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Arc Length in Three Dimensions Question

  1. Feb 4, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Find the arc length of r(t)= <tsin(t), tcost(t), 3t> from 0 to t to 2pi (inclusive)

    2. Relevant equations
    Integral from 0 to 2pi of the magnitude of r'(t) dt

    3. The attempt at a solution
    1. Must find the derivative of the function.
    Using the product rule a few times, the derivative becomes
    <tcos(t)+sin(t), -tsin(t)+cos(t), 3>

    2. Now take the magnitude
    Well, this is a bit lengthy, but it simplifies pretty nice with the double angle trig identity. (I hope what I did is correct, please check my work in this step.

    (tcos(t)+sin(t))^2 = t^2cos^2t+2tsin(t)*cos(t)+sin^2t
    Simplifies to: t^2cos^2t+tsin(2t)+sin^2t

    (-tsin(t)+cos(t))^2 = t^2sin^2t+(-2tsin(t)*cos*(t))+cos^2t
    simplifies to: t^2sin^2t-tsin(2t)+cos^2t

    3^2 = 9

    So, when we add everything the double angles cancel out. The cos^2t+sin^2t is equal to 1. We factor a t^2 out of the t^2cos^2t and t^2sin^2t equations and this becomes t^2*1.

    We are left with 9+1+t^2
    t^2+10, beautiful!

    3. Integrate from 0 to 2pi the square root of (t^2+10) dt

    Well, here's where I am stuck.
    I feel like I've done everything properly up to this point. But this integral cannot be taken easily (at least with calc 3 tools).

    Did I do something wrong? Any hints on the integral?
     
  2. jcsd
  3. Feb 5, 2015 #2

    Dick

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    You use a trig substitution on the integral. Something like t=sqrt(10)*tan(u). Have you done stuff like that?
     
    Last edited: Feb 5, 2015
  4. Feb 5, 2015 #3

    RJLiberator

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    Yes, I have. I will attempt this tomorrow morning and report back. This sounds promising.

    Thank you.
     
  5. Feb 5, 2015 #4

    RJLiberator

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    How's this:

    Let u = sqrt(10)*tan(theta)
    dtheta = sqrt(10)sec^2(theta)

    The integral is from 0 to 2pi the square root of 10+t^2
    then you get the integral of the square root of 10+10tan^2(theta) under the radical
    You can factor the 10 and the tan^2(theta)+1 becomes sec^2(theta) which simplifies to just sec after it is out of the radical

    so now you are left with
    sqrt(10)*sec^2theta/sqrt(10)sec(theta)
    some cancelations make it the integral of sec(theta) from 0 to 2pi

    answer: ln(sec(theta)+tan(theta)) from 0 to 2pi

    1-1=0

    well... that's not good.
     
  6. Feb 5, 2015 #5

    Dick

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    That's pretty sloppy. This might go better if you didn't scramble variables. t=sqrt(10)*tan(u). dt=sqrt(10)*sec^2(u)du. You are going to wind up with a sec^3(u) integral, which is challenging, but you can always look it up. And u doesn't go from 0 to 2pi. t goes from 0 to 2pi.
     
  7. Feb 5, 2015 #6

    RJLiberator

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    Thanks for the hints.

    I was doing the previous (sloppy) work in my head earlier :p.

    Now I've got a pen and paper:

    So after the tangent substitution, my integral becomes
    10*sec^3(u)

    The problem states that we can use table integrals to help solve our problems, so I assume that this is what he means. I apply the table integral of sec^3(u)

    10*[1/2*sec(u)*tan(u)+12*ln|sec(u)+tan(u)|]

    That seems to be correct. Now I just need to figure out the bounds, for some reason I am struggling with this (it's been a while). I'll keep working on it and report back.
     
  8. Feb 5, 2015 #7

    RJLiberator

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    I previously let
    t = sqrt(10)*tan(u)

    to change my bounds, i altered this equation to
    tan^-1(t/sqrt(10)) = u

    I then plugged in 0 and 2pi for t.
    0 works out nicely, however, 2pi does not work out.

    Something seems to be wrong :/
     
  9. Feb 5, 2015 #8

    RJLiberator

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    Ah. I can use the right triangle trigonometry.

    Tan(u) = t/sqrt(10)
    so sec(u)= sqrt(10+t^2)/sqrt(10)

    And then plug in the bounds via this distinction.

    The 2pi however is making the answer messy.

    My final answer is
    10*[pi*sqrt(10+4pi^2)/10+1/2*ln|(sqrt(10+4pi^2)+2pi)/sqrt(10)|
     
  10. Feb 5, 2015 #9

    Dick

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    Looks ok to me.
     
  11. Feb 5, 2015 #10

    RJLiberator

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    Fantastic. That was a great problem. I wasn't expecting it to be that in depth. Thank you for your guidance.
     
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