Question

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

• A. $\tan \theta$
• B. ${\tan }^2\theta$
• C. ${\sec }^2\theta$
• D. $1$

Answer 1

Answer: (C)

${\sec }^2\theta$

We have, $x=a{\cos }^3\theta$ and $y=a{\sin }^3\theta$

Differentiate both w.r.t. $\theta$

We get: $\frac{dx}{d\theta }=-3a{\cos }^2\theta \sin \theta$ and $\frac{dy}{d\theta }=3a{\sin }^2\theta \cos \theta$

Thus $\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{3a{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta }=-\tan \theta$

Therefore $1+{\left(\frac{dy}{dx}\right)}^2=1+{\tan }^2\theta ={\sec }^2\theta$