# If $x=e^{t}$ Sint,$y=l^{t}$ Cost, t is a parameter  then $\frac{d^{2}y}{dx^{2}}$ at (1, 1) os equal to Option 1) $\frac{-1}{2}$ Option 2) $-\frac{1}{4}$ Option 3) 0 Option 4) $\frac{1}{2}$

As learnt in

Differential Equations -

An equation involving independent variable (x), dependent variable (y) and derivative of dependent variable with respect to independent variable
$\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )$

- wherein

eg:

$\frac{d^{2}y}{dx^{2}}- 3\frac{dy}{dx}+5x=0$

$x=e^{t}\sin t\ \Rightarrow \frac{dx}{dt} =e^{t}\sin t+e^{t}\cos t$

$y= e^{t}\cos t\ \Rightarrow \frac{dy}{dt}= e^{t}\cos t- e^{t}\sin t$

$\therefore \frac{dy}{dx} = \frac{\cos t-\sin t}{\cos t+\sin t} =\frac{1-\tan t}{1+\tan t}$

$=\frac{1-\frac{x}{y}}{1+\frac{x}{y}}$

$=\frac{y-x}{y+x}\ \ \ \left [ \because \frac{x}{y}= \tan t \right ]$

$\frac{d^{2}y}{dx^{2}}= \frac{\left ( y+x \right )\left ( \frac{dy}{dv} -1\right )-\left ( y-x \right )\left ( \frac{dy}{dx}+1 \right )}{\left ( y+x \right )^{2}}$

$=\frac{\left ( y+x \right )\left ( \frac{y-x-y-x}{y+x} \right ) -\left ( y-x \right )\left ( \frac{y-x+y+x}{y+x} \right ) }{\left ( y+x \right )^{2}}$

$\Rightarrow \frac{-2x-2y\left ( \frac{y-x}{y+x} \right )}{\left ( y+x \right )^{2}}$

Put

$x=1, \ y=1$

$\therefore \frac{-2\times 1-0}{\left ( 1+1 \right )^{2}}=\frac{-2}{4} =-\frac{1}{2}$

Option 1)

$\frac{-1}{2}$

This option is correct

Option 2)

$-\frac{1}{4}$

This option is incorrect

Option 3)

0

This option is incorrect

Option 4)

$\frac{1}{2}$

This option is incorrect

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