1.3 Polar Coordinates

1. Use $x=1$ and $y=1$ in Equation 1.8:
   
   $$\begin{array}{ccccccc}\begin{array}{ccc}\hfill r^2& =\hfill & x^2+y^2\hfill \\ & =\hfill & 1^2+1^2\hfill \\ \hfill r& =\hfill & \sqrt{2}\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill \text{tan}\theta & =\hfill & \frac{y}{x}\hfill \\ & =\hfill & \frac{1}{1}=1\hfill \\ \hfill \theta & =\hfill & \frac{\pi }{4}.\hfill \end{array}\hfill \end{array}$$
   
   Therefore this point can be represented as $\left(\sqrt{2},\frac{\pi }{4}\right)$ in polar coordinates.
2. Use $x=\mathrm{-3}$ and $y=4$ in Equation 1.8:
   
   $$\begin{array}{ccccccc}\begin{array}{ccc}{r}^2& =& x^2+y^2\\ & =& {\left(\mathrm{-3}\right)}^2+{\left(4\right)}^2\\ r& =& 5\end{array}& & & \text{and}& & & \begin{array}{ccc}\text{tan}\theta & =& \frac{y}{x}\\ & =& -\frac{4}{3}\\ \theta & =& \pi -\arctan \left(\frac{4}{3}\right)\\ & \approx & 2.21.\end{array}\end{array}$$
   
   The point $(-3, 4)$ lies in Quadrant $\text{II}$. Subtract the value of the reference angle, $\text{arctan}\left(\frac{4}{3}\right)$, from $\pi$ to find the radian measure of $\theta$.
   Therefore this point can be represented as $\left(5,2.21\right)$ in polar coordinates.
3. Use $x=0$ and $y=3$ in Equation 1.8:
   
   $$\begin{array}{ccccccc}\begin{array}{ccc}\hfill r^2& =\hfill & x^2+y^2\hfill \\ & =\hfill & {\left(3\right)}^2+{\left(0\right)}^2\hfill \\ & =\hfill & 9+0\hfill \\ \hfill r& =\hfill & 3\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill \text{tan}\theta & =\hfill & \frac{y}{x}\hfill \\ & =\hfill & \frac{3}{0}.\hfill \end{array}\hfill \end{array}$$
   
   Direct application of the second equation leads to division by zero. Graphing the point $\left(0,3\right)$ on the rectangular coordinate system reveals that the point is located on the positive y-axis. The angle between the positive x-axis and the positive y-axis is $\frac{\pi }{2}.$ Therefore this point can be represented as $\left(3,\frac{\pi }{2}\right)$ in polar coordinates.
4. Use $x=5\sqrt{3}$ and $y=\mathrm{-5}$ in Equation 1.8:
   
   $$\begin{array}{ccccccc}\begin{array}{ccc}\hfill r^2& =\hfill & x^2+y^2\hfill \\ & =\hfill & {\left(5\sqrt{3}\right)}^2+{\left(\mathrm{-5}\right)}^2\hfill \\ & =\hfill & 75+25\hfill \\ \hfill r& =\hfill & 10\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill \text{tan}\theta & =\hfill & \frac{y}{x}\hfill \\ & =\hfill & \frac{\mathrm{-5}}{5\sqrt{3}}=-\frac{\sqrt{3}}{3}\hfill \\ \hfill \theta & =\hfill & -\frac{\pi }{6}.\hfill \end{array}\hfill \end{array}$$
   
   Therefore this point can be represented as $\left(10,-\frac{\pi }{6}\right)$ in polar coordinates.
5. Use $r=3$ and $\theta =\frac{\pi }{3}$ in Equation 1.7:
   
   $$\begin{array}{ccccccc}\begin{array}{ccc}\hfill x& =\hfill & r\text{cos}\theta \hfill \\ & =\hfill & 3\text{cos}\left(\frac{\pi }{3}\right)\hfill \\ & =\hfill & 3\left(\frac{1}{2}\right)=\frac{3}{2}\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill y& =\hfill & r\text{sin}\theta \hfill \\ & =\hfill & 3\text{sin}\left(\frac{\pi }{3}\right)\hfill \\ & =\hfill & 3\left(\frac{\sqrt{3}}{2}\right)=\frac{3\sqrt{3}}{2}.\hfill \end{array}\hfill \end{array}$$
   
   Therefore this point can be represented as $\left(\frac{3}{2},\frac{3\sqrt{3}}{2}\right)$ in rectangular coordinates.
6. Use $r=2$ and $\theta =\frac{3\pi }{2}$ in Equation 1.7:
   
   $$\begin{array}{ccccccc}\begin{array}{ccc}\hfill x& =\hfill & r\text{cos}\theta \hfill \\ & =\hfill & 2\text{cos}\left(\frac{3\pi }{2}\right)\hfill \\ & =\hfill & 2\left(0\right)=0\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill y& =\hfill & r\text{sin}\theta \hfill \\ & =\hfill & 2\text{sin}\left(\frac{3\pi }{2}\right)\hfill \\ & =\hfill & 2\left(\mathrm{-1}\right)=\mathrm{-2.}\hfill \end{array}\hfill \end{array}$$
   
