9.3: Polar Coordinates and Functions

1. We draw an angle that begins at the positive \(x\)-axis—the polar axis—and extends \(\pi/4\) radian in the counterclockwise direction. Then we draw a line segment that begins at the pole \((O)\) and extends \(4\) units. This point is labeled \(A\) in Figure 3.

---

2. From the positive \(x\)-axis, we draw an angle that extends \(5 \pi/6\) radians in the counterclockwise direction. The line segment needs to extend \(7\) units from the pole. This point is labeled \(B\) in Figure 3.

---

3. We follow the same procedure in (ii) but rotate the line segment \(\pi\) radians about the pole because \(r \lt 0.\) Hence, we obtain point \(C.\)

We want to eliminate \(r\) and \(\theta\) to write the polar function entirely in terms of \(x\) and \(y.\) It is a good idea to rewrite the trigonometric expression in terms of sine and cosine. We get \[r = \frac{\sin \theta}{\cos^2 \theta} \pd\] Then we use \(\eqref{eq:polar-x-y}\) [\(x = r \cos \theta\) and \(y = r \sin \theta\)] to write \(\sin \theta\) as \(y/r\) and \(\cos^2 \theta\) as \(x^2/r^2 \col\) \[ \ba r &= \frac{y/r}{x^2/r^2} \nl r &= \frac{ry}{x^2} \pd \ea \] We arrive at \[1 = \frac{y}{x^2} \or \boxed{y = x^2}\] Thus, the graph of the polar function \(r = \sec \theta \tan \theta\) is the same as the graph of \(y = x^2\) (Figure 4).

Our goal is to eliminate \(x\) and \(y\) and solve for \(r\) in terms of \(\theta.\) We write \(y/x\) as \(\tan \theta\) and \(x\) as \(r \cos \theta,\) obtaining \[ \ba r \cos \theta - 1 &= \tan^2 \theta \nl r \cos \theta &= \tan^2 \theta + 1 \pd \ea \] Since \(\tan^2 \theta + 1 = \sec^2 \theta,\) we have \[r \cos \theta = \sec^2 \theta \or \boxed{r = \sec^3 \theta}\] The graph of this function is in Figure 5.

The key idea is to obtain values of \(r = 2 \cos \theta\) for some values of \(\theta,\) and plot and connect the corresponding points \((r, \theta).\) We assemble a table with convenient values of \(\theta,\) as follows.

\(\theta\)	\(0\)	\(\ds \frac{\pi}{6}\)	\(\ds \frac{\pi}{4}\)	\(\ds \frac{\pi}{3}\)	\(\ds \frac{\pi}{2}\)	\(\ds \frac{2 \pi}{3}\)	\(\ds \frac{3 \pi}{4}\)	\(\ds \frac{5 \pi}{6}\)	\(\pi\)
\(r\)	\(2\)	\(\sqrt 3\)	\(\sqrt 2\)	\(1\)	\(0\)	\(-1\)	\(-\sqrt 2\)	\(-\sqrt 3\)	\(-2\)

The graph makes a full loop in \(\pi\) radians. For \(\theta \gt \pi,\) the graph of \(r = 2 \cos \theta\) repeats itself. We therefore plot the graph in Figure 8.

The standard form of a circle in Cartesian coordinates is \(\par{x - x_0}^2 + \par{y - y_0}^2 = k^2,\) where the circle is centered at \(\par{x_0, y_0}\) with a radius of \(k.\) We use the conversion formulas in \(\eqref{eq:polar-x-y}\) [\(x = r \cos \theta\) and \(y = r \sin \theta\)] and \(r = \sqrt{x^2 + y^2}\) to convert \(r = 2 \cos \theta\) into \begin{align} r &= 2 \, \frac{x}{r} \nonumber \nl r^2 &= 2x \nonumber \nl x^2 + y^2 &= 2x \nonumber \pd \end{align} We then complete the square to get \[x^2 - 2x + \orange{1} + y^2 = \orange 1 \implies \boxed{(x - 1)^2 + y^2 = 1} \] This circle is centered at \((x, y) =\) \((1, 0)\) with a radius of \(1.\)

