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Arc length of vector function curve

  1. May 20, 2010 #1
    1. The problem statement, all variables and given/known data
    1. Find the length of the curve from t=0 to t=1.
    r(t) = <2t, t^2, (1/3)t^3>

    2. Reparametrize the curve with respect to arc length measured from the point where t=0 in the direction of increasing t.
    r(t) = <e^(2t)cos2t, 2, e^(2t)sin2t>


    2. Relevant equations
    [tex] S = \int{r'(t)} dt [/tex]


    3. The attempt at a solution
    1. I take the definite integral of the absolute value of r'(t).

    r'(t) = <2, 2t, t^2>
    S = integral[sqrt(4 + 4t^2 + t^4)dt] from [0,1]
    This is a pretty complicated integral to solve. It's not that I can't solve it but I think I'm doing something wrong because this is only the second practice question of the section so I don't think it should be that complicated. Our prof has only show us the case for a helix, so the sine and cosine simplify to some number with no "t" terms in the integrand.

    2. Basically, the same problem as number 1. I need to use the arc length function to find s in term of t. But I'm having trouble evaluating the integral for the arc length.
    r'(t) becomes a pretty complicated function by itself but when you square the components, square root them and then take the integral of that from [0, t], it becomes very messy. I don't think I can solve this one. So again, I'm thinking I'm doing something wrong conceptually.
     
    Last edited: May 20, 2010
  2. jcsd
  3. May 20, 2010 #2

    lanedance

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    note that (t^2+2)^2 = ...
     
  4. May 20, 2010 #3

    lanedance

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    for the 2nd, consider some trig inequalities to simpilfy
     
  5. May 20, 2010 #4

    lanedance

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    note just to clarify, though its looks like you're already doing it, but you need the absolute value to show you're intergating a scalar function:
    [tex] s = \int |r'(t)|dt [/tex]
     
  6. May 20, 2010 #5
    Alright, I guess I was too tired. I didn't realize the obvious factor for the 1st question. And the 2nd question isn't hard to integrate at all.

    Thanks.
     
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