Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter P - Review Exercises - Page 38: 53

Answer

(a) Yes; the vertical line test is passed, $y$ is on the $y$-axis, and $t$ is on the $x$-axis (b) Amplitude = 0.25 and Period = 1.1 (c) $\displaystyle f(t)=y=\frac{1}{4}\cos\left(\frac{2\pi}{1.1}t\right)$ (d) See the graph below;

Work Step by Step

(a) The data plotted is a function because no two data points are at the same value of $t$. Therefore, this graph passes the vertical line test and it is a function. $y$ is a function of $t$ since $t$ is on the $x$-axis and $y$ is on the $y$-axis (if $t$ was on the $y$-axis then we'd have to say that $t$ is a function of $y$) (b) The amplitude is the height of the sinusoidal wave and the period is the distance between two successive peaks. The maximum value of the sinusoidal wave is 0.25 so the amplitude is 0.25. From $t = 0$ to $t = 1.1$ the sinusoidal wave completes a full cycle and the graph is back at its original value. Therefore, the period is 1.1. (c) A cosine wave has the following form: $f(t)=A\cos(\omega t)$ where $\omega$ is the angular frequency of the cosine wave and $A$ is the amplitude. We can obtain the angular frequency using the period by using the following fact: $\displaystyle \omega = 2\pi f = \frac{2\pi}{T}$ where $f$ is the regular frequency of the wave, $T$ is the period of the wave, and $f = \displaystyle \frac{1}{T}$. Thus, $\displaystyle \omega = \frac{2\pi}{1.1}$ Our final equation is $f(t) = \displaystyle 0.25\cos\left(\frac{2\pi}{1.1}t\right)$ (d) See graph in solution
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