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Exam Problem

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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 :o

 

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 by ERTW

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 by hal_2011

  • Author

tongue.gif 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. ohmy.gif my bad.

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

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