Question

If $x=e^t\sin t,y=e^t\cos t$ are parametric equations, then $\frac{d^{2}y}{d{x}^2}$ at $(1,1)$ is equal to :

• A. $-\frac{1}{2}$
• B. $-\frac{1}{4}$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $-\frac{1}{2}$

$x=e^t\sin t,$ .... $\left(i\right)$ and $y=e^t\cos t$ .... $\left(ii\right)$ are the parametric equations

At point $(1,1),$ we get

$1=e^t\sin t$ and $1=e^t\cos t$
$\Rightarrow \tan t=1\Rightarrow t=\frac{\pi }{4}$
Now, $\frac{dy}{dt}=e^t(\cos t-\sin t)$
and $\frac{dx}{dt}=e^t(\sin t+\cos t)$

Also, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\cos t-\sin t}{\sin t+\cos t}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}=\frac{d}{dt}\left(\frac{\cos t-\sin t}{\cos t+\sin t}\right)\frac{dt}{dx}$

$=\left|\begin{array}{c}\frac{(\cos t+\sin t)(-\sin t-\cos t)-(\cos t-\sin t)(-\sin +\cos t)}{(\cos t+\sin t{)}^2}\end{array}\right|\frac{dt}{dx}$

$=\left|\begin{array}{c}\frac{-(\cos t+\sin t{)}^2-(\cos t-\sin t{)}^2}{(\cos t+\sin t{)}^2}\end{array}\right|\frac{dt}{dx}$

$=\frac{-2}{(\cos t+\sin t{)}^2}.\frac{1}{e^t(\sin t+\cos t)}$

$=\frac{-2}{(e^t\cos t+e^t\sin t)}.\frac{1}{(\cos t+\sin t{)}^2}$

$=\frac{-2}{x+y}.\frac{1}{(\cos t+\sin t{)}^2}$

$\therefore {\left|\frac{d^{2}y}{d{x}^2}\right|}_{(1,1)}=\frac{-2}{1+1}.\frac{1}{{(\cos \frac{\pi }{4}+\sin \frac{\pi }{4})}^2}=-\frac{1}{2}$