If `x=acos^(3)t,y=asin^(3)t,"show that "(dy)/(dx)=-((y)/(x))^((1)/(3))`

Explanation

Given, ` x = a * cos^(3) t`
Differentiating w.r.t. 't', we get
`(dx)/(dt) = 3a * cos^(2) t * d/(dt) (cos t) `
` = - 3a * cos^(2) t * sin t ` ... (1)
Also ` y = a sin^(3) t`
Differentiating w.r.t. 't', we get
`(dy)/(dt) = 3a sin^(2) t * d/(dt) (sin t) `
` = 3 a sin^(2) t * cos t ` ....(2)
Now ` (dy)/(dx) = (dy//dt)/(dx//dt)`
` = (3a sin^(2) t * cos t)/(-3 a cos^(2) t * sin t)`
[ From (1) and (2)]
` :. " " (dy)/(dx) = (-sin t)/(cos t)` ...(3)
` :' " " x = a cos^(3) t`
` :. " " cos ^(3) t = x/a`
`rArr " " cos t = (x/a) ^(1//3)`
Also ` y = a sin^(3) t`
` :. " " sin^(3) t = y/a`
` rArr " " sin t = (y/a)^(1//3) `
` :. ` Equation (3) reduces to
`(dy)/(dx) = (-(y/a)^(1//3))/((x/a)^(1//3)) = - (y/x)^(1//3)`
` :. " " (dy)/(dx) = - (y/x) ^(1//3)`