1. Dec 22, 2011

aizeltine

1. The problem statement, all variables and given/known data

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

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

3. 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)

2. Dec 22, 2011

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! :-)

Last edited: Dec 22, 2011
3. Dec 22, 2011

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.

4. Dec 22, 2011