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Homework Help: Arc length of vector-valued function; am I starting right?

  1. Mar 14, 2008 #1
    1. The problem statement, all variables and given/known data
    Given: R(t)= <(1/2)t^2, (4/3)t^(3/2), 2*sqrt(3)t>
    Find: Arc length function s(t) where t_0 =0

    2. Relevant equations
    Is this the correct formula?
    ∫[0,t] sqrt( derivative^2 + derivative^2 +derivative^2) dt
    ∫[0,t] sqrt(t^2 + 4t + 12) dt

    3. The attempt at a solution

    ∫[0,t] sqrt(t^2 + 4t + 12) dt

    I am not looking for anyone to do my homework, but before I start a messy integral I just want to know if:
    A) I am starting correctly, and
    B) if their is an easier way to do this (or strategy)

    Last edited: Mar 14, 2008
  2. jcsd
  3. Mar 14, 2008 #2

    Your work is correct so far! Keep going ... what's your next step?
    Last edited: Mar 14, 2008
  4. Mar 14, 2008 #3
    I apreciate your encouraging tone rocophysics; I was about to quit for the night.

    ∫[0,t] sqrt(t^2 + 4t + 12) dt
    ∫[0,t] sqrt((t+2)^2 + 8) dt
    u=t+2 du=1
    ∫[0,t] sqrt((u)^2 + 8) du

    then maybe trigonometric substitution.
    Last edited: Mar 14, 2008
  5. Mar 14, 2008 #4


    Go ahead and do trig sub after this step ...

    [tex]t+2=\sqrt 8\tan\theta[/tex]
    [tex]dt=\sqrt 9\sec^{2}\theta d\theta[/tex]

    But don't forget to change your limits ...
    Last edited: Mar 14, 2008
  6. Mar 15, 2008 #5
    I tried but I know there is at least one mistake somewhere; here's my attempt (each line is numbered 1-14):
    http://www.sudokupuzzles.net/IMG_0030.jpg [Broken]
    Last edited by a moderator: May 3, 2017
  7. Mar 15, 2008 #6
    You forgot about your constant of 8 (line 10), and it should be to both your Integrals, not just one.
    Last edited: Mar 15, 2008
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