A2 June 2024 Paper 2 Q6

OCR ACurrent spec11 marksHyperbolic Functions Polar Coordinates

6 In polar coordinates, the equation of a curve, \(C\), is \(r = 6\sin(2\theta)\sinh\left(\frac{1}{3}\theta\right)\) for \(0 \leqslant \theta \leqslant \frac{1}{2}\pi\).

The pole of the polar coordinate system corresponds to the origin of the cartesian system and the initial line corresponds to the positive \(x\)-axis.

(a) Explain how you can tell that \(C\) comprises a single loop in the first quadrant, passing through the pole. [3]

The incomplete table below shows values of \(r\) for various values of \(\theta\).

\(\theta\)	\(0\)	\(\dfrac{1}{12}\pi\)	\(\dfrac{1}{6}\pi\)	\(\dfrac{1}{4}\pi\)	\(\dfrac{1}{3}\pi\)	\(\dfrac{5}{12}\pi\)	\(\dfrac{1}{2}\pi\)
\(r\)	0	0.262			1.851

(b) Use the copy of the table and the polar coordinate system diagram given below to complete the table and sketch \(C\). [3]

Polar coordinate grid from the Printed Answer Booklet: the pole O, the initial line, quarter-circle arcs r = 0.5, 1.0, 1.5 and 2.0, and half-lines theta = pi/12, pi/6, pi/4, pi/3, 5pi/12 and pi/2

The point on \(C\) which is furthest away from the pole is denoted by \(A\) and the value of \(\theta\) at \(A\) is denoted by \(\phi\).

(c) Show that \(\phi\) satisfies the equation \(\phi = \dfrac{3}{2}\ln\left(\dfrac{6 - \tan 2\phi}{6 + \tan 2\phi}\right)\) [4]

(d) You are given that the relevant solution of the equation given in part (c) is \(\phi = 1.0207\) correct to 5 significant figures.
Find the distance from \(A\) to the pole. Give your answer correct to 3 significant figures. [1]

Mark scheme (a)

Scheme	Marks	AO
\(r = 0\) when \(\theta = 0\) and when \(\theta = \pi/2\) (and the function is continuous so there is a loop).	B1	2.4
Must be in 1^st quadrant since \(\theta\) only takes values between 0 and \(\frac{1}{2}\pi\).	B1	2.1
There are only two values of \(\theta\) for which \(r = 0\), therefore there is only one loop (two loops require three values of \(\theta\)/tangents at the pole).	B1	2.2a
	[3]

Notes

B1: Condone use of “origin” rather than “pole” in explanations.

B1: Must link domain of \(\theta\) with quadrant.

B1: Convincingly explaining why it is a single loop. \(\theta\) values/\(r = 0\) must be linked to loop.
or “There are no other solutions to \(r = 0\) (in the domain)”.

The polar grid with the seven points from the table marked with crosses and joined by a smooth single loop starting and ending at the pole, with its furthest point from the pole between theta = pi/4 and theta = 5pi/12, at r just above 1.6 — Mark scheme (b)

Mark scheme (c)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}r}{\mathrm{d}\theta} = 6\sinh\frac{1}{3}\theta \times \dfrac{\mathrm{d}}{\mathrm{d}\theta}(\sin 2\theta) + 6\sin 2\theta \times \dfrac{\mathrm{d}}{\mathrm{d}\theta}\left(\sinh\frac{1}{3}\theta\right)\)	M1	3.1a
\(\dfrac{\mathrm{d}r}{\mathrm{d}\theta} = 12\cos 2\theta\sinh\frac{1}{3}\theta + 2\sin 2\theta\cosh\frac{1}{3}\theta\)	A1	1.1
\(\therefore\) at \(A\), \(12\cos 2\phi\sinh\frac{1}{3}\phi + 2\sin 2\phi\cosh\frac{1}{3}\phi = 0\) \(\therefore 12\cos 2\phi\sinh\frac{1}{3}\phi = -2\sin 2\phi\cosh\frac{1}{3}\phi\) \(\therefore 6\tanh\frac{1}{3}\phi = -\tan 2\phi\)	M1	1.1
\(\therefore \frac{1}{3}\phi = \tanh^{-1}\left(-\frac{1}{6}\tan 2\phi\right)\) \(= \dfrac{1}{2}\ln\left(\dfrac{1 + \left(-\frac{1}{6}\tan 2\phi\right)}{1 - \left(-\frac{1}{6}\tan 2\phi\right)}\right)\) \(\therefore \phi = \dfrac{3}{2}\ln\left(\dfrac{6 - \tan 2\phi}{6 + \tan 2\phi}\right)\)	A1	2.2a
	[4]

Notes

M1: Attempting to differentiate \(r(\theta)\) using the product rule.
\(a\sinh\frac{1}{3}\theta\cos 2\theta + b\sin 2\theta\cosh\frac{1}{3}\theta\) where \(a\), \(b\) are non-zero constants

A1: All correct (any form)

M1: Setting their derivative to 0 and rearranging to a form containing tan and tanh only.
If using exponential form, must reach tan and reduce to one exponential term.
eg \(\mathrm{e}^{\frac{2}{3}\phi}(6 + \tan 2\phi) = 6 - \tan 2\phi\)

A1: AG so intermediate step must be shown.
Condone use of \(\theta\) rather than \(\phi\) but final answer must be in terms of \(\phi\) (or explicitly stating “\(\theta = \phi\)”).

Scheme	Marks	AO
\(r\ (= 6\sin(2 \times 1.0207) \times \sinh(1.0207/3))\) \(= 1.85\) (3 sf) cao	B1	1.1
	[1]