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Arc length problem.

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the arclength of the function [tex] y=x^2 [/tex] when x is between 0 and 10.

    2. Relevant equations
    Arclength here is [tex] \int_{0}^{10} \sqrt{1+(2x)^2} dx [/tex]
    (It's intended to be the integral from 0 to 10 of the quare root of 1+(2x)^2. My latex skills suck.)

    3. The attempt at a solution
    Trig substitution. 2x= tan (theta)

    Integrate secant theta
    Which is ln(sec(theta)+tan(theta))

    Substituting theta back in for arcsin(2x)
    ln(sec(arctan(2x))+tan(arctan(2x))) evaluated from 10 to 0. Solving for arctan(2x) my final answer is
    ln(sqrt(1+4x^2)+2x) evaluated from 0 to 10. I know this is wrong as it is much to short of an arc length but I don't know where I went wrong. Any clues?






    3. The attempt at a solution
     
    Last edited by a moderator: Feb 16, 2009
  2. jcsd
  3. Feb 16, 2009 #2

    Tom Mattson

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    I've edited your post to fix your TeX.

    That's fine.

    That's not fine. Why would you integrate [itex]\sec (\theta )[/itex]?
     
  4. Feb 16, 2009 #3
    Well, when I use trig substitution here I change 2x to tan(theta). This changes the integrand to sqrt(1+tan^2(theta)). The trig identity says that 1 + tangent squared equals secant squared, and since it is the quare root of that it just becomes secant.
     
  5. Feb 16, 2009 #4

    Tom Mattson

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    But you've forgotten about the [itex]dx[/itex].
     
  6. Feb 16, 2009 #5
    Oh, duh. Wait, why am I having a hard time remembering how to get the dx here? Does dx here equal 1/2*sec^2(theta) d(theta)?
     
  7. Feb 16, 2009 #6

    Tom Mattson

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    Yes, that's right.
     
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