Question

If $x=a{\sec }^3\theta$, $y=a{\tan }^3\theta$ find $\frac{dy}{dx}$ at $\theta =\frac{\pi }{4}$.

Answer 1

We are given, $x=a{\sec }^3\theta$ & $y=a{\tan }^3\theta$.

We have to differentiate $x$ & $y$ wrt $\theta$, we get,

$\frac{dx}{d\theta }=a\left(3{\sec }^2\theta \right).(\sec \theta .\tan \theta )=3a.\tan \theta .{\sec }^3\theta$

& $\frac{dy}{d\theta }=a\left(3{\tan }^2\theta \right).\left({\sec }^2\theta \right)=3a{\tan }^2\theta .{\sec }^2\theta$

[ With Use of chain rule ]

$\therefore \frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}=\frac{3a{\tan }^2\theta .{\sec }^2\theta }{3a\tan \theta .{\sec }^3\theta }$

$\frac{dy}{dx}=\frac{\tan \theta }{\sec \theta }=\frac{\sin \theta }{\cos \theta }\times \cos \theta =\sin \theta$

${\frac{dy}{dx}]}_{\theta =\frac{\pi }{4}}=\sin \theta =\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$