Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.3 Trigonometric Equations II - 6.3 Exercises - Page 281: 63


The least value of $t$ is $0.001$ seconds.

Work Step by Step

$$i=I_{max}\sin(2\pi ft)$$ For $i=40, I_{max}=100, f=60$, we have $$40=100\times\sin(2\pi\times60t)$$ $$40=100\sin(120\pi t)$$ $$\sin(120\pi t)=\frac{40}{100}=\frac{2}{5}$$ $t$ refers to time, so as a rule, $t\in[0,\infty)$ Also $t$ is minimum when $120\pi t$ is minimum, as $120\pi$ is a constant. Therefore, we would be able to find the least of $t$ as we find the first value of $120\pi t$ that would have $\sin(120\pi t)=\frac{2}{5}$ over the interval $[0,2\pi)$. Such a value of $120\pi t$ can be found using the inverse function for sine: $$120\pi t=\sin^{-1}\frac{2}{5}\approx0.4115$$ $$t=\frac{0.4115}{120\pi}$$ $$t\approx0.001 (seconds)$$ Therefore, the least value of $t$ satisfying the given data is $0.001$ seconds.
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