Question

Find the length of the curve y= (e^x+1/e^x-1)

Answer 1

this problems are always cooked up to be easy
but you do need an interval right?

Answer 2

also we need to know if this is
\[\frac{e^x+1}{e^x-1}\] which is what i suspect, or something else

Answer 3

yeah the interval is a<x<b, a>0

Answer 4

and oops i forgot, there is an ln in front of it

Answer 5

do me a favour and write it with the equation editor
that will help a great deal

Answer 6

uh i dont think i can use it through my tablet, sorry

Answer 7

ok let me try
\[\ln\left(\frac{e^x+1}{e^x-1}\right)\] how does that look?

Answer 8

yes thats exactly how it is :)

Answer 9

ok
since i can't compute for beans, lets cheat and find the derivative

Answer 10

oh actually it is not that hard if we rewrite it as
\[\ln(e^x+1)-\ln(e^x-1)\] you get
\[\frac{e^x}{e^x+1}-\frac{e^x}{e^x-1}\] pretty fast, but then you have to actually do the subtraction

Answer 11

so lets cheat and say it is
\[-\frac{e^{2x}}{e^{2x}-1}\]

Answer 12

hmmm but when you take the derivative dont you just get (1/(e^x+1/e^x-1))?

Answer 13

oh no!

Answer 14

you would have to use the chain rule, and it would suck big time, because one you flipped it and got
\[\frac{e^x-1}{e^x+1}\] you would still have to multiply by the derivative of \(\frac{e^x+1}{e^x-1}\) which would require the quotient rule

Answer 15

i.e. it is true that
\[\frac{d}{dx}\ln(x)=\frac{1}{x}\] but
\[\frac{d}{dx}[f(x)]=\frac{f'(x)}{f(x)}\]
you really don't want to do that
always best to "simplify" the log first then take the derivative
clear? (i hope)

Answer 16

@satellite73 You made a mistake in simplification

It is -2*e^x/(e^2x - 1)

Answer 17

course we are still not done
now we have to square that sucker, which is not too bad, it is \[\frac{4e^{2x}}{(e^{2x}-1)^2}\]

Answer 18

@bhaskarbabu so i did
good catch
but the square is right

Answer 19

then we have to add 1 to that

Answer 20

when i said it was easy, i meant they are cooked up so that the integral is easy
the algebra is a bear

Answer 21

yeah the square is right @satellite73

Answer 22

lol thanks god
now we have to add 1 and i bet $4 it is a perfect square when we do it (i hope any ways)

Answer 23

yup i win the $4

Answer 24

http://www.wolframalpha.com/input/?i=1%2B%284+e^%282+x%29%29%2F%28e^%282+x%29-1%29^2

Answer 25

think of all the work some poor underpaid graduate student did to get this problem cooked up so that when you take the derivative
square it
add 1
you get a perfect square! sheer genius

Answer 26

think we lost @TheAni-Trix

Answer 27

lol kinda

Answer 28

oh no, you are back!
ok once all the algebra is done you end up with a perfect square
in fact, wolfram tells me it is \(\cosh^2(x)\) althoug i never would have recognized that

Answer 29

back up like 3 steps

Answer 30

ok we can back up to wherever you like

Answer 31

which step?

Answer 32

to the simplification of the e^x/e^x+1-e^X/e^x-1

Answer 33

ok

Answer 34

spoiler alert, i used wolfram
but we can do it without it
it is a matter of subtracting
\[\frac{e^x}{e^x+1}-\frac{e^x}{e^x-1}=\frac{e^x(e^x-1)-e^x(e^x+1)}{(e^x+1)(e^x-1)}\]

Answer 35

the numerator "simplifies" to \(-2e^x\) and the denominator multiplies out to \(e^{2x}-1\) giving
\[\frac{-2e^x}{e^{2x}-1}\]

Answer 36

hows that look?

Answer 37

how is it -2e^x?

