1.07s Parametric and implicit differentiation

5 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 \(y\).

6 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. Section B (36 marks)

3 Fig. 3 shows the curve defined by the equation \(y = \arcsin ( x - 1 )\), for \(0 \leqslant x \leqslant 2\). \begin{figure}[h] \end{figure}

1. Find \(x\) in terms of \(y\), and show that \(\frac { \mathrm { d } x } { \mathrm {~d} y } = \cos y\).
2. Hence find the exact gradient of the curve at the point where \(x = 1.5\).

3 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 \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\).

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

9 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]{8feffafd-4eba-4968-b4d2-88fa364d6170-4_625_933_589_607} \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 \(9 x ^ { 2 } + y ^ { 2 } = 36 x\).

5 Given that \(y ^ { 3 } = x y - x ^ { 2 }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 2 x } { 3 y ^ { 2 } - x }\).
Hence show that the curve \(y ^ { 3 } = x y - x ^ { 2 }\) has a stationary point when \(x = \frac { 1 } { 8 }\).

9 The curve in Fig. 9.1 has equation \(\sqrt { x } + \sqrt { y } = 1\). \begin{figure}[h] \end{figure}

1. Show that this is part, but not all of the curve \(y = 1 - 2 \sqrt { x } + x\). Sketch the full curve \(y = 1 - 2 \sqrt { x } + x\).
2. Fig.9.2 shows a star shape made up of four parts, one of which is given in part (i) above. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{2f403099-2813-40d8-a9ae-1f7e64d41f80-4_380_681_1197_651} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} For each of the sections of the shape labelled \(\mathrm { A } , \mathrm { B }\) and C , state the equation of the curve and the domain.
3. The shape shown in Fig.9.2 is made into that in Fig. 10.3 by stretching the part of the figure for which \(y > 0\) by a scale factor of 2 . \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{2f403099-2813-40d8-a9ae-1f7e64d41f80-4_405_686_1996_605} \captionsetup{labelformat=empty} \caption{Fig. 9.3}
   \end{figure} Find the area of this shape.

4 A curve has equation \(y ^ { 2 } = 5 x - 4\).
Find the gradient of the curve at the points where \(x = 8\).

5 The equation of a circle is \(x ^ { 2 } + y ^ { 2 } = 25\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { x } { y }\).
2. Hence find the equation of the normal to the circle at the point ( 3,4 ).

1. (i) Differentiate \(x ^ { 3 } \ln x\) with respect to \(x\).
   (ii) Given that

$$x = \frac { y + 1 } { 3 - 2 y }$$ find and simplify an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(y\).

2 Fig. 9 shows the curve with equation \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\). It has an asymptote \(x = a\) and turning point P . \begin{figure}[h] \end{figure}

1. Write down the value of \(a\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 3 x ^ { 2 } } { 3 y ^ { 2 } ( 2 x - 1 ) ^ { 2 } }\). Hence find the coordinates of the turning point P , giving the \(y\)-coordinate to 3 significant figures.
3. Show that the substitution \(u = 2 x - 1\) transforms \(\int \frac { x } { \sqrt [ 3 ] { 2 x - 1 } } \mathrm {~d} x\) to \(\frac { 1 } { 4 } \int \left( u ^ { \frac { 2 } { 3 } } + u ^ { - \frac { 1 } { 3 } } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4.5\).

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 \(y\).
   [0pt]
2. Hence find the exact coordinates of the stationary points of the curve. [You need not determine their nature.]

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).

2 Given that \(y ^ { 3 } = x y - x ^ { 2 }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 2 x } { 3 y ^ { 2 } - x }\).
Hence show that the curve \(y ^ { 3 } = x y - x ^ { 2 }\) has a stationary point when \(x = \frac { 1 } { 8 }\).

4 The equation of a curve is given by \(\mathrm { e } ^ { 2 y } = 1 + \sin x\).

1. By differentiating implicitly, find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
2. Find an expression for \(y\) in terms of \(x\), and differentiate it to verify the result in part (i).

5 Fig. 6 shows the curve \(\mathrm { e } ^ { 2 y } = x ^ { 2 } + y\). \begin{figure}[h] \end{figure}

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x } { 2 \mathrm { e } ^ { 2 y } - 1 }\).
2. Hence find to 3 significant figures the coordinates of the point P , shown in Fig. 6, where the curve has infinite gradient.

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

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 \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\).