Find gradient at given parameter

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 \(t = 0\).

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 \(t = \frac { 1 } { 6 } \pi\).

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.

5. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve \(C\) with parametric equations $$x = 4 \tan t , \quad y = 5 \sqrt { 3 } \sin 2 t , \quad 0 \leqslant t < \frac { \pi } { 2 }$$ The point \(P\) lies on \(C\) and has coordinates \(\left( 4 \sqrt { 3 } , \frac { 15 } { 2 } \right)\).

1. Find the exact value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at the point \(P\). Give your answer as a simplified surd. The point \(Q\) lies on the curve \(C\), where \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\)
2. Find the exact coordinates of the point \(Q\).

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 \(t = 2\).

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 \(y = 8\)

2 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 \(x = 1 + u ^ { 2 } , y = 2 u ^ { 3 }\).

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 \(( 5,16 )\).

A curve \(C\) has parametric equations $$x = 4t + 3, \quad y = 4t + 8 + \frac{5}{2t}, \quad t \neq 0$$

1. Find the value of \(\frac{dy}{dx}\) at the point on \(C\) where \(t = 2\), giving your answer as a fraction in its simplest form. [3]
2. Show that the cartesian equation of the curve \(C\) can be written in the form $$y = \frac{x^2 + ax + b}{x - 3}, \quad x \neq 3$$ where \(a\) and \(b\) are integers to be determined. [3]

A curve \(C\) is defined by the parametric equations $$x = \frac{4 - e^{-6t}}{4}, \quad y = \frac{e^{3t}}{3t}, \quad t \neq 0$$

1. Find the exact value of \(\frac{dy}{dx}\) at the point on \(C\) where \(t = \frac{2}{3}\). [5 marks]
2. Show that \(x = \frac{4 - e^{-6t}}{4}\) can be rearranged into the form \(e^{3t} = \frac{e}{2\sqrt{(1-x)}}\). [2 marks]
3. Hence find the Cartesian equation of \(C\), giving your answer in the form $$y = \frac{e}{f(x)[1 - \ln(f(x))]}$$ [2 marks]

A curve is defined by the parametric equations $$x = t^3 + 2, \quad y = t^2 - 1$$ Find the gradient of the curve at the point where \(t = -2\) [2]

The curve \(C\) has parametric equations $$x = \sin 2\theta \quad y = \cos\text{ec}^3 \theta \quad 0 < \theta < \frac{\pi}{2}$$

1. Find an expression for \(\frac{dy}{dx}\) in terms of \(\theta\) [3]
2. Hence find the exact value of the gradient of the tangent to \(C\) at the point where \(y = 8\) [3]