If $x = ct$ and $y = \frac{c}{t}$,find $\frac{dy}{dt}$ at $t = 2$.

If $x=f(\theta)$ and $y=g(\theta)$,then $\frac{d^2 y}{d x^2}=$

The equation of the tangent to the curves $x = 1 - 3t^2$ and $y = t - 3t^3$ at the point $(-2, 2)$ is...

If $x=a(1-\cos \theta)$ and $y=a(\theta+\sin \theta)$,then $\frac{dy}{dx} =$ . . . . . .

$A$ curve is given by the equations $x = a \cos \theta + \frac{1}{2}b \cos 2\theta$ and $y = a \sin \theta + \frac{1}{2}b \sin 2\theta$. The points for which $\frac{d^2y}{dx^2} = 0$ are given by:

Find the slope of the normal to the curve $x=a \cos ^{3} \theta, y=a \sin ^{3} \theta$ at $\theta=\frac{\pi}{4}$.