1. The graph of the polar function \(r = k\) is a circle of radius \(\abs k\) centered at the origin. Hence, the center is \((x, y)\) \(= (0, 0)\) and the radius is \(4.\)

---

2. The polar function is of the form \(r = k \sin \theta,\) so its graph is centered at a point on the \(y\)-axis. Because the coefficient \(6\) is positive, the graph is centered at a point on the positive \(y\)-axis—namely, \((x, y)\) \(= (0, 3)\)—with a radius of \(3.\)

---

3. For \(r = k \cos \theta,\) the graph is centered on the negative \(x\)-axis if \(k \lt 0.\) The center is \((x, y)\) \(=(-7/2, 0),\) and the radius is \(7/2.\)

1. Since \(\theta\) is fixed, this polar function represents a line. The line's slope \(m\) is \(\tan \theta,\) or \[m = \tan \frac{\pi}{3} = \sqrt 3 \pd\] In Cartesian form a line that passes through the origin with slope \(m\) is \(y = mx.\) Because \(m = \sqrt 3,\) we write the polar function in Cartesian form as \[\boxed{y = x \sqrt 3}\]

---

2. The form \(r = c \csc \theta\) represents the horizontal line \(y = c.\) Here \(c = 4,\) so we write the expression in Cartesian form as \[\boxed{y = 4}\]

---

3. The graph of \(r = c \sec \theta\) is the vertical line \(x = c.\) Thus, we write \(r = -2 \sec \theta\) in Cartesian form as \[\boxed{x = -2}\]

We note that \(r\) is increasing directly with \(\theta.\) Hence, plotting some values of \((r, \theta)\) shows that the graph is a spiral that passes through the pole (Figure 13).

We need to determine how \(r\) varies with \(\theta\) to sketch the polar curve. It is easiest to first plot \(r = 1 + \cos \theta\) in Cartesian coordinates (Figure 14A). We note four key stages.

1. As \(\theta\) increases from \(0\) to \(\pi/2,\) \(r\) decreases from \(2\) to \(1.\) Hence, two points on the polar curve are \((r, \theta) = (2, 0)\) and \((r, \theta) = (1, \pi/2).\) Note the decreasing distance between \((r, \theta)\) and the pole. We connect these points using an arc that extends through these points.
2. Note that \(r = 0\) when \(\theta = \pi.\) Thus, the graph of the polar curve hits the origin when \(\theta = \pi.\) Since \(r\) is decreasing, the polar curve is getting closer to the origin. We therefore draw an arc that connects \((r, \theta)\) \(= \par{1, \pi/2}\) to \((r, \theta)\) \(= \par{0, \pi}.\)
3. As \(\theta\) increases from \(\pi\) to \(3 \pi/2,\) \(r\) increases from \(0\) to \(1.\) Accordingly, on the polar curve we draw an arc connecting \((r, \theta)\) \(= \par{0, \pi}\) to \((r, \theta)\) \(= (1, 3\pi/2).\)
4. The graph of \(r = 1 + \cos \theta\) finishes a complete revolution. So we close the loop on the polar graph.

These remarks allow us to sketch the graph in Figure 14B. If we wished to graph \(r = 1 + \cos \theta\) for \(\theta \gt 2 \pi,\) then we would retrace our path.

Alternatively, we could have noticed that, because cosine is an even function, \[1 + \cos \theta = 1 + \cos(-\theta) \pd\] Hence, the polar graph of \(r = 1 + \cos \theta\) is symmetric about the \(x\)-axis. So we could have traced the upper half of the graph \((0 \leq \theta \leq \pi)\) and mirrored it across the \(x\)-axis to obtain the bottom half of the graph \((\pi \leq \theta \leq 2\pi).\)

To understand the behavior of \(r\) as \(\theta\) changes, we first plot the graph of \(r = 1 + 2 \sin \theta\) in Cartesian coordinates (Figure 15A). Six stages are of interest to us.

