anonymous
  • anonymous
**Calc 2 Help Needed** Find the length of the arc formed by y = (1/8) * (0.5x^2 -16ln(x)) from x=2 to x=6.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Zarkon
  • Zarkon
what seems to be the problem?
Zarkon
  • Zarkon
I assume you know the formula \[\int\limits_{a}^{b}\sqrt{1+(f'(x))^2}dx\]
anonymous
  • anonymous
oh yes.. I do.. I can't figure out the derivative.

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More answers

anonymous
  • anonymous
of y = (1/8) * (0.5x^2 -16ln(x))
Zarkon
  • Zarkon
\[\frac{dy}{dx}=\frac{1}{8}\left(x-\frac{16}{x}\right)\]
anonymous
  • anonymous
Ok so now it is sqrt(1+ 1/8 * (x-16/x)
Zarkon
  • Zarkon
no...look at the formula again.
anonymous
  • anonymous
Ahh forgot ^2
Zarkon
  • Zarkon
correct
anonymous
  • anonymous
so 1+ ((x-16/8)^2
Zarkon
  • Zarkon
no
anonymous
  • anonymous
I can't bring the 1/8 in? Or shouldn't rather.
Zarkon
  • Zarkon
you lost an x
anonymous
  • anonymous
ohhh the ^2 wouldn't allow it.
Zarkon
  • Zarkon
\[\frac{1}{8}\left(x-\frac{16}{x}\right)\] \[=\frac{1}{8}\left(\frac{x^2-16}{x}\right)\] \[=\left(\frac{x^2-16}{8x}\right)\] then 1+square it \[=1+\left(\frac{x^2-16}{8x}\right)^2\]
anonymous
  • anonymous
so how did the top x-16 go to x^2-16?
anonymous
  • anonymous
oh common denominator
Zarkon
  • Zarkon
yes...\[x=\frac{x^2}{x}\] for x not equal to zero
anonymous
  • anonymous
ok so after squaring I have 1+ ((x^2 - 16)^2/64x^2.. Well wolfram did.
Zarkon
  • Zarkon
make into one fraction
anonymous
  • anonymous
with the one?
Zarkon
  • Zarkon
yes
anonymous
  • anonymous
ok so now I have 64x^2 - (x^2 - 16)^2 all over 64x^2
Zarkon
  • Zarkon
why the minus sign
anonymous
  • anonymous
+ I mean.. sorry
Zarkon
  • Zarkon
multiply out the numerator...then factor
anonymous
  • anonymous
so now:( x^4 + 32x^2 + 256) / 64x^2
Zarkon
  • Zarkon
yes...now factor the numerator
anonymous
  • anonymous
the top would factor to (x^2 + 16)^2
Zarkon
  • Zarkon
yes...now take the square root of the whole thing
anonymous
  • anonymous
the sqrt would cancel the ^2
Zarkon
  • Zarkon
yep
anonymous
  • anonymous
so (x+16)/64
Zarkon
  • Zarkon
you lost an x again.
Zarkon
  • Zarkon
also it is not 64 anymore
anonymous
  • anonymous
Whoops.. (x+16)/8x?
Zarkon
  • Zarkon
close \[\frac{x^2+16}{8x}\]
anonymous
  • anonymous
Ohhh that one.. I am the king of mistakes tonight.
Zarkon
  • Zarkon
write as \[\frac{x^2+16}{8x}=\frac{x^2}{8x}+\frac{16}{8x}=\frac{x}{8}+\frac{2}{x}\] and integrate from 2 to 6
anonymous
  • anonymous
2.5
Zarkon
  • Zarkon
no
anonymous
  • anonymous
OMG why did I do that.. ok let me try again.
anonymous
  • anonymous
I keep getting a negative.
Zarkon
  • Zarkon
what did you get for the integral before evaluating the limits?
anonymous
  • anonymous
Oh wow, I need to integrate..
Zarkon
  • Zarkon
:)
anonymous
  • anonymous
Sorry, you probably think I am absolutely dumb.
Zarkon
  • Zarkon
no
anonymous
  • anonymous
1.33333
Zarkon
  • Zarkon
no
anonymous
  • anonymous
After integrating I had x^2/16 + 4/2x
Zarkon
  • Zarkon
\[\int\left(\frac{x}{8}+\frac{2}{x}\right)dx\] \[=\frac{x^2}{16}+2\ln|x|+c\]
anonymous
  • anonymous
ok after wolfram it says the second part should contain a log
anonymous
  • anonymous
4.19721 with a little confidence.
Zarkon
  • Zarkon
yes...which equals \[2(\ln(3)+1)\]
anonymous
  • anonymous
YESSS!!!!!! I can't believe that simple problem turned so nasty!
Zarkon
  • Zarkon
;)
anonymous
  • anonymous
Would you care to help me solve another?
anonymous
  • anonymous
I'll start a new problem of course.
Zarkon
  • Zarkon
hmmm...don't know if I have that much time. post it in a new thread...someone will help...If I have time I'll look
anonymous
  • anonymous
Okay thanks so much.. Were you by any chance a math major?
Zarkon
  • Zarkon
I'm a Math professor
anonymous
  • anonymous
Cool, where at?
Zarkon
  • Zarkon
so yes...I was a math major
Zarkon
  • Zarkon
I don't give out that info :)
anonymous
  • anonymous
Ahhh.. haha, I understand. Well I'm a student at a SEC school. Hopefully you aren't my teacher.
Zarkon
  • Zarkon
I'm not in the SEC
anonymous
  • anonymous
Haha okay.. Well thanks again for the help. This stuff is hard to understand in a lecture hall.
Zarkon
  • Zarkon
good luck
anonymous
  • anonymous
Thank you. Have a great night.

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