Question

x^3 + y^3 = 6xy

At which point is the first quotient of the tangent is horizontal?

Answer 1

what do you mean "first quotient"?

This is the equation for the Folium of Descartes.

3x^2 + 3y^2 y' = 6y + 6xy'
y'(3y^2 - 6x) = 6y - 3x^2
y' = (2y - x^2)/(y^2 - 2x)

Now, how to find x,y eh?

The parametric equations for this curve are

x = 6t/(1+t^3)
y = 6t^2/(1+t^3)

dy/dx = dy/dt / dx/dt
= (2 + 2t^2 - 3t^3)/(1+t^3)^2

dy/dx = 0 when x = 1.16

If my math is right. The derivatives get messy. I'll double-check. Any other takers?

Answer 2

a web search for folium descartes tangent line reveals that the tangent is horizontal at

x = 0
x = 4^(1/3) = 1.587

so, I was off somewhere.

At one site, they work through without parametric equations:


horizontal tangents occur when dy/dx = 0
vertical tangents occur when dx/dy = 0

x^3 + y^3 - 6xy = 0
3x^2 * dx + 3y^2 * dy - 6x * dy - 6y * dx = 0
dx * (3x^2 - 6y) + dy * (3y^2 - 6x) = 0
dy * (3y^2 - 6x) = dx * (6y - 3x^2)
dy * (y^2 - 2x) = dx * (2y - x^2)
dy/dx = (2y - x^2) / (y^2 - 2x)

Horizontal tangent:

2y - x^2 = 0
2y = x^2
y = (1/2) * x^2

x^3 + y^3 - 6xy = 0
x^3 + (1/2) * x^6 - 6x * (1/2) * x^2 = 0
2x^3 + x^6 - 6x^3 = 0
x^6 - 4x^3 = 0
x^3 * (x^3 - 4) = 0
x = 0
x^3 - 4 = 0
x^3 = 4
x = 4^(1/3)

x = 0 , 4^(1/3)