What is the slope of the normal to the curve $x = a(\theta - \sin \theta)$,$y = a(1 - \cos \theta)$ at the point $\theta = \pi / 2$?

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=4t$ and $y=\frac{4}{t}$.

Let $x(t) = 2 \sqrt{2} \cos t \sqrt{\sin 2t}$ and $y(t) = 2 \sqrt{2} \sin t \sqrt{\sin 2t}$,$t \in (0, \frac{\pi}{2})$. Then $\frac{1 + (\frac{dy}{dx})^2}{\frac{d^2y}{dx^2}}$ at $t = \frac{\pi}{4}$ is equal to:

If $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$,then $\sqrt{1 + \left( \frac{dy}{dx} \right)^2} = $

If $x = \frac{t^2}{1+t^5}$ and $y = \frac{2t^3}{1+t^5}$ where $t \neq -1$ is a parameter,then find $\frac{dy}{dx}$.

If $\tan u=\sqrt{\frac{1-x}{1+x}}$ and $\cos v=4 x^{3}-3 x$,then $\frac{d u}{d v}=$