C4 June 2012 Q5
5. The curve \(C\) has equation\[16y^3 + 9x^2y - 54x = 0\]
(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) and \(y\). (5)
(b) Find the coordinates of the points on \(C\) where \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\). (7)
| Scheme | Marks |
|---|---|
| Differentiating implicitly to obtain \(\pm ay^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and/or \(\pm bx^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | M1 |
| \(48y^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + \ldots - 54\ldots\) | A1 |
| \(9x^2y \to 9x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 18xy\) or equivalent | B1 |
| \(\left(48y^2 + 9x^2\right)\dfrac{\mathrm{d}y}{\mathrm{d}x} + 18xy - 54 = 0\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{54 - 18xy}{48y^2 + 9x^2} \quad \left(= \dfrac{18 - 6xy}{16y^2 + 3x^2}\right)\) | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(18 - 6xy = 0\) | M1 |
| Using \(x = \dfrac{3}{y}\) or \(y = \dfrac{3}{x}\) \(16y^3 + 9\left(\dfrac{3}{y}\right)^2 y - 54\left(\dfrac{3}{y}\right) = 0\) or \(16\left(\dfrac{3}{x}\right)^3 + 9x^2\left(\dfrac{3}{x}\right) - 54x = 0\) | M1 |
| Leading to \(16y^4 + 81 - 162 = 0\) or \(16 + x^4 - 2x^4 = 0\) | M1 |
| \(y^4 = \dfrac{81}{16}\) or \(x^4 = 16\) \(y = \dfrac{3}{2}, -\dfrac{3}{2}\) or \(x = 2, -2\) | A1 A1 |
| Substituting either of their values into \(xy = 3\) to obtain a value of the other variable. | M1 |
| \(\left(2, \dfrac{3}{2}\right),\ \left(-2, -\dfrac{3}{2}\right)\) both | A1 |
| (7) | |
| (12 marks) |
Notes
In the printed scheme a bracket joins these method marks: each later M mark is dependent on the M mark before it.