OCR MEI C3 (Core Mathematics 3)

1 Given that \(x ^ { 2 } + x y + y ^ { 2 } = 12\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).

2 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = \sqrt { 4 - x ^ { 2 } }\) for \(- 2 \leqslant x \leqslant 2\).

1. Show that the curve \(y = \sqrt { 4 - x ^ { 2 } }\) is a semicircle of radius 2 , and explain why it is not the whole of this circle. Fig. 9 shows a point \(\mathrm { P } ( a , b )\) on the semicircle. The tangent at P is shown. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{ce82bfc4-90dd-4127-a11c-281cdcca70cf-1_621_934_1046_664} \captionsetup{labelformat=empty} \caption{Fig. 9}
   \end{figure}
2. (A) Use the gradient of OP to find the gradient of the tangent at P in terms of \(a\) and \(b\).
   (B) Differentiate \(\sqrt { 4 - x ^ { 2 } }\) and deduce the value of \(\mathrm { f } ^ { \prime } ( a )\).
   (C) Show that your answers to parts (A) and (B) are equivalent. The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = 3 \mathrm { f } ( x - 2 )\), for \(0 \leqslant x \leqslant 4\).
3. Describe a sequence of two transformations that would map the curve \(y = \mathrm { f } ( x )\) onto the curve \(y = \mathrm { g } ( x )\). Hence sketch the curve \(y = \mathrm { g } ( x )\).
4. Show that if \(y = \mathrm { g } ( x )\) then \(9 x ^ { 2 } + y ^ { 2 } = 36 x\).

3 Fig. 6 shows the triangle OAP , where O is the origin and A is the point \(( 0,3 )\). The point \(\mathrm { P } ( x , 0 )\) moves on the positive \(x\)-axis. The point \(\mathrm { Q } ( 0 , y )\) moves between O and A in such a way that \(\mathrm { AQ } + \mathrm { AP } = 6\). \begin{figure}[h] \end{figure}

1. Write down the length AQ in terms of \(y\). Hence find AP in terms of \(y\), and show that $$( y + 3 ) ^ { 2 } = x ^ { 2 } + 9$$
2. Use this result to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x } { y + 3 }\).
3. When \(x = 4\) and \(y = 2 , \frac { \mathrm {~d} x } { \mathrm {~d} t } = 2\). Calculate \(\frac { \mathrm { d } y } { \mathrm {~d} t }\) at this time.

4 A curve has equation \(2 y ^ { 2 } + y = 9 x ^ { 2 } + 1\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at the point \(\mathrm { A } ( 1,2 )\).
2. Find the coordinates of the points on the curve at which \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\).

5 Given that \(y = ( 1 + 6 x ) ^ { \frac { 1 } { 3 } }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { y ^ { 2 } }\).

6 A curve is defined implicitly by the equation $$y ^ { 3 } = 2 x y + x ^ { 2 }$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 ( x + y ) } { 3 y ^ { 2 } - 2 x }\).
2. Hence write down \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(x\) and \(y\).