Polar coordinates

Given rectangular coordinates $(x, y)$, the corresponding polar coordinates can be found. The value of $r$ is given by $r = \sqrt{x^2 + y^2}$. The value of $\theta$ can be obtained as follows:

From $x = r\cos\theta$ and $y = r\sin\theta$, we get $\tan\theta = \frac{y}{x}$.

Thus, $\theta = \tan^{-1}\left(\frac{y}{x}\right)$.

Note: The arctangent function ($\tan^{-1}$) typically gives angles in the first or fourth quadrants. For points $(x, y)$ in the second or third quadrants, $\pi$ radians (or 180°) must be added to the result. Therefore:

• If $x > 0$: $\theta = \tan^{-1}\left(\frac{y}{x}\right)$

• If $x < 0$: $\theta = \tan^{-1}\left(\frac{y}{x}\right) + \pi$

Example 1

Write the rectangular coordinates $(-7, 0)$ in polar coordinates.

Solution:

$r = \sqrt{(-7)^2 + 0^2} = \sqrt{49} = 7$

$\tan\theta = \frac{0}{-7} = 0$

Since $x < 0$, $\theta = \tan^{-1}(0) + \pi = 0 + \pi = \pi$.

The polar coordinates are $(7, \pi)$.

Example 2

Find the polar coordinates corresponding to the rectangular coordinates $(0, 4)$.

Solution:

$r = \sqrt{0^2 + 4^2} = \sqrt{16} = 4$

$\tan\theta = \frac{4}{0}$, which is undefined. This indicates a vertical line. Since the point is $(0, 4)$, it lies on the positive y-axis.

Thus, $\theta = \frac{\pi}{2}$.

The polar coordinates are $(4, \frac{\pi}{2})$.

Example 3

Transform the polar equation $r = \frac{4}{3 + 5\sin\theta}$ into a rectangular equation. Identify the resulting curve.

Solution:

Given $r = \frac{4}{3 + 5\sin\theta}$, we have $r(3 + 5\sin\theta) = 4$, so $3r + 5r\sin\theta = 4$.

Substituting $y = r\sin\theta$, we get $3r + 5y = 4$, so $3r = 4 - 5y$.

Squaring both sides:

$9r^2 = (4 - 5y)^2$

Substituting $r^2 = x^2 + y^2$:

$9(x^2 + y^2) = 16 - 40y + 25y^2$

$9x^2 + 9y^2 = 16 - 40y + 25y^2$

$9x^2 - 16y^2 + 40y - 16 = 0$

Completing the square for the y terms:

$9x^2 - 16\left(y^2 - \frac{5}{2}y\right) = 16$

$9x^2 - 16\left(y^2 - \frac{5}{2}y + \frac{25}{16}\right) = 16 - 25$

$9x^2 - 16\left(y - \frac{5}{4}\right)^2 = -9$

$\frac{\left(y-\frac{5}{4}\right)^2}{\frac{9}{16}} - \frac{x^2}{1} = 1$

This is the equation of a hyperbola.

Example 4

Transform the rectangular equation $(x^2 + y^2)^3 = 4x^2y^2$ into a polar equation.

Solution:

Substitute $x = r\cos\theta$ and $y = r\sin\theta$, and $x^2+y^2=r^2$:

$(r^2)^3 = 4(r\cos\theta)^2(r\sin\theta)^2$

$r^6 = 4r^4\cos^2\theta\sin^2\theta$

If $r$ is not zero:

$r^2 = 4\cos^2\theta\sin^2\theta$

$r^2 = (2\sin\theta\cos\theta)^2$

$r^2 = (\sin 2\theta)^2$

$r = \sin 2\theta$

Polar equations (e.g., $r = 2\theta$ and $r = 4\sin\theta$) can be graphed. A graph of a polar equation is the set of all points whose coordinates satisfy the equation, typically in the form $r = f(\theta)$.

The form of the polar equation often suggests the curve's nature. Equations like $r = a\sin\theta$, $r = a\cos\theta$, and $r = a$ represent circles.

