Folium of Descartes: Derivative

[ChaoticLlama]

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?

 

[Guest]

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

 

[soroban]

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]

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.