# ARC Length Help!!

• Jun 2nd 2008, 03:55 PM
07 Yamaha R6
ARC Length Help!!
hey, guys and gals,
just wanted to see if any of you can direct me towards the correct way of
approaching this problem.

the problem asks to find the arc length of y=e^2x
now ive obtained the derivative dy/dx to be
2e^2x and when I use the formula for finding the
arclength, I then have the integral:
√(1+(2e^2x )²) dx
next I use
u=√(1+(2e^2x )²)
so that if I square both sides I get:
u²=1+4e^4x then, to get du:
2udu=16e^4x dx

now i noticed that above in "u²=1+4e^4x " i could solve for
e^4x to =(u²-1)/4 so that in the equation,
2udu=16e^4x dx, i could also solve for e^4x dx=1/8udu, then substituting
the e^4x=(u²-1)/4 into e^4x dx=1/8udu to become
(u²-1)/4 dx=1/8udu then solving for dx, I found
dx=u/2(u²-1) du

so my question is what do you do next? I substituted back u and du
into the integral and get u²/2(u²-1) du but I dont know where to go from here. Can somebody please help?

• Jun 2nd 2008, 04:00 PM
Krizalid
Where're the bounds?
• Jun 2nd 2008, 04:18 PM
Mathstud28
Quote:

Originally Posted by 07 Yamaha R6
hey, guys and gals,
just wanted to see if any of you can direct me towards the correct way of
approaching this problem.

the problem asks to find the arc length of y=e^2x
now ive obtained the derivative dy/dx to be
2e^2x and when I use the formula for finding the
arclength, I then have the integral:
√(1+(2e^2x )²) dx
next I use
u=√(1+(2e^2x )²)
so that if I square both sides I get:
u²=1+4e^4x then, to get du:
2udu=16e^4x dx

now i noticed that above in "u²=1+4e^4x " i could solve for
e^4x to =(u²-1)/4 so that in the equation,
2udu=16e^4x dx, i could also solve for e^4x dx=1/8udu, then substituting
the e^4x=(u²-1)/4 into e^4x dx=1/8udu to become
(u²-1)/4 dx=1/8udu then solving for dx, I found
dx=u/2(u²-1) du

so my question is what do you do next? I substituted back u and du
into the integral and get u²/2(u²-1) du but I dont know where to go from here. Can somebody please help?

$\displaystyle \int\sqrt{1+(2e^{2x})^2}dx$

Let $\displaystyle u=\sqrt{1+(2e^{2x})^2}\Rightarrow{\frac{1}{4}\ln(u ^2-1)-\frac{\ln(2)}{2}=x}$

so $\displaystyle dx=\frac{1}{2}\frac{u}{u^2-1}$

So we have

$\displaystyle \frac{1}{2}\int\frac{u^2}{u^2-1}du=\frac{1}{2}\bigg[\int\frac{u^2-1}{u^2-1}du+\int\frac{1}{u^2-1}du\bigg]$

integrating we get

$\displaystyle \frac{1}{2}\bigg[u-arctanh(u)\bigg]+C$

subbing back in we get

$\displaystyle \int\sqrt{1+(2e^{2x})^2}dx=\frac{1}{2}\bigg[\sqrt{1+(2e^{2x})^2}-arctahn\bigg(\sqrt{1+(2e^{2x})^2}\bigg)\bigg]+C$
• Jun 2nd 2008, 04:20 PM
07 Yamaha R6
right, the bounds are from 0 to 1
• Jun 2nd 2008, 04:24 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
$\displaystyle \int\sqrt{1+(2e^{2x})^2}dx$

Let $\displaystyle u=\sqrt{1+(2e^{2x})^2}\Rightarrow{\frac{1}{4}\ln(u ^2-1)-\frac{\ln(2)}{2}=x}$

so $\displaystyle dx=\frac{1}{2}\frac{u}{u^2-1}$

So we have

$\displaystyle \frac{1}{2}\int\frac{u^2}{u^2-1}du=\frac{1}{2}\bigg[\int\frac{u^2-1}{u^2-1}du+\int\frac{1}{u^2-1}du\bigg]$

integrating we get

$\displaystyle \frac{1}{2}\bigg[u-arctanh(u)\bigg]+C$

subbing back in we get

$\displaystyle \int\sqrt{1+(2e^{2x})^2}dx=\frac{1}{2}\bigg[\sqrt{1+(2e^{2x})^2}-arctahn\bigg(\sqrt{1+(2e^{2x})^2}\bigg)\bigg]+C$

