Finding Stationary Points on Implicitly Differentiated Curves

[aanandpatel]

Homework Statement

Find the coordinates of the stationary points on the curve:
x^3 + (3x^2)(y) -2y^3=16

Homework Equations

Stationary points occur when the first derivative of y with respect to x is equal to zero

The Attempt at a Solution

I implicitly differentiated the equation and got
dy/dx = (x^2 + 2xy) / (2y^2 - x^2)

I know I have to make this equal to zero but then I'm not sure how to find the x and y coordinates of the stationary point.

Help would be greatly appreciated :)

 

[ehild]

Hi aanandpatel,

Find y in terms of x from the condition dy/dx=0. Substitute back into the original equation.

ehild

 

[HallsofIvy]

You have two equations,
x^3 + (3x^2)(y) -2y^3=16
and
(x^2 + 2xy) / (2y^2 - x^2)= 0
to solve for x and y. The second equation can easily be solved for y in terms of x since a fraction is equal to 0 if and only if the numerator is 0.

 

[aanandpatel]

Thanks guys - helped a lot! :)