# Homework Help: Arc Length. Simplify expression

1. May 30, 2015

### rossmoesis

1. The problem statement, all variables and given/known data

I'm trying to find Arc Length of F(x) = (e^x + e^-x)/2

0< x < 2]

2. Relevant equations

L = integrate sqrt ( 1 + (dy/dx))^2)

3. The attempt at a solution

(dy/dx)^2 + 1 = 1/4e^2x + 1/4e^-2x + 1/2]

I don't know how to take the square root of the above function so I can be able to take the integral of it to find the arc lengh.

The solution says ]

sqrt (1/4e^2x + 1/4e^-2x + 1/2) = 1/2 (e^-x + e^x)
I'm not sure how they did that. Once I find that out I can do the rest of the problem. Any help or advice would be appreciated. Maybe I'm missing something obvious. Thanks

Last edited by a moderator: May 30, 2015
2. May 30, 2015

### pasmith

$$\frac{e^{2x}}{4} + \frac12 + \frac{e^{-2x}}{4} = \frac{e^{2x}}{4} + 2 \left(\frac{e^x}{2}\right)\left(\frac{e^{-x}}{2}\right) + \frac{e^{-2x}}{4}.$$

Last edited by a moderator: May 30, 2015
3. May 30, 2015

### rossmoesis

sqrt (1/4e^2x + 1/4e^-2x + 1/2) = 1/2 (e^-x + e^x)

I'm not sure how the above math is done. Thanks

4. May 30, 2015

### Ray Vickson

You have $y = \cosh(x)$. There are lots of very useful identities/properties of $\cosh$ (and its companion $\sinh$), which reduce your problem to almost a triviality. See, eg., http://en.wikipedia.org/wiki/Hyperbolic_function .

5. May 30, 2015

### LCKurtz

Do you mean you don't understand how those two expressions are equal?

6. May 30, 2015

### rossmoesis

Thanks for the input. I figured out my issue.

7. May 30, 2015

### rossmoesis

Yes, it was obvious when you factor it out. It just took me a bit to realize that.