Exam Problem

[ERTW]

Hi all, first post

On my calculus exam we had the following question:

 

find the arc length of

 

[latex] \frac{e^x+e^{-x}}{2}[/latex] from zero to a

 

I subbed in hyperbolic cosine and found the arc length to be sinh(a)

Any thoughts? Thanks. I only ask because the arc length was crucial to later questions involving area of rotation and none of my classmates used the hyperbolic cosine substitution

 

My solution (don't laugh):

 

 

[math](\frac{dy}{dx})^2=sinh^2(x)[/math]

 

[math]L = \int_{0}^{a} \sqrt{1+sinh^2(x)}dx[/math]

 

[math]L = \int_{0}^{a} \sqrt{cosh^2(x)}dx[/math]

 

[math]L = \int_{0}^{a} cosh(x)dx[/math]

 

[math]L = sinh(a) - sinh(0) = sinh(a) [/math]

 

edit: e^x + e^-x

Edited April 17, 2011 by ERTW

[Hal.]

I had a look at this problem and I think you're correct . But why is edit: e^x + e^-x near the end of you're post ?

Edited April 17, 2011 by hal_2011

[ERTW]

Initially, I had the identity for sinh at the top (e^x - e^-x)/2

This is my first time with a forum, but i've seen people log their edits, so I figured I should too.

In hindsight, writing that didn't really clarify anything and just led to more confusion. my bad.

Thanks for confirming my solution by the way, that was one question I felt a little shaky on.