C4 January 2012 Q1
1. The curve \(C\) has the equation \(2x + 3y^2 + 3x^2y = 4x^2\).
The point \(P\) on the curve has coordinates \((-1, 1)\).
| Scheme | Marks |
|---|---|
| \(\left\{\cancel{\dfrac{\mathrm{d}y}{\mathrm{d}x}} \times\right\}\quad \underline{2 + 6y\dfrac{\mathrm{d}y}{\mathrm{d}x}} + \underline{\underline{\left(6xy + 3x^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)}} = \underline{8x}\) | M1 A1 B1 |
| \(\left\{\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{8x - 2 - 6xy}{6y + 3x^2}\right\}\) not necessarily required. | |
| At \(P(-1, 1)\), \(\ \mathrm{m}(\mathbf{T}) = \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{8(-1) - 2 - 6(-1)(1)}{6(1) + 3(-1)^2} = -\dfrac{4}{9}\) | dM1 A1 cso |
| (5) |
Notes
M1: Differentiates implicitly to include either \(\pm ky\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(3x^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\). (Ignore \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right)\)).
A1: \((2x + 3y^2) \to \left(\underline{2 + 6y\dfrac{\mathrm{d}y}{\mathrm{d}x}}\right)\) and \((4x^2 \to \underline{8x})\). Note: If an extra “sixth” term appears then award A0.
B1: \(6xy + 3x^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\).
dM1: Substituting \(x = -1\) and \(y = 1\) into an equation involving \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). Allow this mark if either the numerator or denominator of \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{8x - 2 - 6xy}{6y + 3x^2}\) is substituted into or evaluated correctly.
If it is clear, however, that the candidate is intending to substitute \(x = 1\) and \(y = -1\), then award M0.
Candidates who substitute \(x = 1\) and \(y = -1\), will usually achieve \(\mathrm{m}(\mathbf{T}) = -4\)
Note that this mark is dependent on the previous method mark being awarded.
A1: For \(-\dfrac{4}{9}\) or \(-\dfrac{8}{18}\) or \(-0.\dot{4}\) or awrt \(-0.44\)
If the candidate’s solution is not completely correct, then do not give this mark.
Alternative method for part (a): Differentiating with respect to \(y\)
| Scheme | Marks |
|---|---|
| \(\left\{\cancel{\dfrac{\mathrm{d}x}{\mathrm{d}y}} \times\right\}\quad \underline{2\dfrac{\mathrm{d}x}{\mathrm{d}y} + 6y} + \underline{\underline{\left(6xy\dfrac{\mathrm{d}x}{\mathrm{d}y} + 3x^2\right)}} = \underline{8x\dfrac{\mathrm{d}x}{\mathrm{d}y}}\) |
M1: Differentiates implicitly to include either \(2\dfrac{\mathrm{d}x}{\mathrm{d}y}\) or \(6xy\dfrac{\mathrm{d}x}{\mathrm{d}y}\) or \(\pm kx\dfrac{\mathrm{d}x}{\mathrm{d}y}\). (Ignore \(\left(\dfrac{\mathrm{d}x}{\mathrm{d}y} =\right)\)).
A1: \((2x + 3y^2) \to \left(\underline{2\dfrac{\mathrm{d}x}{\mathrm{d}y} + 6y}\right)\) and \(\left(4x^2 \to \underline{8x\dfrac{\mathrm{d}x}{\mathrm{d}y}}\right)\). Note: If an extra “sixth” term appears then award A0.
B1: \(6xy\dfrac{\mathrm{d}x}{\mathrm{d}y} + 3x^2\). (corrected from the printed mark scheme: printed as \(6xy + 3x^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\), the term from the main method)
dM1: Substituting \(x = -1\) and \(y = 1\) into an equation involving \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). Allow this mark if either the numerator or denominator of \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{6y + 3x^2}{8x - 2 - 6xy}\) is substituted into or evaluated correctly.
If it is clear, however, that the candidate is intending to substitute \(x = 1\) and \(y = -1\), then award M0.
Candidates who substitute \(x = 1\) and \(y = -1\), will usually achieve \(\mathrm{m}(\mathbf{T}) = -4\)
Note that this mark is dependent on the previous method mark being awarded.
A1: For \(-\dfrac{4}{9}\) or \(-\dfrac{8}{18}\) or \(-0.\dot{4}\) or awrt \(-0.44\)
If the candidate’s solution is not completely correct, then do not give this mark.
| Scheme | Marks |
|---|---|
| So, \(\mathrm{m}(\mathbf{N}) = \dfrac{-1}{-\frac{4}{9}}\ \left\{= \dfrac{9}{4}\right\}\) | M1 |
| \(\mathbf{N}:\ \ y - 1 = \dfrac{9}{4}(x + 1)\) | M1 |
| \(\mathbf{N}:\ \ 9x - 4y + 13 = 0\) | A1 |
| (3) | |
| (8 marks) |
Notes
M1: Applies \(\mathrm{m}(\mathbf{N}) = -\dfrac{1}{\text{their m}(\mathbf{T})}\).
M1: Uses \(y - 1 = (m_N)(x - -1)\) or finds \(c\) using \(x = -1\) and \(y = 1\) and uses \(y = (m_N)x + \text{"}c\text{"}\),
Where \(m_N = -\dfrac{1}{\text{their m}(\mathbf{T})}\) or \(m_N = \dfrac{1}{\text{their m}(\mathbf{T})}\) or \(m_N = -\text{their m}(\mathbf{T})\).
A1: \(9x - 4y + 13 = 0\) or \(-9x + 4y - 13 = 0\) or \(4y - 9x - 13 = 0\) or \(18x - 8y + 26 = 0\) etc.
Must be “\(= 0\)”. So do not allow \(9x + 13 = 4y\) etc.
Note: \(m_N = -\left(\dfrac{6y + 3x^2}{8x - 2 - 6xy}\right)\) is M0M0 unless a numerical value is then found for \(m_N\).