Question

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

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

Answer 1

Answer: (D)

Given, $x=a \cos ^{3} \,\theta$ and $y=a \sin ^{3} \,\theta$
On differentiating both sides w.r.t. $\theta$, we get
$\frac{d x}{d \theta}=3 a \cos ^{2} \,\theta(-\sin \,\theta)$
and $\frac{d y}{d \,\theta}=3 a \sin ^{2} \,\theta(\cos \,\theta)$
Now, $\frac{d y / d \,\theta}{d x / d \,\theta}=\frac{3 a \sin ^{2} \,\theta(\cos \,\theta)}{3 a \cos ^{2} \,\theta(-\sin \,\theta)}$
$\Rightarrow \frac{d Y}{d x}=-\tan \,\theta$
$ \therefore 1+\left(\frac{d y}{d x}\right)^{2} =1+(-\tan \,\theta)^{2} $
$ =1+\tan ^{2} \,\theta $
$ =\sec ^{2} \,\theta $