C4 June 2005 Q6
6. A curve has parametric equations
\[x = 2\cot t, \quad y = 2\sin^2 t, \quad 0 \lt t \leqslant \frac{\pi}{2}.\]
(a) Find an expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of the parameter \(t\). (4)
(b) Find an equation of the tangent to the curve at the point where \(t = \dfrac{\pi}{4}\). (4)
(c) Find a cartesian equation of the curve in the form \(y = \mathrm{f}(x)\). State the domain on which the curve is defined. (4)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -2\operatorname{cosec}^2 t,\ \dfrac{\mathrm{d}y}{\mathrm{d}t} = 4\sin t\cos t\) both | M1 A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-2\sin t\cos t}{\operatorname{cosec}^2 t} \quad \left(= -2\sin^3 t\cos t\right)\) | M1 A1 |
| (4) |
| Scheme | Marks |
|---|---|
| At \(t = \frac{\pi}{4},\ x = 2,\ y = 1\) both \(x\) and \(y\) | B1 |
| Substitutes \(t = \frac{\pi}{4}\) into an attempt at \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) to obtain gradient \(\left(-\dfrac{1}{2}\right)\) | M1 |
| Equation of tangent is \(y - 1 = -\dfrac{1}{2}(x - 2)\) Accept \(x + 2y = 4\) or any correct equivalent | M1 A1 |
| (4) |
| Scheme | Marks |
|---|---|
| Uses \(1 + \cot^2 t = \operatorname{cosec}^2 t\), or equivalent, to eliminate \(t\) | M1 |
| \(1 + \left(\dfrac{x}{2}\right)^2 = \dfrac{2}{y}\) correctly eliminates \(t\) | A1 |
| \(y = \dfrac{8}{4 + x^2}\) cao | A1 |
| The domain is \(x \geqslant 0\) | B1 |
| (4) | |
| (12 marks) |
An alternative in (c)
| \(\sin t = \left(\dfrac{y}{2}\right)^{\frac{1}{2}};\ \cos t = \dfrac{x}{2}\sin t = \dfrac{x}{2}\left(\dfrac{y}{2}\right)^{\frac{1}{2}}\) | |
| \(\sin^2 t + \cos^2 t = 1 \Rightarrow \dfrac{y}{2} + \dfrac{x^2}{4} \times \dfrac{y}{2} = 1\) | M1 A1 |
| Leading to \(y = \dfrac{8}{4 + x^2}\) | A1 |