# Math Help - Arc length

1. ## Arc length

Hey guys and gals,
I just wanted to see if any one can explain to me how to derive the arc length of y=e^x; 0<x<1.

I know the formula to finding the arc length is the square root of 1 + (dy/dx)^2 in the integrand. my question is how do you go from √(1 + u²)/u du to √(1 + u²)/u². Can some one help out?

Jose

2. Originally Posted by 07 Yamaha R6
Hey guys and gals,
I just wanted to see if any one can explain to me how to derive the arc length of y=e^x; 0<x<1.

I know the formula to finding the arc length is the square root of 1 + (dy/dx)^2 in the integrand. my question is how do you go from √(1 + u²)/u du to √(1 + u²)/u². Can some one help out?

Jose
Are you asking how $\frac{\sqrt{1 + u^2}}{u}$ becomes $\sqrt{\frac{1 + u^2}{u^2}}$ ?

3. ## arc length of y=e^x; 0<x<1.

Yes sir!
I understand we have to use the technique of using the ln to get rid of e^x so that we can find x, but im having trouble understanding that.

4. ## If

your asking how $\frac{\sqrt{u^2+1}}{u}$ becomes $\sqrt{\frac{u^2+1}{u^2}}$ all you do is since powers in this case $\frac{1}{2}$ distribute if you distributed the squareroot of the second one you would get |u| but in this case disregard the ||

5. Originally Posted by mr fantastic
Are you asking how $\frac{\sqrt{1 + u^2}}{u}$ becomes $\sqrt{\frac{1 + u^2}{u^2}}$ ?
Originally Posted by 07 Yamaha R6
Yes sir!
I understand we have to use the technique of using the ln to get rid of e^x so that we can find x, but im having trouble understanding that.
$\frac{\sqrt{1 + u^2}}{u} = \frac{\sqrt{1 + u^2}}{\sqrt{u^2}} = \sqrt{\frac{1 + u^2}{u^2}}$.

6. sorry mr fantastic,

first of all let me thank you for taking the time in helping me out, I appreciate a lot.

second,
I am trying to derive the arc length of y = e^x from 0 to 1

first step is to get dy/dx

that is simply e^x dx

next I have ∫√(1+(e^x)²) dx

then I use u substitution, where u=e^x and du=e^x dx

next I plug the values in and I get ∫√(1+(u)²) du/u or ∫(√(1+(u)²))/u du

now, I have been helped at school and was told that the next step to this would be to go from ∫(√(1+(u)²))/u du to ∫(√(1+(u)²))/u² du,

my question is what did they do to get there and what are the next steps?

7. Originally Posted by 07 Yamaha R6
sorry mr fantastic,

first of all let me thank you for taking the time in helping me out, I appreciate a lot.

second,
I am trying to derive the arc length of y = e^x from 0 to 1

first step is to get dy/dx

that is simply e^x dx

next I have ∫√(1+(e^x)²) dx

then I use u substitution, where u=e^x and du=e^x dx

next I plug the values in and I get ∫√(1+(u)²) du/u or ∫(√(1+(u)²))/u du

now, I have been helped at school and was told that the next step to this would be to go from ∫(√(1+(u)²))/u du to ∫(√(1+(u)²))/u² du,

my question is what did they do to get there and what are the next steps?
I have posted the answer to what they did to get there ......

The substitution u = e^x is not a good one to use. A much better substitution is $u^2 = 1 + e^{2x}$.

Then $2u \frac{du}{dx} = 2 e^{2x} = 2 (u^2 - 1) \Rightarrow \frac{du}{dx} = \frac{u^2 - 1}{u} \Rightarrow dx = \frac{u}{u^2 - 1}$.

Then:

$\int_{0}^{1} \sqrt{1 + e^{2x}} \, dx$ becomes $\int_{\sqrt{2}}^{\sqrt{1 + e^2}} \frac{u^2}{u^2 - 1} \, du = \int_{\sqrt{2}}^{\sqrt{1 + e^2}} \frac{(u^2 - 1) + 1}{u^2 - 1} \, du = \int_{\sqrt{2}}^{\sqrt{1 + e^2}} 1 + \frac{1}{u^2 - 1} \, du$.

8. hey mr fantastic,

thanks again,

that substitution really makes it easier. Just to clarify though, when you substitute u=√(1+e^2x), I see how you have to square u to get u^2=1+e^2x what I dont understand is how you isolated du to be equal to u/u^2-1

Can you clarify?

9. Originally Posted by 07 Yamaha R6
hey mr fantastic,

thanks again,

that substitution really makes it easier. Just to clarify though, when you substitute u=√(1+e^2x), I see how you have to square u to get u^2=1+e^2x what I dont understand is how you isolated du to be equal to u/u^2-1

Can you clarify?
Do you understand where $2u \frac{du}{dx} = 2 e^{2x} = 2 (u^2 - 1)$ has come from?

Then divide both sides by 2u:

$\frac{du}{dx} = \frac{u^2 - 1}{u} \Rightarrow \frac{dx}{du} = \frac{u}{u^2 - 1} \Rightarrow dx = .....$

10. ## Got It!!

Ok, Fantastic,

I got a break through,
feels good too!! anyways, so I see how you got u^2-1. that is, you set e^2x=u^2-1 right?

got it!!!

So, let me ask you this, this would be the same for y=e^3x or y=e^9x? I mean the technique and substitution we used for y=e^x?

Please tell me it is so!

Jose

11. Originally Posted by 07 Yamaha R6
Ok, Fantastic,

I got a break through,
feels good too!! anyways, so I see how you got u^2-1. that is, you set e^2x=u^2-1 right?

got it!!!

So, let me ask you this, this would be the same for y=e^3x or y=e^9x? I mean the technique and substitution we used for y=e^x?

Please tell me it is so!