   Therefore this point can be represented as $\left(0,\mathrm{-2}\right)$ in rectangular coordinates.
7. Use $r=6$ and $\theta =-\frac{5\pi }{6}$ in Equation 1.7:
   
   $$\begin{array}{ccccccc}\begin{array}{ccc}\hfill x& =\hfill & r\text{cos}\theta \hfill \\ & =\hfill & 6\text{cos}\left(-\frac{5\pi }{6}\right)\hfill \\ & =\hfill & 6\left(-\frac{\sqrt{3}}{2}\right)\hfill \\ & =\hfill & \mathrm{-3}\sqrt{3}\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill y& =\hfill & r\text{sin}\theta \hfill \\ & =\hfill & 6\text{sin}\left(-\frac{5\pi }{6}\right)\hfill \\ & =\hfill & 6\left(-\frac{1}{2}\right)\hfill \\ & =\hfill & \mathrm{-3.}\hfill \end{array}\hfill \end{array}$$
   
   Therefore this point can be represented as $\left(\mathrm{-3}\sqrt{3},\mathrm{-3}\right)$ in rectangular coordinates.

The three points are plotted in the following figure.

Figure 1.29 Three points plotted in the polar coordinate system.

Because the function is a multiple of a sine function, it is periodic with period $2\pi ,$ so use values for $\theta$ between 0 and $2\pi .$ The result of steps 1–3 appear in the following table. Figure 1.30 shows the graph based on this table.

Figure 1.30 The graph of the function $r=4\text{sin}\theta$ is a circle.

This is the graph of a circle. The equation $r=4\text{sin}\theta$ can be converted into rectangular coordinates by first multiplying both sides by $r.$ This gives the equation $r^2=4r\text{sin}\theta .$ Next use the facts that $r^2=x^2+y^2$ and $y=r\text{sin}\theta .$ This gives $x^2+y^2=4y.$ To put this equation into standard form, subtract $4y$ from both sides of the equation and complete the square:

This is the equation of a circle with radius 2 and center $\left(0,2\right)$ in the rectangular coordinate system.

1. Take the tangent of both sides. This gives $\text{tan}\theta =\text{tan}(\pi \text{/}3)=\sqrt{3}.$ Since $\text{tan}\theta =y\text{/}x$ we can replace the left-hand side of this equation by $y\text{/}x.$ This gives $y\text{/}x=\sqrt{3},$ which can be rewritten as $y=x\sqrt{3}.$ This is the equation of a straight line passing through the origin with slope $\sqrt{3}.$ In general, any polar equation of the form $\theta =K$ represents a straight line through the pole with slope equal to $\text{tan}K.$
2. First, square both sides of the equation. This gives $r^2=9.$ Next replace $r^2$ with $x^2+y^2.$ This gives the equation $x^2+y^2=9,$ which is the equation of a circle centered at the origin with radius 3. In general, any polar equation of the form $r=k$ where k is a positive constant represents a circle of radius k centered at the origin. (Note: when squaring both sides of an equation it is possible to introduce new points unintentionally. This should always be taken into consideration. However, in this case we do not introduce new points. For example, $\left(\mathrm{-3},\frac{\pi }{3}\right)$ is the same point as $\left(3,\frac{4\pi }{3}\right).)$
3. Multiply both sides of the equation by $r.$ This leads to $r^2=6r\text{cos}\theta -8r\text{sin}\theta .$ Next use the formulas
   
   $$r^2=x^2+y^2,x=r\text{cos}\theta ,y=r\text{sin}\theta .$$
   
   This gives
   
   $$\begin{array}{ccc}\hfill r^2& =\hfill & 6\left(r\text{cos}\theta \right)-8\left(r\text{sin}\theta \right)\hfill \\ \hfill x^2+y^2& =\hfill & 6x-8y.\hfill \end{array}$$
   
   To put this equation into standard form, first move the variables from the right-hand side of the equation to the left-hand side, then complete the square.
   
   $$\begin{array}{ccc}\hfill x^2+y^2& =\hfill & 6x-8y\hfill \\ \hfill x^2-6x+y^2+8y& =\hfill & 0\hfill \\ \hfill \left(x^2-6x\right)+\left(y^2+8y\right)& =\hfill & 0\hfill \\ \hfill \left(x^2-6x+9\right)+\left(y^2+8y+16\right)& =\hfill & 9+16\hfill \\ \hfill {\left(x-3\right)}^2+{\left(y+4\right)}^2& =\hfill & 25.\hfill \end{array}$$
   
   This is the equation of a circle with center at $\left(3,\mathrm{-4}\right)$ and radius 5. Notice that the circle passes through the origin since the center is 5 units away.

