Question

$If\sqrt{(x^2+y^2)}=a.e^{ta{n}^{-1\left(y/x\right)}}a>0,theny"\left(0\right)isequalto$

• A. $\frac{2}{a}{e}^{-\pi /2}$
• B.
  $a{e}^{\pi /2}$
• C. $-\frac{2}{a}{e}^{-\pi /2}$
• D. Does not exist

Answer 1

The correct option is C $-\frac{2}{a}{e}^{-\pi /2}$

$\sqrt{(x^2+y^2)}=a.e^{ta{n}^{-1\left(y/x\right)}}$
$\Rightarrow \frac{1}{2\sqrt{x^2+y^2}}(2x+2y{y}^{\prime })=a.e^{ta{n}^{-1}\left(y/x\right)}\times \frac{1}{1+\frac{y^2}{x^2}}\times \frac{x{y}^{\prime }-y}{x^2}.....\left(i\right)$
$\Rightarrow \frac{x+y{y}^{\prime }}{\sqrt{x^2+y^2}}=\sqrt{(x^2+y^2)}\times \frac{x{y}^{\prime }-y}{x^2+y^2}$
$[fromEq.(i\left)\right]$
$\therefore x+y{y}^{\prime }=x{y}^{\prime }-y\Rightarrow y^{\prime }=\frac{x+y}{x-y}$
$\therefore y"=\frac{2(x{y}^{\prime }-y)}{(x-y{)}^2}$
$y"\left(0\right)=\frac{2(0-y(0\left)\right)}{{\{0-y(0\left)\right\}}^2}=\frac{-2}{a{e}^{\pi /2}}=\frac{-2}{a}{e}^{-\pi /2}$