
#1
Feb2012, 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 :) 



#2
Feb2012, 04:36 AM

HW Helper
Thanks
P: 9,818

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



#3
Feb2012, 06:43 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882

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. 



#4
Feb2112, 04:37 AM

P: 16

Implicit differentiation
Thanks guys  helped a lot! :)



Register to reply 
Related Discussions  
Implicit differentiation help  Calculus & Beyond Homework  3  
Implicit differentiation  Calculus & Beyond Homework  2  
Solve by Implicit Differentiation or Partial Differentiation?  Calculus & Beyond Homework  12  
implicit differentiation  Calculus & Beyond Homework  2  
Implicit differentiation  Calculus & Beyond Homework  3 