Question

**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.

Answer 1

what seems to be the problem?

Answer 2

I assume you know the formula

\[\int\limits_{a}^{b}\sqrt{1+(f'(x))^2}dx\]

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

no...look at the formula again.

Answer 8

Ahh forgot ^2

Answer 9

correct

Answer 10

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

Answer 11

no

Answer 12

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

Answer 13

you lost an x

Answer 14

ohhh the ^2 wouldn't allow it.

Answer 15

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

\[=\frac{1}{8}\left(\frac{x^2-16}{x}\right)\]

\[=\left(\frac{x^2-16}{8x}\right)\]

then 1+square it

\[=1+\left(\frac{x^2-16}{8x}\right)^2\]

Answer 16

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

Answer 17

oh common denominator

Answer 18

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

Answer 19

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

Answer 20

make into one fraction

Answer 21

with the one?

Answer 22

yes

Answer 23

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

Answer 24

why the minus sign

Answer 25

+ I mean.. sorry

Answer 26

multiply out the numerator...then factor

Answer 27

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

Answer 28

yes...now factor the numerator

Answer 29

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

Answer 30

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

Answer 31

the sqrt would cancel the ^2

Answer 32

yep

Answer 33

so (x+16)/64

Answer 34

you lost an x again.

Answer 35

also it is not 64 anymore

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

2.5

Answer 41

no

Answer 42

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

Answer 43

I keep getting a negative.

Answer 44

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

Answer 45

Oh wow, I need to integrate..

Answer 46

:)

Answer 47

Sorry, you probably think I am absolutely dumb.

Answer 48

no

Answer 49

1.33333

Answer 50

no

Answer 51

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

Answer 52

\[\int\left(\frac{x}{8}+\frac{2}{x}\right)dx\]

\[=\frac{x^2}{16}+2\ln|x|+c\]

Answer 53

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

Answer 54

4.19721 with a little confidence.

Answer 55

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

Answer 56

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

Answer 57

;)

Answer 58

Would you care to help me solve another?

Answer 59

I'll start a new problem of course.

Answer 60

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

Answer 61

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

Answer 62

I'm a Math professor

Answer 63

Cool, where at?

Answer 64

so yes...I was a math major

Answer 65

I don't give out that info :)

Answer 66

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

Answer 67

I'm not in the SEC

Answer 68

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

Answer 69

good luck

Answer 70

Thank you. Have a great night.