Equations of the form $r = a \pm b\cos\theta$ and $r = a \pm b\sin\theta$, where $a > 0$ and $b > 0$, produce curves called limacons (snails). Equations of the form $r^2 = a^2\cos 2\theta$ and $r^2 = a^2\sin 2\theta$ are lemniscates.

Steps for sketching polar curves:

1. Select values of $\theta$. Usually multiples of $\frac{\pi}{6}$ (30°).
2. Calculate the corresponding $r$ values.
3. Create a table of values.
4. Sketch the graph of $r$ against $\theta$.
5. Join the points with a smooth curve.

Polar equations of the form $r = a\sin\theta$ and $r = a\cos\theta$

Equations of the form $r = a\sin\theta$ and $r = a\cos\theta$, where $a \ne 0$, represent circles. The value of $a$ determines the diameter.

Example 1

Sketch the graph of $r = 5\sin\theta$.

Solution:

Table of values:

$\theta$	$r = 5\sin\theta$
$0$	$0$
$\frac{\pi}{6}$	$2.5$
$\frac{\pi}{3}$	$4.33$
$\frac{\pi}{2}$	$5$
$\frac{2\pi}{3}$	$4.33$
$\frac{5\pi}{6}$	$2.5$
$\pi$	$0$
$\frac{7\pi}{6}$	$-2.5$

(Insert graph of r = 5sinθ here)

Example 2

Sketch the graph of $r = 1 + 4\cos\theta$.

Solution:

Table of values:

$\theta$	$r = 1 + 4\cos\theta$	$(r, \theta)$
$0$	$5$	$(5, 0)$
$\frac{\pi}{6}$	$1 + 4\left(\frac{\sqrt{3}}{2}\right) \approx 4.46$	$(4.46, \frac{\pi}{6})$
$\frac{\pi}{3}$	$1+4\left(\frac{1}{2}\right)=3$	$(3, \frac{\pi}{3})$
$\frac{\pi}{2}$	$1$	$(1, \frac{\pi}{2})$
$\frac{2\pi}{3}$	$1+4\left(-\frac{1}{2}\right)=-1$	$(-1, \frac{2\pi}{3})$
$\pi$	$-3$	$(-3, \pi)$
$\frac{4\pi}{3}$	$-1$	$(-1, \frac{4\pi}{3})$
$\frac{3\pi}{2}$	$1$	$(1, \frac{3\pi}{2})$
$\frac{5\pi}{3}$	$3$	$(3, \frac{5\pi}{3})$
$\frac{11\pi}{6}$	$4.46$	$(4.46, \frac{11\pi}{6})$
$2\pi$	$5$	$(5, 2\pi)$

Polar equations of the form $r = a \pm b\cos\theta$ and $r = a \pm b\sin\theta$

These equations are called limacons or snails.

Polar equations of the form $r^2 = a^2\cos 2\theta$ and $r^2 = a^2\sin 2\theta$, where $a \ne 0$, produce lemniscates (figure-eight shaped curves).

Example 1

Sketch the graph of the polar equation $r^2 = 2\sin 2\theta$.

Solution:

It's important to note that since we have $r^2$, $r$ can be positive or negative. Also, $r^2$ cannot be negative, so $2\sin 2\theta$ must be greater or equal to zero. This means that $2\theta$ must be in the first and second quadrants (and their coterminal angles), meaning $0 \le 2\theta \le \pi$ or $2\pi \le 2\theta \le 3\pi$ and so on. Therefore $0 \le \theta \le \frac{\pi}{2}$ and $\pi \le \theta \le \frac{3\pi}{2}$. We can then use symmetry to complete the graph.

