C4 June 2016 Q3

3. The curve \(C\) has equation \[2x^2y + 2x + 4y - \cos(\pi y) = 17\]

(a) Use implicit differentiation to find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) and \(y\). (5)

The point \(P\) with coordinates \(\left(3, \dfrac{1}{2}\right)\) lies on \(C\).

The normal to \(C\) at \(P\) meets the \(x\)-axis at the point \(A\).

(b) Find the \(x\) coordinate of \(A\), giving your answer in the form \(\dfrac{a\pi + b}{c\pi + d}\), where \(a\), \(b\), \(c\) and \(d\) are integers to be determined. (4)

Mark scheme (a)

Scheme	Marks
\(2x^2y + 2x + 4y - \cos(\pi y) = 17\)
Way 1
\(\left\{\dfrac{\cancel{\mathrm{d}y}}{\cancel{\mathrm{d}x}} \times\right\}\ \underline{\underline{\left(4xy + 2x^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)}} + \underline{2 + 4\dfrac{\mathrm{d}y}{\mathrm{d}x} + \pi\sin(\pi y)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0}\)	M1 A1 B1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x}\big(2x^2 + 4 + \pi\sin(\pi y)\big) + 4xy + 2 = 0\)	dM1
\(\left\{\dfrac{\mathrm{d}y}{\mathrm{d}x} = \right\}\ \dfrac{-4xy - 2}{2x^2 + 4 + \pi\sin(\pi y)}\) or \(\dfrac{4xy + 2}{-2x^2 - 4 - \pi\sin(\pi y)}\) Correct answer or equivalent	A1 cso
	(5)

Notes

Note: Writing down from no working

\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-4xy - 2}{2x^2 + 4 + \pi\sin(\pi y)}\) or \(\dfrac{4xy + 2}{-2x^2 - 4 - \pi\sin(\pi y)}\) scores M1A1B1M1A1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{4xy + 2}{2x^2 + 4 + \pi\sin(\pi y)}\) scores M1A0B1M1A0

Note: Few candidates will write \(4xy\,\mathrm{d}x + 2x^2\,\mathrm{d}y + 2\,\mathrm{d}x + 4\,\mathrm{d}y + \pi\sin(\pi y)\,\mathrm{d}y = 0\) leading to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-4xy - 2}{2x^2 + 4 + \pi\sin(\pi y)}\) or equivalent. This should get full marks.

M1: Differentiates implicitly to include either \(2x^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(4y \to 4\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(-\cos(\pi y) \to \pm\lambda\sin(\pi y)\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (Ignore \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \right)\)). \(\lambda\) is a constant which can be 1.

1^st A1: \(2x + 4y - \cos(\pi y) = 17 \to 2 + 4\dfrac{\mathrm{d}y}{\mathrm{d}x} + \pi\sin(\pi y)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)

Note: \(4xy + 2x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2 + 4\dfrac{\mathrm{d}y}{\mathrm{d}x} + \pi\sin(\pi y)\dfrac{\mathrm{d}y}{\mathrm{d}x} \to 2x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 4\dfrac{\mathrm{d}y}{\mathrm{d}x} + \pi\sin(\pi y)\dfrac{\mathrm{d}y}{\mathrm{d}x} = -4xy - 2\) will get 1^st A1 (implied) as the "\(= 0\)" can be implied by the rearrangement of their equation.

B1: \(2x^2y \to 4xy + 2x^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

Note: If an extra term appears then award 1^st A0.

dM1: Dependent on the first method mark being awarded.
An attempt to factorise out all the terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) as long as there are at least two terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).
ie. \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\big(2x^2 + 4 + \pi\sin(\pi y)\big) + \ldots = \ldots\)

Note: Writing down an extra \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) and then including it in their factorisation is fine for dM1.

Note: Final A1 cso: If the candidate’s solution is not completely correct, then do not give this mark.

Note: Final A1 isw: You can, however, ignore subsequent working following on from correct solution.

