Question

If $x=aco{s}^3\theta$ and $y=asi{n}^3\theta$, then $\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$ is equal to

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

Answer 1

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

$x=a{\cos }^3\theta$, $y=0{\sin }^3\theta$

diff wrt ${}^{\prime }{\theta }^{\prime }$

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

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

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

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

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

$=|\sec \theta |$