1.07s Parametric and implicit differentiation

6 The equation of a curve is \(x ^ { 3 } + 2 y ^ { 3 } = 3 x y\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - x ^ { 2 } } { 2 y ^ { 2 } - x }\).
2. Find the coordinates of the point, other than the origin, where the curve has a tangent which is parallel to the \(x\)-axis.

4 The parametric equations of a curve are $$x = a ( 2 \theta - \sin 2 \theta ) , \quad y = a ( 1 - \cos 2 \theta )$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \cot \theta\).

3 The equation of a curve is \(x ^ { 3 } - x ^ { 2 } y - y ^ { 3 } = 3\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
2. Find the equation of the tangent to the curve at the point \(( 2,1 )\), giving your answer in the form \(a x + b y + c = 0\).

2 The parametric equations of a curve are $$x = \frac { t } { 2 t + 3 } , \quad y = \mathrm { e } ^ { - 2 t }$$ Find the gradient of the curve at the point for which \(t = 0\).

2 The parametric equations of a curve are $$x = 3 \left( 1 + \sin ^ { 2 } t \right) , \quad y = 2 \cos ^ { 3 } t$$ Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\), simplifying your answer as far as possible.

8 \includegraphics[max width=\textwidth, alt={}, center]{6025cf1d-525e-4f12-9517-f20ef5fff2fa-3_698_1006_758_571} The diagram shows the curve with parametric equations $$x = \sin t + \cos t , \quad y = \sin ^ { 3 } t + \cos ^ { 3 } t$$ for \(\frac { 1 } { 4 } \pi < t < \frac { 5 } { 4 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 3 \sin t \cos t\).
2. Find the gradient of the curve at the origin.
3. Find the values of \(t\) for which the gradient of the curve is 1 , giving your answers correct to 2 significant figures.

7 The equation of a curve is \(\ln ( x y ) - y ^ { 3 } = 1\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y } { x \left( 3 y ^ { 3 } - 1 \right) }\).
2. Find the coordinates of the point where the tangent to the curve is parallel to the \(y\)-axis, giving each coordinate correct to 3 significant figures.

3 The parametric equations of a curve are $$x = \frac { 4 t } { 2 t + 3 } , \quad y = 2 \ln ( 2 t + 3 )$$

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\), simplifying your answer.
2. Find the gradient of the curve at the point for which \(x = 1\).

4 The parametric equations of a curve are $$x = \mathrm { e } ^ { - t } \cos t , \quad y = \mathrm { e } ^ { - t } \sin t$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \tan \left( t - \frac { 1 } { 4 } \pi \right)\).

4 A curve has equation \(3 \mathrm { e } ^ { 2 x } y + \mathrm { e } ^ { x } y ^ { 3 } = 14\). Find the gradient of the curve at the point \(( 0,2 )\).

4 The parametric equations of a curve are $$x = \frac { 1 } { \cos ^ { 3 } t } , \quad y = \tan ^ { 3 } t$$ where \(0 \leqslant t < \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sin t\).
2. Hence show that the equation of the tangent to the curve at the point with parameter \(t\) is \(y = x \sin t - \tan t\).

2 A curve is defined for \(0 < \theta < \frac { 1 } { 2 } \pi\) by the parametric equations $$x = \tan \theta , \quad y = 2 \cos ^ { 2 } \theta \sin \theta$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 \cos ^ { 5 } \theta - 4 \cos ^ { 3 } \theta\).

4 A curve has parametric equations $$x = t ^ { 2 } + 3 t + 1 , \quad y = t ^ { 4 } + 1$$ The point \(P\) on the curve has parameter \(p\). It is given that the gradient of the curve at \(P\) is 4 .

1. Show that \(p = \sqrt [ 3 ] { } ( 2 p + 3 )\).
2. Verify by calculation that the value of \(p\) lies between 1.8 and 2.0.
3. Use an iterative formula based on the equation in part (i) to find the value of \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

4 The equation of a curve is \(x y ( x - 6 y ) = 9 a ^ { 3 }\), where \(a\) is a non-zero constant. Show that there is only one point on the curve at which the tangent is parallel to the \(x\)-axis, and find the coordinates of this point.

