Question

find the length: f(x) = x^2-1/8 ln x, 1 11 years ago

Answer 1

take the derivative
square it
then add one
those are the first three steps, mostly algebra
usually the problem is cooked up to be easy

Answer 2

is it
\[f(x)=\frac{x^2-1}{8}\]?

Answer 3

no its f(x) = x^2 - (1/8) ln x

Answer 4

oh ok

Answer 5

then the derivative is \(2x-\frac{1}{8x}\) i guess right?

Answer 6

yeah

Answer 7

then square it

Answer 8

4x^2+1/64x^2-1/2

Answer 9

that is the annoying algebra part
ok then add one

Answer 10

it would be +1/2 instead of -1/2

Answer 11

then write as a single fraction

Answer 12

uhhhhh

Answer 13

lol

Answer 14

gotta add
no way around it

Answer 15

4x^2 + 1/64x^2 +1/2 ?!

Answer 16

the denominator will be \(64x^2\) the only denominator you see
the numerator will be
\[4x^2\times 62x^2+32x^2+1\]

Answer 17

uhhhhhhhhhhh that blew my mind lol

Answer 18

nay it is how you add is all
what should blow your mind is the next step

Answer 19

uhh whats the next step O_o

Answer 20

the miracle that the numerator is actually a perfect square
you have to admire the work that when it to this problem
the numerator is ....
\[(16x^2+1)^2\]

Answer 21

btw the algebra to add is not that bad at all, i am sure you know how to add
to recognize it as a perfect square in the ingeneos part

Answer 22

yeah i get it till here now

Answer 23

however you spell it
and it has to be a perfect square because the next step it to take the square root!!

Answer 24

then multiply by 2 pi x

Answer 25

your final job is to take the square root, and integrate
the square root of a square is easy enough
you get
\[\int_a^b\frac{16x^2+1}{8x}dx\]

Answer 26

nah it is the arc length right?

Answer 27

yes

Answer 28

oh yeah i thought it was surface area sorry

Answer 29

thank god
it is
\[\int_a^b\sqrt{1+(f')^2}dx\]

Answer 30

imagine all the work that went in to making
\[1+(f')^2\] a perfect square
i tried it once and it took the better part of an hour

Answer 31

oh damnnnnn. i just hope he don't put any on the exam

Answer 32

tomorrow

Answer 33

maybe a real easy one
or just "set up but don't evaluate"
anyway i hope the steps are clear
the integral is now routine
good luck on your exam

Answer 34

so i take the integral of that the one you sent

Answer 35

thank you so much for the help.

Answer 36

yeah that is it
if forgot the limits

Answer 37

break it in to two parts
the first is easy, the second is the log, that is why one limit had an \(e\) in it i guess

Answer 38

yw

Answer 39

take 8x out ?

Answer 40

?

Answer 41

oh sorry i was thinking something else lol

Answer 42

\[\frac{16x^2+1}{8x}=2x+\frac{1}{8x}\]

Answer 43

take 8x as u

Answer 44

integral is therefore
\[x^2+\frac{1}{8}\ln(x)\]

Answer 45

you are thinking way way too hard

Answer 46

oh its already in dx form

Answer 47

so the answer is the one you gave?

Answer 48

that is the integral, yes
convince yourself it was a real real easy one
then evaluate etc
good luck

Answer 49

now just plug the values?

Answer 50

1 and e and f(B)-f(A)

Answer 51

wait where did the 8x came from in the denominator

Answer 52

\(f(x) = x^{2}-(1/8) ln x,\;1<x<e\)
\(f'(x) = 2x - 1/(8x) = \dfrac{16x^2 - 1}{8x},\;1<x<e\)