Folium of Descartes: Derivative

ChaoticLlama

Junior Member
I just had a calculus test today where one of the questions asked to find the derivative of the Folium of Descartes:

x³ + y³ = 3xy

I found the derivative to be:

dy/dx = (y - x²) / (y² - x)

Then one of the parts for this question read:
Find all points in the first quadrant where the tangent is horizontal.

My Work:

Set dy/dx = 0
The slope of the tangent is 0 when the numerator is equal to 0, therefore

0 = y - x²

y = x²

For all values of x and y which satisfy the relation y = x² for x>0 the slope of the tangent is 0.

Am I correct?

G

Guest

Guest
Looks correct , however also need to consider what happens when y² - x =0 as this has an impact.

soroban

Elite Member
Hello, ChaoticLlama!

I just had a calculus test today where one of the questions asked to find the derivative
of the Folium of Descartes: $$\displaystyle x^3\,+\,y^3\:=\:3xy$$

I found the derivative to be: $$\displaystyle \frac{dy}{dx} \:= \:\frac{y - x^2}{y^2 - x}$$

Then one of the parts for this question read:
. . "Find all points in the first quadrant where the tangent is horizontal."

My Work:

Set $$\displaystyle \frac{dy}{dx}\,=\,0$$
The slope of the tangent is 0 when the numerator is equal to 0,
. . therefore: $$\displaystyle y\,-\,x^2\:=\:0\qquad\Rightarrow\qquad y\,=\,x^2$$

For all values of x and y which satisfy the relation $$\displaystyle y\,=\,x^2$$ for $$\displaystyle x > 0$$
. . the slope of the tangent is 0.

Am I correct? . . . . well, yes and no
Yes, your work is correct . . . up to that point.

However, you are probably expected to find the <u>exact</u> <u>points</u>.

You found that: $$\displaystyle y\,=\,x^2.$$

Substitute into the original equation: . $$\displaystyle x^3\,+\,(x^2)^3 \:= \:3x(x^2)$$

We get: . $$\displaystyle x^6\,-\,2x^3 \:= \:0$$

Factor: . $$\displaystyle x^3(x^3 - 2)\:= \:0$$

And we have roots: . $$\displaystyle x\:=\:0,\;\sqrt[3]{2}$$

The y-values are: . $$\displaystyle y\:=\:0,\;\sqrt[3]{4}$$

At $$\displaystyle (0,0)$$, the slope is indeterminate (as apm warned us).
. . But the Folium of Descartes has a "loop"; it cross itself at the origin
. . where there is both a horizontal <u>and</u> a vertical tangent.
And there is another horizontal tangent at: .$$\displaystyle (\sqrt[3]{2},\,\sqrt[3]{4})$$

[IMHO: a very intricate problem for a test.]

ChaoticLlama

Junior Member
Thank you for your insight into this problem.

My teacher has a tendancy to push us very hard on tests (It's for the preparation for the AP exam).

As far as i know, only 1 person has gotten the answer that you spoke of.