Way 2

Scheme	Marks
\(\left\{\dfrac{\cancel{\mathrm{d}x}}{\cancel{\mathrm{d}y}} \times\right\}\ \underline{\underline{\left(4xy\dfrac{\mathrm{d}x}{\mathrm{d}y} + 2x^2\right)}} + \underline{2\dfrac{\mathrm{d}x}{\mathrm{d}y} + 4 + \pi\sin(\pi y) = 0}\)	M1 A1 B1
\(\dfrac{\mathrm{d}x}{\mathrm{d}y}(4xy + 2) + 2x^2 + 4 + \pi\sin(\pi y) = 0\)	dM1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-4xy - 2}{2x^2 + 4 + \pi\sin(\pi y)}\) or \(\dfrac{4xy + 2}{-2x^2 - 4 - \pi\sin(\pi y)}\) Correct answer or equivalent	A1 cso
	(5)

Way 2: Apply the mark scheme for Way 2 in the same way as Way 1.

Mark scheme (b)

Scheme	Marks
At \(\left(3, \dfrac{1}{2}\right)\), \(m_T = \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-4(3)(\frac{1}{2}) - 2}{2(3)^2 + 4 + \pi\sin\left(\frac{1}{2}\pi\right)}\ \left\{= \dfrac{-8}{22 + \pi}\right\}\) Substituting \(x = 3\) & \(y = \dfrac{1}{2}\) into an equation involving \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)	M1
\(m_N = \dfrac{22 + \pi}{8}\) Applying \(m_N = \dfrac{-1}{m_T}\) to find a numerical \(m_N\). Can be implied by later working	M1
\(\bullet\ y - \dfrac{1}{2} = \left(\dfrac{22 + \pi}{8}\right)(x - 3)\) \(\bullet\ \dfrac{1}{2} = \left(\dfrac{22 + \pi}{8}\right)(3) + c \Rightarrow c = \dfrac{1}{2} - \dfrac{66 + 3\pi}{8}\) \(\Rightarrow y = \left(\dfrac{22 + \pi}{8}\right)x + \dfrac{1}{2} - \dfrac{66 + 3\pi}{8}\) Cuts \(x\)-axis \(\Rightarrow y = 0\) \(\Rightarrow -\dfrac{1}{2} = \left(\dfrac{22 + \pi}{8}\right)(x - 3)\) \(y - \dfrac{1}{2} = m_N(x - 3)\) or \(y = m_N x + c\) where \(\dfrac{1}{2} = (\text{their } m_N)3 + c\) with a numerical \(m_N\ (\neq m_T)\) where \(m_N\) is in terms of \(\pi\) and sets \(y = 0\) in their normal equation.	dM1
So, \(\left\{x = \dfrac{-4}{22 + \pi} + 3 \Rightarrow\right\}\ x = \dfrac{3\pi + 62}{\pi + 22}\) \(\dfrac{3\pi + 62}{\pi + 22}\) or \(\dfrac{6\pi + 124}{2\pi + 44}\) or \(\dfrac{62 + 3\pi}{22 + \pi}\)	A1 o.e.
	(4)
	(9 marks)

Notes

1^st M1: M1 can be gained by seeing at least one example of substituting \(x = 3\) and at least one example of substituting \(y = \dfrac{1}{2}\). E.g. "\(-4xy\)" \(\to\) "\(-6\)" in their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) would be sufficient for M1, unless it is clear that they are instead applying \(x = \dfrac{1}{2}\), \(y = 3\).

3^rd M1: is dependent on the first M1.

Note: The 2^nd M1 mark can be implied by later working.
Eg. Award 2^nd M1 3^rd M1 for \(\dfrac{\frac{1}{2}}{3 - x} = \dfrac{-1}{\text{their } m_T}\)

Note: We can accept \(\sin\pi\) or \(\sin\left(\dfrac{\pi}{2}\right)\) as a numerical value for the 2^nd M1 mark.
But, \(\sin\pi\) by itself or \(\sin\left(\dfrac{\pi}{2}\right)\) by itself are not allowed as being in terms of \(\pi\) for the 3^rd M1 mark.
The 3^rd M1 can be accessed for terms containing \(\pi\sin\left(\dfrac{\pi}{2}\right)\).