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Question
If x = a cos t and y = b sin t, then find `(d^2y)/(dx^2)`.
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Solution
If x = a cos t, y = b sin t
`dx/(dt)` = – a sin t
`dy/(dt)` = b cos t
`dy/dx = (dy/(dt))/(dx/(dt))`
= `(b cos t)/(-a sin t)`
= `(-b)/a cot t`
`(d^2y)/(dx^2) = b/a "cosec"^2t xx (dt)/dx`
= `b/a "cosec"^2t xx (-1)/(a sin t)`
= `-b/a^2 "cosec"^3t`.
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