Question

Find the arc length of y=2sqrt(x), with its starting point at (0,0)

Answer 1

No arclength... unless you know where it ends... it has to end somewhere, right?

Answer 2

It's just an integration from 0 to x.

I'm stuck at the part of getting the integration done.

\[\int\limits_{0}^{x} \sqrt(1+1/x) dx \]

Answer 3

Maybe we shouldn't use x in the integrand...
\[\Large \int\limits_0^x\sqrt{1+\frac1t}dt\]

Answer 4

Anyway, currently, I'm stumped too... try integration by parts?

Answer 5

I'm thinking trig sub\[x=\cot^2\theta\]

Answer 6

Well, let's see where that goes :>

Answer 7

i get
\[t+1-1/2t^2lnt+1/4t^2\] from 0 to x with integration by parts,

but I think it's nowhere close to the answer I have, which is

\[\sqrt{x+x^2}+\ln(\sqrt(x)+\sqrt(1+x))\]

Answer 8

oh nvm ive omitted the square root during my process, so i guess i'm out of luck :(

Answer 9

trig sub worked out for me I think...\[x=\cot^2\theta\implies dx=-2\cot\theta\csc^2\theta d\theta=-2{\cos\theta\over\sin^3\theta}d\theta\]

Answer 10

\[1+\cot^2\theta=\csc^2\]so our integral becomes...
let me know if you see it from here

Answer 11

\[\int\csc\theta\left(-2{\cos\theta\over\sin^3\theta}\right)d\theta=\cdots\]

Answer 12

oh I messed up, it should have been tan in the radical

Answer 13

yup i felt omething was weird too :)

Answer 14

\[x=\cot^2\theta\implies dx=-2\cot\theta\csc^2\theta d\theta=-2{\cos\theta\over\sin^3\theta}d\theta\]\[\int\sqrt{1+\tan^2\theta}\left(-2{\cos\theta\over\sin^2\theta}\right)d\theta=\int\sec\theta\left(-2{\cos\theta\over\sin^2\theta}\right)d\theta\]

Answer 15

I think that's better...

Answer 16

\[=-2\int\csc^2\theta d\theta\]even easier

Answer 17

so the result of the integration would be then -2cot(theta),
but since we have x=cot^2 theta, how should we convert this back in terms of x?

Answer 18

gotta draw the triangle

Answer 19

i get sqrt(x)?

Answer 20

I dunno, I'm just taking it as it comes... I didn't work out the final answer or anything

Answer 21

I have feeling there was a way better way to do this :P

Answer 22

hmm.. i guess we have usd other method or something since the derived solution looks far off from the actual solution I have, but thanks a bunch for helping me out :) i guess Ima leave this question for now :P

Answer 23

dang, sorry, it's been a while since I've done this. may I ask what the actual solution is?

Answer 24

oh I see you wrote it above

Answer 25

its all good :) i will tackle this later on i guess :)

Answer 26

how about\[u=1+\frac1x\implies du=-\frac1{x^2}dx\]\[dx=-x^2du=-\frac1{(u-1)^2}du\]

Answer 27

I would use \[u=\sqrt{1+\frac{1}{t}}\]

but that's just me :)

Answer 28

Well @grit0614 that's you're answer, trust me.
I'll try it out on my own too, good luck!

Answer 29

using my substitution you will get to a point where you will need to use partial fractions.

Answer 30

\[\int\limits_{0}^{x} \ -sqrt(u)*(1/(u-1)^2) du \]

Wow this is getting nasty :) appreciate it. I will give it a try myself as well :)

Answer 31

1) try my substitution
2) remember that your limits of integration change with u-sub

Answer 32

I will try partial fraction as well :) thanks for all your help