Find dy/dx at a point

4 A curve has parametric equations $$x = \ln ( 2 t + 6 ) - \ln t , \quad y = t \ln t$$

1. Find the value of \(t\) at the point \(P\) on the curve for which \(x = \ln 4\).
2. Find the exact gradient of the curve at \(P\).

3 The parametric equations of a curve are $$x = \mathrm { e } ^ { 2 t } \cos 4 t , \quad y = 3 \sin 2 t$$ Find the gradient of the curve at the point for which \(t = 0\).

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.

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 \(t\).
2. Find the exact gradient of the curve at the point where the curve crosses the \(y\)-axis.

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 .

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 The parametric equations of a curve are $$x = 2 \cos 2 \theta + 3 \sin \theta , \quad y = 3 \cos \theta$$ for \(0 < \theta < \frac { 1 } { 2 } \pi\).

1. Find the gradient of the curve at the point for which \(\theta = 1\) radian.
2. Find the value of \(\sin \theta\) at the point on the curve where the tangent is parallel to the \(y\)-axis.

3 The parametric equations of a curve are $$x = \ln ( 2 t + 3 ) , \quad y = \frac { 3 t + 2 } { 2 t + 3 }$$ Find the gradient of the curve at the point where it crosses the \(y\)-axis.

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.

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 \(t = 0\).

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 \(x = 1\).

7 The parametric equations of a curve are $$x = 3 \sin 2 \theta , \quad y = 1 + 2 \tan 2 \theta$$ for \(0 \leqslant \theta < \frac { 1 } { 4 } \pi\).

1. Find the exact gradient of the curve at the point for which \(\theta = \frac { 1 } { 6 } \pi\).
2. Find the value of \(\theta\) at the point where the gradient of the curve is 2 , giving the value correct to 3 significant figures.

2 The parametric equations of a curve are $$x = ( \ln t ) ^ { 2 } , \quad y = \mathrm { e } ^ { 2 - t ^ { 2 } }$$ for \(t > 0\).
Find the gradient of the curve at the point where \(t = \mathrm { e }\), simplifying your answer.

4. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with parametric equations $$x = \tan t , \quad y = \sin 2 t , \quad - \frac { \pi } { 2 } < t < \frac { \pi } { 2 } .$$

1. Find the gradient of the curve at the point \(P\) where \(t = \frac { \pi } { 3 }\).
2. Find an equation of the normal to the curve at \(P\).
3. Find an equation of the normal to the curve at the point \(Q\) where \(t = \frac { \pi } { 4 }\).

6 \includegraphics[max width=\textwidth, alt={}, center]{798da17d-0af5-4aa6-b731-564642dc28d5-3_766_611_251_703} The diagram shows the curve with parametric equations $$x = a \sin \theta , \quad y = a \theta \cos \theta$$ where \(a\) is a positive constant and \(- \pi \leqslant \theta \leqslant \pi\). The curve meets the positive \(y\)-axis at \(A\) and the positive \(x\)-axis at \(B\).

1. Write down the value of \(\theta\) corresponding to the origin, and state the coordinates of \(A\) and \(B\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 - \theta \tan \theta\), and hence find the equation of the tangent to the curve at the origin.

7 Fig. 7 shows the curve with parametric equations $$x = \cos \theta , y = \sin \theta - \frac { 1 } { 8 } \sin 2 \theta , 0 \leqslant \theta < 2 \pi$$ The curve crosses the \(x\)-axis at points \(\mathrm { A } ( 1,0 )\) and \(\mathrm { B } ( - 1,0 )\), and the positive \(y\)-axis at C . D is the maximum point of the curve, and E is the minimum point. The solid of revolution formed when this curve is rotated through \(360 ^ { \circ }\) about the \(x\)-axis is used to model the shape of an egg. \begin{figure}[h] \end{figure}

1. Show that, at the point \(\mathrm { A } , \theta = 0\). Write down the value of \(\theta\) at the point B , and find the coordinates of C .
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Hence show that, at the point D, $$2 \cos ^ { 2 } \theta - 4 \cos \theta - 1 = 0 .$$
3. Solve this equation, and hence find the \(y\)-coordinate of D , giving your answer correct to 2 decimal places. The cartesian equation of the curve (for \(0 \leqslant \theta \leqslant \pi\) ) is $$y = \frac { 1 } { 4 } ( 4 - x ) \sqrt { 1 - x ^ { 2 } } .$$
4. Show that the volume of the solid of revolution of this curve about the \(x\)-axis is given by $$\frac { 1 } { 16 } \pi \int _ { - 1 } ^ { 1 } \left( 16 - 8 x - 15 x ^ { 2 } + 8 x ^ { 3 } - x ^ { 4 } \right) \mathrm { d } x .$$ Evaluate this integral.

3 A curve has parametric equations $$x = 2 \sin \theta , \quad y = \cos 2 \theta$$

1. Find the exact coordinates and the gradient of the curve at the point with parameter \(\theta = \frac { 1 } { 3 } \pi\).
2. Find \(y\) in terms of \(x\).

6 A curve has parametric equations $$x = a t ^ { 3 } , \quad y = \frac { a } { 1 + t ^ { 2 } }$$ where \(a\) is a constant.
Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { - 2 } { 3 t \left( 1 + t ^ { 2 } \right) ^ { 2 } }\).
Hence find the gradient of the curve at the point \(\left( a , \frac { 1 } { 2 } a \right)\).

7 A curve has parametric equations \(x = 1 + u ^ { 2 } , y = 2 u ^ { 3 }\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(u\).
2. Hence find the gradient of the curve at the point with coordinates \(( 5,16 )\).

4 The parametric equations of a curve are $$x = \sin \theta , \quad y = \sin 2 \theta , \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi$$

1. Find the exact value of the gradient of the curve at the point where \(\theta = \frac { 1 } { 6 } \pi\).
2. Show that the cartesian equation of the curve is \(y ^ { 2 } = 4 x ^ { 2 } - 4 x ^ { 4 }\).

3 A curve is defined parametrically by the equations $$x = t - \ln t , \quad y = t + \ln t \quad ( t > 0 )$$ Find the gradient of the curve at the point where \(t = 2\).

5 A curve has parametric equations $$x = a t ^ { 3 } , \quad y = \frac { a } { 1 + t ^ { 2 } }$$ where \(a\) is a constant.
Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { - 2 } { 3 t \left( 1 + t ^ { 2 } \right) ^ { 2 } }\).
Hence find the gradient of the curve at the point \(\left( a , \frac { 1 } { 2 } a \right)\).

4 A curve has parametric equations $$x = 2 \sin \theta , \quad y = \cos 2 \theta$$

1. Find the exact coordinates and the gradient of the curve at the point with parameter \(\theta = \frac { 1 } { 3 } \pi\).
2. Find \(y\) in terms of \(x\).

1. The curve \(C\) has parametric equations

$$x = \sin 2 \theta \quad y = \operatorname { cosec } ^ { 3 } \theta \quad 0 < \theta < \frac { \pi } { 2 }$$

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\)
2. Hence find the exact value of the gradient of the tangent to \(C\) at the point where \(y = 8\)