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Question
If `x = acos^3t`, `y = asin^3 t`,
Show that `(dy)/(dx) =- (y/x)^(1/3)`
Solution
We have `(dy)/(dx) = "dy/dx"/"dx/dt" , "dx/dt" != 0` ...(1)
Now `y = asin^3t = a(sint)^3 => sint = (y/a) ^(1/3)`
`:. dy/dx = a d/dt (sint)^3 = a.3(sint)^2 d/dt (sin t)`
`= 3asin^2 t cost` ... (2)
Also, `x = acos^3t = a(cost)^3 => cost = (x/a)^(1/3)`
`:. dx/dt = a.3cos^2t d/dt (cost)`
`= 3acos^2t(-sint)`
`= -3acos^2t sin t` ...3
From (1), (2) and (3),
`dy/dx = (3asin^2t cost)/(-3acos^2tsint) = -sint/cost = -(y/x)^(1/3)`
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