Calculus 1: Tangent Line to x³ + y³ = 6xy at Point (3, 3) (Implicit Differentiation)

Exact Midterm Question

Find the equation of the tangent line to the curve defined by $x^3 + y^3 = 6xy$ (the Folium of Descartes) at the point $(3, 3)$.

Step-by-Step Mathematical Solution

1. Differentiate both sides with respect to $x$:

$$ \frac{d}{dx}\left[x^3 + y^3\right] = \frac{d}{dx}\left[6xy\right] $$

1. Apply Power Rule, Chain Rule on $y^3$, and Product Rule on $6xy$:

$$ 3x^2 + 3y^2\frac{dy}{dx} = 6\left(y + x\frac{dy}{dx}\right) $$

$$ 3x^2 + 3y^2 y' = 6y + 6x y' $$

1. Isolate $y'$ ($\frac{dy}{dx}$):

$$ 3y^2 y' - 6x y' = 6y - 3x^2 \implies y' = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x} $$

1. Evaluate Slope $m$ at $(3, 3)$:

$$ m = \frac{2(3) - (3)^2}{(3)^2 - 2(3)} = \frac{6 - 9}{9 - 6} = \frac{-3}{3} = -1 $$

1. Write Point-Slope Tangent Line Equation:

$$ y - 3 = -1(x - 3) \implies \mathbf{y = -x + 6} $$