7.3 Polar Coordinates

Learning Objectives

  • Locate points in a plane using polar coordinates.
  • Convert points between rectangular and polar coordinates.
  • Sketch polar curves with given equations.
  • Convert equations between rectangular and polar coordinates.
  • Identify symmetry in polar curves and equations.

The rectangular coordinate system (or Cartesian plane) provides a means of mapping points to ordered pairs and ordered pairs to points. This is called a one-to-one mapping from points in the plane to ordered pairs. The polar coordinate system provides an alternative method of mapping points to ordered pairs. In this section we see that in some circumstances, polar coordinates can be more useful than rectangular coordinates.

Defining Polar Coordinates

To find the coordinates of a point in the polar coordinate system, consider Figure 1 below. The point \ds P has Cartesian coordinates \ds \left(x,y\right). Consider the line segment connecting the origin to the point \ds P. Its length is equal to the distance from the origin to \ds P and we denote it by \ds r. We also denote the angle between the positive \ds x-axis and the line segment by \ds \theta . Then (r,\theta) are the polar coordinates of P

A point P(x, y) is given in the first quadrant with lines drawn to indicate its x and y values. There is a line from the origin to P(x, y) marked r and this line make an angle θ with the x axis.
Figure 1. An arbitrary point in the Cartesian plane.

Each point \ds \left(x,y\right) in the Cartesian coordinate system can be represented as an ordered pair \ds \left(r,\theta \right) in the polar coordinate system. The first coordinate is called the radial coordinate and the second coordinate is called the angular coordinate. In the polar coordinate system, the origin is called the pole and the positive x-axis is the polar axis. Note that every point in the Cartesian plane has two values (hence the term ordered pair) associated with it. In the polar coordinate system, each point also has two values associated with it: \ds r and \ds \theta .
Using right-triangle trigonometry, we obtain the following equations:

\ds \text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) =\frac{x}{r},\ \phantom{\rule{0.2em}{0ex}}\text{and so}\ \phantom{\rule{0.2em}{0ex}}x=r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta)

\ds \text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) =\frac{y}{r},\phantom{\rule{0.2em}{0ex}}\ \text{and so}\phantom{\rule{0.2em}{0ex}}\ y=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) .

Furthermore,

\ds {r}^{2}={x}^{2}+{y}^{2}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{tan}\phantom{\rule{0.2em}{0ex}}(\theta) =\frac{y}{x}.

These formulas can be used to convert from rectangular to polar or from polar to rectangular coordinates.

Note that the equation \ds \text{tan}\phantom{\rule{0.2em}{0ex}}(\theta) =\frac yx has an infinite number of solutions for any ordered pair \ds \left(x,y\right). However, if we restrict the solutions \theta to values in (-\pi,\pi], then we can assign a unique solution to the quadrant in which the point \ds \left(x,y\right) is located.

Converting Points between Coordinate Systems

Given a point \ds P in the plane with Cartesian coordinates \ds \left(x,y\right) and polar coordinates \ds \left(r,\theta \right), the following conversion formulas hold true:

(*)\quad\ds x=r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) ,

(**)\quad\ds {r}^{2}={x}^{2}+{y}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{tan}\phantom{\rule{0.2em}{0ex}}(\theta) =\frac{y}{x}.

Converting between Rectangular and Polar Coordinates

Convert the following rectangular coordinates into polar coordinates.

  1. \ds \left(1,1\right)
  2. \ds \left(-3,4\right)
  3. \ds \left(0,3\right)
  4. \ds \left(5\sqrt{3},-5\right)

Convert the following polar coordinates into rectangular coordinates.

  1. \ds \left(3,\frac{\pi}3\right)
  2. \ds \left(2,\frac{3\pi}2\right)
  3. \ds \left(6,-\frac{5\pi}6\right)

