# How do you find the length of the curve x=e^t+e^-t, y=5-2t, where 0<=t<=3 ?

Aug 20, 2014

The answer is ${e}^{3} - {e}^{- 3}$.

Recall that the arclength for parametric curves is:

$L = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

So,

$\frac{\mathrm{dx}}{\mathrm{dt}} = {e}^{t} - {e}^{- t}$
$\frac{\mathrm{dy}}{\mathrm{dt}} = - 2$

Now substituting:

$L = {\int}_{0}^{3} \sqrt{{\left({e}^{t} - {e}^{- t}\right)}^{2} + {\left(- 2\right)}^{2}} \mathrm{dt}$
$= {\int}_{0}^{3} \sqrt{{e}^{2 t} - 2 + {e}^{- 2 t} + 4} \mathrm{dt}$ expand
$= {\int}_{0}^{3} \sqrt{{e}^{2 t} + 2 + {e}^{- 2 t}} \mathrm{dt}$ simplify
$= {\int}_{0}^{3} \sqrt{{\left({e}^{t} + {e}^{- t}\right)}^{2}} \mathrm{dt}$ factor
$= {\int}_{0}^{3} \left({e}^{t} + {e}^{- t}\right) \mathrm{dt}$ simplify
$= {e}^{t} - {e}^{- t} {|}_{0}^{3}$ integrate
$= {e}^{3} - {e}^{- 3} - \left({e}^{0} - {e}^{0}\right)$ evaluate
$= {e}^{3} - {e}^{- 3}$

Note that there aren't many questions that can be solved algebraically. Please note the pattern of this problem because most algebraic solutions have this form.