6 The equation of a curve is \(x ^ { 3 } y - 3 x y ^ { 3 } = 2 a ^ { 4 }\), where \(a\) is a non-zero constant.

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 x ^ { 2 } y - 3 y ^ { 3 } } { 9 x y ^ { 2 } - x ^ { 3 } }\).
2. Hence show that there are only two points on the curve at which the tangent is parallel to the \(x\)-axis and find the coordinates of these points.

4 The parametric equations of a curve are $$x = 2 \sin \theta + \sin 2 \theta , \quad y = 2 \cos \theta + \cos 2 \theta$$ where \(0 < \theta < \pi\).

1. Obtain an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\).
2. Hence find the exact coordinates of the point on the curve at which the tangent is parallel to the \(y\)-axis.

3 The parametric equations of a curve are $$x = 2 t + \sin 2 t , \quad y = \ln ( 1 - \cos 2 t )$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \operatorname { cosec } 2 t\).

5 The equation of a curve is \(2 x ^ { 2 } y - x y ^ { 2 } = a ^ { 3 }\), where \(a\) is a positive constant. Show that there is only one point on the curve at which the tangent is parallel to the \(x\)-axis and find the \(y\)-coordinate of this point.

3 Find the equation of the normal to the curve $$x ^ { 2 } \ln y + 2 x + 5 y = 11$$ at the point \(( 3,1 )\).

6 \includegraphics[max width=\textwidth, alt={}, center]{f5e0b088-73db-405b-a832-aa01d9fcba64-08_396_716_260_712} The diagram shows the curve with parametric equations $$x = 3 t - 6 \mathrm { e } ^ { - 2 t } , \quad y = 4 t ^ { 2 } \mathrm { e } ^ { - t }$$ for \(0 \leqslant t \leqslant 2\). At the point \(P\) on the curve, the \(y\)-coordinate is 1 .

1. Show that the value of \(t\) at the point \(P\) satisfies the equation \(t = \frac { 1 } { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } t }\).
2. Use the iterative formula \(t _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } t _ { n } }\) with \(t _ { 1 } = 0.7\) to find the value of \(t\) at \(P\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.
3. Find the gradient of the curve at \(P\), giving the answer correct to 2 significant figures.

6 \includegraphics[max width=\textwidth, alt={}, center]{0d15e5a1-d05f-48bc-8613-198804ff605c-08_396_716_260_712} The diagram shows the curve with parametric equations $$x = 3 t - 6 \mathrm { e } ^ { - 2 t } , \quad y = 4 t ^ { 2 } \mathrm { e } ^ { - t }$$ for \(0 \leqslant t \leqslant 2\). At the point \(P\) on the curve, the \(y\)-coordinate is 1 .

1. Show that the value of \(t\) at the point \(P\) satisfies the equation \(t = \frac { 1 } { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } t }\).
2. Use the iterative formula \(t _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } t _ { n } }\) with \(t _ { 1 } = 0.7\) to find the value of \(t\) at \(P\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.
3. Find the gradient of the curve at \(P\), giving the answer correct to 2 significant figures.

7 The equation of a curve is \(2 x ^ { 3 } + y ^ { 3 } = 24\).

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\), and show that the gradient of the curve is never positive.
2. Find the coordinates of the two points on the curve at which the gradient is - 2 .

4 Find the gradient of the curve $$x ^ { 2 } \sin y + \cos 3 y = 4$$ at the point \(\left( 2 , \frac { 1 } { 2 } \pi \right)\).

7 The parametric equations of a curve are $$x = 2 t - \sin 2 t , \quad y = 5 t + \cos 2 t$$ for \(0 \leqslant t \leqslant \frac { 1 } { 2 } \pi\). At the point \(P\) on the curve, the gradient of the curve is 2 .

1. Show that the value of the parameter at \(P\) satisfies the equation \(2 \sin 2 t - 4 \cos 2 t = 1\).
2. By first expressing \(2 \sin 2 t - 4 \cos 2 t\) in the form \(R \sin ( 2 t - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), find the coordinates of \(P\). Give each coordinate correct to 3 significant figures.
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7 The equation of a curve is $$2 x ^ { 2 } + 3 y ^ { 2 } - 2 x y = 10 .$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 2 x } { 3 y - x }\).
2. Find the coordinates of the points on the curve where the tangent is parallel to the \(x\)-axis.