극좌표와 극방정식의 그래프

$r=7/(\sin\theta-4\cos\theta)$

$r=5\sec \theta$

$r=5\csc \theta$

$r=\cos\theta/\sin^2\theta$

$r=1/\sqrt[3]{\cos^2\theta\sin\theta}$

$r=\sqrt{7}$

$r=-2\cos\theta$

$P(\sqrt{3}/2,\pi/6)$, $P(0,\pi/2)$, $P(-\sqrt{3}/2,5\pi/6)$

$P(1,0)$, $P(0,\pi/2)=P(0,\pi/4)$, $P(-1/2,2\pi/3)$

$P(0,0)=P(0,\pi/2)$, $P(\sqrt{2},\pi/4)$

$P(\sqrt{3}/2,\pi/3)=P(-\sqrt{3}/2,4\pi/3)$, $P(\sqrt{3}/2,2\pi/3)=P(-\sqrt{3}/2,5\pi/3)$, $P(0,\pi/2)$

$P(0,0)=P(0,\pi/6)$, $P(\sqrt{2}/2,\pi/12)$, $P(-\sqrt{2}/2,5\pi/12)$, $P(\sqrt{2}/2,3\pi/4)$

$P(3/2,\pi/3)$, $P(3/2,-\pi/3)$

For all points, $r=1$; $\theta = \pi/12,\ 5\pi/12,\ 7\pi/12,\ 11\pi/12,\ 13\pi/12,\ 17\pi/12,\ 19\pi/12,\ 23\pi/12$.

$P(0,0)=P(0,3\pi/2)$, $P(1+\sqrt{2}/2,3\pi/4)$, $P(1-\sqrt{2}/2,7\pi/4)$

Answers will vary. If $m$ and $n$ do not have any common factors, then an interval of $2n\pi$ is needed to sketch the entire graph.

Answers will vary.

The left hand sketch below contains the points, $(x_1,y_1)$, $(x_3,y_3)$, $(x_5,y_5)$, that are on the axes. The right hand sketch below contains the points, $(x_2,y_2)$, $(x_4,y_4)$, that are not on the axes.

Recall that the polar coordinates $r$, $\theta$ are related to the cartesian coordinates $x$, $y$, by $x=r\cos\theta$, $y=r\sin\theta$. So $r=\sqrt{x^2+y^2}$ and $\tan\theta=\frac{y}{x}$ (assuming that $x\ne 0$) and

$$\begin{aligned} (x_1,y_1) &= (3,0) &&\implies &r_1=3,\ \tan\theta_1=0 &&\implies \theta_1&=0 \text{ as $(x_1,y_1)$ is on the positive $x$-axis} \\ (x_2,y_2) &= (1,1) &&\implies &r_2=\sqrt{2},\ \tan\theta_2=1 &&\implies \theta_2&=\frac{\pi}{4} \text{ as $(x_2,y_2)$ is in the first octant} \\ (x_3,y_3) &= (0,1) &&\implies &r_3=1,\ \cos\theta_3=0 &&\implies \theta_3&=\frac{\pi}{2} \text{ as $(x_3,y_3)$ is on the positive $y$-axis} \\ (x_4,y_4) &= (-1,1) &&\implies &r_4=\sqrt{2},\ \tan\theta_4=-1 &&\implies \theta_4&=\frac{3\pi}{4} \text{ as $(x_4,y_4)$ is in the third octant} \\ (x_5,y_5) &= (-2,0) &&\implies &r_5=2,\ \tan\theta_5=0 &&\implies \theta_5&=\pi \text{ as $(x_5,y_5)$ is on the negative $x$-axis} \end{aligned}$$

Note that the distance from the point $\big(r\cos\theta\,,\,r\sin\theta\big)$ to the origin is

$$\sqrt{r^2\cos^2\theta + r^2\sin^2\theta} =\sqrt{r^2} =|r|$$

Thus $r$ can be either the distance to the origin or minus the distance to the origin.

(a) The distance from $(-2,0)$ to the origin is $2$. So either $r=2$ or $r=-2$.

• If $r=2$, then $\theta$ must obey

$$\begin{aligned} (-2,0) = \big(2\cos\theta\,,\,2\sin\theta\big) &\iff \sin\theta=0,\ \cos\theta=-1 \\ &\iff \theta= n\pi,\ n\text{ integer },\ \cos\theta=-1 \\ &\iff \theta= n\pi,\ n\text{ odd integer } \end{aligned}$$

• If $r=-2$, then $\theta$ must obey

$$\begin{aligned} (-2,0) = \big(-2\cos\theta\,,\,-2\sin\theta\big) &\iff \sin\theta=0,\ \cos\theta=1 \\ &\iff \theta= n\pi,\ n\text{ integer },\ \cos\theta=1 \\ &\iff \theta= n\pi,\ n\text{ even integer } \end{aligned}$$

In the figure on the left below, the blue half-line is the set of all points with polar coordinates $\theta=\pi$, $r>0$ and the pink half-line is the set of all points with polar coordinates $\theta=\pi$, $r<0$. In the figure on the right below, the blue half-line is the set of all points with polar coordinates $\theta=0$, $r>0$ and the pink half-line is the set of all points with polar coordinates $\theta=0$, $r<0$.

(b) The distance from $(1,1)$ to the origin is $\sqrt{2}$. So either $r=\sqrt{2}$ or $r=-\sqrt{2}$.

• If $r=\sqrt{2}$, then $\theta$ must obey

$$\begin{aligned} (1,1) = \big(\sqrt{2}\,\cos\theta\,,\,\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta={1}/{\sqrt{2}} \\ &\iff \theta= {\pi}/{4} + 2n\pi,\ n\text{ integer } \end{aligned}$$

• If $r=-\sqrt{2}$, then $\theta$ must obey

$$\begin{aligned} (1,1) = \big(-\sqrt{2}\,\cos\theta\,,\,-\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta=-{1}/{\sqrt{2}} \\ &\iff \theta= {5\pi}/{4} + 2n\pi,\ n\text{ integer } \end{aligned}$$

In the figure on the left below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}$, $r>0$ and the pink half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}$, $r<0$. In the figure on the right below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}$, $r>0$ and the pink half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}$, $r<0$.

(c) The distance from $(-1,-1)$ to the origin is $\sqrt{2}$. So either $r=\sqrt{2}$ or $r=-\sqrt{2}$.

• If $r=\sqrt{2}$, then $\theta$ must obey

$$\begin{aligned} (-1,-1) = \big(\sqrt{2}\,\cos\theta\,,\,\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta=-{1}/{\sqrt{2}} \\ &\iff \theta= {5\pi}/{4} + 2n\pi,\ n\text{ integer } \end{aligned}$$

• If $r=-\sqrt{2}$, then $\theta$ must obey

$$\begin{aligned} (-1,-1) = \big(-\sqrt{2}\,\cos\theta\,,\,-\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta={1}/{\sqrt{2}} \\ &\iff \theta= {\pi}/{4} + 2n\pi,\ n\text{ integer } \end{aligned}$$

In the figure on the left below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}$, $r>0$ and the pink half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}$, $r<0$. In the figure on the right below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}$, $r>0$ and the pink half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}$, $r<0$.