How can I find the arc length of y=(1/4)((e^(2x)+e^(-2x))), & the x intervals are [0,2]

Question

anonymous · Answer

The equation given was $$l=\int\limits_{a}^{b}\sqrt{1+[f'(x)]^2} dx$$

tkhunny · Answer

Would it help to observe that $\dfrac{1}{4}\left(e^{2x}+e^{-2x}\right) = \dfrac{1}{2}\cosh(2x)$?  Derivatives and things might be easier.

anonymous · Answer

@tkhunny the cosh(2x)/2 insight would honestly not help us here in anyway. The best way to compute the integrals involved in the arc length of smooth curves is to use a graphing calculator since they often get very complicated or do it by hand if things simplify to such to an extent..

anonymous · Answer

How did you get that? $$\frac{ 1 }{ 2 }\cosh(2x)?$$
& what's h?

anonymous · Answer

@nelly_416 don't worry about that, @tkhunny is referring to the hyperbolic cosine also known as cosh(x) but we don't need it here and neither will it help us much.

tkhunny · Answer

It's a definition.  Look up the definition of cosh(x).

anonymous · Answer

alright, thank you

anonymous · Answer

btw @tkhunny is dat u in the pic?

anonymous · Answer

oh nvm, now that i look at the larger version of the pic lol...

tkhunny · Answer

Yes.  It was one of my last operas, some 4-5 years ago.

anonymous · Answer

oh so it is u lmfao. i actually thought it was some scientist's pic from 300 years ago..lol

tkhunny · Answer

Well, that is about the proper setting.  I guess costume and makeup did a good job.

@nelly_416 Trust me on this one.  This problem is designed so that it works nicely with the transformation I gave you.  Do the transformation.  Find the derivative.  Do the algebra and you will be pleasantly surprised.

In general, these arc length things do get pretty ugly and numerical methods are the way to go.  This one, however, is a thing of beauty.  Do the algebra!  Trust me on this.

tkhunny · Answer

$\dfrac{d}{dx}\dfrac{1}{2}\cosh(2x) = sinh(2x)$

$1 + \sinh^{2}(2x) = \cosh^{2}(2x)$

anonymous · Answer

Will do, thank you very much for your help.