Question

Find the arc length of the curve:

y=(1/2)((e^x)+(e^-x))

Answer 1

I know that

(1/2)((e^x)+(e^-x)) = coshx

but I am not sure what to do after that.

Answer 2

if i recall correctly the arc length is given by
\[\int_a^b\sqrt{1+(f'(x)})^2dx\]

Answer 3

Could you help me with the derivation? d/dx of e^x is just e^x, right? So would e^-x be e^-x?

Answer 4

no it would be \(-e^{-x}\) by the chain rule

Answer 5

so your derivative would be \[\frac{e^x-e^{-x}}{2}\]

Answer 6

or with hyperbolic function, you have cosh and the derivative is sinh

Answer 7

and by coincidence, \(1+\sinh^2(x)=\cosh^2(x)\) i believe

Answer 8

Yes, got the same thing. The hyperbolic method seems like it would be easier, however I am not very well versed in them.

Answer 9

i think that is the gimmick here, because these problems are always cooked up to give you something you can actually integrate

Answer 10

so you take the derivative of cosh, get sinh, then square and add one and you get \(\cosh^2(x)\)

Answer 11

take the square root and you get \(\cosh(x)\) again

Answer 12

i assume you have limits of integration too right?

Answer 13

Yes, the arc is from x=0, and x=1.