Implicit differentiation


by aanandpatel
Tags: implicit diff., stationary point
aanandpatel
aanandpatel is offline
#1
Feb20-12, 03:55 AM
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 :)
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ehild
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#2
Feb20-12, 04:36 AM
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Hi aanandpatel,

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

ehild
HallsofIvy
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#3
Feb20-12, 06:43 AM
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You have two equations,
[tex]x^3 + (3x^2)(y) -2y^3=16[/tex]
and
[tex](x^2 + 2xy) / (2y^2 - x^2)= 0[/tex]
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
aanandpatel is offline
#4
Feb21-12, 04:37 AM
P: 16

Implicit differentiation


Thanks guys - helped a lot! :)


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