1. As \(\theta\) increases from \(0\) to \(\pi/2,\) \(r\) increases from \(1\) to \(3.\) Thus, the polar graph of \(r = 1 + 2 \sin \theta\) passes through the points \((r, \theta) = (1, 0)\) and \((r, \theta) = (3, \pi/2).\) We therefore draw an arc that connects these points.
2. As \(\theta\) increases from \(\pi/2\) to \(\pi,\) \(r\) decreases from \(3\) to \(1.\) So the polar curve passes through the points \((r, \theta) = (3, \pi/2)\) and \((r, \theta) = (1, \pi).\)
3. The Cartesian graph of \(r = 1 + 2 \sin \theta\) has a zero when \(\theta = 7 \pi/6.\) The polar graph therefore crosses the pole when \(\theta = 7 \pi/6.\) So we connect the point \((r, \theta) = (1, \pi)\) to \((r, \theta) = (0, 7 \pi/6).\)
4. As \(\theta\) increases from \(7 \pi/6\) to \(3 \pi/2,\) \(r\) decreases from \(0\) to \(-1.\) Accordingly, the polar curve passes through \((r, \theta) = (-1, 3\pi/2).\) Since \(r \lt 0,\) this point is a reflection of the point \((1, 3 \pi/2)\) by \(\pi\) radians—giving us a point on the positive \(y\)-axis.
5. Note that \(r = 0\) when \(\theta = 11 \pi/6.\) So we draw an arc connecting the polar curve from \((r, \theta) = (-1, 3\pi/2)\) to the pole at \(\theta = 11 \pi/6.\) Steps IV and V trace an inner loop of the graph.
6. Because \((r, \theta) = (1, 2 \pi)\) represents the same point as \((r, \theta) = (1, 0),\) we close the graph of the polar curve.

We therefore attain the polar curve in Figure 15B.

We first plot the graph of \(r = \cos 2 \theta\) in Cartesian coordinates (Figure 17A). We analyze eight stages.

1. As \(\theta\) increases from \(0\) to \(\pi/4,\) \(r\) decreases from \(1\) to \(0.\) Hence, on the polar curve \(r = \cos 2 \theta\) we plot two points: \((r, \theta) = (1, 0)\) and \((r, \theta) = (0, \pi/4).\) The arc connecting these points forms half a petal.
2. For \(\theta \in (\pi/4, \pi/2),\) \(r\) is negative and drops to \(-1\) when \(\theta = \pi/2.\) The points on the polar graph therefore lie on the opposite side of the first quadrant—in the third quadrant, forming a half-petal that begins at the origin and terminates at the ray \(\theta = 3 \pi/2.\)
3. As \(\theta\) increases from \(\pi/2\) to \(3 \pi/4,\) \(r\) increases from \(-1\) to \(0.\) Since \(r\) is negative in this interval, the points on the polar curve reside in the fourth quadrant instead of in the second quadrant. These points trace a half-petal that starts at the ray \(\theta = 3\pi/2\) and ends at the origin. Stages II and III form a full petal of the graph on the negative \(y\)-axis.
4. As \(\theta\) increases from \(3 \pi/4\) to \(\pi,\) \(r\) increases from \(0\) to \(1.\) Thus, points on the polar graph of \(r = \cos 2 \theta\) trace a half-petal in the second quadrant that begins at the origin and terminates at the ray \(\theta = \pi.\)
5. As \(\theta\) increases from \(\pi\) to \(5 \pi/4,\) \(r\) decreases from \(1\) to \(0.\) Accordingly, in the third quadrant the polar graph of \(r = \cos 2 \theta\) traces a half-petal that begins at the ray \(\theta = \pi\) and ends at the origin. Stages IV and V therefore form a full petal that resides on the negative \(x\)-axis.
6. The polar graph passes through \((r, \theta) = (0, 5 \pi/4)\) and \((r, \theta) = (-1, 3\pi/2).\) Since \(r \lt 0,\) instead of the points tracing an arc in the third quadrant, they form an arc in the first quadrant that begins at the origin and ends at the positive \(y\)-axis.
7. Now \(r\) increases from \(-1\) to \(0\) as \(\theta\) increases from \(3\pi/2\) to \(7 \pi/4.\) Because \(r \lt 0,\) the arc of the polar graph—instead of residing in the fourth quadrant—traces out a half-petal in the second quadrant. Stages VI and VII trace a full petal that lies on the positive \(y\)-axis.
8. The polar graph passes through the points \((r, \theta) = (0, 7\pi/4)\) and \((r, \theta) = (1, 2\pi).\) Hence, this last arc culminates the petal that lies on the positive \(x\)-axis.

