Folium of Descartes derivative at $0$

Question

With regard to this curve: $$3xy=x^3+y^3$$ I understand that $\frac{dy}{dx}$ is not defined at $(0,0)$, but, there must be some more information right as there are $2$ tangent lines. I know my question is not very specific but if anyone can elaborate on the derivative as we approach $(0,0)$. Thanks

Answer 1

If I understand your question correctly, you understand that the Folium of Descarte has certain 'tangent-like' directions at the origin $(0,0)$. However (since there are two directions), we cannot identify them by computing the derivative $\frac{dy}{dx}$, so it would be nice to have some other way of finding them.

One way is to use polar coordinates. In polar coordinates, $x = r\cos \theta$ and $y = r\sin \theta$. There, our problem comes into focus: the origin $(x,y) = (0,0)$ can be represented as $r = 0$ with no restriction on $\theta$. However, if we write down the equation in polar, we find $$r^2 \sin \theta \cos \theta = r^3 \left(\sin^3 \theta + \cos^3 \theta\right)$$ Notice that the left-hand side has two powers of $r$, while the right-hand side has three. This means that, to keep the equation balanced, as $r \to 0$, $\sin \theta \cos \theta$ must go to $0$. The only way for this to happen is if $\theta = \frac{\pi}{2} n$ for some natural number $n$. From this, we can conclude that the 'tangent-like' directions at $(0,0)$ are vertical and horizontal, which agrees with the image Wikipedia has of the curve.

Answer 2

Here's an alternative approach, which usually shows up in differential and algebraic geometry. It's called "blowing up" the origin.

Introduce the slope coordinate $m$ by $y=mx$ and rewrite the equation. You have $$3xy=x^3+y^3 \iff 3x(mx) = x^3+(mx)^3 \iff 3mx^2 = x^3(1+m^3).$$ Dividing out the $x^2$, we obtain $3m = x(1+m^3)$. When $x=0$ we get $m=0$ and we see that one branch of the curve comes in with slope $0$. Now make the reciprocal substitution (to see what happens with infinite slope) $x=\ell y$. Similarly we end up with $3\ell = y(1+\ell^3)$, and, when $y=0$, we find that $\ell=0$, so there is also a branch of the curve at the origin with infinite slope.

Comment: By the way, polar coordinates is itself a blow-up of the origin, as we get the whole circle of possible directions $\theta\in [0,2\pi)$ when $r=0$.