0

Background: Given $\alpha : I \rightarrow R$, a simple curve of class $C^0$, do Carmo defines $\alpha$ to have a weak tangent at $t=t_0 \in I$ if the line determined by $\alpha (t_0 +h)$ and $\alpha (t_0)$ has a limit position when $h \rightarrow 0$. Also, he defines $\alpha$ to have a strong tangent at $t=t_0$ if the line determined by $\alpha (t_0 +h)$ and $\alpha (t_0 + k)$ has a limit position when $h,k \rightarrow 0$.

Question: What does it mean for a line to have a limit position exactly? Also, $\alpha (t) = (t^2, t^3)$ seems to be an example of a curve having a weak tangent but not a strong tangent. Do these definitions have relevance to the fact that $\alpha$ is not regular $\alpha '(t) = 0$ at point $t=0$?

Thanks!

T J. Kim
  • 610
  • Which book of him? – edm Apr 14 '18 at 18:51
  • It's from the second edition of differential geometry of curves and surfaces – T J. Kim Apr 14 '18 at 18:51
  • Well, without anything too fancy, a line in the plane has an equation $y=mx+b$ for some real numbers $m$ and $b$. If you have lines $L_t: y=m_tx+b_t$, they'll have a limit $y=mx+b$ precisely when $m_t\to m$ and $b_t\to b$. (We'll have to handle vertical lines separately, but you can figure that out.) – Ted Shifrin Apr 14 '18 at 18:55
  • 1
    Yes, if the parametrization has non-vanishing derivative at $t=t_0$, then you'll get a strong tangent. – Ted Shifrin Apr 14 '18 at 19:15
  • @Ted Shifrin Thanks. For the example of $\alpha (t) = (t^3,t^2)$, I took the slope of the line joining $\alpha(t_0)$ and $\alpha (t_0+h)$ to be $\frac{y(t_0 +h)-y(t_0)}{x(t_0 +h) - x(t_0)} = \frac{h^2}{h^3}$, which approaches infinity as $h \rightarrow 0$; but doesn't this mean the line would have no limit? But this particular curve should have a weak tangent, I'm not sure why. – T J. Kim Apr 14 '18 at 19:23
  • Remember that I said you have to handle vertical lines separately. When you're going to have a vertical line as a limit, do $x=ky+c$ instead. – Ted Shifrin Apr 14 '18 at 19:25
  • Do you mean that we must check $\frac{dx}{dy}$ also if $\frac{dy}{dx}$ does not converge to a limit as $h\rightarrow 0$? Wouldn't the former always converge if the latter diverges? – T J. Kim Apr 14 '18 at 19:31
  • 1
    That last question was a bit vague. Of course, if $|\frac{dy}{dx}|\to\infty$, then $\frac{dx}{dy} \to 0$, but you can have other ways that $\frac{dy}{dx}$ fails to have a finite limit. – Ted Shifrin Apr 14 '18 at 19:49
  • Ah, I see what you mean. Thanks! – T J. Kim Apr 14 '18 at 19:52
  • 1
    I believe this is strongly related --- pun intended --- to the notion of "strong derivative" I discuss in my answer to “Strong” derivative of a monotone function. In particular, see pp. 104-105 in The Elementary Differential Geometry of Plane Curves by Ralph Howard Fowler (1920). In early literature there was often no distinction made between tangents obtained as arbitrary approaching chord limits (continued) – Dave L. Renfro Apr 14 '18 at 20:17
  • 1
    and chord limits with one endpoint fixed. Fowler's book is the earliest non-specialist publication I know of which draws attention to this distinction. – Dave L. Renfro Apr 14 '18 at 20:18

0 Answers0