Question

Find $\frac{dy}{dx}$

$\frac{dx}{dt}=ap\cos pt,\frac{dy}{dt}=-bp\sin pt$

Answer 1

We have,

$\frac{dx}{dt}=ap\cos pt$ ……. (1)

$\frac{dy}{dt}=-bp\sin pt$ …….. (2)

On dividing equation (2) and (1), we get

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-bp\sin pt}{ap\cos pt}$

$\frac{dy}{dx}=-\frac{b\tan pt}{a}$

Hence, this is the answer.