Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 10 - Parametric Equations and Polar Coordinates - 10.2 Calculus with Parametric Curves - 10.2 Exercises - Page 696: 49

Answer

$L$ $\approx$ $612.3053$

Work Step by Step

$x$ = $t-e^t$ $y$ = $t+e^t$ $\frac{dx}{dt}$ = $1-e^t$ $\frac{dy}{dt}$ = $1+e^t$ $\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2$ = $(1-e^t)^2+(1+e^t)^2$ = $2+2e^{2t}$ $L$ = $\int_{-6}^{6}\sqrt {2+2e^{2t}}dt$ $f(t)$ = $\sqrt {2+2e^{2t}}$ Simpson's rule with $n$ = $6$ $Δt$ = $\frac{6-(-6)}{6}$ = $2$ $L$ = $\frac{2}{3}[f(-6)+4f(-4)+2f(-2)+4f(0)+2f(2)+4f(4)+f(6)]$ $L$ $\approx$ $612.3053$
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