Note

Polar coordinates are appropriate when distance from the origin or angle plays a role. For example, the gravitational force exerted on a planet by the sun depends only on the distance \(r\) from the sun and is conveniently described in polar coordinates.

EXAMPLE 1 From Polar to Rectangular Coordinates

Find the rectangular coordinates of point \(Q\) in Figure 11.40.

Solution The point \(Q= (r,\theta) = \big(3,\frac{5\pi}6\big)\) has rectangular coordinates: \begin{align*} x &= r\cos\theta = 3\cos\left(\frac{5\pi}6\right) =3\left(-\frac{\sqrt{3}}2\right) = -\frac{3\sqrt{3}}2\\ y &= r\sin\theta =3\sin\left(\frac{5\pi}6\right) =3\left(\frac{1}2\right) = \frac32 \end{align*}

EXAMPLE 2 From Rectangular to Polar Coordinates

Find the polar coordinates of point \(P\) in Figure 11.41.

Figure 11.41: The polar coordinates of \(P\) satisfy \(r=\sqrt{3^2+2^2}\) and \(\tan\theta=\frac23\).

Solution Since \(P=(x,y)=(3,2)\), \begin{align*} r&=\sqrt{x^2+y^2}=\sqrt{3^2+2^2}=\sqrt{13}\approx 3.6\\ \tan\theta &= \frac{y}{x} = \frac23 \end{align*} and because \(P\) lies in the first quadrant, \begin{align*} \theta &= \tan^{-1} \frac{y}x = \tan^{-1} \frac23 \approx 0.588 \end{align*} Thus, \(P\) has polar coordinates \((r,\theta) \approx (3.6,0.588)\).

Definition

By definition, \[ -\frac{\pi}2<\tan^{-1} x < \frac{\pi}2 \] If \(r>0\), a coordinate \(\theta\) of \(P=(x,y)\) is \[ \theta = \begin{cases} \tan^{-1}\frac{y}x & \text{ if \(x > 0\)} \\ \tan^{-1}\frac{y}x +\pi & \text{ if \(x<0\)}\\ \pm \frac{\pi}{2}& \text{ if \(x = 0\)} \end{cases} \]}

EXAMPLE 3 Choosing \(\theta\) Correctly

Find two polar representations of \(P = (-1,1)\), one with \(r>0\) and one with \(r<0\).

Solution The point \(P=(x,y) = (-1,1)\) has polar coordinates \((r,\theta)\), where \[ r=\sqrt{(-1)^2+1^2}=\sqrt 2,\qquad \tan\theta=\tan\frac{y}{x}=-1 \] However, \(\theta\) is not given by \[ \tan^{-1}\frac{y}x = \tan^{-1}\left(\frac1{-1}\right) = -\frac{\pi}4 \] because \(\theta = -\frac{\pi}4\) this would place \(P\) in the fourth quadrant (Figure 11.44). Since \(P\) is in the second quadrant, the correct angle is \[ \theta =\tan^{-1}\frac{y}x +\pi = -\frac{\pi}4 + \pi = \frac{3\pi}4 \] If we wish to use the negative radial coordinate \(r= -\sqrt 2\), then the angle becomes \(\theta= -\frac{\pi}4\) or \(\frac{7\pi}4\). Thus, \[ P = \left(\sqrt 2,\frac{3\pi}4\right)\qquad \textrm{or} \qquad \left(-\sqrt 2,\frac{7\pi}4\right) \]

EXAMPLE 4 Line Through the Origin

Find the polar equation of the line through the origin of slope \(\frac32\) (Figure 11.46).

Solution A line of slope \(m\) makes an angle \(\theta_0\) with the \(x\)-axis, where \(m=\tan\theta_0\). In our case, \(\theta_0 = \tan^{-1}\frac32 \approx 0.98\). The equation of the line is \(\theta= \tan^{-1} \frac32\) or \(\theta \approx 0.98\).

EXAMPLE 5 Line Not Passing Through \(O\)

Show that \begin{equation} \boxed{\bbox[#fef7e5,5pt]{r=d\sec(\theta-\alpha)}}\tag{1} \end{equation} is the polar equation of the line \(L\) whose point closest to the origin is \(P_0 = (d,\alpha)\).

Figure 11.47: \(P_0\) is the point on \(L\) closest to the origin.

Solution The point \(P_0\) is obtained by dropping a perpendicular from the origin to \(L\) (Figure 11.48), and if \(P = (r,\theta)\) is any point on \(L\) other than \(P_0\), then \(\triangle \textit{OPP}_0\) is a right triangle. Therefore, \({d}/{r} = \cos(\theta - \alpha)\), or \(r = d\sec(\theta - \alpha)\), as claimed.

EXAMPLE 6

Find the polar equation of the line \(L\) tangent to the circle \(r=4\) at the point with polar coordinates \(P_0=\big(4,\frac{\pi}3\big)\).

Solution The point on \(L\) closest to the origin is \(P_0\) itself (Figure 11.48). Therefore, we take \((d,\alpha)=\big(4,\frac{\pi}3\big)\) in Eq. (1) to obtain the equation \(r=4\sec\big(\theta-\frac{\pi}3\big)\).

Figure 11.48: The tangent line has equation \(r=4\sec\left(\theta-\frac{\pi}3\right)\).

EXAMPLE 7 Converting to Rectangular Coordinates

Identify the curve with polar equation \(r=2a\cos\theta\) (\(a\) a constant).

Solution Multiply the equation by \(r\) to obtain \(r^2=2ar\cos\theta\). Because \(r^2=x^2+y^2\) and \(x=r\cos\theta\), this equation becomes \[ x^2+y^2 = 2ax\qquad\textrm{or}\qquad x^2-2ax+y^2= 0 \] Then complete the square to obtain \((x-a)^2+y^2=a^2\). This is the equation of the circle of radius \(a\) and center \((a,0)\) (Figure 11.49).

EXAMPLE 8 Symmetry About the \(x\)-Axis

Sketch the limaçon curve \newline \(r = 2\cos\theta-1\).

Solution Since \(\cos\theta\) is periodic, it suffices to plot points for \(-\pi\le \theta\le \pi\).

Step 1. Plot points

To get started, we plot points \(A\)–\(G\) on a grid and join them by a smooth curve (Figure 11.51).

Step 2. Analyze \(r\) as a function of \(\theta\)

For a better understanding, it is helpful to graph \(r\) as a function of \(\theta\) in rectangular coordinates. Figure 11.52 shows that

We conclude:

• The graph begins at point \(A\) in Figure 11.52 and moves in toward the origin as \(\theta\) varies from \(0\) to \(\frac{\pi}3\).
• Since \(r\) is negative for \(\frac{\pi}3\le\theta\le \pi\), the curve continues into the third and fourth quadrants (rather than into the first and second quadrants), moving toward the point \(G = (-3,\pi)\) in Figure 11.52.

Figure 11.51: Plotting \(r = 2\cos\theta-1\) using a grid.

Step 3. Use symmetry

Since \(r(\theta)=r(-\theta)\), the curve is symmetric with respect to the \(x\)-axis. So the part of the curve with \(-\pi\le\theta\le 0\) is obtained by reflection through the \(x\)-axis as in Figure 11.52.