Calculus 8th Edition

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

Chapter 16 - Vector Calculus - 16.2 Line Integrals - 16.2 Exercises - Page 1126: 42



Work Step by Step

Work done$W=\int_C F\cdot dr=\int_0^{1} \dfrac{k}{(4+26t^2)^{3/2}}\lt 2,t,5t \gt \cdot \lt 0, 1,5 \gt dt$ $=\int_0^{1} \dfrac{k(t+25t)}{(4+26t^2)^{3/2}}dt$ $=(\dfrac{1}{2}) \int_0^{1} \dfrac{k \times (52t)}{(4+26t^2)^{3/2}}dt$ Use $4+26 t^2=t ;dt=52 t dt$ $=(k/2) \int_4^{30} \dfrac{dt}{t^{3/2}}$ $=[\dfrac{k}{2}]\dfrac{-2}{p^{1/2}}]_4^{30}$ $=k[\dfrac{1}{2}-\dfrac{1}{\sqrt{30}}]$
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