Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 16 - Vector Calculus - 16.2 Exercises - Page 1098: 42

Answer

$W=k[\dfrac{1}{2}-\dfrac{1}{\sqrt{30}}]$

Work Step by Step

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