Solution

  1. Use \ds x=1 and \ds y=1 in (**):

    \ds \begin{array}{ccccccc}\begin{array}{ccc}\hfill {r}^{2}&\ds =\hfill &\ds {x}^{2}+{y}^{2}\hfill \\[5mm]\ds &\ds =\hfill &\ds {1}^{2}+{1}^{2}\hfill \\[5mm]\ds \hfill r&\ds =\hfill &\ds \sqrt{2}\hfill \end{array}\hfill &\ds &\ds &\ds \text{and}\hfill &\ds &\ds &\ds \begin{array}{ccc}\hfill \text{tan}\phantom{\rule{0.2em}{0ex}}(\theta) &\ds =\hfill &\ds \frac{y}{x}\hfill \\[5mm]\ds &\ds =\hfill &\ds \frac{1}{1}=1\hfill \\[5mm]\ds \hfill \theta &\ds =\hfill &\ds \frac{\pi }{4},\hfill \end{array}\hfill \end{array}

    where we chose the angle of \ds\frac{\pi}4 since the point (1,1) belongs to the first quadrant and hence \ds\theta\in\left[0,\frac{\pi}2\right].
     Therefore this point can be represented as \ds \left(\sqrt{2},\frac{\pi }{4}\right) in polar coordinates.

  2. Use \ds x=-3 and \ds y=4 in (**):

    \ds \begin{array}{ccccccc}\begin{array}{ccc}\hfill {r}^{2}&\ds =\hfill &\ds {x}^{2}+{y}^{2}\hfill \\[5mm]\ds &\ds =\hfill &\ds {\left(-3\right)}^{2}+{\left(4\right)}^{2}\hfill \\[5mm]\ds \hfill r&\ds =\hfill &\ds 5\hfill \end{array}\hfill &\ds &\ds &\ds \text{and}\hfill &\ds &\ds &\ds \begin{array}{ccc}\hfill \text{tan}\phantom{\rule{0.2em}{0ex}}(\theta) &\ds =\hfill &\ds \frac{y}{x}\hfill \\[5mm]\ds &\ds =\hfill &\ds -\frac{4}{3}\hfill \\[5mm]\ds \hfill \theta &\ds =\hfill &\ds \pi+ \text{arctan}\left(-\frac{4}{3}\right)\hfill \\[5mm]\ds &\ds = \hfill &\ds\pi-\arctan\left(\frac43\right),\hfill \end{array}\hfill \end{array}

    since (-3,4) belongs to the second quadrant and hence \ds\theta\in\left[\frac{\pi}2,\pi\right]. So the polar coordinates of this point are \ds \left(5,\pi-\arctan\left(\frac43\right)\right).

  3. Use \ds x=0 and \ds y=3 in (**):

    \ds \begin{array}{ccccccc}\begin{array}{ccc}\hfill {r}^{2}&\ds =\hfill &\ds {x}^{2}+{y}^{2}\hfill \\[5mm]\ds &\ds =\hfill &\ds {\left(3\right)}^{2}+{\left(0\right)}^{2}\hfill \\[5mm]\ds &\ds =\hfill &\ds 9+0\hfill \\[5mm]\ds \hfill r&\ds =\hfill &\ds 3\hfill \end{array}\hfill &\ds &\ds &\ds \text{and}\hfill &\ds &\ds &\ds \begin{array}{ccc}\hfill \text{tan}\phantom{\rule{0.2em}{0ex}}(\theta) &\ds =\hfill &\ds \frac{y}{x}\hfill \\[5mm]\ds &\ds =\hfill &\ds \frac{3}{0}.\hfill \end{array}\hfill \end{array}

    Direct application of the second equation leads to division by zero. Graphing the point \ds \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 \ds \frac{\pi }{2}. Therefore this point can be represented as \ds \left(3,\frac{\pi }{2}\right) in polar coordinates.
  4. Use \ds x=5\sqrt{3} and \ds y=-5 in (**):

    \ds \begin{array}{ccccccc}\begin{array}{ccc}\hfill {r}^{2}&\ds =\hfill &\ds {x}^{2}+{y}^{2}\hfill \\[5mm]\ds &\ds =\hfill &\ds {\left(5\sqrt{3}\right)}^{2}+{\left(-5\right)}^{2}\hfill \\[5mm]\ds &\ds =\hfill &\ds 75+25\hfill \\[5mm]\ds \hfill r&\ds =\hfill &\ds 10\hfill \end{array}\hfill &\ds &\ds &\ds \text{and}\hfill &\ds &\ds &\ds \begin{array}{ccc}\hfill \text{tan}\phantom{\rule{0.2em}{0ex}}(\theta) &\ds =\hfill &\ds \frac{y}{x}\hfill \\[5mm]\ds &\ds =\hfill &\ds \frac{-5}{5\sqrt{3}}=-\frac{\sqrt{3}}{3}\hfill \\[5mm]\ds \hfill \theta &\ds =\hfill &\ds -\frac{\pi }{6},\hfill \end{array}\hfill \end{array}

