if `sqrt(x^2+y^2)=ae^(tan^-1 (y/x)) , a > 0, (y(0) > 0)` then `y"(0)` equals

Options

A. `(20)/(a) e^(-pi//2)`
B. `- (20)/(a) e^(pi//2)`
C. `- (2)/(a) e^(-pi //2)`
D. None of these

Answer

C

Explanation

Given , ` sqrt(x^(2) + y^(2) = ae^(tan^(-1))((y)/(x)) ` when
When , ` x = 0 , y gt 0 rArr y = ae^(pi//2)`
On taking log on both the sides of the given equation , we get
`(1)/(2) log (x^(2) + y^(2) )= log a + log ^(-1)((y)/(x))`
On differentiating both the sides w.r.t.x, we get
`(1)/(2) xx(2x + 2yy')/(x^(2) + y^(2)) = (1)/(1 + ((y)/(x))^(2)) xx (xy'-y)/(x^(2))`
` rArr x + yy' = xy' - y ` ..(i)
On differentiating again both sides w.r.t.x we get
` 1 + (y')^(2) + yy'' = xy'' + y' - y'`
` rarr 1 + (y')^(2) = (x - y) y''`
` rArr y'' = (1 + (y')^(2))/(x-y)`
when x = 0 , we get from Eq. (i) ,` y' = - 1 `
`y''(0) = (1 + (-1)^(2) )/(- ae^(pi//2)) = (-2)/(a)e^(-pi//2)` .