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If `X = Acos^3t`, `Y = Asin^3 T`, Show that `(Dy)/(Dx) =- (Y/X)^(1/3)` - Mathematics and Statistics

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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)` 

  Is there an error in this question or solution?

APPEARS IN

 2017-2018 (March) (with solutions)
Question 5.1.2 | 3.00 marks
Solution If `X = Acos^3t`, `Y = Asin^3 T`, Show that `(Dy)/(Dx) =- (Y/X)^(1/3)` Concept: Derivatives of Functions in Parametric Forms.
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