Question

Find $\frac{dy}{dx}$, if x and y are connected parametrically by the equations given in questions without eliminating the parameter.

$x=a(\theta -sin\theta ),y=a(1+cos\theta )$

Answer 1

Given, $x=a(\theta -sin\theta ),y=a(1+cos\theta )$

Differentiating w.r.t. $\theta$ , we get

$\therefore \frac{dx}{d\theta }=\frac{d}{d\theta }a(\theta -sin\theta )=a(1-cos\theta )and\frac{dy}{d\theta }=\frac{d}{d\theta }a(1+cos\theta )=a(0-sin\theta )\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}=\frac{-asin\theta }{a(1-cos\theta )}=\frac{-2sin\frac{\theta }{2}cos\frac{\theta }{2}}{2si{n}^2\left(\frac{\theta }{x}\right)}=-cot\left(\frac{\theta }{2}\right)\left[\begin{array}{c}\because sin\theta =2sin\frac{\theta }{2}cos\frac{\theta }{2}\\ 1-cos\theta =2si{n}^2\frac{\theta }{2}\end{array}\right]$