# Arc length - please check my work

• Mar 9th 2009, 05:33 PM
mollymcf2009
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:

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

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

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

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

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

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

$\displaystyle = \int\limits^{\frac{1}{8}}_{0} 1 - \frac{1}{x+1} + \frac{1}{x-1}$

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

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

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

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

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

Am I combining those log correctly? I'm drawing a blank.
Thanks!! Molly
• Mar 9th 2009, 05:36 PM
Zizoo
Yeah, that looks good except that ln(9/8) + ln(7/8) = ln(9/8 * 7/8) = ln(63/64)
• Mar 9th 2009, 05:47 PM
GaloisTheory1
Quote:

Originally Posted by mollymcf2009
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:

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

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

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

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

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

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

$\displaystyle = \int\limits^{\frac{1}{8}}_{0} 1 - \frac{1}{x+1} + \frac{1}{x-1}$

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

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

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

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

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

Am I combining those log correctly? I'm drawing a blank.
Thanks!! Molly

no

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

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

$\displaystyle =\ln\frac{7}{ 9}$
• Mar 9th 2009, 06:01 PM
mollymcf2009
Quote:

Originally Posted by Zizoo
Yeah, that looks good except that ln(9/8) + ln(7/8) = ln(9/8 * 7/8) = ln(63/64)

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

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

Quote:

Originally Posted by GaloisTheory1
no

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

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

$\displaystyle =\ln\frac{7}{ 9}$

Ok, so because of the negative in front of the first one, I divide. Got it! THanks!

Well, I thought I had it.

$\displaystyle \frac{1}{8} + ln(\frac{7}{9})$ is not right in webassign. I don't get it.
• Mar 9th 2009, 06:57 PM
Mentia
That looks correct. Perhaps try less simplified:

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

or the strange:
$\displaystyle ln(\frac{7e^{\frac{1}{8}}}{9})$