consider the graph of the polar function r = f(θ), where f(θ) = 1 + 2sinθ, in the polar coordinate system…

consider the graph of the polar function r = f(θ), where f(θ) = 1 + 2sinθ, in the polar coordinate system for 0 ≤ θ ≤ 2π. which of the following statements is true about the distance between the point with polar coordinates (f(θ),θ) and the origin? a the distance is increasing for 0 ≤ θ ≤ π/2, because f(θ) is positive and increasing on the interval. b the distance is increasing for 3π/2 ≤ θ ≤ 11π/6, because f(θ) is negative and increasing on the interval. c the distance is decreasing for 0 ≤ θ ≤ π/2, because f(θ) is positive and decreasing on the interval. d the distance is decreasing for 3π/2 ≤ θ ≤ 11π/6, because f(θ) is negative and decreasing on the interval.

consider the graph of the polar function r = f(θ), where f(θ) = 1 + 2sinθ, in the polar coordinate system for 0 ≤ θ ≤ 2π. which of the following statements is true about the distance between the point with polar coordinates (f(θ),θ) and the origin? a the distance is increasing for 0 ≤ θ ≤ π/2, because f(θ) is positive and increasing on the interval. b the distance is increasing for 3π/2 ≤ θ ≤ 11π/6, because f(θ) is negative and increasing on the interval. c the distance is decreasing for 0 ≤ θ ≤ π/2, because f(θ) is positive and decreasing on the interval. d the distance is decreasing for 3π/2 ≤ θ ≤ 11π/6, because f(θ) is negative and decreasing on the interval.

Answer

Explanation:

Step1: Recall polar - distance property

In polar coordinates, the distance between a point $(r,\theta)$ and the origin is given by $r$. Here $r = f(\theta)=1 + 2\sin\theta$.

Step2: Analyze the derivative of $f(\theta)$

First, find the derivative of $f(\theta)$: $f'(\theta)=2\cos\theta$.

Step3: Analyze the sign of $f'(\theta)$ on different intervals

  • For $0\leq\theta\leq\frac{\pi}{2}$, $\cos\theta\geq0$, so $f'(\theta)=2\cos\theta\geq0$. Also, $f(\theta)=1 + 2\sin\theta\geq1$ (since $\sin\theta\geq0$ for $0\leq\theta\leq\frac{\pi}{2}$). When $f'(\theta)>0$, the function $f(\theta)$ is increasing. The distance $r = f(\theta)$ from the point $(f(\theta),\theta)$ to the origin is increasing for $0\leq\theta\leq\frac{\pi}{2}$ because $f(\theta)$ is positive and increasing on this interval.
  • For $\frac{3\pi}{2}\leq\theta\leq\frac{11\pi}{6}$, $\cos\theta\geq0$, $f(\theta)=1 + 2\sin\theta$. When $\theta=\frac{3\pi}{2}$, $f(\frac{3\pi}{2})=1+2\times(- 1)=-1$, and as $\theta$ increases from $\frac{3\pi}{2}$ to $\frac{11\pi}{6}$, $\sin\theta$ increases from - 1 to $-\frac{1}{2}$, so $f(\theta)$ is negative but increasing. However, the distance from the point to the origin is $|f(\theta)|$. When $f(\theta)$ is negative and increasing, the distance (which is non - negative) is actually decreasing.

Answer:

A. The distance is increasing for $0\leq\theta\leq\frac{\pi}{2}$, because $f(\theta)$ is positive and increasing on the interval.