Question

If $x=a{\cos }^3\theta$ and $y=a{\sin }^3\theta$, $1+{(dy/dx)}^2$ is equal to:

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

Answer 1

The correct option is D

${\sec }^2\theta$

Explanation for the correct option.

Step 1: Differentiate $x\mathrm{and}y$ w.r.t $\theta$

$\begin{array}{rcl}x& =& a{\cos }^3\theta \\ & \Rightarrow & \frac{dx}{d\theta }=-3a{\cos }^2\theta \sin \theta \end{array}$

$\begin{array}{rcl}y& =& a{\sin }^3\theta \\ & \Rightarrow & \frac{dy}{d\theta }=3a{\sin }^2\theta \cos \theta \end{array}$

Step 2: Find $\frac{dy}{dx}$

$$\begin{gathered} \frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx} \\ =\frac{\begin{array}{rcl}& & 3a{\sin }^2\theta \cos \theta \end{array}}{\begin{array}{rcl}& & -3a{\cos }^2\theta \sin \theta \end{array}} \\ =-\tan \theta \end{gathered}$$

Step 3: Find $1+{(dy/dx)}^2$

$\begin{array}{rcl}1+{(dy/dx)}^2& =& 1+{\left(-\tan \theta \right)}^2\\ & =& {\sec }^2\theta \end{array}$

Hence, option D is correct.