Question

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

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

Answer 1

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

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

$\therefore \frac{dx}{d\theta }=a\frac{d}{d\theta }(cos\theta +\theta sin\theta )=a\left\{\frac{d}{d\theta }\right(cos\theta )+\frac{d}{d\theta }(\theta sin\theta \left)\right\}=a\{-sin\theta +(\theta cos\theta +sin\theta .1\left)\right\}=a\theta cos\theta \left(Usingproductrulein\frac{d}{d\theta }\right(\theta sin\theta \left)\right)and\frac{dy}{d\theta }=a\frac{d}{d\theta }(sin\theta -\theta cos\theta )=a\left\{\frac{d}{d\theta }\right(sin\theta )-\frac{d}{d\theta }(\theta cos\theta \left)\right\}=a[cos\theta -\{\theta (-sin\theta )+cos\theta .1\left\}\right]=a\theta sin\theta \left(Usingproductrulein\frac{d}{d\theta }\right(\theta cos\theta \left)\right)\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{a\theta sin\theta }{a\theta cos\theta }=tan\theta$