Questions: Plot the point (1, 5π/6), given in polar coordinates, and find other polar coordinates (r, θ) of the point for which the following are true. (a) r>0, -2π ≤ θ<0 (b) r<0, 0 ≤ θ<2π (c) r>0, 2π ≤ θ<4π

Solution Steps

The given polar coordinates are $(1, \frac{5\pi}{6})$. We can convert these to rectangular coordinates $(x,y)$ using the formulas $x = r\cos\theta$ and $y = r\sin\theta$. In this case, $r=1$ and $\theta = \frac{5\pi}{6}$, so we have: $x = 1\cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}$ $y = 1\sin(\frac{5\pi}{6}) = \frac{1}{2}$

So the rectangular coordinates are $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$.

The point $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$ lies in the second quadrant. Among the given options, only option B shows a point in the second quadrant.

(a) r > 0, -2π ≤ θ < 0: We need a positive $r$ and a negative $\theta$. We can achieve this by subtracting $2\pi$ from the original $\theta$: $\theta = \frac{5\pi}{6} - 2\pi = \frac{5\pi - 12\pi}{6} = -\frac{7\pi}{6}$. So the polar coordinates are $(1, -\frac{7\pi}{6})$.

(b) r < 0, 0 ≤ θ < 2π: We need a negative $r$ and a positive $\theta$. We can achieve this by adding $\pi$ to the original $\theta$ and making $r$ negative: $\theta = \frac{5\pi}{6} + \pi = \frac{5\pi + 6\pi}{6} = \frac{11\pi}{6}$. So the polar coordinates are $(-1, \frac{11\pi}{6})$.

(c) r > 0, 2π ≤ θ < 4π: We need a positive $r$ and a $\theta$ between $2\pi$ and $4\pi$. We can achieve this by adding $2\pi$ to the original $\theta$: $\theta = \frac{5\pi}{6} + 2\pi = \frac{5\pi + 12\pi}{6} = \frac{17\pi}{6}$. So the polar coordinates are $(1, \frac{17\pi}{6})$.

Final Answer