Question

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

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

Answer 1

Answer: (D)

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

$=\frac{2}{e^{\pi }}$