1.07s Parametric and implicit differentiation

6 A curve has parametric equations $$x = \frac { \mathrm { e } ^ { 2 t } - 2 } { \mathrm { e } ^ { 2 t } + 1 } , \quad y = \mathrm { e } ^ { 3 t } + 1$$

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\). \includegraphics[max width=\textwidth, alt={}, center]{dcc483e9-630e-4f02-ad8c-4a27c0720fc6-10_2718_42_107_2007} \includegraphics[max width=\textwidth, alt={}, center]{dcc483e9-630e-4f02-ad8c-4a27c0720fc6-11_2725_35_99_20}
2. Find the exact gradient of the curve at the point where the curve crosses the \(y\)-axis.

3 A curve has equation \(6 \mathrm { e } ^ { - x } y ^ { 2 } + \mathrm { e } ^ { 2 x } - 12 y + 7 = 0\).
Find the gradient of the curve at the point \(( \ln 3,2 )\).

6 The parametric equations of a curve are $$x = \mathrm { e } ^ { 2 t } , \quad y = 4 t \mathrm { e } ^ { t } .$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 ( t + 1 ) } { \mathrm { e } ^ { t } }\).
2. Find the equation of the normal to the curve at the point where \(t = 0\).

7 The parametric equations of a curve are $$x = t + 2 \ln t , \quad y = 2 t - \ln t$$ where \(t\) takes all positive values.

1. Express \(\frac { d y } { d x }\) in terms of \(t\).
2. Find the equation of the tangent to the curve at the point where \(t = 1\).
3. The curve has one stationary point. Show that the \(y\)-coordinate of this point is \(1 + \ln 2\) and determine whether this point is a maximum or a minimum.

7 The parametric equations of a curve are $$x = 2 \theta - \sin 2 \theta , \quad y = 2 - \cos 2 \theta$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \cot \theta\).
2. Find the equation of the tangent to the curve at the point where \(\theta = \frac { 1 } { 4 } \pi\).
3. For the part of the curve where \(0 < \theta < 2 \pi\), find the coordinates of the points where the tangent is parallel to the \(x\)-axis.

6 The parametric equations of a curve are $$x = 2 t + \ln t , \quad y = t + \frac { 4 } { t }$$ where \(t\) takes all positive values.

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { t ^ { 2 } - 4 } { t ( 2 t + 1 ) }\).
2. Find the equation of the tangent to the curve at the point where \(t = 1\).
3. The curve has one stationary point. Find the \(y\)-coordinate of this point, and determine whether this point is a maximum or a minimum.

5

1. By differentiating \(\frac { 1 } { \cos \theta }\), show that if \(y = \sec \theta\) then \(\frac { \mathrm { d } y } { \mathrm {~d} \theta } = \sec \theta \tan \theta\).
2. The parametric equations of a curve are $$x = 1 + \tan \theta , \quad y = \sec \theta$$ for \(- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi\). Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sin \theta\).
3. Find the coordinates of the point on the curve at which the gradient of the curve is \(\frac { 1 } { 2 }\).

5 The equation of a curve is \(3 x ^ { 2 } + 2 x y + y ^ { 2 } = 6\). It is given that there are two points on the curve where the tangent is parallel to the \(x\)-axis.

1. Show by differentiation that, at these points, \(y = - 3 x\).
2. Hence find the coordinates of the two points.

3 The parametric equations of a curve are $$x = 3 t + \ln ( t - 1 ) , \quad y = t ^ { 2 } + 1 , \quad \text { for } t > 1$$

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Find the coordinates of the only point on the curve at which the gradient of the curve is equal to 1 .

7 The equation of a curve is $$x ^ { 2 } + y ^ { 2 } - 4 x y + 3 = 0$$

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

4 The parametric equations of a curve are $$x = 4 \sin \theta , \quad y = 3 - 2 \cos 2 \theta$$ where \(- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi\). Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\), simplifying your answer as far as possible.

6 The equation of a curve is $$x ^ { 2 } y + y ^ { 2 } = 6 x$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 - 2 x y } { x ^ { 2 } + 2 y }\).
2. Find the equation of the tangent to the curve at the point with coordinates ( 1,2 ), giving your answer in the form \(a x + b y + c = 0\).

2 A curve has parametric equations $$x = 3 t + \sin 2 t , \quad y = 4 + 2 \cos 2 t$$ Find the exact gradient of the curve at the point for which \(t = \frac { 1 } { 6 } \pi\).

