C4 June 2009 Q4
4. The curve \(C\) has the equation \(y\mathrm{e}^{-2x} = 2x + y^2\).
(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) and \(y\). (5)
The point \(P\) on \(C\) has coordinates \((0, 1)\).
(b) Find the equation of the normal to \(C\) at \(P\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. (4)
| Scheme | Marks |
|---|---|
| \(\mathrm{e}^{-2x}\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y\mathrm{e}^{-2x} = 2 + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) A1 correct RHS | M1 A1 |
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(y\mathrm{e}^{-2x}\right) = \mathrm{e}^{-2x}\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y\mathrm{e}^{-2x}\) | B1 |
| \(\left(\mathrm{e}^{-2x} - 2y\right)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2 + 2y\mathrm{e}^{-2x}\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2 + 2y\mathrm{e}^{-2x}}{\mathrm{e}^{-2x} - 2y}\) | A1 |
| (5) |
Alternative
Alternative for (a) differentiating implicitly with respect to \(y\).
| \(\mathrm{e}^{-2x} - 2y\mathrm{e}^{-2x}\dfrac{\mathrm{d}x}{\mathrm{d}y} = 2\dfrac{\mathrm{d}x}{\mathrm{d}y} + 2y\) A1 correct RHS | M1 A1 |
| \(\dfrac{\mathrm{d}}{\mathrm{d}y}\left(y\mathrm{e}^{-2x}\right) = \mathrm{e}^{-2x} - 2y\mathrm{e}^{-2x}\dfrac{\mathrm{d}x}{\mathrm{d}y}\) | B1 |
| \(\left(2 + 2y\mathrm{e}^{-2x}\right)\dfrac{\mathrm{d}x}{\mathrm{d}y} = \mathrm{e}^{-2x} - 2y\) | M1 |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{\mathrm{e}^{-2x} - 2y}{2 + 2y\mathrm{e}^{-2x}}\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2 + 2y\mathrm{e}^{-2x}}{\mathrm{e}^{-2x} - 2y}\) | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| At \(P\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2 + 2\mathrm{e}^0}{\mathrm{e}^0 - 2} = -4\) | M1 |
| Using \(mm' = -1\) \(m' = \dfrac{1}{4}\) | M1 |
| \(y - 1 = \dfrac{1}{4}(x - 0)\) | M1 |
| \(x - 4y + 4 = 0\) or any integer multiple | A1 |
| (4) | |
| (9 marks) |