{d}}x}}{{{\text{d}}t}} \times \frac{{{\text{d}}\theta }}{{{\text{d}}x}}$ (M1)

$3 tan θ = x$ $3\,{\text{tan}}\,\theta = x$ A1

EITHER

$3 se c^{2} θ = \frac{d x}{d θ}$ $3\,{\text{se}}{{\text{c}}^2}\,\theta = \frac{{{\text{d}}x}}{{{\text{d}}\theta }}$ A1

$\frac{d θ}{d t} = \frac{24}{3 se c^{2} θ}$ $\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{24}}{{3\,{\text{se}}{{\text{c}}^2}\,\theta }}$

attempt to substitute for $θ = 0$ $\theta = 0$ into their differential equation M1

$θ = arctan (\frac{x}{3})$ $\theta = {\text{arctan}}\left( {\frac{x}{3}} \right)$

$\frac{d θ}{d x} = \frac{1}{3} \times \frac{1}{1 + \frac{x^{2}}{9}}$ $\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{1}{3} \times \frac{1}{{1 + \frac{{{x^2}}}{9}}}$ A1

$\frac{d θ}{d t} = 24 \times \frac{1}{3 (1 + \frac{x^{2}}{9})}$ $\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = 24 \times \frac{1}{{3\left( {1 + \frac{{{x^2}}}{9}} \right)}}$

attempt to substitute for $x = 0$ $x = 0$ into their differential equation M1

THEN

$\frac{d θ}{d t} = \frac{24}{3} = 8$ $\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{24}}{3} = 8$ (rad s⁻¹) A1

Note: Accept −8 rad s⁻¹.

METHOD 2

$\frac{d x}{d t} = 24$ $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 24$ (or −24) (A1)

$3 tan θ = x$ $3\,{\text{tan}}\,\theta = x$ A1

attempt to differentiate implicitly with respect to $t$ $t$ M1

$3 se c^{2} θ \times \frac{d θ}{d t} = \frac{d x}{d t}$ $3\,{\text{se}}{{\text{c}}^2}\,\theta \times \frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{{\text{d}}x}}{{{\text{d}}t}}$ A1

$\frac{d θ}{d t} = \frac{24}{3 se c^{2} θ}$ $\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{24}}{{3\,{\text{se}}{{\text{c}}^2}\,\theta }}$

attempt to substitute for $θ = 0$ $\theta = 0$ into their differential equation M1

$\frac{d θ}{d t} = \frac{24}{3} = 8$ $\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{24}}{3} = 8$ (rad s⁻¹) A1

Note: Accept −8 rad s⁻¹.

Note: Can be done by consideration of CX, use of Pythagoras.

METHOD 3

let the position of the car be at time $t$ $t$ be $d - 24 t$ $d - 24t$ from O (A1)

$tan θ = \frac{d - 24 t}{3} (= \frac{d}{3} - 8 t)$ ${\text{tan}}\,\theta = \frac{{d - 24t}}{3}\left( { = \frac{d}{3} - 8t} \right)$ M1

Note: For $tan θ = \frac{24 t}{3}$ ${\text{tan}}\,\theta = \frac{{24t}}{3}$ award A0M1 and follow through.

EITHER

attempt to differentiate implicitly with respect to $t$ $t$ M1

$se c^{2} θ \frac{d θ}{d t} = - 8$ ${\text{se}}{{\text{c}}^2}\,\theta \frac{{{\text{d}}\theta }}{{{\text{d}}t}} = - 8$ A1

attempt to substitute for $θ = 0$ $\theta = 0$ into their differential equation M1

$θ = arctan (\frac{d}{3} - 8 t)$ $\theta = {\text{arctan}}\left( {\frac{d}{3} - 8t} \right)$ M1

$\frac{d θ}{d t} = \frac{8}{1 + {(\frac{d}{3} - 8 t)}^{2}}$ $\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{8}{{1 + {{\left( {\frac{d}{3} - 8t} \right)}^2}}}$ A1

at O, $t = \frac{d}{24}$ $t = \frac{d}{{24}}$ A1

THEN

$\frac{d θ}{d t} = - 8$ $\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = - 8$ A1

[6 marks]

Examiners report

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Syllabus sections

Topic 5—Calculus » SL 5.1—Introduction of differential calculus

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