A2 October 2020 Paper 1 Q11

OCR ACurrent spec8 marksPolar Coordinates

11 A curve has cartesian equation \(x^3 + y^3 = 2xy\).

\(C\) is the portion of the curve for which \(x \geqslant 0\) and \(y \geqslant 0\). The equation of \(C\) in polar form is given by \(r = \mathrm{f}(\theta)\) for \(0 \leqslant \theta \leqslant \frac{1}{2}\pi\).

(a) Find \(\mathrm{f}(\theta)\). [2]

(b) Find an expression for \(\mathrm{f}\left(\frac{1}{2}\pi - \theta\right)\), giving your answer in terms of \(\sin\theta\) and \(\cos\theta\). [2]

(d) Find the value of \(r\) when \(\theta = \frac{1}{4}\pi\). [1]

(e) By finding values of \(\theta\) when \(r = 0\), show that \(C\) has a loop. [2]

Mark scheme (a)

Scheme	Marks	AO
\(x = r\cos\theta, y = r\sin\theta \Rightarrow (r\cos\theta)^3 + (r\sin\theta)^3 = 2r\cos\theta . r\sin\theta\) \(\Rightarrow r\left(\cos^3\theta + \sin^3\theta\right) = 2\cos\theta\sin\theta\)	M1	3.1a
\(\Rightarrow r = \dfrac{2\cos\theta\sin\theta}{\cos^3\theta + \sin^3\theta}\) oe	A1	1.1
	[2]

Notes

M1: Substitution
May see “or \(r = 0\)” but not required.

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{f}\left(\dfrac{1}{2}\pi - \theta\right) = \dfrac{2\cos\left(\frac{1}{2}\pi - \theta\right)\sin\left(\frac{1}{2}\pi - \theta\right)}{\cos^3\left(\frac{1}{2}\pi - \theta\right) + \sin^3\left(\frac{1}{2}\pi - \theta\right)}\)	M1	1.1a
\(= \dfrac{2\sin\theta\cos\theta}{\sin^3\theta + \cos^3\theta}\)	A1	1.1
	[2]

Notes

M1: Correct substitution into their \(\mathrm{f}(\theta)\)

Scheme	Marks	AO
So the line of symmetry is \(\theta = \dfrac{\pi}{4}\)	B1	2.2a
	[1]

Scheme	Marks	AO
\(r = \mathrm{f}\left(\tfrac{1}{4}\pi\right) = \sqrt{2}\)	B1	1.1
	[1]

Mark scheme (e)

Scheme	Marks	AO
\(r = 0\) when \(\theta = 0\). \(r = 0\) also when \(\theta = \dfrac{\pi}{2}\)	B1	3.1a
In range \(0 \lt \theta \lt \dfrac{\pi}{2}\), \(r \gt 0\) and is continuous So there is a loop	B1	2.4
	[2]

Notes

B1: For both, ignore extras.

B1: Conclusion – both statements for \(r\) need to be mentioned