5 A curve has equation \(x ^ { 2 } + 2 y ^ { 2 } + 5 x + 6 y = 10\). Find the equation of the tangent to the curve at the point \(( 2 , - 1 )\). Give your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.
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6 A curve has parametric equations $$x = \frac { 1 } { ( 2 t + 1 ) ^ { 2 } } , \quad y = \sqrt { } ( t + 2 )$$ The point \(P\) on the curve has parameter \(p\) and it is given that the gradient of the curve at \(P\) is - 1 .

1. Show that \(p = ( p + 2 ) ^ { \frac { 1 } { 6 } } - \frac { 1 } { 2 }\).
2. Use an iterative process based on the equation in part (i) to find the value of \(p\) correct to 3 decimal places. Use a starting value of 0.7 and show the result of each iteration to 5 decimal places.

5 The parametric equations of a curve are $$x = \ln ( t + 1 ) , \quad y = \mathrm { e } ^ { 2 t } + 2 t$$

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Find the equation of the normal to the curve at the point for which \(t = 0\). Give your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.

5 The parametric equations of a curve are $$x = \mathrm { e } ^ { 2 t } , \quad y = 4 t \mathrm { e } ^ { t }$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 ( t + 1 ) } { \mathrm { e } ^ { t } }\).
2. Find the equation of the normal to the curve at the point where \(t = 0\).

5 The equation of a curve is $$x ^ { 2 } - 2 x ^ { 2 } y + 3 y = 9$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x - 4 x y } { 2 x ^ { 2 } - 3 }\).
2. Find the equation of the normal to the curve at the point where \(x = 2\), giving your answer in the form \(a x + b y + c = 0\).

7 The equation of a curve is $$2 x ^ { 2 } + 3 x y + y ^ { 2 } = 3$$

1. Find the equation of the tangent to the curve at the point \(( 2 , - 1 )\), giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.
2. Show that the curve has no stationary points.

4 The parametric equations of a curve are $$x = 2 \ln ( t + 1 ) , \quad y = 4 \mathrm { e } ^ { t }$$ Find the equation of the tangent to the curve at the point for which \(t = 0\). Give your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.

7 The equation of a curve is $$y ^ { 3 } + 4 x y = 16$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 4 y } { 3 y ^ { 2 } + 4 x }\).
2. Show that the curve has no stationary points.
3. Find the coordinates of the point on the curve where the tangent is parallel to the \(y\)-axis.

5 A curve is defined by the parametric equations $$x = 2 \tan \theta , \quad y = 3 \sin 2 \theta$$ for \(0 \leqslant \theta < \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 \cos ^ { 4 } \theta - 3 \cos ^ { 2 } \theta\).
2. Find the coordinates of the stationary point.
3. Find the gradient of the curve at the point \(\left( 2 \sqrt { } 3 , \frac { 3 } { 2 } \sqrt { } 3 \right)\).

7 \includegraphics[max width=\textwidth, alt={}, center]{a07e6d2f-ded1-4c62-957b-41fb94b46a2d-3_423_837_1352_651} The diagram shows the curve with parametric equations $$x = 2 - \cos t , \quad y = 1 + 3 \cos 2 t$$ for \(0 < t < \pi\). The minimum point is \(M\) and the curve crosses the \(x\)-axis at points \(P\) and \(Q\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 12 \cos t\).
2. Find the coordinates of \(M\).
3. Find the gradient of the curve at \(P\) and at \(Q\).

7 \includegraphics[max width=\textwidth, alt={}, center]{f85c4010-17b1-441c-ae8a-e77573d1b0c3-3_423_837_1352_651} The diagram shows the curve with parametric equations $$x = 2 - \cos t , \quad y = 1 + 3 \cos 2 t$$ for \(0 < t < \pi\). The minimum point is \(M\) and the curve crosses the \(x\)-axis at points \(P\) and \(Q\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 12 \cos t\).
2. Find the coordinates of \(M\).
3. Find the gradient of the curve at \(P\) and at \(Q\).

7 The parametric equations of a curve are $$x = t ^ { 3 } + 6 t + 1 , \quad y = t ^ { 4 } - 2 t ^ { 3 } + 4 t ^ { 2 } - 12 t + 5$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and use division to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) can be written in the form \(a t + b\), where \(a\) and \(b\) are constants to be found.
2. The straight line \(x - 2 y + 9 = 0\) is the normal to the curve at the point \(P\). Find the coordinates of \(P\).