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Homework Help: Parameterised Curve Proof Part 1

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose that [itex]\vec r(t)[/itex] is a parameterised curve defined for [itex]a \le t\le b[/itex] and

    [itex] \displaystyle s(t)=\int_{a}^{t}\left \| d \vec r (t) \right \|dt[/itex] is the arc length function measured from r(a)

    a) Prove that s'(t) = || dr(t)||

    How do I start this? It is easy to see that differentiating both sides will yield the proof but I dont know how to go about t. Any clues?

    Note I have this also posted at MHF with no replies
    http://www.mathhelpforum.com/math-help/f57/parameterised-curve-proof-part-1-a-191196.html"
     
    Last edited by a moderator: Apr 26, 2017
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  3. Nov 5, 2011 #2

    LCKurtz

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    Look up Leibniz rule in your calculus book. It will tell you how do differentiate an integral as a function of the upper limit.
     
  4. Nov 5, 2011 #3

    I like Serena

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    Hi bugatti79! :smile:

    Funny that you'd get no reply at MHF.
    I'm not familiar with MHF, but can you compare PF to MHF?
    And what is that avatar?
    http://www.mathhelpforum.com/math-help/avatars/bugatti79.gif [Broken]
     
    Last edited by a moderator: May 5, 2017
  5. Nov 5, 2011 #4
    The MHF website was down for a few days so there are not many people checking the forum as the minute. I think its a very good website, very responsive.
    Its just some fractals jpeg I got off the web. :-)
     
    Last edited by a moderator: May 5, 2017
  6. Nov 5, 2011 #5
    My interpretation of Leibniz rule for this problem is not correct which is

    [itex]\displaystyle \frac{d}{dt} \int_{u(x)=a}^{v(x)=t} || dr(t)||dt=f(v(x))\frac{dv}{dx}-f(u(x))\frac{du}{dx}[/itex]............?
     
  7. Nov 5, 2011 #6

    I like Serena

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    Hmm, you did not apply the rule correctly...

    But I'll give you another way to look at it.

    Suppose D(t) is the anti-derivative of [itex]||d\vec r(t)||[/itex].
    Then D'(t)=[itex]||d\vec r(t)||[/itex].

    Furthermore s(t)=D(t)-D(a).
    Can you take the derivative on both sides with respect to t?
     
  8. Nov 5, 2011 #7

    LCKurtz

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    [tex]\frac d {dt}\int_a^t f(u)\,du = f(t)[/tex]
     
  9. Nov 11, 2011 #8
    No, I dont know how to differentiate this [itex]\displaystyle s(t)=\int_{a}^{t}\left \| d \vec r (t) \right \|dt

    [/itex] on the RHS...? THe LHS is just s'(t)
     
  10. Nov 11, 2011 #9

    LCKurtz

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    Write it like this:
    [tex]s(t)=\int_{a}^{t}\left \| \vec r' (u) \right \|du
    [/tex]
    and look at post #7.
     
  11. Nov 11, 2011 #10
    What is throwing me off is how to integrate an implicit function and put in the the limits and then differentiate it.

    Ie, one cant write f(u)^2/2|^t_a etc....I dont believe this is correct.


    I can see how integrating the function yields 2 terms..one a function of t and the other a function of a. Then differentiating wrt t takes the term a function of a go to 0 etc.
     
  12. Nov 11, 2011 #11

    LCKurtz

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    Of course that isn't correct.
    Suppose the antiderivative of f is F, whatever formula that may be. Then the integral gives you F(t) - F(a). What happens if you differentiate that?
     
  13. Nov 11, 2011 #12

    I like Serena

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    If you take the derivative on both sides you get:
    s'(t)=D'(t)-0
    And because of our definition of D we have: D'(t)=[itex]||d\vec r(t)||[/itex].
    So:
    [itex]s'(t)=||d\vec r(t)||[/itex].
     
  14. Nov 11, 2011 #13
    You are just left with s'(t)=F'(t) where the RHS is ||r'(t)|| etc. But how do you prove it? THat is the original question.
     
  15. Nov 11, 2011 #14

    I like Serena

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    This is the proof.
    Note that you only used the definitions of the anti-derivative and the integral.
     
  16. Nov 11, 2011 #15

    LCKurtz

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    If what you are asking is how to prove Leibnitz Rule then, like I said in my very first post, look in your calculus book for a proof. Although ILS and I have both outlined a version of a proof in this thread.
     
  17. Nov 18, 2011 #16
    OK, the next bit.

    Assume dr (t) is never 0. Then ds(t) >0 for all t so we can in principle express t in terms of s and then obtain the new parameterisation r_1 (s)= r(t(s)) where 0 <= s<=L(=\int_{a}^{b} ||dr(t)||dt).

    a) Prove ||dr(s)||=1.

    My attempt.

    r_1(s)=dr(t).dt(s). We know s'(t)= ||dr(t)|| therefore t'(s)=1/||dr(t)||

    then r_1(s)=dr(t)/||dr(t)||=1

    Correct?
     
  18. Nov 18, 2011 #17

    I like Serena

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    You have your integral wrong.
    It should be \int_{a}^{b} ||dr(t)||.
    That is, without the dt.
    There can (effectively) be only 1 infinitesimal in an integral.

    This cannot be true, since the left hand side (LHS) is an infinitesimal, while the right hand side (RHS) is not.

    Perhaps you meant: ||dr(s)||=ds?


    I'm afraid you have another infinitesimal problem here.
    No infinitesimals on the LHS, and a product of 2 at the RHS is simply not possible.

    You should define r_1 in terms of r though.
    This would be:
    r_1(s) = r(t(s))


    Good!
    You will want to use this.


    I'm afraid you have a vector at the LHS, and in the middle, but a scalar at the RHS.
    Not possible.




    Perhaps you can start with:
    r_1(s) = r(t(s))

    Convert first both sides to vector-infinitesimals.
    And then take the norm of the vector-infinitesimals.
     
  19. Nov 19, 2011 #18
    it says in the handout ||dr_1(s)||=1 for all s....


     
  20. Nov 19, 2011 #19

    HallsofIvy

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    Since, for this problem, the lower limit is a constant and the upper limit is t, you don't really need the full Leibniz rule. Just the "Fundamental Theorem of Calculus:
    The derivative of
    [tex]\frac{d}{dx}\int_a^x f(t) dt= f(x)[/tex]
     
  21. Nov 19, 2011 #20
    Yes, I have that thanks. Its the next bit I wanna work on

    Assume r' (t) is never 0. Then s'(t) >0 for all t so we can in principle express t in terms of s and then obtain the new parameterisation r_1 (s)= r(t(s)) where 0 <= s<=L(=\int_{a}^{b} ||r'(t)||).

    a) Prove ||r_1(s)||=1 for all s

    My attempt.

    [itex]r_1(s)=\frac{dr}{dt}*\frac{dt}{ds}[/itex].

    We know s'(t)= ||r'(t)|| therefore t'(s)=1/||r'(t)||

    then r_1(s)=r'(t)/||r'(t)||=1....?
     
    Last edited by a moderator: Nov 22, 2011
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