A biconditional statement for arc length of a function

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Homework Help Overview

The discussion revolves around the conditions under which a parameterization of a curve in \(\mathbb{R}^2\) is considered a natural parameterization. The original poster presents a biconditional statement involving the arc length of a curve defined by a parameterization \(\gamma\) and seeks to prove it.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to demonstrate the equivalence of the parameterization and the arc length condition through integration and change of variables, questioning the correctness of their approach.
  • Some participants question the validity of the original statement, suggesting it may only hold for natural parameterizations rather than arbitrary ones.
  • Others assert the correctness of the statement as presented in the text, indicating a potential disagreement on the interpretation of parameterization types.

Discussion Status

Contextual Notes

There is a noted lack of clarity regarding the distinction between general parameterizations and natural parameterizations, which is central to the discussion. The original poster also mentions an oversight in their initial post, indicating that additional context may be relevant.

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] <br /> \int ^{s}_{a} \left| \gamma ' (t) \right| dt = s - a.<br /> [/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] <br /> \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.<br /> [/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?
 
Last edited:
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stripes said:
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] <br /> \int ^{s}_{a} \left| \gamma ' (t) \right| dt = s - a.<br /> [/itex]

This statement is incorrect. Is it supposed to be "natural parametrization" rather than just "parametrization?
 
I do believe that the statement I am to prove is correct. I have copied it directly out of my text. My instructor says my proof is complete. I'm not sure what you meant.

And it is natural parametrization.
 
The statement is correct only if it is about natural parametrization. It is not correct for an arbitrary parametrization.
 
Oh my bad. I failed to mention that in my
 

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