consider the graph of the polar function r = f(θ), where f(θ) = 4sin(2θ), in the polar coordinate system. on…

consider the graph of the polar function r = f(θ), where f(θ) = 4sin(2θ), in the polar coordinate system. on the interval 0 ≤ θ ≤ 2π, which of the following is true about the graph of r = f(θ)? a for the input values θ = π/4, θ = 3π/4, θ = 5π/4, and θ = 7π/4, the function r = f(θ) has extrema that correspond to points that are farthest from the origin. b for the input values θ = π/4, θ = 3π/4, θ = 5π/4, and θ = 7π/4, the function r = f(θ) has extrema. however, only the points corresponding to θ = π/4 and θ = 5π/4 are farthest from the origin. c for the input values θ = π/2 and θ = 3π/2, the function r = f(θ) has extrema that correspond to points that are farthest from the origin. d for the input values θ = π/2 and θ = 3π/2, the function r = f(θ) has extrema. however, only the point corresponding to θ = π/2 is farthest from the origin.

consider the graph of the polar function r = f(θ), where f(θ) = 4sin(2θ), in the polar coordinate system. on the interval 0 ≤ θ ≤ 2π, which of the following is true about the graph of r = f(θ)? a for the input values θ = π/4, θ = 3π/4, θ = 5π/4, and θ = 7π/4, the function r = f(θ) has extrema that correspond to points that are farthest from the origin. b for the input values θ = π/4, θ = 3π/4, θ = 5π/4, and θ = 7π/4, the function r = f(θ) has extrema. however, only the points corresponding to θ = π/4 and θ = 5π/4 are farthest from the origin. c for the input values θ = π/2 and θ = 3π/2, the function r = f(θ) has extrema that correspond to points that are farthest from the origin. d for the input values θ = π/2 and θ = 3π/2, the function r = f(θ) has extrema. however, only the point corresponding to θ = π/2 is farthest from the origin.

Answer

Explanation:

Step1: Recall the property of polar - function distance from origin

In polar coordinates, the distance of a point from the origin is given by (r = f(\theta)). We want to find the maxima of (r = 4\sin(2\theta)) on the interval (0\leq\theta\leq2\pi). The maximum value of the sine - function (y = \sin(u)) is 1. We set (u = 2\theta) and find when (\sin(2\theta)=1) or (\sin(2\theta)= - 1).

Step2: Solve for (\theta) when (\sin(2\theta)=\pm1)

If (\sin(2\theta)=1), then (2\theta=\frac{\pi}{2}+2k\pi), (k\in\mathbb{Z}), so (\theta=\frac{\pi}{4}+k\pi). If (\sin(2\theta)= - 1), then (2\theta=\frac{3\pi}{2}+2k\pi), (k\in\mathbb{Z}), so (\theta=\frac{3\pi}{4}+k\pi). For (k = 0), (\theta=\frac{\pi}{4},\frac{3\pi}{4}); for (k = 1), (\theta=\frac{\pi}{4}+\pi=\frac{5\pi}{4},\frac{3\pi}{4}+\pi=\frac{7\pi}{4}) in the interval (0\leq\theta\leq2\pi). When (\theta=\frac{\pi}{4},\frac{5\pi}{4}), (r = 4\sin(2\times\frac{\pi}{4})=4\sin(\frac{\pi}{2}) = 4) and (r = 4\sin(2\times\frac{5\pi}{4})=4\sin(\frac{5\pi}{2}) = 4). When (\theta=\frac{3\pi}{4},\frac{7\pi}{4}), (r = 4\sin(2\times\frac{3\pi}{4})=4\sin(\frac{3\pi}{2})=-4) and (r = 4\sin(2\times\frac{7\pi}{4})=4\sin(\frac{7\pi}{2})=-4). The absolute - value of (r) represents the distance from the origin, and (|r| = 4) for (\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}). So the function (r = f(\theta)) has extrema at (\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}), and all these points are at a distance of 4 from the origin (farthest from the origin since the maximum value of (|r|) for this function is 4).

Answer:

A. For the input values (\theta=\frac{\pi}{4},\theta=\frac{3\pi}{4},\theta=\frac{5\pi}{4}), and (\theta=\frac{7\pi}{4}), the function (r = f(\theta)) has extrema that correspond to points that are farthest from the origin.