The angle [latex]\frac{\pi }{2}[/latex] 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 [latex]\frac{\pi }{2}[/latex] direction, as shown in Figure 3.

Figure 3

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

1. Plot the point [latex]\left(2,\frac{\pi }{6}\right)[/latex] by moving [latex]\frac{\pi }{6}[/latex] 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;
2. Move [latex]\frac{\pi }{6}[/latex] 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 [latex]\left(2,\frac{\pi }{6}\right)[/latex] shown in Figure 4(b).

Figure 4

Use the equivalent relationships.

[latex]\begin{align} &x=r\cos \theta \\ &x=3\cos \frac{\pi }{2}=0 \\ &y=r\sin \theta \\ &y=3\sin \frac{\pi }{2}=3 \end{align}[/latex]

The rectangular coordinates are [latex]\left(0,3\right)[/latex].

Figure 6

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

[latex]\begin{align}&x=r\cos \theta \\ &x=-2\cos \left(0\right)=-2 \\ \text{ } \\ &y=r\sin \theta \\ &y=-2\sin \left(0\right)=0 \end{align}[/latex]

The rectangular coordinates are also [latex]\left(-2,0\right)[/latex].

Figure 7

We see that the original point [latex]\left(3,3\right)[/latex] is in the first quadrant. To find [latex]\theta[/latex], use the formula [latex]\tan \theta =\frac{y}{x}[/latex]. This gives

[latex]\begin{align}&\tan \theta =\frac{3}{3} \\ &\tan \theta =1 \\ &{\tan }^{-1}\left(1\right)=\frac{\pi }{4} \end{align}[/latex]

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

[latex]\begin{align} r&=\sqrt{{3}^{2}+{3}^{2}} \\ r&=\sqrt{9+9} \\ r&=\sqrt{18}=3\sqrt{2} \end{align}[/latex]

So, [latex]r=3\sqrt{2}[/latex] and [latex]\theta \text{=}\frac{\pi }{4}[/latex], giving us the polar point [latex]\left(3\sqrt{2},\frac{\pi }{4}\right)[/latex].

Figure 9

Analysis of the Solution

There are other sets of polar coordinates that will be the same as our first solution. For example, the points [latex]\left(-3\sqrt{2},\frac{5\pi }{4}\right)[/latex] and [latex]\left(3\sqrt{2},-\frac{7\pi }{4}\right)[/latex] will coincide with the original solution of [latex]\left(3\sqrt{2},\frac{\pi }{4}\right)[/latex]. The point [latex]\left(-3\sqrt{2},\frac{5\pi }{4}\right)[/latex] indicates a move further counterclockwise by [latex]\pi[/latex], which is directly opposite [latex]\frac{\pi }{4}[/latex]. The radius is expressed as [latex]-3\sqrt{2}[/latex]. However, the angle [latex]\frac{5\pi }{4}[/latex] is located in the third quadrant and, as [latex]r[/latex] is negative, we extend the directed line segment in the opposite direction, into the first quadrant. This is the same point as [latex]\left(3\sqrt{2},\frac{\pi }{4}\right)[/latex]. The point [latex]\left(3\sqrt{2},-\frac{7\pi }{4}\right)[/latex] is a move further clockwise by [latex]-\frac{7\pi }{4}[/latex], from [latex]\frac{\pi }{4}[/latex]. The radius, [latex]3\sqrt{2}[/latex], is the same.

The conversion is

[latex]\begin{gathered}r=2\sec \theta \\ r=\frac{2}{\cos \theta } \\ r\cos \theta =2 \\ x=2 \end{gathered}[/latex]

Notice that the equation [latex]r=2\sec \theta[/latex] drawn on the polar grid is clearly the same as the vertical line [latex]x=2[/latex] drawn on the rectangular grid. Just as [latex]x=c[/latex] is the standard form for a vertical line in rectangular form, [latex]r=c\sec \theta[/latex] is the standard form for a vertical line in polar form.

