Find parameter value given gradient condition

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 \(\frac { 1 } { 2 }\).

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 .

4 The parametric equations of a curve are $$x = t + \cos t , \quad y = \ln ( 1 + \sin t )$$ where \(- \frac { 1 } { 2 } \pi < t < \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec t\).
2. Hence find the \(x\)-coordinates of the points on the curve at which the gradient is equal to 3 . Give your answers correct to 3 significant figures.

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

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\tan \theta\).
2. Find the exact \(y\)-coordinate of the point on the curve at which the gradient of the normal is equal to 1 . \includegraphics[max width=\textwidth, alt={}, center]{b00cefad-7c3c-4672-b309-f19aafab8b01-08_378_689_260_726} The diagram shows a semicircle with centre \(O\), radius \(r\) and diameter \(A B\). The point \(P\) on its circumference is such that the area of the minor segment on \(A P\) is equal to half the area of the minor segment on \(B P\). The angle \(A O P\) is \(x\) radians.

4 The parametric equations of a curve are $$x = 1 - \mathrm { e } ^ { - t } , \quad y = \mathrm { e } ^ { t } + \mathrm { e } ^ { - t }$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { 2 t } - 1\).
2. Hence find the exact value of \(t\) at the point on the curve at which the gradient is 2 .

4 The parametric equations of a curve are $$x = \ln ( 1 - 2 t ) , \quad y = \frac { 2 } { t } , \quad \text { for } t < 0$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 - 2 t } { t ^ { 2 } }\).
2. Find the exact coordinates of the only point on the curve at which the gradient is 3 .

5 The parametric equations of a curve are $$x = \cos 2 \theta - \cos \theta , \quad y = 4 \sin ^ { 2 } \theta$$ for \(0 \leqslant \theta \leqslant \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 8 \cos \theta } { 1 - 4 \cos \theta }\).
2. Find the coordinates of the point on the curve at which the gradient is - 4 .

6 The parametric equations of a curve are $$x = \sqrt { t } + 3 , \quad y = \ln t$$ for \(t > 0\).

1. Obtain a simplified expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Hence find the exact coordinates of the point on the curve at which the gradient of the normal is - 2 .

6 A curve has parametric equations $$x = 9 t - \ln ( 9 t ) , \quad y = t ^ { 3 } - \ln \left( t ^ { 3 } \right)$$ Show that there is only one value of \(t\) for which \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3\) and state that value.

5 The parametric equations of a curve are $$x = 2 + 3 \sin \theta \text { and } y = 1 - 2 \cos \theta \text { for } 0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$$

1. Find the coordinates of the point on the curve where the gradient is \(\frac { 1 } { 2 }\).
2. Find the cartesian equation of the curve.

10 In this question you must show detailed reasoning. Fig. 10 shows the curve given parametrically by the equations \(\mathrm { x } = \frac { 1 } { \mathrm { t } ^ { 2 } } , \mathrm { y } = \frac { 1 } { \mathrm { t } ^ { 3 } } - \frac { 1 } { \mathrm { t } }\), for \(t > 0\). \begin{figure}[h] \end{figure}

1. Show that \(\frac { d y } { d x } = \frac { 3 - t ^ { 2 } } { 2 t }\).
2. Find the coordinates of the point on the curve at which the tangent to the curve is parallel to the line \(4 \mathrm { y } + \mathrm { x } = 1\).
3. Find the cartesian equation of the curve. Give your answer in factorised form.

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

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Find the exact value of \(t\) at the point on the curve where the gradient is 2 .

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

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Show that \(t = - \ln 2\) at the point on the curve where the gradient is 3 .

The parametric equations of a curve are $$x = t - \tan t, \quad y = \ln(\cos t),$$ for \(-\frac{1}{4}\pi < t < \frac{1}{4}\pi\).

1. Show that \(\frac{dy}{dx} = \cot t\). [5]
2. Hence find the \(x\)-coordinate of the point on the curve at which the gradient is equal to 2. Give your answer correct to 3 significant figures. [2]