# Implicit differentiation

by aanandpatel
Tags: implicit diff., stationary point
 P: 16 1. The problem statement, all variables and given/known data Find the coordinates of the stationary points on the curve: x^3 + (3x^2)(y) -2y^3=16 2. Relevant equations Stationary points occur when the first derivative of y with respect to x is equal to zero 3. 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 :)
 HW Helper Thanks P: 9,675 Hi aanandpatel, Find y in terms of x from the condition dy/dx=0. Substitute back into the original equation. ehild
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,705 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.
P: 16

## Implicit differentiation

Thanks guys - helped a lot! :)

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