C4 June 2014 Q1

Notes

Note: Writing down \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3x^2 + 2y - 1}{3y^2 - 2x}\) or \(\dfrac{1 - 3x^2 - 2y}{2x - 3y^2}\) from no working is full marks.

Note: Writing down \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3x^2 + 2y - 1}{2x - 3y^2}\) or \(\dfrac{1 - 3x^2 - 2y}{3y^2 - 2x}\) from no working is M1A0B0M1A0

Note: Few candidates will write \(3x^2 + 2y + 2x\,\mathrm{d}y - 1 - 3y^2\mathrm{d}y = 0\) leading to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3x^2 + 2y - 1}{3y^2 - 2x}\), o.e.
This should get full marks.

M1: Differentiates implicitly to include either \(2x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(-y^3 \to \pm ky^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\). (Ignore \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right)\)).

A1: \(x^3 \to 3x^2\) and \(-x - y^3 - 20 = 0 \to -1 - 3y^2\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)

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

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

Note: \(3x^2 + 2y + 2x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 1 - 3y^2\dfrac{\mathrm{d}y}{\mathrm{d}x} \to 3x^2 + 2y - 1 = 3y^2\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2x\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
will get 1^st A1 (implied) as the "\(= 0\)" can be implied by rearrangement of their equation.

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. \(\ldots + \left(2x - 3y^2\right)\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\)

Note: Placing an extra \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) at the beginning and then including it in their factorisation is fine for dM1.

A1: For \(\dfrac{1 - 2y - 3x^2}{2x - 3y^2}\) or equivalent. Eg: \(\dfrac{3x^2 + 2y - 1}{3y^2 - 2x}\)
cso: If the candidate’s solution is not completely correct, then do not give this mark.
isw: You can, however, ignore subsequent working following on from correct solution.

Alternative method for part (a)

Alternative method for part (a): Differentiating with respect to \(y\)

M1: Differentiates implicitly to include either \(2y\dfrac{\mathrm{d}x}{\mathrm{d}y}\) or \(x^3 \to \pm kx^2\dfrac{\mathrm{d}x}{\mathrm{d}y}\) or \(-x \to -\dfrac{\mathrm{d}x}{\mathrm{d}y}\)
(Ignore \(\left(\dfrac{\mathrm{d}x}{\mathrm{d}y} =\right)\)).

A1: \(x^3 \to 3x^2\dfrac{\mathrm{d}x}{\mathrm{d}y}\) and \(-x - y^3 - 20 = 0 \to -\dfrac{\mathrm{d}x}{\mathrm{d}y} - 3y^2 = 0\)

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

dM1: dependent on the first method mark being awarded.
An attempt to factorise out all the terms in \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) as long as there are at least two terms in \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\).

A1: For \(\dfrac{1 - 2y - 3x^2}{2x - 3y^2}\) or equivalent. Eg: \(\dfrac{3x^2 + 2y - 1}{3y^2 - 2x}\)
cso: If the candidate’s solution is not completely correct, then do not give this mark.

Notes

M1: Some attempt to substitute both \(x = 3\) and \(y = -2\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) which contains both \(x\) and \(y\) to find \(m_T\) and

• either applies \(y - -2 = (\text{their } m_T)(x - 3)\), where \(m_T\) is a numerical value.
• or finds \(c\) by solving \((-2) = (\text{their } m_T)(3) + c\), where \(m_T\) is a numerical value.

Note: Using a changed gradient (i.e. applying \(\dfrac{-1}{\text{their } \frac{\mathrm{d}y}{\mathrm{d}x}}\) or \(\dfrac{1}{\text{their } \frac{\mathrm{d}y}{\mathrm{d}x}}\) is M0).

A1: Accept any integer multiple of \(11x - 3y - 39 = 0\) or \(11x - 39 - 3y = 0\) or \(-11x + 3y + 39 = 0\), where their tangent equation is equal to 0.

cso: A correct solution is required from a correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).

isw: You can ignore subsequent working following a correct solution.