C3 January 2007 Q3

3. The curve \(C\) has equation\[x = 2\sin y.\]

(a) Show that the point \(P\left(\sqrt{2}, \dfrac{\pi}{4}\right)\) lies on \(C\). (1)

(b) Show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sqrt{2}}\) at \(P\). (4)

(c) Find an equation of the normal to \(C\) at \(P\). Give your answer in the form \(y = mx + c\), where \(m\) and \(c\) are exact constants. (4)

Mark scheme (a)

Scheme	Marks
\(y = \dfrac{\pi}{4} \Rightarrow x = 2\sin\dfrac{\pi}{4} = 2 \times \dfrac{1}{\sqrt{2}} = \sqrt{2} \Rightarrow P \in C\)	B1
	(1)

Accept equivalent (reversed) arguments. In any method it must be clear that \(\sin\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}\) or exact equivalent is used.

Mark scheme (b)

Scheme	Marks
\(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 2\cos y\) or \(1 = 2\cos y\dfrac{\mathrm{d}y}{\mathrm{d}x}\)	M1 A1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2\cos y}\) May be awarded after substitution	M1
\(y = \dfrac{\pi}{4} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sqrt{2}}\) * cso	A1
	(4)