Table of values:

$\theta$	$2\theta$	$\sin 2\theta$	$r^2 = 2\sin 2\theta$	$r$
$0$	$0$	$0$	$0$	$0$
$\frac{\pi}{12}$ (15°)	$\frac{\pi}{6}$ (30°)	$0.5$	$1$	$\pm 1$
$\frac{\pi}{6}$ (30°)	$\frac{\pi}{3}$ (60°)	$\frac{\sqrt{3}}{2} \approx 0.866$	$\sqrt{3} \approx 1.73$	$\pm\sqrt[4]{3} \approx \pm 1.32$
$\frac{\pi}{4}$ (45°)	$\frac{\pi}{2}$ (90°)	$1$	$2$	$\pm\sqrt{2} \approx \pm 1.41$
$\frac{\pi}{3}$ (60°)	$\frac{2\pi}{3}$ (120°)	$\frac{\sqrt{3}}{2} \approx 0.866$	$\sqrt{3} \approx 1.73$	$\pm\sqrt[4]{3} \approx \pm 1.32$
$\frac{5\pi}{12}$ (75°)	$\frac{5\pi}{6}$ (150°)	$0.5$	$1$	$\pm 1$
$\frac{\pi}{2}$ (90°)	$\pi$ (180°)	$0$	$0$	$0$
$\frac{7\pi}{6}$ (210°)	$\frac{7\pi}{3}$ (420°) or $60°$	$\frac{\sqrt{3}}{2}$	$\sqrt{3}$	$\pm\sqrt[4]{3}$
$\frac{5\pi}{4}$ (225°)	$\frac{5\pi}{2}$ (450°) or $90°$	$1$	$2$	$\pm\sqrt{2}$
$\frac{4\pi}{3}$ (240°)	$\frac{8\pi}{3}$ (480°) or $120°$	$\frac{\sqrt{3}}{2}$	$\sqrt{3}$	$\pm\sqrt[4]{3}$
$\frac{11\pi}{6}$ (330°)	$\frac{11\pi}{3}$ (660°) or $300°$	$-\frac{\sqrt{3}}{2}$	$-\sqrt{3}$	No Real Solution

The graph of the polar equation $r^2 = 2\sin 2\theta$ is presented in as follows

Polar equations of the form $r = a\sin n\theta$ and $r = a\cos n\theta$, where $a \ne 0$ and $n$ is an integer greater than 1, produce graphs known as roses. These graphs have loops resembling leaves or petals. The number of leaves depends on $n$:

If $n$ is even, the rose has $2n$ leaves.

If $n$ is odd, the rose has $n$ leaves.

Example 1

Draw the graph of $r = 10\cos 2\theta$.

Solution:

Table of values:

$\theta$	$10\cos 2\theta$	$(r, \theta)$
$0°$	$10$	$(10, 0°)$
$30°$	$5$	$(5, 30°)$
$60°$	$-5$	$(-5, 60°)$
$90°$	$-10$	$(-10, 90°)$
$120°$	$-5$	$(-5, 120°)$
$150°$	$5$	$(5, 150°)$
$180°$	$10$	$(10, 180°)$
$210°$	$5$	$(5, 210°)$
$240°$	$-5$	$(-5, 240°)$
$270°$	$-10$	$(-10, 270°)$
$300°$	$-5$	$(-5, 300°)$
$330°$	$5$	$(5, 330°)$
$360°$	$10$	$(10, 360°)$

The graph of $r = 10\cos 2\theta$

Example 2

Sketch the graph of $r = 5\sin 3\theta$.

Solution:

Table of values:

$\theta$	$5\sin 3\theta$	$(r, \theta)$
$0°$	$0$	$(0, 0°)$
$30°$	$5$	$(5, 30°)$
$60°$	$0$	$(0, 60°)$
$90°$	$-5$	$(-5, 90°)$
$120°$	$0$	$(0, 120°)$
$150°$	$5$	$(5, 150°)$
$180°$	$0$	$(0, 180°)$
$210°$	$-5$	$(-5, 210°)$
$240°$	$0$	$(0, 240°)$
$270°$	$5$	$(5, 270°)$
$300°$	$0$	$(0, 300°)$
$330°$	$-5$	$(-5, 330°)$
$360°$	$0$	$(0, 360°)$

The graph of $r = 5\sin 3\theta$