If `x=2cost-cos2t ,\ \ y=2sint-sin2t` , find `(d^2y)/(dx^2)` at `t=pi/2` .

Options

A. `1//2`
B. `5//2`
C. `-3//2`
D. 2

Answer

C

Explanation

We have `x=2cost-cos2t`
and `y=2sint-sin2t`
`therefore" "(dx)/(dt)=-2sint+2sin2t`
`=-2(sint-sin2t)`
And `(dy)/(dt)=2cos t-2cos 2t`
`=2(cost-cos2t)`
`therefore" "(dy)/(dx)=(cos2t-cost)/(sint-sint)`
`(sint-sin2t)(-2sin 2t+sint)`
`therefore" "(d^(2)y)/(dx^(2))=(-(cos2t-cost)(cost-2 cos 2t))/((sint-sin2t)^(2))xx(dt)/(dx)`
`therefore" "(d^(2)y)/(dx^(2)):|_(t=(pi)/(2))=(1(1)-(-1)(+2))/((1)^(2))xx(1)/(-2(1))`
`=-(3)/(2)`