4
$\begingroup$

If we take the implicit derivative of $x^3+x^2-y^2=0$, we find that $\frac{dy}{dx}=\frac{3x^2+2x}{2y}.$ So, the slope of the tangent line should be undefined at any point where y is 0.

To me, the tangent line to the graph of the equation at $x=0$ should not have an undefined slope. It appears that the function is not differentiable at the point $x = 0$.

How do we know if the implicit derivative is undefined due to an actual vertical slope or due to the equation being non-differentiable at that point?

$\endgroup$
  • $\begingroup$ Consider $\lim_{x\to x_0}f’(x)$. Like $\lim_{ x\to 0} (\sqrt{|x|})’=\infty$ and $\lim_{x\to 0}(e^{-1/x^2})’=0$. Both functions are non-differentiable at zero. The first one can be understood as ‘non-differentiable due to vertical slope’. For the second one, although the limit exists, it is still non-differentiable at zero; in fact, there is an essential singularity. $\endgroup$ – Szeto Jun 13 '18 at 23:30
6
$\begingroup$

It looks like you have to be mores specific about what you mean about when $y=0$

plot of $x^3+x^2-y^2=0$

At $(-1,0)$ you have a vertical tangent line $x=-1$ with undefined slope.

At $(0,0)$ you are correct there is not a uniquely defined tangent line. However you can make a uniquely defined tangent line of you parameterize the curve. That is both $x$ and $y$ would be functions of some parameter for example $t$ then the curve will be drawn $(x(t),y(t))$.

In this way of looking at the curve, the self intersection will be two unique $t$-values and you can draw a distinct tangent line for each one and it looks like the slope will actually be defined in that case, but you are adding extra information with the parameterization.

Edit Also I might add that not just any parameterization will work you'd need to pick a differentiable one, that is $\frac{dx}{dt}$ and $\frac{dy}{dt}$ both exist and are not simultaneously zero. This forces no sharp turns along the path you follow. So at the offending point $(0,0)$ you'd need to make sure it follows to the opposite quadrant as $t$ gets larger.


Just to complete what I was saying above I found a differentiable parameterization of the curve

$$ \begin{split} x&=\tan^2 t -1\\ y&=\tan^3 t- \tan t \end{split} $$

Which has derivatives

$$ \begin{split} \frac{dx}{dt}&=2\tan t\sec^2 t \\ \frac{dy}{dt}&=3\tan^2 t \sec^2 t- \sec^2 t \end{split} $$

On the interval $-\frac{\pi}{2} < t < \frac{\pi}{2}$

And the $(0,0)$ point on the graph occurs at the $t$-values $t=-\frac{\pi}{4}$ and $t=\frac{\pi}{4}$

$$ \begin{split} \frac{dx}{dt}|_{t=\frac{\pi}{4}}&=4 \\ \frac{dy}{dt}|_{t=\frac{\pi}{4}}&=4\\ \frac{dy}{dx}|_{t=\frac{\pi}{4}}&=\frac{dy/dy|_{t=\frac{\pi}{4}}}{dx/dt|_{t=\frac{\pi}{4}}}=1 \end{split} $$

$$ \begin{split} \frac{dx}{dt}|_{t=-\frac{\pi}{4}}&=-4 \\ \frac{dy}{dt}|_{t=-\frac{\pi}{4}}&=4\\ \frac{dy}{dx}|_{t=-\frac{\pi}{4}}&=\frac{dy/dy|_{t=-\frac{\pi}{4}}}{dx/dt|_{t=-\frac{\pi}{4}}}=-1 \end{split} $$

I'm fairly sure this curve is well-studied, but don't know the name of it and just felt like deriving something like this myself :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.