Question

Find length of curve x = ln (t), y = sqrt(t+1) 1<=t<=2

Answer 1

first you differentiate y and x

Answer 2

and then put it in this for square root of (1+(dx?dy)^2)

Answer 3

and do the integral since you are given the boundries

Answer 4

good try but not quite

Answer 5

We're actually dealing with parametric curves at this stage, so I assume it's the sqrt of (dx/dt)^2 + (dy/dt)^2 dt.
I can't seem to simplify the algebra.

Answer 6

do you like algebra

Answer 7

Yes, for the most part. I suspect that I have done something wrong because the answer looks worse than the original problem.

Answer 8

you r right

Answer 9

i have 2 go kay kay

Answer 10

well it was the same whether use the equation but Kix001 equation is most accurate because it a parametraization

Answer 11

(dx/dt)^2=1/t^2
(dy/dt)^2=1/(t+1)
\[\int\limits_{1}^{2}{\sqrt{t^2+t+1\over t^2(t+1)}}dt\]seems to be where I'm stuck at...

Answer 12

\[\int_{1}^{2} \sqrt{((\log{t})\prime)^2+((\sqrt{t+1)}\prime)^{2}}\]
so
\[\int_{0}^{1} \sqrt{\frac{1}{t^2}+\frac{1}{(2\sqrt{t+1)^2}}}\]
\[\int_{0}^{1} \sqrt{\frac{1}{t^2}+\frac{1}{4t+1}}\]
\[\int_{0}^{1} \sqrt{\frac{t^2+4t+1}{4t^4+t^2}} =0.680816 \;\;\; Aprox.\]

Answer 13

Made a mistake, never mind the above.

Answer 14

I did lose a two in my answer, so I'm at\[\int\limits\limits_{1}^{2}{\sqrt{t^2+4t+4\over4t^2(t+1)}}dt=\int\limits\limits_{1}^{2}{(t+2) \over2t \sqrt{t+1}}dt\]umm...

Answer 15

Same here. I have the same answer, but don't know how to solve it further.

Answer 16

same as me

Answer 17

Maybe\[\int\limits_{1}^{2}{1\over2\sqrt{t+1}}+{1\over t \sqrt{t+1}}dt\]now about that second term... I'm thinking either integration by parts, or it might be a form in some table. Let me check...

Answer 18

but that is how did you get the second term

Answer 19

how? by what?

Answer 20

http://www.wolframalpha.com/input/?i=integral+1%2F%28t+sqrt%28t%2B1%29%29 but I can't get it to show steps...

Answer 21

oh, you asked how I split up the fraction?

Answer 22

anyway, Wolframs answer works, but it involves two substitutions.\[\sqrt{t+1}+\ln(1-\sqrt{t+1})-\ln(1+\sqrt{t+1})\]evaluated from 1 to 2 is the answer.

Answer 23

Ouch, did you get that via integration by parts?

Answer 24

go to Wolfram on the link and look at "show steps" in the upper right:
http://www.wolframalpha.com/input/?i=integral+1%2F%28t+sqrt%28t%2B1%29%29
Like I said, two tricky substitutions were used here. There's probably another way.

Answer 25

Thanks.

Answer 26

hey

Answer 27

if you are here let me know