Quote:

Originally Posted by 07 Yamaha R6
right, the bounds are from 0 to 1

$\displaystyle \int_0^{1}\sqrt{1+(2e^{2x})^2}dx=\frac{1}{2}\bigg[\sqrt{1+(2e^{2x})^2}-arctanh\bigg(\sqrt{1+(2e^{2x})^2}\bigg)\bigg]\bigg|_0^{1}\approx{6.5}$
• Jun 2nd 2008, 04:24 PM
galactus
Yes, where are the bounds?. But we can proceed and then you can use them.

One possible sub is to let $\displaystyle u=4e^{4x}$

Then, $\displaystyle x=\frac{ln(u/4)}{4}, \;\ dx=\frac{1}{4u}du$

$\displaystyle \frac{1}{4}\int\frac{\sqrt{1+u}}{u}du$

Now, a sub could be $\displaystyle w=\sqrt{1+u}, \;\ w^{2}=1+u, \;\ 2wdw=du$

Making the subs and expanding gives:

$\displaystyle \frac{1}{2}+\frac{1}{4(w-1)}-\frac{1}{4(w+1)}$

Now, it is just a matter of ln.

This is one possible approach.

Resubbing we get $\displaystyle \frac{ln(\sqrt{1+4e^{4x}}-1)}{2}+\frac{\sqrt{1+4e^{4x}}}{2}-x$
• Jun 2nd 2008, 04:30 PM
Mathstud28
Quote:

Originally Posted by galactus
Yes, where are the bounds?. But we can proceed and then you can use them.

One possible sub is to let $\displaystyle u=4e^{4x}$

Then, $\displaystyle x=\frac{ln(u/4)}{4}, \;\ dx=\frac{1}{4u}du$

$\displaystyle \frac{1}{4}\int\frac{\sqrt{1+u}}{u}du$

Now, a sub could be $\displaystyle w=\sqrt{1+u}, \;\ w^{2}=1+u, \;\ 2wdw=du$

Making the subs and expanding gives:

$\displaystyle \frac{1}{2}+\frac{1}{4(w-1)}-\frac{1}{4(w+1)}$

Now, it is just a matter of ln.

This is one possible approach.

Resubbing we get $\displaystyle \frac{ln(\sqrt{1+4e^{4x}}-1)}{2}+\frac{\sqrt{1+4e^{4x}}}{2}-x$

Just to satisfy my curiousity, I have always when seeing a nested exponential defined u as the whole quantity due to the simplification qualities used when solving for x (because of ln(u(x))), but no matter where I look no one seems to do it that way,

Is my way either invalid, overcomplicated, or just uncommon?
• Jun 2nd 2008, 04:32 PM
07 Yamaha R6
mathstud,
hey thanks for the help bra! thats right, my problem was
not taking the 1/2 out of the integral, thats why I couldnt see
the "solution" to the 1/u²-1. The only question was that
isnt 1/u²-1=1/(2a)*ln|(u-a)/(u+a)| + C ?

07 Yamaha R6
• Jun 2nd 2008, 04:35 PM
Mathstud28
Quote:

Originally Posted by 07 Yamaha R6
mathstud,
hey thanks for the help bra! thats right, my problem was
not taking the 1/2 out of the integral, thats why I couldnt see
the "solution" to the 1/u²-1. The only question was that
isnt 1/u²-1=1/(2a)*ln|(u-a)/(u+a)| + C ?

07 Yamaha R6

I did not put $\displaystyle arctan(x)$, I put $\displaystyle arctanH(x)$, which is hyperbolic arctangent, it is actually equal to what you said,

$\displaystyle arctanh(x)=\frac{1}{2}\ln\bigg(\frac{1+x}{1-x}\bigg)$
• Jun 2nd 2008, 04:42 PM
galactus
Whatever works.
• Jun 2nd 2008, 04:44 PM
Jhevon
Quote:

Originally Posted by galactus
Whatever works.

I found that post funny for some reason (Speechless)