C3 June 2015 Q5

5. The point \(P\) lies on the curve with equation\[x = (4y - \sin 2y)^2\]Given that \(P\) has \((x, y)\) coordinates \(\left(p, \dfrac{\pi}{2}\right)\), where \(p\) is a constant,

(a) find the exact value of \(p\). (1)

The tangent to the curve at \(P\) cuts the \(y\)-axis at the point \(A\).

(b) Use calculus to find the coordinates of \(A\). (6)

Scheme	Marks
\(p = 4\pi^2\) or \((2\pi)^2\)	B1
	(1)

Mark scheme (b)

Scheme	Marks
\(x = (4y - \sin 2y)^2 \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}y} = 2(4y - \sin 2y)(4 - 2\cos 2y)\)	M1A1
Sub \(y = \dfrac{\pi}{2}\) into \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 2(4y - \sin 2y)(4 - 2\cos 2y)\)
\(\Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}y} = 24\pi\) \((= 75.4)\) / \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{24\pi}\ (= 0.013)\)	M1
Equation of tangent \(y - \dfrac{\pi}{2} = \dfrac{1}{24\pi}\left(x - 4\pi^2\right)\)	M1
Using \(y - \dfrac{\pi}{2} = \dfrac{1}{24\pi}\left(x - 4\pi^2\right)\) with \(x = 0 \Rightarrow y = \dfrac{\pi}{3}\) cso	M1, A1
	(6)
	(7 marks)

Alt (b) I

Scheme	Marks
\(x = (4y - \sin 2y)^2 \Rightarrow x^{0.5} = 4y - \sin 2y\)
\(\Rightarrow 0.5x^{-0.5}\dfrac{\mathrm{d}x}{\mathrm{d}y} = 4 - 2\cos 2y\)	M1A1

Alt (b) II

Scheme	Marks
\(x = \left(16y^2 - 8y\sin 2y + \sin^2 2y\right)\)
\(\Rightarrow 1 = 32y\dfrac{\mathrm{d}y}{\mathrm{d}x} - 8\sin 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} - 16y\cos 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} + 4\sin 2y\cos 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) Or \(1\,\mathrm{d}x = 32y\,\mathrm{d}y - 8\sin 2y\,\mathrm{d}y - 16y\cos 2y\,\mathrm{d}y + 4\sin 2y\cos 2y\,\mathrm{d}y\)	M1A1

Notes

M1: Uses the chain rule of differentiation to get a form \(A(4y - \sin 2y)(B \pm C\cos 2y)\), \(A, B, C \neq 0\) on the right hand side
Alternatively attempts to expand and then differentiate using product rule and chain rule to a form \(x = \left(16y^2 - 8y\sin 2y + \sin^2 2y\right) \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}y} = Py \pm Q\sin 2y \pm Ry\cos 2y \pm S\sin 2y\cos 2y\) \(P, Q, R, S \neq 0\)
A second method is to take the square root first. To score the method look for a differentiated expression of the form \(Px^{-0.5}\ldots = 4 - Q\cos 2y\)
A third method is to multiply out and use implicit differentiation. Look for the correct terms, condoning errors on just the constants.

A1: \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 2(4y - \sin 2y)(4 - 2\cos 2y)\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2(4y - \sin 2y)(4 - 2\cos 2y)}\) with both sides correct. The lhs may be seen elsewhere if clearly linked to the rhs.
In the alternative \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 32y - 8\sin 2y - 16y\cos 2y + 4\sin 2y\cos 2y\)

M1: Sub \(y = \dfrac{\pi}{2}\) into their \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) or inverted \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\). Evidence could be minimal, eg \(y = \dfrac{\pi}{2} \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}y} = \ldots\)
It is not dependent upon the previous M1 but it must be a changed \(x = (4y - \sin 2y)^2\)

M1: Score for a correct method for finding the equation of the tangent at \(\left(\text{'}4\pi^2\text{'}, \dfrac{\pi}{2}\right)\).
Allow for \(y - \dfrac{\pi}{2} = \dfrac{1}{\text{their numerical}\left(\mathrm{d}x/\mathrm{d}y\right)}\left(x - \text{their } 4\pi^2\right)\)
Allow for \(\left(y - \dfrac{\pi}{2}\right)\text{their numerical}\left(\mathrm{d}x/\mathrm{d}y\right) = \left(x - \text{their } 4\pi^2\right)\)
Even allow for \(y - \dfrac{\pi}{2} = \dfrac{1}{\text{their numerical}\left(\mathrm{d}x/\mathrm{d}y\right)}(x - p)\)
It is possible to score this by stating the equation \(y = \dfrac{1}{24\pi}x + c\) as long as \(\left(\text{'}4\pi^2\text{'}, \dfrac{\pi}{2}\right)\) is used in a subsequent line.

M1: Score for writing their equation in the form \(y = mx + c\) and stating the value of '\(c\)'
Or setting \(x = 0\) in their \(y - \dfrac{\pi}{2} = \dfrac{1}{24\pi}\left(x - 4\pi^2\right)\) and solving for \(y\).
Alternatively using the gradient of the line segment \(AP\) = gradient of tangent.
Look for \(\dfrac{\frac{\pi}{2} - y}{4\pi^2} = \dfrac{1}{24\pi} \Rightarrow y = ..\) Such a method scores the previous M mark as well.
At this stage all of the constants must be numerical. It is not dependent and it is possible to score this using the "incorrect" gradient.

A1: cso \(y = \dfrac{\pi}{3}\). You do not have to see \(\left(0, \dfrac{\pi}{3}\right)\)