# What is the arc length of teh curve given by r(t)=(2t , 2e^(-t) , e^(t) ) in the interval t in [0,3]?

Jul 10, 2018

1+e^3-2e^-3#

#### Explanation:

The parametric equation for the curve is

$x \left(t\right) = 2 t q \quad \implies \mathrm{dx} = 2 \mathrm{dt}$
$y \left(t\right) = 2 {e}^{-} t \implies \mathrm{dy} = - 2 {e}^{-} t \mathrm{dt}$
$z \left(t\right) = {e}^{t} \setminus \quad \implies \mathrm{dz} = {e}^{t} \mathrm{dt}$

Thus

$\mathrm{ds} = \sqrt{{\mathrm{dx}}^{2} + {\mathrm{dy}}^{2} + {\mathrm{dz}}^{2}}$
$q \quad = \sqrt{4 + 4 {e}^{- 2 t} + {e}^{2 t}} \mathrm{dt}$
$q \quad = \sqrt{{\left({e}^{t} + 2 {e}^{-} t\right)}^{2}} \mathrm{dt}$
$q \quad = \left({e}^{t} + 2 {e}^{-} t\right) \mathrm{dt}$

Hence, the required length is

$\int \mathrm{ds} = {\int}_{0}^{3} \left({e}^{t} + 2 {e}^{-} t\right) \mathrm{dt}$
$q \quad \quad = {\left[{e}^{t} - 2 {e}^{-} t\right]}_{0}^{3}$
$\quad q \quad = \left({e}^{3} - 2 {e}^{-} 3\right) - \left({e}^{0} - 2 {e}^{0}\right)$
$q \quad \quad = 1 + {e}^{3} - 2 {e}^{-} 3$