Choose the correct alternative in the following:

Solution:

We are given that

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

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

Now, we know

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

Now,

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\frac{\mathrm{d}}{\mathrm{d} \theta} \mathrm{a} \cdot \cos ^{3} \theta$

$=-3 a \cos ^{2} \theta \sin \theta$ (Using Chain Rule)

Again

$\frac{d y}{d \theta}=\frac{d}{d \theta} a \cdot \sin ^{3} \theta$

$=3 a \sin ^{2} \theta \cos \theta$ (Using Chain Rule)

Now, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}=\frac{3 \mathrm{a} \sin ^{2} \theta \cos \theta}{-3 \mathrm{a} \cos ^{2} \theta \sin \theta}$

By Simplifying we get,

$\frac{d y}{d x}=-\tan \theta$

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

$\therefore \sqrt{1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}}=|\sec \theta|=$ (D)