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If x=2cost-cos2t ,\ \ y=2sint-sin2t , fi...

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

A

`1//2`

B

`5//2`

C

`-3//2`

D

2

Text Solution

Verified by Experts

The correct Answer is:
C
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)`
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