Arc Length. Simplify expression

[rossmoesis]

Homework Statement

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

0< x < 2]

Homework Equations

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

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

 

[pasmith]

Homework Statement

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

Homework Equations

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

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

\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}.

 

[rossmoesis]

\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}.

Thanks for the quick reply but I'm not able to follow your work.

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

 

[Ray Vickson]

Homework Statement

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

0< x < 2

Homework Equations

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

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

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 .

 

[LCKurtz]

\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}.

Thanks for the quick reply but I'm not able to follow your work.

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

 

[rossmoesis]

Thanks for the input. I figured out my issue.

 

[rossmoesis]

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

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