Solve the problem.
-A generator produces an alternating current according to the equation $\mathrm { I } = 80 \sin 11 \pi \mathrm { t }$, where $t$ is time in seconds and $I$ is the current in amperes. What is the smallest time $t$ such that $I = 40$ ?

Question

Quizplus · Accepted Answer

To find the smallest time $t$ such that $I = 40$, we set the equation $40 = 80 \sin 11 \pi t$ and solve for $t$. Dividing both sides by 80 gives $\sin 11 \pi t = \frac{1}{2}$. The smallest positive solution for $t$ when $\sin 11 \pi t = \frac{1}{2}$ occurs at $11 \pi t = \frac{\pi}{6}$, solving for $t$ gives $t = \frac{1}{66}$ sec.

Solve the Problem I=80sin⁡11πt\mathrm { I } = 80 \sin 11 \pi \mathrm { t }I=80sin11πt

Solve the Problem $\mathrm { I } = 80 \sin 11 \pi \mathrm { t }$