Question

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

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

$0$

$x=\cos t\Rightarrow \frac{dx}{dt}=-\sin t$
$y=\log t\Rightarrow \frac{dy}{dt}=\frac{1}{t}$
$\therefore \frac{dy}{dx}=-\frac{1}{t\sin t}$
$\Rightarrow \frac{d^{2}y}{{dx}^2}=\frac{-0-(-\sin t+t\cos t)}{{\left(t\sin t\right)}^2}\times -\sin t$
$=\frac{-\sin t-t\cos t}{t^2\sin t}$
$\therefore \frac{d^{2}y}{{dx}^2}+{\left(\frac{dy}{dt}\right)}^2=0$