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Math Help - Arc length query

  1. #1
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    Exclamation Arc length query

    I have been trying to find the archlength of y^2=4x, where 0 ≤ x ≤ 2.

    I first solved y=\sqrt{4x}.

    Then I solved for y'.
    y'=\frac{4x^\frac{-1}{2}}{2}*4=2(4x)^\frac{-1}{2}.

    The formula I have to find arclength is \int_a^b \sqrt{1+(y')^2} dx.

    So I plugged in and got this:

    \int_0^2 \sqrt{1+(2(4x)^\frac{-1}{2})^2}=\int_0^2 \sqrt{1+(4(4x)^{-1})}=\int_0^2 \sqrt{1+\frac{4}{4x}}=\int_0^2 \sqrt{1+\frac{1}{x}}.

    I evaluated \frac{1}{x}=(\frac{1}{\sqrt{x}})^2, because my book offers a formula for the form \sqrt{a^2+u^2}.

    The formula was

    My a=1 and my u=\frac{1}{\sqrt{x}}, so I plugged those into the formula.

    What I got out was this:

    \int_0^2 \sqrt{1^2+\frac{1}{\sqrt{x}}^2} du=\frac{\frac{1}{\sqrt{x}}}{2}\sqrt{1^2+\frac{1}{  \sqrt{x}}^2}+\frac{1^2}{2}\ln(\frac{1}{\sqrt{x}}+\  sqrt{1^2+\frac{1}{\sqrt{x}}^2})=
    \frac{1}{2\sqrt{x}}\sqrt{1+\frac{1}{\sqrt{x}}^2}+\  frac{1}{2}\ln(\frac{1}{\sqrt{x}}+\sqrt{1+(\frac{1}  {\sqrt{x}})^2})

    This is what I got to. Now, I need to evaluate it from 0 to 2. But when I plug in 2 for the value of x, my answer is undefined. Did I do something wrong? Is there a different way to do this? Please...PLEASE help!
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  2. #2
    Senior Member
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    Alternatively, you can set the independent variable to y and integrate from 0 to 2\sqrt{2}.

    Edit: You'd multiply it by 2 to get both sides' worth.
    Last edited by Scott H; October 17th 2009 at 07:42 PM.
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  3. #3
    Senior Member apcalculus's Avatar
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    http://www.wolframalpha.com/input/?i...1%2B(1/x))^0.5

    click on 'Show Steps'

    I hope it helps.
    Last edited by mr fantastic; October 17th 2009 at 03:15 PM. Reason: Added url tags
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