Question

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

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

Answer 1

The correct option is D $|\sec \theta |$

From given we have,

$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$

$\frac{dy}{d\theta }=\frac{d}{d\theta }\left({\sin }^3\theta \right)=3{\sin }^2\theta \cos \theta$

$\frac{dx}{d\theta }=\frac{d}{d\theta }\left({\cos }^3\theta \right)=-3{\cos }^2\theta \sin \theta$

$\therefore \frac{dy}{dx}=\frac{3{\sin }^2\theta \cos \theta }{-3{\cos }^2\theta \sin \theta }$

$=-\frac{\sin \theta }{\cos \theta }=-\tan \theta$

$\therefore \sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$

$=\sqrt{1+{(-\tan \theta )}^2}$

$=\sqrt{1+{\tan }^2\theta }$

$=|\sec \theta |$