If `x=e^tsint,y=e^tcost` then `(d^2y)/(dx^2)` at `x=pi` is

Options

A. `2 e^(pi)`
B. `(1)/(2) e^(pi)`
C. `(1)/(2e^(pi))`
D. `(2)/(e^(pi))`

Answer

D

Explanation

Since , ` x = e^(t) sin t " and " y = e^(t) cos t `
` rArr (dx)/(dt) = e^(t) cos t + sin t e^(t)`
and `(dx)/(dt) = -e^(t) sin t+ e^(t)cos t`
` therefore (dy)/(dx) = (dy//dt)/(dx//dt)= ((cos t - sin t))/((cos t + sin t)) `
`(d^(2) y)/(dx^(2)) = ([{:((cos t + sin t)(-sin t - cos t)),(- (cos t- sin t)(- sin t + cos t)):}])/((cos t + sin t))(dt)/(dx) `
` = (- (sin t + cos t)^(2) - (cos t - sin t)^(2))/((cos t + sin t)^(2)) xx(1)/(e^(t) (cos t + sin t))`
` = - (2) /(e^(t) (cos t + sin t)^(2)) `
` rArr ((d^(2) y)/(dx^(2)))_((x= pi) ) = (-2)/(e^(pi) (cos pi + sin pi)^(3))= (2)/(e^(pi))` .