Question: Show that $\alpha(t)=(t^3,t^2)$, $t\in \Bbb R$, has a weak tangent but not a strong tangent at $t=0$.

Definitions from this answer:

(Weak tangent) $\alpha: I \to \Bbb R^3$ has a weak tangent at $t_0 \in I$, if the line determined by $\alpha(t_0 + h)$ and $\alpha(t_0)$ has a limit position when $h \to 0$.

(Strong tangent) $\alpha: I \to \Bbb R^3$ has a strong tangent at $t_0 \in I$, if the line determined by $\alpha(t_0 + h)$ and $\alpha(t_0 + k)$ has a limit position when $h \to 0$ and $k \to 0$.

My query:

I'm not really clear what argument to use to demonstrate this. The weak tangent is the line joining $\alpha(t_0)$ and $\alpha(t_0+h)$, which is $$ (\lambda(x(t_0+h)-x(t_0))+x(t_0), \lambda(y(t_0+h)-y(t_0))+y(t_0)) $$ If $t_0=0$ then $x(t_0)=0$ so this becomes $$ (\lambda h^3, \lambda h^2) $$ The strong tangent is $$ (\lambda(x(t_0+h)-x(t_0+k))+x(t_0+k), \lambda(y(t_0+h)-y(t_0+k))+y(t_0+k)) $$ $$ =(\lambda (h^3-k^3)+k^3,\lambda (h^2-k^2)+k^2) $$ As $h,k\rightarrow0$ this seems badly defined. But how can I make this argument precise?

Also, what is the intuitive meaning of the strong and weak tangents?

[This is exercise 1-3-7 of Differential Geometry of Curves and Surfaces by Do Carmo.]


2 Answers 2


Weak tangent

Notice that the slope of $(\lambda h^3, \lambda h^2)$ tends to $\infty$ as $h\to 0$, meaning that the line direction approaches vertical. Since the line always passes through $(0,0)$, this means it has a limiting position (the $y$ axis).

Strong tangent

If it exists, it has to be the same as the weak tangent, because if the double limit exists, iterated limit "$k\to 0$ then $h\to 0$" exists and is equal to it. However, approaching via $h=-k$ you will find that the lines stay horizontal.

Intuitive meaning

  • Strong tangent: if you walk along the curve and someone is walking along the tangent line with the same speed, you can spend some time walking together and holding hands.

  • Weak tangent: looks like strong tangent at first, but at the point of tangency there is a break-up and someone goes away in the opposite direction.

  • 1
    $\begingroup$ I'm working the same problem and was confused in the same way. Looking at the slope is OK in $\mathbb{R}^2$, but doesn't easily generalize to $\mathbb{R}^n$. I think do Carmo has hidden a rigorous definition in the solutions, where he suggests that $\lim_{(h,k) \to \vec{0}} (\alpha(t_0+h) - \alpha(t_0+k))/(h-k)$ should exist and not be the zero vector. I believe one could equivalently assert that $\lim_{(h,k) \to \vec{0}} (\alpha(t_0+h) - \alpha(t_0+k))/\lvert \alpha(t_0+h) - \alpha(t_0+k) \rvert$ exists and is not zero. $\endgroup$ Mar 13, 2019 at 20:18
  • $\begingroup$ Upon further analysis, it looks like neither of those definitions work. I do, however, think it suffices to normalize the vector in the first limit. $\endgroup$ Mar 13, 2019 at 20:28
  • $\begingroup$ Sorry, I can't see why the strong tangent stay horizontal via $h=-k$, in this case the slope won't be $k^2/k^3(1-2\lambda) = 1/k(1-2\lambda)$, which will also aproach $\infty$? $\endgroup$ Aug 25, 2021 at 16:46
  • 1
    $\begingroup$ @AnalyticHarmony It is easier to see this by drawing the picture. The (trace of the) curve $\alpha$ is symmetric about the $y$-axis. The line joining $\alpha(0+k)$ and $\alpha(0+h)=\alpha(0-k)$ is the horizontal line $y=k^2$, for every $k\neq 0$. Thus the limiting line as $k\to 0$ also stays horizontal. $\endgroup$ Sep 15, 2021 at 5:41

Maybe one comment is worth here. Observe that the curve is $y = x^{2/3}$ and $dy/dx = (2/3)x^{-1/3}$ goes to $\infty$ when $x \rightarrow 0$, so the derivative does not exist at $x=0$ and $\alpha(t)$ is not regular at $t=0$. The same result should be obtained by the definition of weak and strong tangent if they exist. As mentioned before, the slope of the line determined by $\alpha(t_0+h)$ and $\alpha(t_0)$ goes to $\infty$ when $h\rightarrow 0$. So the weak tangent exists. But this does not happen for any $h$ and $k$ in the case of the strong tangent (the slope is $0$ if you choose $h=-k$).


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