Parametric differentiation

1. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

A curve \(C\) has equation $$x = \sin ^ { 2 } 4 y \quad 0 \leqslant y \leqslant \frac { \pi } { 8 } \quad 0 \leqslant x \leqslant 1$$ The point \(P\) with \(x\) coordinate \(\frac { 1 } { 4 }\) lies on \(C\)

1. Find the exact \(y\) coordinate of \(P\)
2. Find \(\frac { \mathrm { d } x } { \mathrm {~d} y }\)
3. Hence show that \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) can be written in the form $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { q + r ( x + s ) ^ { 2 } } }$$ where \(q , r\) and \(s\) are constants to be found. Using the answer to part (c),
4. 1. state the \(x\) coordinate of the point where the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) is a minimum,
   2. state the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at this point.

1. Given that

$$x = \sec ^ { 2 } 3 y , \quad 0 < y < \frac { \pi } { 6 }$$

1. find \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(y\).
2. Hence show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 6 x ( x - 1 ) ^ { \frac { 1 } { 2 } } }$$
3. Find an expression for \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) in terms of \(x\). Give your answer in its simplest form.

4

1. Given that \(x = ( 4 t + 9 ) ^ { \frac { 1 } { 2 } }\) and \(y = 6 \mathrm { e } ^ { \frac { 1 } { 2 } x + 1 }\), find expressions for \(\frac { \mathrm { d } x } { \mathrm {~d} t }\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Hence find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} t }\) when \(t = 4\), giving your answer correct to 3 significant figures.