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.]