Folium of Descartes. In geometry, the folium of Descartes is an algebraic curve
ID: 2881944 • Letter: F
Question
Folium of Descartes. In geometry, the folium of Descartes is an algebraic curve defined by the equation x^3 + y^3 - 3axy = 0. The curve was first proposed by Descartes in 1638 as a challenge to Fermat's method for finding tangent lines to curves. Fermat solved the problem easily, something at the time, Descartes could not do. In today's calculus, we say the slope of the tangent line can be found using implicit differentiation. In this problem, we will consider the curve above for a = 2, i.e. x^3 + y^3 - 6xy = 0. Find y' by implicit differentiation. Show that the point (3, 3) is on the curve. Find the tangent line to the curve passing through the point (3, 3). At what point on the curve is the tangent line horizontal?Explanation / Answer
Given equation is
x^3 + y^3 - 6xy =0
x^3 + y^3 = 6xy ------(1)
1) Use implicit differentiation to find dy/dx
First, let us compute y. Differentiating both sides of equation
d[x^3 + y^3]/dx = d[6xy]/dx
3x^2 + 3y^2 dy/dx = 6y + 6x dy/dx
y' = dy/ dx = (2y - x2)/ (y2 - 2x)
2) Show that the point (3,3) is on the curve
Now we need to show that (3,3) is on the line (ie that it satisfies the eq)
x^3 + y^3 = 6xy
3^3 + 3^3 = 6*3*3
27 + 27 = 45
54 = 54
3)
In order to find an equation of the line that is tangent to the curve at this point, we must
compute the derivative at point (3,3)
y'= dy/dx = (2*3 - 3^2)/(3^2 - 2*3)
y'= dy/dx = (6-9)/(9-6)
y'= dy/dx = -1 ( the slope of the tangent line)
Since we know a point on the line and the slope of the line, it is now easy to
find an equation of the line
y - 3 = -1(x - 3)
=> y = -x + 6
4)
let us determine the points on the curve that have a horizontal
tangent line. For a point to have a horizontal tangent line, we must have
y' = dy/ dx = (2y - x2)/ (y2 - 2x) = 0
=> (2y - x2)= (y2 - 2x) * 0 =0
2y - x2 = 0 or y = x2 /2
Substituting y = x2/2 into equation (1)
x3 +(x2 /2)3 = 6x (x2/2) => x6 = 16x3
if we simplify,
x6 = 16x3 = x6 - 16x3 = x3(x3 - 16) = 0, and either x = 0 or x3 = 16.
Thus, we have horizontal tangent lines when x = 0 or x = (16)1/3