Question

If $x=\cos t$ and $y=\ln t$ then the value of $\frac{ d ^2 y }{ dx ^2}+\left(\frac{ dy }{ dx }\right)^2$ at $t =\frac{\pi}{2}$ is equal to

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (B)

$ x=\cos t$
$\frac{d x}{d t}=-\sin t$
$y=\ln t$
$\frac{d y}{d t}=\frac{1}{t} $
$\frac{d y}{d x}=\frac{-1}{t \sin t} \Rightarrow\left(\frac{d y}{d x}\right)_{t=\pi / 2}^2=\frac{4}{\pi^2}$
$\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{-1}{t \sin t}\right)=\frac{d}{d t}\left(\frac{-1}{t \sin t}\right) \frac{d t}{d x} $
$=\frac{1}{(t \sin t)^2}(t \cos t+\sin t)\left(\frac{-1}{\sin t}\right) $
$\frac{d^2 y}{d x^2}=\frac{-t \cos t-\sin t}{t^2 \sin ^3 t}$
$\left.\frac{d^2 y}{d x^2}\right|_{t=\pi / 2}=\frac{-4}{\pi^2} $
$\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0$ .