1 The curve \(C\) is defined parametrically by $$x = 2 \cos ^ { 3 } t \quad \text { and } \quad y = 2 \sin ^ { 3 } t , \quad \text { for } 0 < t < \frac { 1 } { 2 } \pi .$$ Show that, at the point with parameter \(t\), $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { 1 } { 6 } \sec ^ { 4 } t \operatorname { cosec } t$$

## Question 1:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\dot{x} = -6\cos^2 t \sin t$, $\dot{y} = 6\sin^2 t \cos t$ | B1 | |
| $\Rightarrow \frac{dy}{dx} = -\tan t$ | B1 | (OE) |
| $\frac{d^2y}{dx^2} = -\sec^2 t \times \frac{-1}{6\cos^2 t \sin t} = \frac{1}{6}\sec^4 t \cosec t$ | M1A1 | AG |
| **Total: 4 marks** | [4] | |

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1 The curve $C$ is defined parametrically by

$$x = 2 \cos ^ { 3 } t \quad \text { and } \quad y = 2 \sin ^ { 3 } t , \quad \text { for } 0 < t < \frac { 1 } { 2 } \pi .$$

Show that, at the point with parameter $t$,

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { 1 } { 6 } \sec ^ { 4 } t \operatorname { cosec } t$$

\hfill \mbox{\textit{CAIE FP1 2015 Q1}}