Question

If $ x = e^t \, \sin\, t, y=e^t\, \cos\,t$, then $\frac{d^{2}y}{dx^{2}}$ 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}{dx^{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}{dx^{2}}\right)_{{\left(x=\pi\right)}}$
$ = \frac{-2}{e^{\pi} \left(\cos\, \pi +\sin \,\pi\right)^{3}}$

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