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Math Help - Arc length - please check my work

  1. #1
    Senior Member mollymcf2009's Avatar
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    Arc length - please check my work

    I think I did this correctly, but Webassign does not like my answer. Can someone please check my work? Thanks!! Molly

    Find the arc length for:

    y = ln(1-x^2) on 0 \leq x \leq \frac{1}{8}

    L = \int\limits^{\frac{1}{8}}_{0} \sqrt{1 + \frac{4x^2}{x^4 - 2x^2 + 1}}dx

    = \int\limits^{\frac{1}{8}}_{0} \sqrt{\frac{x^4 + 2x^2 + 1}{x^4 - 2x^2 + 1}}

    = \int\limits^{\frac{1}{8}}_{0} \sqrt{\frac{(x^2+1)(x^2-1)}{(x^2-1)(x^2-1)}}

    = \int\limits^{\frac{1}{8}}_{0} \frac{(x^2+1)}{(x^2-1)}

    <br />
= \int\limits^{\frac{1}{8}}_{0} 1 + \int\limits^{\frac{1}{8}}_{0} \frac{2}{x^2-1}

    <br />
= \int\limits^{\frac{1}{8}}_{0} 1 - \frac{1}{x+1} + \frac{1}{x-1}

    = \biggl[ x - ln |x+1| + ln|x-1| \biggr]^{\frac{1}{8}}_0

    * If it looks good up to here, I think I may be botching the evaluation:

    = \biggl[\frac{1}{8} - ln(\frac{1}{8} + 1) + ln(\frac{1}{8} - 1)\biggr]

    = \frac{1}{8} - ln\frac{9}{8} + ln\frac{7}{8}

    = \frac{1}{8} - ln\frac{63}{8}

    Am I combining those log correctly? I'm drawing a blank.
    Thanks!! Molly
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  2. #2
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    Yeah, that looks good except that ln(9/8) + ln(7/8) = ln(9/8 * 7/8) = ln(63/64)
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  3. #3
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    Quote Originally Posted by mollymcf2009 View Post
    I think I did this correctly, but Webassign does not like my answer. Can someone please check my work? Thanks!! Molly

    Find the arc length for:

    y = ln(1-x^2) on 0 \leq x \leq \frac{1}{8}

    L = \int\limits^{\frac{1}{8}}_{0} \sqrt{1 + \frac{4x^2}{x^4 - 2x^2 + 1}}dx

    = \int\limits^{\frac{1}{8}}_{0} \sqrt{\frac{x^4 + 2x^2 + 1}{x^4 - 2x^2 + 1}}

    = \int\limits^{\frac{1}{8}}_{0} \sqrt{\frac{(x^2+1)(x^2-1)}{(x^2-1)(x^2-1)}}

    = \int\limits^{\frac{1}{8}}_{0} \frac{(x^2+1)}{(x^2-1)}

    <br />
= \int\limits^{\frac{1}{8}}_{0} 1 + \int\limits^{\frac{1}{8}}_{0} \frac{2}{x^2-1}

    <br />
= \int\limits^{\frac{1}{8}}_{0} 1 - \frac{1}{x+1} + \frac{1}{x-1}

    = \biggl[ x - ln |x+1| + ln|x-1| \biggr]^{\frac{1}{8}}_0

    * If it looks good up to here, I think I may be botching the evaluation:

    = \biggl[\frac{1}{8} - ln(\frac{1}{8} + 1) + ln(\frac{1}{8} - 1)\biggr]

    = \frac{1}{8} - ln\frac{9}{8} + ln\frac{7}{8}

    = \frac{1}{8} - ln\frac{63}{8}

    Am I combining those log correctly? I'm drawing a blank.
    Thanks!! Molly
    no

    - \ln\frac{9}{8} + \ln\frac{7}{8}

    =\ln\frac{7 \cdot 8}{8 \cdot 9}

    =\ln\frac{7}{ 9}
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  4. #4
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by Zizoo View Post
    Yeah, that looks good except that ln(9/8) + ln(7/8) = ln(9/8 * 7/8) = ln(63/64)

    \frac{1}{8} - ln(\frac{63}{64}) is still not correct in webassign.

    I think I just have a negative missing or something. Any ideas?

    Quote Originally Posted by GaloisTheory1 View Post
    no

    - \ln\frac{9}{8} + \ln\frac{7}{8}

    =\ln\frac{7 \cdot 8}{8 \cdot 9}

    =\ln\frac{7}{ 9}
    Ok, so because of the negative in front of the first one, I divide. Got it! THanks!

    Well, I thought I had it.

    \frac{1}{8} + ln(\frac{7}{9}) is not right in webassign. I don't get it.
    Last edited by Krizalid; March 9th 2009 at 06:14 PM.
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  5. #5
    Member Mentia's Avatar
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    That looks correct. Perhaps try less simplified:

    \frac{1}{8}+ln(7)-ln(9)

    or the strange:
    <br />
ln(\frac{7e^{\frac{1}{8}}}{9})
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