Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Section 16.2 - Line Integrals - 16.2 Exercise - Page 1086: 42



Work Step by Step

The work done is given by $W=\int_C F\cdot dr=\int_0^{1} \dfrac{k}{(4+26t^2)^{3/2}}\lt 2,t,5t \gt \cdot \lt 0, 1,5 \gt dt=\int_0^{1} \dfrac{k(t+25t)}{(4+26t^2)^{3/2}}dt=(\dfrac{1}{2}) \int_0^{1} \dfrac{k(52t)}{(4+26t^2)^{3/2}}dt$ Suppose $4+26 t^2=a ; 52 t dt =da$ $W=(k/2) \int_4^{30} \dfrac{dp}{p^{3/2}}=[\dfrac{k}{2}]\dfrac{-2}{p^{1/2}}]_4^{30}$ Work done, $W=[\dfrac{k}{2}]\dfrac{-2}{p^{1/2}}]_4^{30}=k[\dfrac{1}{2}-\dfrac{1}{\sqrt{30}}]$
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