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Proof: Arc Length parameter

  1. Mar 14, 2014 #1
    1. The problem statement, all variables and given/known

    If C is a smooth curve given by
    r(s)= x(s)i + y(s)j + z(s)k

    Where s is the arc length parameter. Then

    ||r'(s)|| = 1.

    My professor has stated that this is true for all cases the magnitude of r'(s) will always equal 1. But he wants me to PROVE it. ( of course not with an example)

    2. Relevant equations

    r(s)= x(s)i + y(s)j + z(s)k

    r'(s)= x'(s)i + y'(s)j + z'(s)k

    3. Attempt at the solution

    To be quite honest, usually with math problems I will have some sort of attempt to try and solve it. But when it comes to proofs... I seem to get stuck.

    Well I know I'm trying to prove that. ||r'(s)|| = 1

    So the magnitude of r'(s)
    Will be given by
    SQRT[ x'(s)^2 + y'(s)^2 + z'(s)^2 ]

    After this I don't know what I can do to make it equal 1. Any help will be greatly appreciated! ImageUploadedByPhysics Forums1394824295.975449.jpg
     
  2. jcsd
  3. Mar 14, 2014 #2

    micromass

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    You know ##s## is the arc-length parameter, so what does that mean?
     
  4. Mar 14, 2014 #3
    Um, That it is measuring the length along the Curve ..
     
  5. Mar 14, 2014 #4

    micromass

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    Can you express that mathematically?
     
  6. Mar 14, 2014 #5
    Unfortunately no:/
     
  7. Mar 14, 2014 #6

    micromass

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    How do you calculate the arc length of a curve?
     
  8. Mar 17, 2014 #7

    by integrating from a to b
    On the following
    sqrt[ x'(t)^2 + y'(t)^2 + z'(t)^2 ] dt
     
  9. Mar 17, 2014 #8

    micromass

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    Right, so the length from ##0## to ##s## is defined as

    [tex]\int_0^s \sqrt{(x^\prime(t))^2 + (y^\prime(t))^2 + (z^\prime(t))^2}dt[/tex]

    by definition of arc-lenght parameter, this is equal to ##s##. So

    [tex]s = \int_0^s \sqrt{(x^\prime(t))^2 + (y^\prime(t))^2 + (z^\prime(t))^2}dt[/tex]

    Now differentiate both sides.
     
  10. Mar 17, 2014 #9
    Will differentiating on the side with the integral wipe out the integral completely? Or would I have to integrate and then differentiate?
     
  11. Mar 17, 2014 #10

    HallsofIvy

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    Differentiating on the side with the integral, use the fundamental theorem of Calculus:
    [tex]\frac{d}{ds}\left(\int_a^s f(t)dt\right)= f(s)[/tex]
     
  12. Mar 17, 2014 #11

    Ok so by using that theorem, to find the derivative of an integral. I have to plug my upper limit back into the function.

    Which would be
    SQRT[ x'(s)^2 + y'(s)^2 + z'(s)^2 ]

    How does this prove that ||r'(t)|| always =1 where t is any parameter?
     
  13. Mar 17, 2014 #12

    LCKurtz

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    Here's a hint for an alternate (easier) way. Use the chain rule for$$
    \frac d {ds} \vec r(t) = r'(t)\cdot (?)$$by the chain rule. Work that out and calculate its length. You don't have to consider components.
     
  14. Mar 18, 2014 #13

    pasmith

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    So far you have
    [tex]
    s = \int_0^s \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}\,dt
    [/tex]
    and have differentiated with respect to [itex]s[/itex] to find
    [tex]
    1 = \sqrt{x'(s)^2 + y'(s)^2 + z'(s)^2}.
    [/tex]
    But the right hand side is exactly [itex]\|r'(s)\|[/itex]!

    It doesn't, becuase it's not true that [itex]\|r'(t)\| = 1[/itex] for any parameter: consider the line from [itex](0,0,0)[/itex] to [itex](1,1,1)[/itex] with the parametrization
    [tex]
    (x,y,z) = (t,t,t),\qquad 0 \leq t \leq 1
    [/tex]
    and compare it with the parametrization
    [tex]
    (x,y,z) = (2u,2u,2u),\qquad 0 \leq u \leq \tfrac 12.
    [/tex]
    The result [itex]\|r'(s)\| = 1[/itex] holds only when [itex]s[/itex] is arclength.
     
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