Question

Find a point on the curve x^3+y^3=11xy other than the origin at which the tangent line is horizontal.

Answer 1

Compute the derivative dy/dx using the power, product, and chain rules. Given

x³ + y³ = 11xy

differentiate both sides with respect to x to get

3x² + 3y² dy/dx = 11y + 11x dy/dx

Solve for dy/dx :

(3y² - 11x) dy/dx = 11y - 3x²

dy/dx = (11y - 3x²)/(3y² - 11x)

The tangent line to the curve is horizontal when the slope dy/dx = 0; this happens when

11y - 3x² = 0

or

y = 3/11 x²

(provided that 3y² - 11x ≠ 0)

Substitute y into into the original equation:

x³ + (3/11 x²)³ = 11x (3/11 x²)

x³ + (3/11)³ x⁶ = 3x³

(3/11)³ x⁶ - 2x³ = 0

x³ ((3/11)³ x³ - 2) = 0

One (actually three) of the solutions is x = 0, which corresponds to the origin (0,0). This leaves us with

(3/11)³ x³ - 2 = 0

(3/11 x)³ - 2 = 0

(3/11 x)³ = 2

3/11 x = ³√2

x = (11•³√2)/3

Solving for y gives

y = 3/11 x²

y = 3/11 ((11•³√2)/3)²

y = (11•³√4)/3

So the only other point where the tangent line is horizontal is ((11•³√2)/3, (11•³√4)/3).