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The trace of the parametrized curve is

$$\alpha(t)=(t,\cosh t),\ t\in\mathbb R$$

is called catenary.

I want to show the curvature of the catenary is

$$k(t)=\frac{1}{\cosh^2t}$$

Before finding the curvature I need to parametrize it by arc length. Do Carmo in his classical Differential Geometry book makes the following remark about it on page 22:

since the catenary is defined for every $t\in \mathbb R$, I'm having trouble to know the value of $t_0$, can it be anything?

user42912
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  • yes; $0$ would be a convenient choice – Nick Pavlov Oct 30 '17 at 11:22
  • Do you have go through to the arc length? If you use this you can get the result you're looking for relatively straightforward – caverac Oct 30 '17 at 11:27
  • @NickPavlov can it be whatever I want? – user42912 Oct 30 '17 at 11:41
  • "yes" was the answer to your actual question, "can it be anything?" – Nick Pavlov Oct 30 '17 at 11:46
  • @NickPavlov what about the case $t<0$? – user42912 Oct 30 '17 at 11:49
  • no matter what $t_0$ you choose, there will always be some $t < t_0$, but that's fine, for those the corresponding arc length will also be negative (they are in "the past") – Nick Pavlov Oct 30 '17 at 11:51
  • @NickPavlov yes, they are negative, but what does it have to do with my actual question? I need a new parametrization which deals with every $t$. If I take $t_0=0$ the arc length parametrization is valid only for $t>0$ – user42912 Oct 30 '17 at 11:55
  • no, it will be valid for every $t$. The function $s(t)$ is from $\mathbb{R}$ to itself, it takes care of all $t$'s – Nick Pavlov Oct 30 '17 at 11:58
  • if you insist that "arc length" can only be non-negative (if I am correctly understanding the source of your discomfort), note that for the purposes of re-parametrization, arc length is also a "directed" quantity, that is from your chosen base point $\alpha(t_0)$ you can go forward ($s > 0$) as well as back ($s < 0$). – Nick Pavlov Oct 30 '17 at 12:05
  • @NickPavlov I think I got it, we can take any $t_0\in I$. Thank you – user42912 Oct 30 '17 at 13:07
  • In fact, you don’t need a paramerization to arc length, cf. https://math.stackexchange.com/questions/2202579/intuitive-definition-for-curvature/2202754#2202754 – Michael Hoppe Oct 30 '17 at 15:12

1 Answers1

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You can always use the chain rule to do curvature/torsion computations for curves that are not arclength parametrized (and, of course, there are other formulas as well). This has been discussed in numerous MSE posts.

However, note that for your catenary, $s(t) = \displaystyle\int_0^t \sqrt{1+\sinh^2 u}du = \int_0^t \cosh u\,du = \sinh t$. And, using the quadratic formula (and the definition of $\sinh$), you can easily solve the equation $s=\sinh t$ for $t$ as a function of $s$ and reparametrize the curve. ... Again, whether you should want to is another matter.

Ted Shifrin
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