Answer 38

\[e^x(e^x-1)-e^x(e^x+1)\\
=e^{2x}-e^x-e^{2x}-e^x=-2e^x\]

Answer 39

ok with that?

Answer 40

okay yeah i got it :) whats next?

Answer 41

square it
that part is easy no wolfram needed
you get \[\left(\frac{dy}{dx}\right)^2=\frac{4e^{2x}}{(e^{2x}-1)^2}\]

Answer 42

ok for the next step?

Answer 43

hope it is clear what we are aiming for
\[\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx\]

Answer 44

we got \[\left(\frac{dy}{dx}\right)^2=\frac{4e^{2x}}{(e^{2x}-1)^2}\] now we need
\[1+\left(\frac{dy}{dx}\right)^2=1+\frac{4e^{2x}}{(e^{2x}-1)^2}\]

Answer 45

@satellite73 You are awesome!! I can solve this but I cannot explain like you. You are doing agreat job!! keep it up!! :)

Answer 46

which is
\[\frac{(2e^{2x}-1)^2+4e^{2x}}{(e^{2x}-1)^2}\]

Answer 47

@bhaskarbabu thanks!

Answer 48

ready for more algebra?

Answer 49

lol yes so far im understanding

Answer 50

lol you're back!

Answer 51

ok when you multiply all that stuff out in the numerator, you are going to get
\[e^{4x}+2e^{2x}+1\] which is luckily a perfect square i.e. it is
\[(e^{2x}+1)^2\]

Answer 52

so now
\[1+(\frac{dy}{dx})^2=\frac{(e^{2x}+1)^2}{(e^{2x}-1)^2}\]

Answer 53

that makes \[\sqrt{1+(\frac{dy}{dx})^2}=\frac{e^{2x}+1}{e^{2x}-1}\] as the square root of the square makes it go away

Answer 54

see
cooked up to work out nice

Answer 55

okay give me a sec to look over all this and see if i understand it all

Answer 56

okay give me a sec to look over all this and see if i understand it all

Answer 57

take your time

Answer 58

almost all of it is algebra, fraction algebra

Answer 59

okay i got it up to the last point

Answer 60

are you good to here \[1+(\frac{dy}{dx})^2=\frac{(e^{2x}+1)^2}{(e^{2x}-1)^2}\]?

Answer 61

yeah lol i understood all of tha

Answer 62

and then take the square root and get
\[\int\sqrt{1+(\frac{d}{dx})^2}=\int \frac{e^{2x}+1}{e^{2x}-1}dx\]

Answer 63

there are lots of ways to do that integral
i will let you do it unless you still need help, but it can't be that hard after all that work

Answer 64

okay yeah i got that too

Answer 65

then you are done right?

Answer 66

um yeah i got how to do it but can you give me the answer so i can see if i did

Answer 67

it right

Answer 68

i guess there are several ways, but partial fractions is probably best

Answer 69

actually i would do this
\[\int \frac{e^{2x}+1}{e^{2x}-1}dx=\int \frac{e^{2x}-1+2}{e^{2x}-1}dx=\int 1dx+\int\frac{2}{e^{2x}-1}dx\]

Answer 70

a neat trick easier algebra than division

Answer 71

hmmm i see what you did but i dont think i could come up with that on my own

Answer 72

yeah it is a nice trick to know
but maybe it doesn't really help that much in this case
of course you do the mental u-sub and think of it as
\[\frac{1}{2}\int \frac{e^u+1}{e^u-1}du\]

Answer 73

actually it is easier than i am making out
break in to two pieces
\[\int \frac{e^u}{e^u-1}du+\int \frac{1}{e^u-1}du\] then the first one is obviously
\[\ln(e^u-1)\] by substitution

Answer 74

the second one you can make \(z=e^u-1, z+1=e^u, \ln(z+1)=u, \frac{1}{z+1}dz=du\) and solve
\[\int \frac{1}{(z+1)z}dz\] using partial fractions

Answer 75

actually you get the same answer using either method
you decide which is easier

Answer 76

lol yeah im more used to using the other way with u