Chapter 10 - Parametric Equations and Polar Coordinates - 10.2 Calculus with Parametric Curves - 10.2 Exercises - Page 696: 37

$L$ $\approx$ $3.1416$

$x$ = $t+e^{-t}$ $y$ = $t-e^{-t}$ $\frac{dx}{dt}$ = $1-e^{-t}$ $\frac{dy}{dt}$ = $1+e^{-t}$ So $\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}$ = $(1-e^{-t})^{2}+(1+e^{-t})^{2}$ = $2+2e^{-2t}$ Thus $L$ = $\int_a^b{\sqrt {\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}}dt$ $L$ = $\int_0^2\sqrt {2+2e^{-2t}}dt$ $L$ $\approx$ $3.1416$