Question

EXAMPLE 2 (a) Find y' if x3 + y3 = 18xy. (b) Find the tangent to the folium of Descartes x3 + y3 = 18xy at the point (9, 9). (c) At what point in the first quadrant is the tangent line horizontal?

Answer 1

(a) Via implicit differentiation, we get

$x^3+y^3=18xy\implies 3x^2+3y^2y'=18y+18xy'$

Solve for $y'$:

$y'=\dfrac{18y-3x^2}{3y^2-18x}=\dfrac{6y-x^2}{y^2-6x}$

(b) Find the slope of the tangent line at (9, 9) by plugging in x = y = 9 into the equation above:

$y'=\dfrac{6\cdot9-9^2}{9^2-6\cdot9}=-1$

Use the point-slope formula to find the equation of the line:

$y-9=-1(x-9)\implies y=-x+18$

(c) The tangent line is horizontal when its slope is 0, so solve $y'=0$:

$\dfrac{6y-x^2}{y^2-6x}=0\implies6y-x^2=0\implies y=\dfrac{x^2}6$

Now substitute y in the equation for the folium to solve for x :

$x^3+\left(\dfrac{x^2}6\right)^3=18x\cdot\dfrac{x^2}6$

$x^3+\dfrac{x^6}{6^3}=3x^3$

$\dfrac{x^6}{6^3}-2x^3=0$

$x^3\left(\left(\dfrac x6\right)^3-2\right)=0$

$\implies x=0\text{ or }x=6\sqrt[3]{2}$

x = 0 corresponds to y = 0 (plug x = 0 into the folium equation to see why), i.e. the origin. If you don't consider the origin to belong to the first quadrant, then we only keep

$x=6\sqrt[3]{2}\implies y=6\sqrt[3]{4}$