OCR MEI C3 (Core Mathematics 3)

1 A curve has implicit equation \(y ^ { 2 } + 2 x \ln y = x ^ { 2 }\).
Verify that the point \(( 1,1 )\) lies on the curve, and find the gradient of the curve at this point.

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

1. By differentiating implicitly, find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
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2. Hence find the exact coordinates of the stationary points of the curve. [You need not determine their nature.]

3 Given that \(y = \ln \left( \sqrt { \frac { 2 x - 1 } { 2 x + 1 } } \right)\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 x - 1 } - \frac { 1 } { 2 x + 1 }\).

4 Fig. 7 shows the curve \(x ^ { 3 } + y ^ { 3 } = 3 x y\). The point P is a turning point of the curve. \begin{figure}[h] \end{figure}

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - x ^ { 2 } } { y ^ { 2 } - x }\).
2. Hence find the exact \(x\)-coordinate of P .

5 Find the gradient at the point \(( 0 , \ln 2 )\) on the curve with equation \(\mathrm { e } ^ { 2 y } = 5 - \mathrm { e } ^ { - x }\).

6 A curve is defined by the equation \(( x + y ) ^ { 2 } = 4 x\). The point \(( 1,1 )\) lies on this curve.
By differentiating implicitly, show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { x + y } - 1\).
Hence verify that the curve has a stationary point at \(( 1,1 )\).

7 A curve is defined by the equation \(\sin 2 x + \cos y = \sqrt { 3 }\).

1. Verify that the point \(\mathrm { P } \left( \frac { 1 } { 6 } \pi , \frac { 1 } { 6 } \pi \right)\) lies on the curve.
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at the point P .

8

1. Given that \(y = \sqrt [ 3 ] { 1 + 3 x ^ { 2 } }\), use the chain rule to find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\).
2. Given that \(y ^ { 3 } = 1 + 3 x ^ { 2 }\), use implicit differentiation to find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Show that this result is equivalent to the result in part (i).