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Arc Length

  1. Sep 3, 2016 #1
    1. The problem statement, all variables and given/known data

    Consider the path r(t) = <10t,5t2,5ln(t) defined for t >0. Find the length of the curve between (10,5,0) and (20,20,5ln(2))

    2. Relevant equations

    L= ∫ab |r'(t)|dt

    3. The attempt at a solution
    r'(t) = <10, 10t, 5/t>

    t values are 1 and 2 based on the x values for the points listed

    √(102 + (10t)2 + (5/t)2)
    √(100 + 100t2 + 25/t2)
    √(25) ⋅ √(4 + 4t2 + 1/t2)
    5√(4 + 4t2 + 1/t2)


    This is where I am getting stuck. Is there an additional step that is missing?
     
  2. jcsd
  3. Sep 3, 2016 #2

    Ray Vickson

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    Yes. You need to figure out the initial ##t_1## final value ##t_2## of ##t##, then do the integral
    $$\text{Length} = \int_{t_1}^{t_2} 5 \sqrt{ \displaystyle 4 + 4 t^2 + \frac{1}{t^2}} \, dt $$
     
  4. Sep 3, 2016 #3

    LCKurtz

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    Hint: If you re-arrange the expression under the square root, you may recognize it as a perfect square.
     
  5. Sep 3, 2016 #4
    Based on the two points listed, I found that t1 = 1 and t2=2. So next I would then substitute in the t values and solve? I attempted that previously, but I did not get the correct answer.

    $$\text{Length} = \int_{1}^{2} 5 \sqrt{ \displaystyle 4 + 4 (2^2 -1^2) + \frac{1}{2^2-1^2}} \, dt $$
     
  6. Sep 3, 2016 #5

    Ray Vickson

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    If ##f(t) = 5 \sqrt{4 + 4t^2 + 1/t^2}##, then ##\int_{t_1}^{t_2} f(t) \, dt = F(t_2) - F(t_1)##, where ##F(t) = \int f(t) \, dt## is the indefinite integral of ##f(t)##. This is not anything like what you wrote!
     
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