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.]


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.

  • $\begingroup$ 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. $\endgroup$ – terrygarcia Mar 13 '19 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. $\endgroup$ – terrygarcia Mar 13 '19 at 20:28

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 $h=-k$).


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.