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x=tcost ,y=t+sintdot Then (d^(2x))/(dy^2...

`x=tcost ,y=t+sintdot` Then `(d^(2x))/(dy^2)a tt=pi/2` is `(pi+4)/2` (b) `-(pi+4)/2` `-2` (d) none of these

A

`(pi+4)/(2)`

B

`-(pi+4)/(2)`

C

`-2`

D

none of these

Text Solution

Verified by Experts

`(dx)/(dy)=((dx)/(dt))/((dy)/(dt))=(cos t - t sin t)/(1+ cos t)`
`therefore" "(d^(2)x)/(dy^(2))=((d)/(dt)((dx)/(dy)))/((dy)/(dt))`
`=(((-2 sin t- t cos t)(1+ cos t )-(cos t - t sin t)(- sin t))/((1+cos t)^(2)))/(1+cos t)`
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