16 mark for review a microphone detects sound based on the distance and direction of the sound from the…

16 mark for review a microphone detects sound based on the distance and direction of the sound from the microphone. the location of the microphone is considered the origin of the polar coordinate system. the microphone detects sound horizontally, measured in feet (ft), within the graph of the polar function $r = f(\theta)$, where $f(\theta)=1 + sin\theta$. for $\frac{pi}{6}leq\thetaleq\frac{pi}{3}$, what is the maximum distance from the microphone that sound will be detected? a 1.500 ft b 1.707 ft c 1.866 ft d 2.000 ft

16 mark for review a microphone detects sound based on the distance and direction of the sound from the microphone. the location of the microphone is considered the origin of the polar coordinate system. the microphone detects sound horizontally, measured in feet (ft), within the graph of the polar function $r = f(\theta)$, where $f(\theta)=1 + sin\theta$. for $\frac{pi}{6}leq\thetaleq\frac{pi}{3}$, what is the maximum distance from the microphone that sound will be detected? a 1.500 ft b 1.707 ft c 1.866 ft d 2.000 ft

Answer

Explanation:

Step1: Analyze the sine - function range

The function is $r = f(\theta)=1 + \sin\theta$, and we know that the sine function $y = \sin\theta$ has a range. For $\frac{\pi}{6}\leq\theta\leq\frac{\pi}{3}$, we need to find the maximum value of $\sin\theta$ in this interval. The sine function $y = \sin\theta$ is an increasing function on the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, and $\frac{\pi}{6}\leq\theta\leq\frac{\pi}{3}\subseteq\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

Step2: Evaluate $\sin\theta$ at the end - points

When $\theta=\frac{\pi}{6}$, $\sin\theta=\sin\frac{\pi}{6}=\frac{1}{2}$. When $\theta = \frac{\pi}{3}$, $\sin\theta=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}\approx0.866$.

Step3: Find the maximum of $r = 1+\sin\theta$

Substitute the maximum value of $\sin\theta$ in the interval $\left[\frac{\pi}{6},\frac{\pi}{3}\right]$ into $r = 1+\sin\theta$. Since the maximum value of $\sin\theta$ in the interval $\frac{\pi}{6}\leq\theta\leq\frac{\pi}{3}$ is $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$, then $r = 1+\sin\theta$ has a maximum value of $r_{max}=1+\frac{\sqrt{3}}{2}\approx1 + 0.866=1.866$.

Answer:

C. 1.866 ft