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Homework Help: Arc Length of e^5x from 0 to ln(4)

  1. Feb 22, 2010 #1
    1. The problem statement, all variables and given/known data


    Note... I used wolfram alpha to get the answer, I did not get it myself... So I still need help. The answer shown is correct, so you'll know if you got it.

    2. Relevant equations

    Integral [0, ln(4)] sqrt(1+(dy/dx)^2)

    3. The attempt at a solution

    2pi Integral [0, ln(4)] y*sqrt(1+(dy/dx)^2)

    2pi Integral [0, ln(4)] (e^5x)*sqrt(1+5e^5x^2)dx

    u = 5e^5x
    du = 25e^5x dx
    dx = du/25e^5x

    2pi Integral [0, ln(4)] (e^5x)*sqrt(1+u^2)du/25e^5x

    2pi Integral [0, ln(4)] sqrt(1+u^2)du

    u = tan(t)

    2pi/25 Integral [0, ln(4)] sqrt(1+tan^2(t))du

    2pi/25 Integral [0, ln(4)] sqrt(sec^2(t))du

    2pi/25 Integral [0, ln(4)] sec(t)du

    du = sec^2(t)dt
    dt = du*cos^2(t)

    2pi/25 Integral [0, ln(4)] cos^2(t)/cos(t)dt

    2pi/25 Integral [0, ln(4)] cos(t)dt

    Edit bounds...

    [arctan(5), arctan(5e^(5*ln(4)))]

    Then get ****ed over with an answer of .0048...

    What did I do wrong.
    Last edited: Feb 22, 2010
  2. jcsd
  3. Feb 22, 2010 #2
    the question asks the surface area and not the arc length. Perhaps this is your mistake
  4. Feb 23, 2010 #3
    I have never heard of wolfram alpha before...very interesting. What did you type in to get the answer to your question? I can't figure it out...
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