the polar function r = f(θ), where f(θ) = 1 + 2 cos θ, is graphed in the polar coordinate system for 0 ≤ θ ≤…

the polar function r = f(θ), where f(θ) = 1 + 2 cos θ, is graphed in the polar coordinate system for 0 ≤ θ ≤ 2π. on which of the following intervals of θ is the distance between the point with polar coordinates (f(θ), θ) and the origin decreasing? a (0, 2.094) only b (2.094, 4.189) c (0, 2.094) and (3.142, 4.189) d (2.094, 3.142) and (4.189, 6.283)

the polar function r = f(θ), where f(θ) = 1 + 2 cos θ, is graphed in the polar coordinate system for 0 ≤ θ ≤ 2π. on which of the following intervals of θ is the distance between the point with polar coordinates (f(θ), θ) and the origin decreasing? a (0, 2.094) only b (2.094, 4.189) c (0, 2.094) and (3.142, 4.189) d (2.094, 3.142) and (4.189, 6.283)

Answer

Explanation:

Step1: Recall the distance in polar - coordinates

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

Step2: Find the derivative of $r$ with respect to $\theta$

Differentiate $r = 1+2\cos\theta$ with respect to $\theta$. Using the derivative formula $\frac{d}{d\theta}\cos\theta=-\sin\theta$, we get $r'=- 2\sin\theta$.

Step3: Determine when $r$ is decreasing

A function $y = r(\theta)$ is decreasing when $r'(\theta)<0$. So we set $-2\sin\theta<0$, which simplifies to $\sin\theta>0$.

Step4: Solve the inequality for $\theta$ in the given domain

We know that $\sin\theta>0$ for $0 <\theta<\pi$ in the domain $0\leq\theta\leq2\pi$. Also, we can find the critical - points of $r = 1 + 2\cos\theta$ by setting $r'=-2\sin\theta = 0$, so $\theta = 0,\pi,2\pi$. We can also test some values. For example, when $\theta = 0$, $r=1 + 2\cos(0)=3$; when $\theta=\frac{\pi}{2}$, $r = 1+2\cos(\frac{\pi}{2})=1$; when $\theta=\pi$, $r=1 + 2\cos(\pi)=-1$; when $\theta=\frac{3\pi}{2}$, $r=1+2\cos(\frac{3\pi}{2})=1$. We want to find where $r$ is decreasing. We know that $r = 1 + 2\cos\theta$. The function $r$ is decreasing when $\sin\theta>0$. If we consider the values in radians, $\sin\theta>0$ for $0<\theta<\pi$. Converting the values in the options to radians (approximate values: $2.094\approx\frac{2\pi}{3}$, $4.189\approx\frac{4\pi}{3}$, $3.142\approx\pi$, $6.283\approx2\pi$), the function $r = 1 + 2\cos\theta$ is decreasing when $\sin\theta>0$. The function $r = 1+2\cos\theta$ is decreasing on the interval $(\frac{2\pi}{3},\pi)$ and $(\frac{4\pi}{3},2\pi)$. In decimal form, approximately on the interval $(2.094,3.142)$ and $(4.189,6.283)$.

Answer:

D. $(2.094,3.142)$ and $(4.189,6.283)$