Calculus: Early Transcendentals 9th Edition

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

Chapter 3 - Section 3.7 - Rates of Change in the Natural and Social Sciences - 3.7 Exercises - Page 236: 23

Answer

$F = \frac{m_0~a}{(1-\frac{v^2}{c^2})^{3/2}}$

Work Step by Step

$F = (\frac{d}{dt})(m~v)$ $F = (\frac{d}{dt})(\frac{m_0~v}{\sqrt{1-\frac{v^2}{c^2}}})$ $F = \frac{m_0~\frac{dv}{dt}~\sqrt{1-\frac{v^2}{c^2}}-\frac{1}{2}(1-\frac{v^2}{c^2})^{-1/2}~(-\frac{2v}{c^2})(\frac{dv}{dt})~(m_0~v)}{1-\frac{v^2}{c^2}}$ $F = m_0~\frac{dv}{dt}\cdot ~\frac{\sqrt{1-\frac{v^2}{c^2}}+(1-\frac{v^2}{c^2})^{-1/2}~(\frac{v^2}{c^2})}{1-\frac{v^2}{c^2}}$ $F = m_0~\frac{dv}{dt}\cdot ~\frac{\sqrt{1-\frac{v^2}{c^2}}+(1-\frac{v^2}{c^2})^{-1/2}~(\frac{v^2}{c^2})}{1-\frac{v^2}{c^2}}\cdot \frac{\sqrt{1-\frac{v^2}{c^2}}}{\sqrt{1-\frac{v^2}{c^2}}}$ $F = m_0~a\cdot ~\frac{(1-\frac{v^2}{c^2})+\frac{v^2}{c^2}}{(1-\frac{v^2}{c^2})^{3/2}}$ $F = m_0~a\cdot ~\frac{1}{(1-\frac{v^2}{c^2})^{3/2}}$ $F = \frac{m_0~a}{(1-\frac{v^2}{c^2})^{3/2}}$
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