Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 1 - First-Order Differential Equations - 1.1 Differential Equations Everywhere - Problems - Page 12: 23


$m_{1}=\frac{m_{2}-\tan a}{1+m_{2} \tan a}$

Work Step by Step

From the given figure: $a_{1}=a_{2}-a$ $m_{1}$ is the slope of trajactory desired $m_{2}$ is the slope of trajactory given $m_{1}=\tan{a_{1}}$ and $m_{2}=\tan{a_{2}}$ Now, $\tan{a_{1}}=\tan{(a_{2}-a)}$ Thus, $m_{1}=\frac{\tan a_{2}-\tan a}{1+\tan{a_{2}\tan a}}$ $m_{1}=\frac{m_{2}-\tan a}{1+m_{2} \tan a}$ Thus, the final answer: $m_{1}=\frac{m_{2}-\tan a}{1+m_{2} \tan a}$
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