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
Folium of Descartes derivative at $0$
analytic geometrycalculusimplicit-differentiation
Best Answer
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.