    since the point (5\sqrt3,-5) is in the fourth quadrant and hence \ds\theta\in\left[-\frac{\pi}2,0\right].
    Therefore this point can be represented as \ds \left(10,-\frac{\pi }{6}\right) in polar coordinates.
  5. Use \ds r=3 and \ds \theta =\frac{\pi }{3} in (*)

    \ds \begin{array}{ccccccc}\begin{array}{ccc}\hfill x&\ds =\hfill &\ds r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \hfill \\[5mm]\ds &\ds =\hfill &\ds 3\phantom{\rule{0.2em}{0ex}}\text{cos}\left(\frac{\pi }{3}\right)\hfill \\[5mm]\ds &\ds =\hfill &\ds 3\left(\frac{1}{2}\right)=\frac{3}{2}\hfill \end{array}\hfill &\ds &\ds &\ds \text{and}\hfill &\ds &\ds &\ds \begin{array}{ccc}\hfill y&\ds =\hfill &\ds r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \hfill \\[5mm]\ds &\ds =\hfill &\ds 3\phantom{\rule{0.2em}{0ex}}\text{sin}\left(\frac{\pi }{3}\right)\hfill \\[5mm]\ds &\ds =\hfill &\ds 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 \ds \left(\frac{3}{2},\phantom{\rule{0.2em}{0ex}}\frac{3\sqrt{3}}{2}\right) in rectangular coordinates.
  6. Use \ds r=2 and \ds \theta =\frac{3\pi }{2} in (*):

    \ds \begin{array}{ccccccc}\begin{array}{ccc}\hfill x&\ds =\hfill &\ds r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \hfill \\[5mm]\ds &\ds =\hfill &\ds 2\phantom{\rule{0.2em}{0ex}}\text{cos}\left(\frac{3\pi }{2}\right)\hfill \\[5mm]\ds &\ds =\hfill &\ds 2\left(0\right)=0\hfill \end{array}\hfill &\ds &\ds &\ds \text{and}\hfill &\ds &\ds &\ds \begin{array}{ccc}\hfill y&\ds =\hfill &\ds r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \hfill \\[5mm]\ds &\ds =\hfill &\ds 2\phantom{\rule{0.2em}{0ex}}\text{sin}\left(\frac{3\pi }{2}\right)\hfill \\[5mm]\ds &\ds =\hfill &\ds 2\left(-1\right)=-2.\hfill \end{array}\hfill \end{array}

    Therefore this point can be represented as \ds \left(0,-2\right) in rectangular coordinates.
  7. Use \ds r=6 and \ds \theta =-\frac{5\pi }{6} in (*):

    \ds \begin{array}{ccccccc}\begin{array}{ccc}\hfill x&\ds =\hfill &\ds r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \hfill \\[5mm]\ds &\ds =\hfill &\ds 6\phantom{\rule{0.2em}{0ex}}\text{cos}\left(-\frac{5\pi }{6}\right)\hfill \\[5mm]\ds &\ds =\hfill &\ds 6\left(-\frac{\sqrt{3}}{2}\right)\hfill \\[5mm]\ds &\ds =\hfill &\ds -3\sqrt{3}\hfill \end{array}\hfill &\ds &\ds &\ds \text{and}\hfill &\ds &\ds &\ds \begin{array}{ccc}\hfill y&\ds =\hfill &\ds r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \hfill \\[5mm]\ds &\ds =\hfill &\ds 6\phantom{\rule{0.2em}{0ex}}\text{sin}\left(-\frac{5\pi }{6}\right)\hfill \\[5mm]\ds &\ds =\hfill &\ds 6\left(-\frac{1}{2}\right)\hfill \\[5mm]\ds &\ds =\hfill &\ds -3.\hfill \end{array}\hfill \end{array}

    Therefore this point can be represented as \ds \left(-3\sqrt{3},-3\right) in rectangular coordinates.

