Question

find the arc length of x=e^t+e^(-t) y=5-2t for the interval [0,pi]

Answer 1

I know i need to do this:\[\int\limits_{0}^{\pi}ds\]

Answer 2

\[ds=\sqrt{(dx/dt )^{2}+(dy/dt)^2}\]

Answer 3

but im getting like (e^t-e^-t)^2+4 under the radical..how do i simplify this by hand?

Answer 4

or is that off completely?

Answer 5

It can be done, but it requires a trick:\[(e^t-e^-t)^2+4=e ^{2t}-2e^te ^{-t}+e ^{-2t}+4\]\[=e ^{2t}-2+e ^{-2t}+4=e ^{2t}+2+e ^{-2t}\]Factor this last one again:\[=(e^t+e ^{-t})^2\]Now you have a square under the root sign, so you can happily go on integrating...

Answer 6

thanks, i was messing up with the -2e^te^-t part..i was just canceling it and forgetting the two

Answer 7

YW