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: 65


The least possible value of $t$ is $0.04$ seconds.

Work Step by Step

$$i=I_{max}\sin(2\pi ft)$$ For $i=I_{max},f=60$, we have $$i=i\sin(2\pi\times60t)$$ (since $i=I_{max}$, we can replace $I_{max}$ with $i$) $$i=i\sin(120\pi t)$$ Because $I_{max}$ is the maximum current, $I_{max}\ne0$. And as $i=I_{max}$, $i\ne0$. So now we can divide both sides of the equation by $i$: $$\sin(120\pi t)=1$$ $t$ 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 $\sin(120\pi t)=1$ over the interval $[0,2\pi)$. Such a value of $120\pi t$ can be found using the inverse function for sine: $$120\pi t=\sin^{-1}1$$ $$120\pi t=\frac{\pi}{2}$$ $$t=\frac{\pi}{2\times120\pi}$$ $$t=\frac{1}{240}\approx0.04(seconds)$$ Therefore, the least possible value of $t$ satisfying the given data is $0.04$ seconds.
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