Convert \ds \left(-8,-8\right) from rectangular into polar coordinates and \ds \left(4,\frac{2\pi }{3}\right) — from polar coordinates into rectangular coordinates.

Answer

\ds \left(8\sqrt{2},-\frac{3\pi }{4}\right) and \ds \left(-2,2\sqrt{3}\right)

Hint

Make sure to check the quadrant when calculating \ds \theta .

The polar representation of a point is not unique. For example, the polar coordinates \ds \left(2,\frac{\pi }{3}\right) and \ds \left(2,\frac{7\pi }{3}\right) both represent the point \ds \left(1,\sqrt{3}\right) in the rectangular system. We can also allow \ds r to be negative. For example, the point with polar coordinates \ds \left(-2,\frac{4\pi }{3}\right) also represents the point \ds \left(1,\sqrt{3}\right) in the rectangular system, as we can see by using (*):

\ds \begin{array}{ccccccc}\begin{array}{ccc}\hfill x&\ds =\hfill &\ds r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \hfill \\[5mm]\ds &\ds =\hfill &\ds -2\phantom{\rule{0.2em}{0ex}}\text{cos}\left(\frac{4\pi }{3}\right)\hfill \\[5mm]\ds &\ds =\hfill &\ds -2\left(-\frac{1}{2}\right)=1\hfill \end{array}\hfill &\ds &\ds &\ds \text{and}\hfill &\ds &\ds &\ds \begin{array}{ccc}\hfill y&\ds =\hfill &\ds r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \hfill \\[5mm]\ds &\ds =\hfill &\ds -2\phantom{\rule{0.2em}{0ex}}\text{sin}\left(\frac{4\pi }{3}\right)\hfill \\[5mm]\ds &\ds =\hfill &\ds -2\left(-\frac{\sqrt{3}}{2}\right)=\sqrt{3}.\hfill \end{array}\hfill \end{array}

(Geometrically, when we plot a point with a negative radial coordinate, we measure the distance of |r| along the halfline that is in the opposite direction to the one that makes the angle of \theta with the positive x-axis, so basically the minus reverses the direction, the same way as with angles.)

Every point in the plane has an infinite number of representations in polar coordinates. However, each point in the plane has only one representation in the rectangular coordinate system.

Note that the polar representation of a point in the plane also has a visual interpretation. In particular, \ds r is the directed distance that the point lies from the origin, and \ds \theta measures the angle that the line segment from the origin to the point makes with the positive \ds x-axis. Positive angles are measured in a counterclockwise direction and negative angles are measured in a clockwise direction. The polar coordinate system appears in the following figure.

A series of concentric circles is drawn with spokes indicating different values between 0 and 2π in increments of π/12. The first quadrant starts with 0 where the x axis would be, then the next spoke is marked π/12, then π/6, π/4, π/3, 5π/12, π/2, and so on into the second, third, and fourth quadrants. The polar axis is noted near the former x axis line.
Figure 2. The polar coordinate system.

The line segment starting from the center of the graph going to the right (called the positive x-axis in the Cartesian system) is the polar axis. The center point is the pole, or origin, of the coordinate system, and corresponds to \ds r=0. The innermost circle shown in Figure 2 above contains all points a distance of 1 unit from the pole, and is represented by the equation \ds r=1. Then \ds r=2 is the set of points 2 units from the pole, and so on. The line segments emanating from the pole correspond to fixed angles. To plot a point in the polar coordinate system, start with the angle. If the angle is positive, then measure the angle from the polar axis in a counterclockwise direction. If it is negative, then measure it clockwise. If the value of \ds r is positive, move that distance along the terminal ray of the angle. If it is negative, move along the ray that is opposite to the terminal ray of the given angle.

Plotting Points in the Polar Plane

Plot each of the following points on the polar plane.

  1. \ds \left(2,\frac{\pi }{4}\right)
  2. \ds \left(-3,\frac{2\pi }{3}\right)
  3. \ds \left(4,\frac{5\pi }{4}\right)

Solution

The three points are plotted in the following figure.

Three points are marked on a polar coordinate plane, specifically (2, π/4) in the first quadrant, (4, 5π/4) in the third quadrant, and (−3, 2π/3) in the fourth quadrant.
Figure 3. Three points plotted in the polar coordinate system.

