Question

If $x=e^t\sin t,y=e^t\cos t$, then $\frac{d^{2}y}{{dx}^2}$ at $t=\pi$ is

• A. ${2e}^{\pi }$
• B. $\frac{1}{2}{e}^{\pi }$
• C. $\frac{1}{{2e}^{\pi }}$
• D. $\frac{2}{e^{\pi }}$

Answer 1

The correct option is D $\frac{2}{e^{\pi }}$

Since, $x=e^t\sin t$ and $y=e^t\cos t$
On differentiating w.r.t. t respectively, we get
$\frac{dx}{dt}=e^t\cos t+\sin t{e}^t$
and $\frac{dy}{dt}=-e^t\sin t+e^t\cos t$
$\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{e^t(\cos t-\sin t)}{e^t(\cos t+\sin t)}$
$=\frac{\cos t-\sin t}{\cos t+\sin t}$
Again differentiating, we get
$\left[(\cos t+\sin t)(\cos t-\sin t)\right]$
$\frac{d^{2}y}{{dx}^2}=\frac{-(\cos t-\sin t)(-\sin t+\cos t)}{{(\cos t+\sin t)}^2}\frac{dt}{dx}$
$=\frac{{-(\sin t+\cos t)}^2-{(\cos t-\sin t)}^2}{{(\cos t+\sin t)}^2}\times \frac{1}{e^t(\cos t+\sin t)}$
$=-\left[\frac{{sin}^{2}t+{cos}^{2}t+2\sin t\cos t+{cos}^{2}t+{sin}^{2}t-2\sin t\cos t}{{e^t(\cos t+\sin t)}^3}\right]$
$=-\frac{2}{{e^t(\cos t+\sin t)}^3}$
$\Rightarrow {\left(\frac{d^{2}y}{{dx}^2}\right)}_{(t=\pi )}=\frac{2}{e^{\pi }{(\cos \pi +\sin \pi )}^3}$
$=\frac{2}{e^{\pi }}$