As the point P travels around the spiral in a counterclockwise direction, its distance d from the origin increases. Assume that the distance d is a constant multiple k of the angle $\theta$ that the line segment OP makes with the positive r-axis. Therefore $d\left(P,O\right)=k\theta ,$ where $O$ is the origin. Now use the distance formula and some trigonometry:

Although this equation describes the spiral, it is not possible to solve it directly for either x or y. However, if we use polar coordinates, the equation becomes much simpler. In particular, $d\left(P,O\right)=r,$ and $\theta$ is the second coordinate. Therefore the equation for the spiral becomes $r=k\theta .$ Note that when $\theta =0$ we also have $r=0,$ so the spiral emanates from the origin. We can remove this restriction by adding a constant to the equation. Then the equation for the spiral becomes $r=a+k\theta$ for arbitrary constants $a$ and $k.$ This is referred to as an Archimedean spiral, after the Greek mathematician Archimedes.

Another type of spiral is the logarithmic spiral, described by the function $r=a·b^{\theta }.$ A graph of the function $r=1.2\left({1.25}^{\theta }\right)$ is given in Figure 1.36. This spiral describes the shell shape of the chambered nautilus.

Figure 1.36 A logarithmic spiral is similar to the shape of the chambered nautilus shell. (credit: modification of work by Jitze Couperus, Flickr)

Suppose the point $\left(r,\theta \right)$ is on the graph of $r=3\text{sin}\left(2\theta \right).$

1. To test for symmetry about the polar axis, first try replacing $\theta$ with $\text{−}\theta .$ This gives $r=3\text{sin}\left(2\left(\text{−}\theta \right)\right)=\mathrm{-3}\text{sin}\left(2\theta \right).$ Since this changes the original equation, this test is not satisfied. However, returning to the original equation and replacing $r$ with $\text{−}r$ and $\theta$ with $\pi -\theta$ yields
   
   $$\begin{array}{}\\ \text{−}r=3\text{sin}\left(2\left(\pi -\theta \right)\right)\hfill \\ \text{−}r=3\text{sin}\left(2\pi -2\theta \right)\hfill \\ \text{−}r=3\text{sin}\left(\mathrm{-2}\theta \right)\hfill \\ \text{−}r=\mathrm{-3}\text{sin}2\theta .\hfill \end{array}$$
   
   Multiplying both sides of this equation by $\mathrm{-1}$ gives $r=3\text{sin}2\theta ,$ which is the original equation. This demonstrates that the graph is symmetric with respect to the polar axis.
2. To test for symmetry with respect to the pole, first replace $r$ with $\text{−}r,$ which yields $\text{−}r=3\text{sin}\left(2\theta \right).$ Multiplying both sides by −1 gives $r=\mathrm{-3}\text{sin}\left(2\theta \right),$ which does not agree with the original equation. Therefore the equation does not pass the test for this symmetry. However, returning to the original equation and replacing $\theta$ with $\theta +\pi$ gives
   
   $$\begin{array}{cc}\hfill r& =3\text{sin}\left(2\left(\theta +\pi \right)\right)\hfill \\ & =3\text{sin}\left(2\theta +2\pi \right)\hfill \\ & =3\left(\text{sin}2\theta \text{cos}2\pi +\text{cos}2\theta \text{sin}2\pi \right)\hfill \\ & =3\text{sin}2\theta .\hfill \end{array}$$
   
   Since this agrees with the original equation, the graph is symmetric about the pole.
3. To test for symmetry with respect to the vertical line $\theta =\frac{\pi }{2},$ first replace both $r$ with $\text{−}r$ and $\theta$ with $\text{−}\theta .$
   
   $$\begin{array}{}\\ \text{−}r=3\text{sin}\left(2\left(\text{−}\theta \right)\right)\hfill \\ \text{−}r=3\text{sin}\left(\mathrm{-2}\theta \right)\hfill \\ \text{−}r=\mathrm{-3}\text{sin}2\theta .\hfill \end{array}$$
   
   Multiplying both sides of this equation by $\mathrm{-1}$ gives $r=3\text{sin}2\theta ,$ which is the original equation. Therefore the graph is symmetric about the vertical line $\theta =\frac{\pi }{2}.$

This graph has symmetry with respect to the polar axis, the origin, and the vertical line going through the pole. To graph the function, tabulate values of $\theta$ between 0 and $\pi \text{/}2$ and then reflect the resulting graph.

This gives one petal of the rose, as shown in the following graph.

Figure 1.37 The graph of the equation between $\theta =0$ and $\theta =\pi \text{/}2.$

Reflecting this image into the other three quadrants gives the entire graph as shown.

Figure 1.38 The entire graph of the equation is called a four-petaled rose.