Section 8.3Polar Coordinates

The angle \(\frac{\pi }{2}\) is found by sweeping in a counterclockwise direction 90° from the polar axis. The point is located at a length of 3 units from the pole in the \(\frac{\pi }{2}\) direction, as shown in Figure 3.

We know that \(\frac{\pi }{6}\) is located in the first quadrant. However, \(r=-2.\) We can approach plotting a point with a negative \(r\) in two ways:

• Plot the point \((2,\frac{\pi }{6})\) by moving \(\frac{\pi }{6}\) in the counterclockwise direction and extending a directed line segment 2 units into the first quadrant. Then retrace the directed line segment back through the pole, and continue 2 units into the third quadrant;
• Move \(\frac{\pi }{6}\) in the counterclockwise direction, and draw the directed line segment from the pole 2 units in the negative direction, into the third quadrant.

See Figure 4(a). Compare this to the graph of the polar coordinate \((2,\frac{\pi }{6})\) shown in Figure 4(b).

Use the equivalent relationships.

\[\begin{array}{l}x=rcos\,θ \\ x=3cos\,\frac{\pi }{2}=0 \\ y=rsin\,θ \\ y=3sin\,\frac{\pi }{2}=3\end{array}\]

The rectangular coordinates are \((0,3).\) See Figure 6.

See Figure 7. Writing the polar coordinates as rectangular, we have

\[\begin{array}{l}x=rcos\,θ \\ x=-2cos(0)=-2 \\ \\ y=rsin\,θ \\ y=-2sin(0)=0\end{array}\]

The rectangular coordinates are also \((-2,0).\)

We see that the original point \((3,3)\) is in the first quadrant. To find \(θ,\) use the formula \(tan\,θ=\frac{y}{x}.\) This gives

\[\begin{array}{l}tan\,θ=\frac{3}{3} \\ tan\,θ=1 \\ θ={tan}^{-1}(1) \\ θ=\frac{\pi }{4}\end{array}\]

To find \(r,\) we substitute the values for \(x\) and \(y\) into the formula \(r=\sqrt{{x}^{2}+{y}^{2}}.\) We know that \(r\) must be positive, as \(\frac{\pi }{4}\) is in the first quadrant. Thus

\[\begin{array}{l}r=\sqrt{{3}^{2}+{3}^{2}} \\ r=\sqrt{9+9} \\ r=\sqrt{18}=3\sqrt{2}\end{array}\]

So, \(r=3\sqrt{2}\,\) and \(θ\text{=}\frac{\pi }{4},\) giving us the polar point \((3\sqrt{2},\frac{\pi }{4}).\) See Figure 9.

The goal is to eliminate \(x\) and \(y\) from the equation and introduce \(r\) and \(θ.\) Ideally, we would write the equation \(r\) as a function of \(θ.\) To obtain the polar form, we will use the relationships between \((x,y)\) and \((r,θ).\) Since \(x=rcos\,θ\) and \(y=rsin\,θ,\) we can substitute and solve for \(r.\)

\[\begin{array}{ll}\,{(rcos\,θ)}^{2}+{(rsin\,θ)}^{2}=9 & \\ \,{r}^{2}{cos}^{2}θ+{r}^{2}{sin}^{2}θ=9 & \\ {r}^{2}({cos}^{2}θ+{sin}^{2}θ)=9 & \\ {r}^{2}(1)=9 & {\text{Substitute cos}}^{2}θ+{sin}^{2}θ=1. \\ \,r=\pm 3\begin{array}{llll} & & & \end{array} & \text{Use the square root property}.\end{array}\]

Thus, \({x}^{2}+{y}^{2}=9,r=3,\) and \(r=-3\) should generate the same graph. See Figure 10.

To graph a circle in rectangular form, we must first solve for \(y.\)

\[\begin{array}{l}\begin{array}{l} \\ {x}^{2}+{y}^{2}=9\end{array} \\ \,{y}^{2}=9-{x}^{2} \\ \,y=\pm \sqrt{9-{x}^{2}}\end{array}\]

Note that this is two separate functions, since a circle fails the vertical line test. Therefore, we need to enter the positive and negative square roots into the calculator separately, as two equations in the form \({Y}_{1}=\sqrt{9-{x}^{2}}\) and \({Y}_{2}=-\sqrt{9-{x}^{2}}.\) Press GRAPH.

This equation appears similar to the previous example, but it requires different steps to convert the equation.

