Find Points on 2x^3 + 2y^3 -9xy = 0 Curve

[aizeltine]

Homework Statement

Find all points on the curve 2x^3 + 2y^3 -9xy= 0 where you will have a horizontal and vertical tangent lines.

Homework Equations

2x^3 + 2y^3 -9xy= 0

The Attempt at a Solution

6x^2+6y^2dy/dx - (9y +9x dy/dx)=0
6x^2 + 6y^2dy/dx - 9y -9x dy/dx=0
6y^2dy/dx-9xdy/dx=9y-6x^2
dy/dx= (9y-6x^2)/(6y^2-9x)

 

[Underhill]

2x³ + 2y³ - 9xy = 0

To find the lines tangent to this curve, take the derivative:

6x² + 6y²\frac{dy}{dx} - 9y - 9x\frac{dy}{dx} = 0

2x² - 3y = 3x\frac{dy}{dx} - 2y²\frac{dy}{dx}

(2x² - 3y) / (3x - 2y²) = \frac{dy}{dx}

It is now merely a matter of setting the denominator equal to zero to find the vertical tangent lines, and the numerator equal to zero to find the horizontal tangent lines.

Remember, when the denominator of the derivative equals zero, the derivative goes to infinity, and therefore the slope it represents also goes to infinity. When the derivative equals zero, the slope it represents on the original curve is also zero - which means it is horizontal.

You're on the right track - now finish the job! :-)

 

[HallsofIvy]

aizeltine, the only difference between your derivative and Underhills is that he has factored a "3" out of both numerator and denominator and canceled.

 

[micromass]

This is already being discussed at https://www.physicsforums.com/showthread.php?t=562060