Finding Stationary Points on Implicitly Differentiated Curves

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Homework Help Overview

The discussion revolves around finding the coordinates of stationary points on the implicitly defined curve given by the equation x^3 + (3x^2)(y) - 2y^3 = 16. The problem involves implicit differentiation and the conditions for stationary points.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to implicitly differentiate the equation and expresses uncertainty about how to find the coordinates of the stationary points after setting the derivative to zero. Some participants suggest finding y in terms of x from the condition dy/dx=0 and substituting back into the original equation.

Discussion Status

Participants are exploring the relationship between the two equations derived from the implicit differentiation and the original equation. Guidance has been offered regarding solving for y in terms of x, but there is no explicit consensus on the next steps.

Contextual Notes

The original poster indicates a lack of clarity on how to proceed after differentiating, which may suggest constraints in their understanding of implicit differentiation or the conditions for stationary points.

aanandpatel
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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 :)
 
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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
 
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.
 
Thanks guys - helped a lot! :)
 

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