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Suppose you have a parametric curve $r(t) = (x(t),y(t))$. From my understanding, we typically require a smoothness condition that its derivative is not equal to the zero vector for all $t$ in $r(t)$'s domain.

Suppose we ignore that condition. Ignoring some potential uninteresting edge cases, it is easy to find two polynomials $x(t)$, $y(t)$ such that there exists a $c$ so that $c$ is a non-repeated root of $x'$ and $y'$. If we then define "the slope of the line tangent to the curve $r(t)$" as $m = \lim_{t\rightarrow c} \frac{y'(t)}{x'(t)}$ then this limit will exist and the equation of the tangent line can be written as $$y = m(x-x(c)) + y(c).$$

It seems to me that this is the "right" way to try to define a "tangent line" and you could relax the smoothness condition to allow the case when $(x'(c),y'(c)) = (0,0)$ but the limit of $m$ exists at $c$.

My question is then, what am I actually doing here when I apply this procedure? Is this definition of a tangent line reasonable and does this slope really tell me anything? I suppose as a follow up, is this equation of the tangent line unique and independent of the parameterization?

JessicaK
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  • By l'Hopital's rule, you would get $m = \lim_{t\to c} \frac{y(t) - y(c)}{x(t) - x(c)}$ in most cases, except for some where the curve crosses the vertical line $x = x(c)$ infinitely often for $t$ near $c$. Why not use that as your definition? (Might allow a slightly increased generality.) – Daniel Schepler Jan 18 '23 at 01:31
  • So for example, to me the canonical example of a curve with a singularity is $y^2 = x^3$, which can be parameterized by $x(t) = t^2, y(t) = t^3$. So at $t = 0$ either definition would give $m=0$ as the slope of the tangent line, which seems reasonable given the picture of the curve as having a "cusp" at $(0,0)$ which graphically approaches $(0, 0)$ horizontally. – Daniel Schepler Jan 18 '23 at 01:33
  • @MarianoSuárez-Álvarez I have tried to draw a few pictures. It seems reasonable in that if you consider an astroid with a cusp at (1,0), you can see that the "slope" of the tangent line approaches 0 from both sides in the limit. Are the pictures ever a reasonable tangent line? I would think that the answer is no. But if I had to choose from a family of candidate tangent lines which ones I would guess have the most useful properties, it would be the one I chose. – JessicaK Jan 18 '23 at 03:31

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Suppose the curve is $t\mapsto(x(t),y(t))$. The equation of the tangent line at a smooth point $t$ is $$y'(t)\cdot(X-x(t))-x'(t)\cdot(Y-y(t))=0.$$ Let us now suppose that $x'(t_0)=y'(t_0)=0$ but that the limit $\lim\limits_{t\to t_0}y'(t)/x'(t)$ exists and has value $m$. For all $t$ for which it $x'(t)$ is not zero we can rewrite the equation of the tangent line at $t$ in the form $$\frac{y'(t)}{x'(t)}\cdot(X-x(t))-(Y-y(t))=0.$$ This tells us that if we take $t\to t_0$ then the tangent line converges to the line with equation $$m\cdot(X-x(t))-(Y-y(t))=0.$$ So this line is the limit (in an appropriate sense) of the tangent lines at nearby points. It is strictly not the tangent line, but surely deserves to be called that.

A typical example is the curve $t\mapsto(t^2,t^3)$, which looks as follows: enter image description here Clearly the tangents at points near the cusp "converge" to the horizontal line.


$\def\RR{\mathbb{R}}$ This construction makes sense. One way to see this is to construct a space $\mathcal L$ whose points are the lines in the plane and put a sensible topology on it. Then if $\gamma:\RR\to\RR^2$ is a curve and $S$ is the subset of $\RR$ of the smooth points, then we can define a function $\bar\gamma:S\to\mathcal L$ that maps each point in $S$ to the line tangent to the curve at $\gamma(t)$.

Sometimes there is a unique way to extend the function $\bar\gamma$ to a continuous function $\bar{\bar\gamma}:\RR\to\mathcal L$, and in that case it makes sense to say that $\bar{\bar\gamma}(s)$ is the tangent line to $\gamma$ at the singular point $\gamma(s)$.

This construction is very standard in algebtraic geometry. There we consider curves in the projective plane $\mathbb P^2$, and use the fact that the set of lines in that plane is in a natural way another projective plane, the dual projective plane, which we usually write $\mathbb P^{2*}$. The advantage of working in the projective plane is that it is compact, so taking limits is easier.

  • I've never seen the equation of a tangent line written in the form $y'(t) (X-x(t)) - x'(t) (Y-y(t)) = 0$ before and that helps clear up a lot of the uncertainty. Thank you. I still can't help but be a little annoyed that the phrase used is "tangent line" without an adjective like calling it "weakly tangent". – JessicaK Jan 18 '23 at 03:39
  • Mathematics makes moves forward by changing the meaning of words in such a way that true statements using those words keep beeing true. – Mariano Suárez-Álvarez Jan 18 '23 at 03:40
  • Ah I have just seen your edit with the idea of extending $\gamma$. That completely satisfies me now. – JessicaK Jan 18 '23 at 03:42
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    Notice that the image of $\bar{\bar\gamma}$ is a curve in the space of lines. It is called the dual curve to the original one. If you google for that you will see that it is a beautiful and very well studied subject! (and probably many nice pictures) – Mariano Suárez-Álvarez Jan 18 '23 at 03:46