x^3 + y^3 = 6xy
At which point is the first quotient of the tangent is horizontal?
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?
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)
To find the point at which the first quotient of the tangent is horizontal, we need to find the derivative of the given equation and set it equal to zero. Let's start step by step.
Step 1: Differentiate the equation with respect to x:
d/dx(x^3 + y^3) = d/dx(6xy)
Using the power rule of differentiation, we get:
3x^2 + 3y^2(dy/dx) = 6y + 6xy'
Step 2: Solve for dy/dx:
3y^2(dy/dx) = 6y - 3x^2 - 6xy'
Rearranging the equation, we have:
3y^2(dy/dx) + 6xy' = 6y - 3x^2
Step 3: Express dy/dx in terms of y and x:
dy/dx = (6y - 3x^2) / (3y^2 + 6x)
Step 4: Set dy/dx equal to zero to find the point(s) where the first quotient of the tangent is horizontal:
(6y - 3x^2) / (3y^2 + 6x) = 0
Since the numerator is zero, we have:
6y - 3x^2 = 0
Step 5: Solve for y:
6y = 3x^2
y = (3x^2) / 6
y = x^2 / 2
So, the first quotient of the tangent is horizontal at the point (x, y) = (x, x^2/2).
To find the point where the first quotient of the tangent is horizontal, we need to find the derivative of the equation and set it equal to zero. Let's start by finding the first derivatives of x and y with respect to the variable t.
Differentiating x^3 + y^3 = 6xy implicitly with respect to t, we get:
3x^2(dx/dt) + 3y^2(dy/dt) = 6(dx/dt)(y) + 6x(dy/dt)
Now, let's solve this equation for dy/dx (the first quotient of the tangent), by dividing through by dx/dt:
3x^2(dx/dt)/(dx/dt) + 3y^2(dy/dt)/(dx/dt) = 6(dx/dt)(y)/(dx/dt) + 6x(dy/dt)/(dx/dt)
Simplifying, we have:
3x^2 + 3y^2(dy/dx) = 6y + 6x(dy/dx)
Now, set dy/dx equal to zero (to find the horizontal tangent):
3y^2(dy/dx) - 6x(dy/dx) = -3x^2 - 6y
Factoring out dy/dx:
(dy/dx)(3y^2 - 6x) = -3x^2 - 6y
Setting this expression equal to zero:
(3y^2 - 6x)(dy/dx) = -3x^2 - 6y
We can solve this equation to find the values of x and y at the point where the first quotient of the tangent is horizontal.