Chapter 10 - Parametric Equations and Polar Coordinates - 10.2 Calculus with Parametric Curves - 10.2 Exercises - Page 696: 42

$L$ = $e^{2}+1$

$x$ = $e^{t}-t$ $y$ = $4e^{\frac{t}{2}}$ $\frac{dx}{dt}$ = $e^{t}-1$ $\frac{dy}{dt}$ = $2e^{\frac{t}{2}}$ so $\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2$ = $(e^{t}-1)^2+4e^{t}$ = $e^{2t}-2e^{t}+1+4e^{t}$ = $e^{2t}+2e^{t}+1$ = $(e^{t}+1)^2$ $L$ = $\int_0^2{\sqrt {(e^{t}+1)^2}}dt$ $L$ = $\int_0^2(e^{t}+1)dt$ $L$ = $[e^{t}+t]_0^2$ $L$ = $e^{2}+2-1$ $L$ = $e^{2}+1$