**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.

- anonymous

- schrodinger

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- Zarkon

what seems to be the problem?

- Zarkon

I assume you know the formula
\[\int\limits_{a}^{b}\sqrt{1+(f'(x))^2}dx\]

- anonymous

oh yes.. I do.. I can't figure out the derivative.

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## More answers

- anonymous

of y = (1/8) * (0.5x^2 -16ln(x))

- Zarkon

\[\frac{dy}{dx}=\frac{1}{8}\left(x-\frac{16}{x}\right)\]

- anonymous

Ok so now it is sqrt(1+ 1/8 * (x-16/x)

- Zarkon

no...look at the formula again.

- anonymous

Ahh forgot ^2

- Zarkon

correct

- anonymous

so 1+ ((x-16/8)^2

- Zarkon

no

- anonymous

I can't bring the 1/8 in? Or shouldn't rather.

- Zarkon

you lost an x

- anonymous

ohhh the ^2 wouldn't allow it.

- 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

so how did the top x-16 go to x^2-16?

- anonymous

oh common denominator

- Zarkon

yes...\[x=\frac{x^2}{x}\] for x not equal to zero

- anonymous

ok so after squaring I have 1+ ((x^2 - 16)^2/64x^2.. Well wolfram did.

- Zarkon

make into one fraction

- anonymous

with the one?

- Zarkon

yes

- anonymous

ok so now I have 64x^2 - (x^2 - 16)^2 all over 64x^2

- Zarkon

why the minus sign

- anonymous

+ I mean.. sorry

- Zarkon

multiply out the numerator...then factor

- anonymous

so now:( x^4 + 32x^2 + 256) / 64x^2

- Zarkon

yes...now factor the numerator

- anonymous

the top would factor to (x^2 + 16)^2

- Zarkon

yes...now take the square root of the whole thing

- anonymous

the sqrt would cancel the ^2

- Zarkon

yep

- anonymous

so (x+16)/64

- Zarkon

you lost an x again.

- Zarkon

also it is not 64 anymore

- anonymous

Whoops.. (x+16)/8x?

- Zarkon

close \[\frac{x^2+16}{8x}\]

- anonymous

Ohhh that one.. I am the king of mistakes tonight.

- 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

2.5

- Zarkon

no

- anonymous

OMG why did I do that.. ok let me try again.

- anonymous

I keep getting a negative.

- Zarkon

what did you get for the integral before evaluating the limits?

- anonymous

Oh wow, I need to integrate..

- Zarkon

:)

- anonymous

Sorry, you probably think I am absolutely dumb.

- Zarkon

no

- anonymous

1.33333

- Zarkon

no

- anonymous

After integrating I had x^2/16 + 4/2x

- Zarkon

\[\int\left(\frac{x}{8}+\frac{2}{x}\right)dx\]
\[=\frac{x^2}{16}+2\ln|x|+c\]

- anonymous

ok after wolfram it says the second part should contain a log

- anonymous

4.19721 with a little confidence.

- Zarkon

yes...which equals \[2(\ln(3)+1)\]

- anonymous

YESSS!!!!!! I can't believe that simple problem turned so nasty!

- Zarkon

;)

- anonymous

Would you care to help me solve another?

- anonymous

I'll start a new problem of course.

- 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

Okay thanks so much.. Were you by any chance a math major?

- Zarkon

I'm a Math professor

- anonymous

Cool, where at?

- Zarkon

so yes...I was a math major

- Zarkon

I don't give out that info :)

- anonymous

Ahhh.. haha, I understand. Well I'm a student at a SEC school. Hopefully you aren't my teacher.

- Zarkon

I'm not in the SEC

- anonymous

Haha okay.. Well thanks again for the help. This stuff is hard to understand in a lecture hall.

- Zarkon

good luck

- anonymous

Thank you. Have a great night.

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