# Weak tangent but not a strong tangent

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

(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

### 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.

• 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. – terrygarcia Mar 13 '19 at 20:18
• 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. – 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 $$h=-k$$).