Question

If $x=4 \cos ^{3} \theta$ and $y=3 \sin ^{2} \theta$
then $\frac{d^{2} y}{d x^{2}}$ at $\theta=\frac{\pi}{4}$, is

• A. $\frac{1}{3}$
• B. $\frac{1}{6}$
• C. $\frac{-1}{6}$
• D. $\frac{-1}{3}$

Answer 1

Answer: (B)

Given, $x=4 \cos ^{3} \theta .....(i)$
and $y=3 \sin ^{2} \theta .......(ii)$
and $y=3 \sin ^{2} \theta \ldots$
Differentiating Eq. (i) and Eq. (ii) w. r. t. $t \theta$
$ \frac{d x}{d \theta} =4 \times 3 \cos ^{2} \theta(-\sin \theta) $
$=-12 \cos ^{2} \theta \sin \theta \,......(iii)$
and $\frac{d y}{d \theta}=3 \times 2 \sin \theta \cos \theta$
$=6 \sin \theta \cos \theta \,.....(iv)$
Now, $ \frac{d y}{d x} =\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} $
$=-\frac{1}{2} \sec \theta[\text { by Eqs. (iii) and (iv)] }$
So, $\frac{d^{2} y}{d x^{2}}=-\frac{1}{2} \sec \theta \tan \theta . \frac{d \theta}{d x} $
$=-\frac{1}{2} \times \sec \theta \tan \theta \times \frac{-1}{12 \cos ^{2} \theta \sin \theta} [$ by Eq. (iii) $]$
$=\frac{1}{24} \times \frac{1}{\cos \theta} \times \frac{\sin \theta}{\cos \theta} \times \frac{1}{\cos ^{2} \theta \sin \theta}$
$=\frac{1}{24(\cos \theta)^{4}}=\frac{1}{24\left(\cos \frac{\pi}{4}\right)^{4}} \left(\because \theta=\frac{\pi}{4}\right)$
$=\frac{1}{24} \times \frac{1}{\left(\frac{1}{\sqrt{2}}\right)^{4}}=\frac{1}{24} \times 4=\frac{1}{6}$