Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - 10.2 Exercises - Page 682: 48

Answer

$e^2+1$

Work Step by Step

Recall: The integral which represents the length of the parametric curve for $a\leq t\leq b$ is given by $L=\int_a^b\sqrt{(dx/dt)^2+(dy/dt)^2}dt$. Given: $x=e^t-t$ and $y=4e^{t/2}$ where $0\leq t\leq 2$ Find $dx/dt$ and $dy/dt$: $dx/dt=e^t-1$ and $dy/dt=2e^{t/2}$ Find the length of the given curve: $L=\int_0^2\sqrt{(e^t-1)^2+(2e^{t/2})^2}dt$ $L=\int_0^2\sqrt{(e^{t})^2-2e^t+1+4e^t}dt$ $L=\int_0^2\sqrt{(e^t)^2+2e^t+1}dt$ $L=\int_0^2\sqrt{(e^t+1)^2}dt$ $L=\int_0^2e^t+1 dt$ $L=[e^t+t]_0^2$ $L=(e^2+2)-(e^0+0)$ $L=e^2+2-1$ $L=e^2+1$ Thus, the length of the curve is $e^2+1$.
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