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Show that the arc length $L(\gamma)$ of a curve $\gamma$ is unchanged if $\gamma$ is reparametrized

Can you help me please?

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    What have you tried? Do you know the definitions of arc-length and parameterization?2012-05-09
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    Remember integration by substitution?2012-05-09
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    yes, I remember2012-05-09
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    But I have trouble understanding the definition of arc length2012-05-09
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    It's the length of a line. Like integration is the area under a curve. Arc length is a length of a line. It is easy for a straight line. For a curve line, you approximate the arc length by using lots of little straight lines which, together, approximate the curve. Taking the sum of infinitely many, infinitesimally small straight lines, we get the actual "arc length" of the line.2012-05-09
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    Okay, would you at least be able to explain why $$\sqrt{\left(\frac{\mathrm d}{\mathrm dt}f(w(t))\right)^2+\left(\frac{\mathrm d}{\mathrm dt}g(w(t))\right)^2}=\sqrt{w^\prime(t)^2 (f^\prime(w(t))^2+g^\prime(w(t))^2)}$$?2012-05-09
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    To add to Adam Rubinson's comment, locally the curve can be approximated by $t\gamma'(t)+\gamma(t)$, which means its speed is locally $|\gamma'(t)|$, which means that over a small interval $dt$ the curve travels a distance $|\gamma'(t)|dt$. So the total distance traveled is the sum: $\int |\gamma'(t)|dt$.2012-05-09

3 Answers 3

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Let $\gamma = \gamma(t)$

Then $$L(\gamma) = \int_0^t |\frac{d\gamma}{dt}|dt$$

If $\gamma$ is reparametrized so $\gamma = \gamma(s(t))$ then $\displaystyle\gamma' = \frac{d\gamma}{ds}\frac{ds}{dt}$ by the chain rule so $$L(\gamma) = \int_0^s \left|\frac{d\gamma}{ds}\frac{ds}{dt}\right|ds$$

Can you show that $L(\gamma)$ is the same computed either way?

Hopefully this helps, this is my first post so go easy on me!

Edit: Be careful on the limits of integration, it will depend on the domain of the curve.

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The way I learned it was to let the curve $\gamma$ be described by two functions, $X$ and $Y$ such that $Y(t)=X[u(t)]$ for $c\le t\le d$. (Here $u$ defines the change of parameter.) Also assume that $u'(t)>0$ for $c\le t \le d$. Now prove that

$$\int_{u(c)}^{u(d)} \! ||X'(h)||\,\mathrm d h=\int_c^d \! || Y'(t)|| \, \mathrm dt\,.$$

(If this is the part you're stuck on, let me know and I'll try to think of another hint.)

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Here is another way to look at the problem:

In order to define the length of a curve $$\gamma:\ t\mapsto{\bf x}(t)\quad(a\leq t\leq b)$$ one considers partitions $$T:\quad a=t_0 of the parameter interval $[a,b]$. Such a partition $T$ induces a "piecewise linear approximation" $\gamma_T$ of $\gamma$ whose length is given by $$L(\gamma_T):=\sum_{k=1}^N\bigl|{\bf x}(t_k)-{\bf x}(t_{k-1})\bigr|\ .$$ One then defines $$L(\gamma):=\sup_{T\in{\cal T}} L(\gamma_T)\ ,$$ where ${\cal T}$ denotes the set of all partitions of $[a,b]$. Given the geometric interpretation of this definition it is obvious that $L(\gamma)$ stays the same when $\gamma$ is reparametrized.

(It is a theorem that the above $L(\gamma)$ equals $\int_a^b\bigl|\dot{\bf x}(t)\bigr|\ dt$ when $\gamma$ is continuously differentiable.)