Arc Length. Simplify expression

  • Thread starter rossmoesis
  • Start date
  • #1

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
 
Last edited by a moderator:

Answers and Replies

  • #2
pasmith
Homework Helper
1,857
513

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
[tex]\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}.[/tex]
 
Last edited by a moderator:
  • #3
[tex]\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}.[/tex]

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
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

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 .
 
  • #5
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,555
766
[tex]\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}.[/tex]
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?
 
  • #6
Thanks for the input. I figured out my issue.
 
  • #7
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.
 

Related Threads on Arc Length. Simplify expression

  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
4
Views
730
  • Last Post
2
Replies
25
Views
2K
  • Last Post
Replies
6
Views
894
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
1
Views
962
Top