Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 4 - Graphs of the Circular Functions - Section 4.5 Harmonic Motion - 4.5 Exercises - Page 180: 22

Answer

The points of intersection of $y_1$ and $y_2$ are: $( \frac{\pi}{2}+2\pi~n, e^{-\frac{\pi}{2}-2\pi~n})$ The t-coordinates of the points of intersection of $y_1$ and $y_2$ are the same as the x-coordinates where the graph of $~~y = sin~x~~$ has a value of $1$ The points of intersection of $y_1$ and $y_3$ are: $( \frac{3\pi}{2}+2\pi~n, -e^{-\frac{3\pi}{2}-2\pi~n})$ The t-coordinates of the points of intersection of $y_1$ and $y_3$ are the same as the x-coordinates where the graph of $~~y = sin~x~~$ has a value of $-1$

Work Step by Step

$y_1 = e^{-t}~sin~t$ $y_2 = e^{-t}$ We can find the t-coordinates of the points of intersection: $y_1 = y_2$ $e^{-t}~sin~t = e^{-t}$ $sin~t = 1$ $t = \frac{\pi}{2}+2\pi~n,~~$ where $n$ is an integer The points of intersection of $y_1$ and $y_2$ are: $( \frac{\pi}{2}+2\pi~n, e^{-\frac{\pi}{2}-2\pi~n})$ The t-coordinates of the points of intersection of $y_1$ and $y_2$ are the same as the x-coordinates where the graph of $~~y = sin~x~~$ has a value of $1$ $y_1 = e^{-t}~sin~t$ $y_3 = -e^{-t}$ We can find the t-coordinates of the points of intersection: $y_1 = y_3$ $e^{-t}~sin~t = -e^{-t}$ $sin~t = -1$ $t = \frac{3\pi}{2}+2\pi~n,~~$ where $n$ is an integer The points of intersection of $y_1$ and $y_3$ are: $( \frac{3\pi}{2}+2\pi~n, -e^{-\frac{3\pi}{2}-2\pi~n})$ The t-coordinates of the points of intersection of $y_1$ and $y_3$ are the same as the x-coordinates where the graph of $~~y = sin~x~~$ has a value of $-1$
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