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If `x=a cos^3 t " and "y=a sin ^3 t "then find " dy/dx`

Answer» `x=a cos ^3 t `
`dx/dt=a.d/dt(cos t)^3=-3a cos^2 t. d/dt(cos t )`
and `y=a sin^3 t`
`rArr dy/dt =a. d/dt(sin t )^3=a.3 sin ^2 t .d/dtsin t `
`= 3a sin ^2 t cos t`
`Now dy/dx=(dy//dt)/(dx//dt)=(3asin^2 tcost )/(-3 a cos^2 t.sin t)=-tan t `.


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