Question

Let $y=e^{\sin t}$ and $x=e^{\cos t}$. Then Slope of normal to the curve $y=f\left(x\right)$ at $t=\frac{\pi }{4}$ is

Answer 1

Given : $y=e^{\sin t}$
On differentiating both sides w.r.t. $t$
$\frac{dy}{dt}=e^{\sin t}\cdot \cos t$
and $x=e^{\cos t}$
On differentiating both sides w.r.t. $t$
$\frac{dx}{dt}=e^{\cos t}\cdot (-\sin t)$
$\therefore \frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}=\frac{\cos t\cdot e^{\sin t}}{-\sin t\cdot e^{\cos t}}$
At $t=\frac{\pi }{4}$
$\frac{dy}{dx}{|}_{(t=\pi /4)}=-1(\because \sin t=\cos t)$
Since,
slope of normal $=-\frac{1}{\text{slope of tangent}}$
$\therefore$ slope of normal $=1$