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If x=acos^3theta and y=asin^3theta, then...

If `x=acos^3theta` and `y=asin^3theta,` then find the value of `(d^2y)/(dx^2)` at `theta=pi/6dot`

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Given
`x=acos^3theta` and `y=asin^3theta,`
Therefore
`(dy)/(dx)=((dy)/(d theta))/((dx)/(d theta))=(3bsin^2theta,costheta)/(-3acos^2theta,sintheta)`
`(dy)/(dx)=(-b/a)tantheta`
Therefore,
`(d^2y)/(dx^2)=d/(d theta)((dy)/(dx))(d theta)/(dx)`
...
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