We can still follow the same procedures we have already learned and make the following substitutions:

\[\begin{array}{ll}{r}^{2}=6y & \text{Use }{x}^{2}+{y}^{2}={r}^{2}. \\ {r}^{2}=6rsin\,θ & \text{Substitute}\,y=rsin\,θ. \\ {r}^{2}-6rsin\,θ=0 & \text{Set equal to 0}. \\ r(r-6sin\,θ)=0 & \text{Factor and solve}. \\ r=0 & \text{We reject }r=0,\,\text{as it only represents one point, }(0,0). \\ r=6sin\,θ\begin{array}{llll} & & & \end{array} & \end{array}\]

Therefore, the equations \({x}^{2}+{y}^{2}=6y\) and \(r=6sin\,θ\) should give us the same graph. See Figure 11.

The Cartesian or rectangular equation is plotted on the rectangular grid, and the polar equation is plotted on the polar grid. Clearly, the graphs are identical.

We will use the relationships \(x=rcos\,θ\) and \(y=rsin\,θ.\)

\[\begin{array}{ll}\,y=3x+2 & \\ \,rsin\,θ=3rcos\,θ+2 & \\ rsin\,θ-3rcos\,θ=2 & \\ r(sin\,θ-3cos\,θ)=2 & \text{Isolate }r. \\ r=\frac{2}{sin\,θ-3cos\,θ}\begin{array}{llll} & & & \end{array} & \text{Solve for }r.\end{array}\]

The conversion is

\[\begin{array}{lll}r & = & 2sec\,θ \\ r & = & \frac{2}{cos\,θ} \\ r\,cos\,θ & = & 2 \\ x & = & 2\end{array}\]

Notice that the equation \(r=2sec\,θ\) drawn on the polar grid is clearly the same as the vertical line \(x=2\) drawn on the rectangular grid (see Figure 12). Just as \(x=c\) is the standard form for a vertical line in rectangular form, \(r=csec\,θ\) is the standard form for a vertical line in polar form.

A similar discussion would demonstrate that the graph of the function \(r=2csc\,θ\) will be the horizontal line \(y=2.\) In fact, \(r=ccsc\,θ\) is the standard form for a horizontal line in polar form, corresponding to the rectangular form \(y=c.\)

The goal is to eliminate \(θ\) and \(r,\) and introduce \(x\) and \(y.\) We clear the fraction, and then use substitution. In order to replace \(r\) with \(x\) and \(y,\) we must use the expression \({x}^{2}+{y}^{2}={r}^{2}.\)

\[\begin{array}{llll}r & = & \frac{3}{1-2\,cos\,θ} & \\ r(1-2(\frac{x}{r})) & = & 3 & \\ r-2x & = & 3 & \text{Use}\,cos\,θ=\frac{x}{r}\,to\,\text{eliminate}\,θ \\ r-2x & = & 3 & \\ r & = & 3+2x & \text{Isolate}\,r \\ {r}^{2} & = & {(3+2x)}^{2} & \text{Square}\,\text{both}\,\text{sides} \\ {x}^{2}+{y}^{2} & = & {(3+2x)}^{2} & \text{Use}\,{x}^{2}+{y}^{2}={r}^{2}\end{array}\]

The Cartesian equation is \({x}^{2}+{y}^{2}={(3+2x)}^{2}.\) However, to graph it, especially using a graphing calculator or computer program, we want to isolate \(y.\)

\[\begin{array}{l}{x}^{2}+{y}^{2}={(3+2x)}^{2} \\ \,{y}^{2}={(3+2x)}^{2}-{x}^{2} \\ \,y=\pm \sqrt{{(3+2x)}^{2}-{x}^{2}}\end{array}\]

When our entire equation has been changed from \(r\) and \(θ\) to \(x\) and \(y,\) we can stop, unless asked to solve for \(y\) or simplify. See Figure 13.

The “hour-glass” shape of the graph is called a hyperbola. Hyperbolas have many interesting geometric features and applications, which we will investigate further in Analytic Geometry.

\[\begin{array}{ll}\,r=sin(2θ) & \text{Use the double angle identity for sine}. \\ \,r=2sin\,θcos\,θ\begin{array}{llll} & & & \end{array} & \text{Use }cos\,θ=\frac{x}{r}\text{ and }sin\,θ=\frac{y}{r}. \\ \,r=2(\frac{x}{r})(\frac{y}{r}) & \text{Simplify}. \\ \,r=\frac{2xy}{{r}^{2}} & \text{ Multiply both sides by }{r}^{2}. \\ \,{r}^{3}=2xy & \\ {(\sqrt{{x}^{2}+{y}^{2}})}^{3}=2xy & \text{As}\,{x}^{2}+{y}^{2}={r}^{2},r=\sqrt{{x}^{2}+{y}^{2}}.\end{array}\]

This equation can also be written as

\[{({x}^{2}+{y}^{2})}^{\frac{3}{2}}=2xy\,\text{or}\,{x}^{2}+{y}^{2}={(2xy)}^{\frac{2}{3}}\]