Figure 12. (a) Polar grid (b) Rectangular coordinate system

A similar discussion would demonstrate that the graph of the function [latex]r=2\csc \theta[/latex] will be the horizontal line [latex]y=2[/latex]. In fact, [latex]r=c\csc \theta[/latex] is the standard form for a horizontal line in polar form, corresponding to the rectangular form [latex]y=c[/latex].

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

[latex]\begin{align}&r=\frac{3}{1 - 2\cos \theta } \\ &r\left(1 - 2\cos \theta \right)=3 \\ &r\left(1 - 2\left(\frac{x}{r}\right)\right)=3&& \text{Use }\cos \theta =\frac{x}{r}\text{ to eliminate }\theta . \\ &r - 2x=3 \\ &r=3+2x&& \text{Isolate }r. \\ &{r}^{2}={\left(3+2x\right)}^{2}&& \text{Square both sides}. \\ &{x}^{2}+{y}^{2}={\left(3+2x\right)}^{2}&& \text{Use }{x}^{2}+{y}^{2}={r}^{2}. \end{align}[/latex]

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

[latex]\begin{gathered}{x}^{2}+{y}^{2}={\left(3+2x\right)}^{2} \\ {y}^{2}={\left(3+2x\right)}^{2}-{x}^{2} \\ y=\pm \sqrt{{\left(3+2x\right)}^{2}-{x}^{2}} \end{gathered}[/latex]

When our entire equation has been changed from [latex]r[/latex] and [latex]\theta[/latex] to [latex]x[/latex] and [latex]y[/latex], we can stop, unless asked to solve for [latex]y[/latex] or simplify.

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.

Analysis of the Solution

In this example, the right side of the equation can be expanded and the equation simplified further, as shown above. However, the equation cannot be written as a single function in Cartesian form. We may wish to write the rectangular equation in the hyperbola’s standard form. To do this, we can start with the initial equation.

[latex]\begin{align}&{x}^{2}+{y}^{2}={\left(3+2x\right)}^{2} \\ &{x}^{2}+{y}^{2}-{\left(3+2x\right)}^{2}=0 \\ &{x}^{2}+{y}^{2}-\left(9+12x+4{x}^{2}\right)=0 \\ &{x}^{2}+{y}^{2}-9 - 12x - 4{x}^{2}=0 \\ &-3{x}^{2}-12x+{y}^{2}=9&& \text{Multiply through by }-1. \\ &3{x}^{2}+12x-{y}^{2}=-9 \\ &3\left({x}^{2}+4x+\right)-{y}^{2}=-9&& \text{Organize terms to complete the square for}x. \\ &3\left({x}^{2}+4x+4\right)-{y}^{2}=-9+12 \\ &3{\left(x+2\right)}^{2}-{y}^{2}=3 \\ &{\left(x+2\right)}^{2}-\frac{{y}^{2}}{3}=1 \end{align}[/latex]

[latex]\begin{align}&r=\sin \left(2\theta \right)&& \text{Use the double angle identity for sine}. \\ &r=2\sin \theta \cos \theta&& \text{Use }\cos \theta =\frac{x}{r}\text{ and }\sin \theta =\frac{y}{r}. \\ &r=2\left(\frac{x}{r}\right)\left(\frac{y}{r}\right)&& \text{Simplify}. \\ &r=\frac{2xy}{{r}^{2}}&& \text{ Multiply both sides by }{r}^{2}. \\ &{r}^{3}=2xy \\ &{\left(\sqrt{{x}^{2}+{y}^{2}}\right)}^{3}=2xy&& \text{As }{x}^{2}+{y}^{2}={r}^{2},r=\sqrt{{x}^{2}+{y}^{2}}. \end{align}[/latex]

This equation can also be written as

[latex]{\left({x}^{2}+{y}^{2}\right)}^{\frac{3}{2}}=2xy\text{ or }{x}^{2}+{y}^{2}={\left(2xy\right)}^{\frac{2}{3}}[/latex]