We therefore sketch the graph in Figure 17B.

Alternatively, a simpler option is to exploit the symmetry of the graph: The curve is symmetric about the \(x\)-axis because \[\cos 2 \theta \equalsCheck \cos(-2 \theta) \pd\] It is also symmetric about the \(y\)-axis since \[\cos 2 \theta \equalsCheck \cos[2(\pi - \theta)] \pd\] From these remarks, we can sketch the half-petal in stage I, reflect a copy across the \(x\)-axis (obtaining VIII), and then reflect both halves across the \(y\)-axis (to get IV and V). Doing so gives us the two petals that lie on the \(x\)-axis. To obtain the vertical petals, we can sketch stage II of the polar graph. Reflecting a copy of this half-petal across the \(y\)-axis gives III, and reflecting both halves across the \(x\)-axis yields portions VII and VI. This entire procedure covers all values of \(\theta\) and therefore produces the full graph of \(r = \cos 2 \theta.\) When we apply symmetry, we must ensure that our strategy covers all values of \(\theta;\) otherwise, we may miss portions of the graph.

First graphing \(r = \sin 3 \theta\) in Cartesian coordinates (Figure 18A), we remark the following properties.

1. The polar graph passes through the origin when \(\theta = 0\) since \(r = 0.\) Then \(r\) increases to \(1\) when \(\theta = \pi/6,\) before falling to \(0\) when \(\theta = \pi/3.\) Accordingly, for the polar curve \(r = \sin 3 \theta\) we trace a petal in the first quadrant that intersects the origin at \(\theta = 0\) and \(\theta = \pi/3,\) with a vertex occurring when \(\theta = \pi/6.\)
2. For \(\pi/3 \lt \theta \lt 2 \pi/3,\) \(r\) is negative. The polar graph therefore enters the third and fourth quadrants. For \(\pi/3 \leq \theta \leq \pi/2\) we trace a half-petal in the third quadrant, and for \(\pi/2 \leq \theta \leq 2 \pi/3\) we draw the other half of the petal in the fourth quadrant. Doing so yields a petal along the negative \(y\)-axis.
3. For \(2 \pi/3 \lt \theta \lt \pi,\) \(r\) is positive. Hence, in the second quadrant we draw a petal that strikes the origin when \(\theta = 2 \pi/3\) and \(\theta = \pi.\)

This analysis yields the graph in Figure 18B.

We dissect the equation into two parts: \begin{equation} r = \sqrt 2 \sqrt{\cos 2 \theta} \and r = - \sqrt 2 \sqrt{\cos 2 \theta} \pd \label{eq:ex-lemniscate-parts} \end{equation} Since \(\cos 2 \theta\) cannot be negative in these equations, \(\sqrt{\cos 2 \theta}\) is defined for \(0 \leq \theta \leq \pi/4\) and \(3 \pi/4 \leq \theta \leq \pi.\) We sketch the graph of each part of \(\eqref{eq:ex-lemniscate-parts}\) and combine the curves. The lemniscate passes through the Cartesian points \((x, y)\) \(= (-\sqrt 2, 0)\) and \((x, y)\) \(= (\sqrt 2, 0).\) At \(\theta = \pi/4\) and \(\theta = 3 \pi/4,\) both graphs hit the pole. The graph of \(r = \sqrt 2 \sqrt{\cos 2 \theta}\) forms the upper half of the lemniscate, and the graph of \(r = -\sqrt 2 \sqrt{\cos 2 \theta}\) forms the bottom half. (See Figure 19.)