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stripes
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Homework Statement
Show that [itex]\gamma : [a, b] \rightarrow \Re^{2}[/itex] is a parameterization of [itex]\Gamma[/itex] if and only if the length of the curve from [itex]\gamma(a)[/itex] to [itex]\gamma(s)[/itex] is [itex]s - a[/itex]; i.e.,
[itex]
\int ^{s}_{a} \left| \gamma ' (t) \right| dt = s - a.
[/itex]
Homework Equations
The Attempt at a Solution
Part 1; show [itex]\left| \gamma ' (s) \right| = 1 \Rightarrow \int ^{s}_{a} \left| \gamma ' (t) \right| dt = s - a.[/itex]
We have
[itex]
\ell = \int ^{b}_{a} \left| \gamma ' (t) \right| dt = \int ^{b}_{a} \left| \gamma ' (s(t)) s'(t) \right| dt = \int ^{b}_{a} \left| \gamma ' (s(t))\right| \left| s'(t) \right| dt = \int ^{\ell}_{0} \left| \gamma ' (s) \right| ds = \int ^{\ell}_{0} 1 ds = s(\ell) - s(0) = s - a.
[/itex]
My question here is: did I do the change of variables correctly? Specifically in the limits of integration?
Now how do I prove the converse; i.e., how do I show that [itex]\int ^{s}_{a} \left| \gamma ' (t) \right| dt = s - a \Rightarrow \left| \gamma ' (s) \right| = 1 ?[/itex]
[itex]\int ^{s}_{a} \left| \gamma ' (t) \right| dt [/itex]
[itex]= \int ^{s}_{a} \sqrt{x'(t)^{2} + y'(t)^{2} } dt [/itex]
[itex]= \int ^{s}_{a} \sqrt{ (x'(s(t)) s'(t) )^{2} + (y'(s(t)) s'(t) )^{2} } dt [/itex]
[itex]= \int ^{s}_{a} \sqrt{ s'(t)^{2} (x'(s)^{2} + y'(s)^{2}) } dt [/itex]
[itex]= \int ^{s}_{a} s'(t) \sqrt{ (x'(s)^{2} + y'(s)^{2}) } dt [/itex]
[itex]= \int ^{s}_{t=a} \sqrt{ (x'(s)^{2} + y'(s)^{2}) } ds[/itex]
[itex]= \int ^{s}_{t=a} \left| \gamma' (s) \right| ds = s - a[/itex]
[itex]= \frac{d}{ds}(\int ^{s}_{t=a} \left| \gamma' (s) \right| ds)[/itex]
[itex] = \frac{d}{ds}(s - a)[/itex]
[itex] = 1 - 0 = 1[/itex]
so [itex]1 = \left| \gamma' (s) \right| .[/itex]
or is that garbage?
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