Question:
If $x=3 \tan t$ and $y=3 \sec t$, then the value of
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$ at $\mathrm{t}=\frac{\pi}{4}$, is:
Correct Option: , 4
Solution:
$\frac{\mathrm{dx}}{\mathrm{dt}}=3 \sec ^{2} \mathrm{t}$
$\frac{\mathrm{dy}}{\mathrm{dt}}=3 \sec t \tan t$
$\frac{d y}{d x}=\frac{\tan t}{\sec t}=\sin t$
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\cos \mathrm{t} \frac{\mathrm{dt}}{\mathrm{dx}}$
$=\frac{\cos t}{3 \sec ^{2} t}=\frac{\cos ^{3} t}{3}=\frac{1}{3.2 \sqrt{2}}=\frac{1}{6 \sqrt{2}}$