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