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Homework Help: Arc length in 3-D

  1. Jul 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider the path f{r}(t) = (8t, 4t^2, 4log(t) ) defined for t > 0.
    Find the length of the curve between the points (8, 4, 0) and (24, 36, 4log(3)).

    2. Relevant equations

    [tex]\int[/tex]|r' (t)|dt

    3. The attempt at a solution

    r(t)=(8t, 4t^2, 4log(t))
    r'(t)=(8, 8t, 4/(ln(10)t))
    |r' (t)|=[tex]\sqrt{8^2+(8t)^2+(1.737177927/t)^2}[/tex]
    |r' (t)|=[tex]\sqrt{64+64t^2+3.01778715219/t^2}[/tex]

    At Point (8,4,0) t=1 and at Point (24, 36, 4log(3)) t=3
    Therefore the integral is from 1 to 3

    from here, the integral of [tex]\int[/tex][tex]\sqrt{64+64t^2+3.01778715219/t^2}[/tex]
    is to complex to do by hand, so with matlab and TI-89 I am getting an answer of 36.106527, which according to the assignment is wrong. Am I going about this problem wrong? what should I do?
    Last edited: Jul 13, 2009
  2. jcsd
  3. Jul 13, 2009 #2


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    Staff Emeritus
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    Gold Member

    Are you sure that by log(t) it's not referring to base e logarithm? Some books use log and ln as the same thing. Also, you have a 1/x when you should have a 1/t in the third coordinate
  4. Jul 13, 2009 #3
    I am sure it is log(t) and not ln(t). That would make it a bit easier... Thanks for the variable mistake catch... I fixed it...
  5. Jul 13, 2009 #4


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    Homework Helper

    I get 36.10652942031572. I don't think you are doing anything fundamentally wrong. Why are you sure it's not ln(x)?
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