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Length of a curve problem

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the length of the curve r(t) = <2t3/2 , cos 2t, sin 2t>, for 0<= t <=1

    2. Relevant equations

    L = [tex]\int\sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2}dt[/tex]

    3. The attempt at a solution

    (dx/dt)2 = (3t1/2)2
    (dy/dt)2 = (-2 sin(2t))2
    (dz/dt)2 = (2 cos(2t))2

    [tex]\int\sqrt{9t+4sin^2 (2t) + 4 cos^2 (2t)}dt[/tex]

    and since sin^2 + cos^2 = 1, that reduces to:


    I found a site that had an identity for integrals with square roots, and this resembles number 6 on the list:

    so using that identity, I get
    [tex]2\sqrt{(9t+4)^3} / 27 [/tex] evaluated from 0 to 1

    Is my approach correct, and if so, should I just keep these integrals on hand, or do I need to memorize all those forms?
    Last edited: Dec 11, 2008
  2. jcsd
  3. Dec 11, 2008 #2
    you should know how to calculate this integral.
    Exchanging variables, you can define u=9t+4. then du = 9dt. = > dt = du/9
    int {sqrt(9t+4)}dt = int {sqrt(u)/9}du = (1/9)*(u^(3/2))/(3/2).
    Then stick u = 9t+4 back and use the upper and lower limits.

    So, to answer your question - it's not that you should know this integral by heart, but you should be able to solve it using basic techniques.
    Hope that was clear, and that it answered your question.
  4. Dec 11, 2008 #3
    I don't see any problem. You should be familiar with a method of integration called u-substitution that would do the trick. If you're in a Calc I class, then you might learn this soon.
  5. Dec 11, 2008 #4
    doh! of course. I don't know why I couldn't see it as a u substitution problem. Thanks all for the responses.
  6. Dec 11, 2008 #5
    You're welcomed :)
    By the way - off topic, but - could you tell me how I write formulas here? Do I need some kind of a program, or is it a code, or what?
    New here...
  7. Dec 11, 2008 #6
    At first it's really tedious, but once you start to get the hang of it, it's not that bad. When you're posting, there is a series of controls at the top of the box. the sigma will bring up a clickable interface so you can choose different symbols, and the subscript and superscript symbols are pretty useful too. once you get the hang of using it, it's pretty easy to just type in the code as you go along. just don't forget to close your tags, otherwise the output won't look right. : )
  8. Dec 11, 2008 #7
    Thanks man. Have some Grog :)

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