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Parametric Arclength

  1. Aug 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the arc length of a curve given parametrically from t = 0 to t = 1.
    Curve given by x = 4t^2, y = 2t


    2. Relevant equations

    [I think] parametric arclength =
    integral from t = b to t = a of sqrt( (dx/dt)^2 + (dy/dt)^2)dt


    3. The attempt at a solution

    dx/dt = 8t, and dy/dt = 2
    Then I did S from 0 to 1 of sqrt((8t)^2) + (2)^2)dt

    I then drew a triangle for trigonometric substitution, with the tangent case.
    Let t = (1/4)tan(σ)
    Let dt = (1/4)sec^2(σ) dσ
    Let sqrt(64t^2 + 4) = 4sec(σ)
    Then S[ sqrt( (8t)^2 + (2)^2) ]dt becomes
    S of 4sec(σ) * (1/4)sec^2(σ) dσ
    which equals S[ sec^3(σ) ] dσ

    Now I solved that with integration by parts, letting u = sec(σ), and dv = sec^2(σ)
    so that du = sec(σ)tan(σ)dσ and v = tan(σ)
    now S[ sec^3(σ) ]dσ = sec(σ)tan(σ) - S[ tan^3(σ)sec(σ) ]dσ

    that last little integral = S[ tan^3(σ)sec(σ) ]dσ = S[ ((sec^2(σ) - 1)*sec(σ)tan(σ)]dσ]
    =(1/3)sec^3(σ) - sec(σ)
    so the entire S[ sec^3(σ) ]dσ = sec(σ)tan(σ) - (1/3)sec^3(σ) + sec(σ)
    so (4/3)S = sec(σ) + sec(σ)*tan(σ)
    so S = (3/4)[sec(σ)tan(σ) + sec(σ)]

    Now I'm replacing this value with the value given by my triangle for sec(σ) and tan(σ)
    so that S = (3/4)[sqrt(64t^2 + 4)*4t + sqrt(64t^2 + 4)] evaluated from t = 0 to t = 1
    which would be
    (3)sqrt(68) + (3/4)sqrt(68)
    minus
    (6/4) + 2
    =
    (15/4)sqrt(68) - (14/4) as my final answer.

    Could someone please tell me if I'm doing anything wrong? Maybe if I'm going about this entirely the wrong way? Unfortunately I don't even know the right answer, but even if I can't get the right answer, as long as I understand how to get there, I feel okay.

    I would really appreciate it, and I'm sorry about the typing, I'm not very good at typing these problems out. So much thanks,
    Sassy
     
  2. jcsd
  3. Aug 16, 2013 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that's correct.


    3. The attempt at a solution

    dx/dt = 8t, and dy/dt = 2
    Then I did S from 0 to 1 of sqrt((8t)^2) + (2)^2)dt

    I then drew a triangle for trigonometric substitution, with the tangent case.
    Let t = (1/4)tan(σ)

    Let dt = (1/4)sec^2(σ) dσ[/quote]
    No big deal but I wouldn't use the word "let" here. You already said "let t= (1/4)tan(σ) and then dt= (1/4)sec^2(σ) follows from that.

    Same point

    How did you get this? with [itex]du= sec(\sigma)tan(\sigma)d\sigma)dx[itex] and [itex]v= tan(\sigma)[/itex], as you say, that last integrand is [itex]sec(\sigma)tan^2(\sigma)d\sigma[/itex], not [itex]tan^3(\sigma)sec(\sigma)d\sigma[/itex].

    With [itex]tan^2(\sigma) sec(\sigma)d\sigma[/itex], I would proceed by putting everything in terms of sine and cosine:
    [tex]\int tan^2(\sigma) sec(\sigma)d\sigma= \int \frac{sin^2(\sigma)}{cos^2(\sigma)}\frac{1}{cos(\sigma)}d\sigma= \int \frac{sin^2(\sigma)}{cos^3(\sigma)}d\sigma[/tex]

    With that odd power of cosine, I would multiply both numerator and denominator by [/itex]cos(\sigma)[/itex] to get
    [tex]\int \frac{sin^2(\sigma)}{cos^4(\sigma)}cos(\sigma)d\sigma= \int\frac{sin^2(\sigma)}{(1- sin^2(\sigma))^2} cos(\sigma)d\sigma[/tex]
    and use the substitution [itex]u= sin(\sigma)[/itex], [itex]du= cos(\sigma)d\sigma[/itex] so that the integral becomes
    [tex]\int \frac{u^2}{(1- u^2)^2}du[/tex]

     
  4. Aug 16, 2013 #3

    verty

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    Homework Helper

    Should this be 2sec(σ)?

    Then also, let t = 1/4 sinh(σ) for a shorter solution, although this is an example of why I disliked integration, it is sometimes very arcane.
     
  5. Aug 16, 2013 #4
    Okay great thank you very much. So I took it from where you left off, HallsofIvy, and I see what you mean that yes, I did mess up with the integral and it should have been tan^2(σ)sec(σ) d(σ).
    Picking up where you left off, I suppose I would then need partial fractions?
    Wow, this has been quite the problem! Looks deceivingly simple.
     
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