Plot \ds \left(4,\frac{5\pi }{3}\right) and \ds \left(-3,-\frac{7\pi }{2}\right) on the polar plane.

Answer

 

Two points are marked on a polar coordinate plane, specifically (−3, −7π/2) on the y axis and (4, 5π/3) in the fourth quadrant.

Polar Curves

Now that we know how to plot points in the polar coordinate system, we can discuss how to plot curves. In the rectangular coordinate system, we can graph a function \ds y=f\left(x\right) and create a curve in the Cartesian plane. In a similar fashion, we can graph a curve that is generated by a function \ds r=f\left(\theta \right).

The general idea behind graphing a function in polar coordinates is the same as graphing a function in rectangular coordinates. Start with a list of values for the independent variable \ds \left(\theta in this case) and calculate the corresponding values of the dependent variable \ds r. This process generates a list of ordered pairs, which can be plotted in the polar coordinate system. Finally, connect the points, and take advantage of any patterns that may appear. The function may be periodic, for example, which indicates that only a limited number of values for the independent variable are needed.

Problem-Solving Strategy: Plotting a Curve in Polar Coordinates

  1. Create a table with two columns. The first column is for \ds \theta , and the second column is for \ds r.
  2. Create a list of values for \ds \theta .
  3. Calculate the corresponding \ds r values for each \ds \theta .
  4. Plot each ordered pair \ds \left(r,\theta \right) on the coordinate axes.
  5. Connect the points and look for a pattern.

Graphing a Function in Polar Coordinates

Graph the curve defined by the function \ds r=4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) . Identify the curve and rewrite the equation in rectangular coordinates.

Solution

Because the function is a multiple of a sine function, it is periodic with period \ds 2\pi , so use values for \ds \theta between 0 and \ds 2\pi . The result of steps 1–3 appear in the following table.

\ds \theta \ds r=4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \ds \theta \ds r=4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta)
0 0 \ds \pi 0
\ds \frac{\pi }{6} \ds 2 \ds \frac{7\pi }{6} \ds -2
\ds \frac{\pi }{4} \ds 2\sqrt{2}\approx 2.8 \ds \frac{5\pi }{4} \ds -2\sqrt{2}\approx -2.8
\ds \frac{\pi }{3} \ds 2\sqrt{3}\approx 3.4 \ds \frac{4\pi }{3} \ds -2\sqrt{3}\approx -3.4
\ds \frac{\pi }{2} \ds 4 \ds \frac{3\pi }{2} \ds 4
\ds \frac{2\pi }{3} \ds 2\sqrt{3}\approx 3.4 \ds \frac{5\pi }{3} \ds -2\sqrt{3}\approx -3.4
\ds \frac{3\pi }{4} \ds 2\sqrt{2}\approx 2.8 \ds \frac{7\pi }{4} \ds -2\sqrt{2}\approx -2.8
\ds \frac{5\pi }{6} \ds 2 \ds \frac{11\pi }{6} \ds -2
\ds 2\pi 0
Figure 4 below shows the graph based on this table.
On the polar coordinate plane, a circle is drawn with radius 2. It touches the origin, (2 times the square root of 2, π/4), (4, π/2), and (2 times the square root of 2, 3π/4).
Figure 4. The graph of the function \ds r=4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) is a circle.

This is the graph of a circle. The equation \ds r=4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) can be converted into rectangular coordinates by first multiplying both sides by \ds r. This gives the equation \ds {r}^{2}=4r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) . Next use the facts that \ds {r}^{2}={x}^{2}+{y}^{2} and \ds y=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) . This gives \ds {x}^{2}+{y}^{2}=4y. To put this equation into standard form, subtract \ds 4y from both sides of the equation and complete the square:

\ds \begin{array}{ccc}\hfill {x}^{2}+{y}^{2}-4y&\ds =\hfill &\ds 0\hfill \\[5mm]\ds \hfill {x}^{2}+\left({y}^{2}-4y\right)&\ds =\hfill &\ds 0\hfill \\[5mm]\ds \hfill {x}^{2}+\left({y}^{2}-4y+4\right)&\ds =\hfill &\ds 0+4\hfill \\[5mm]\ds \hfill {x}^{2}+{\left(y-2\right)}^{2}&\ds =\hfill &\ds 4.\hfill \end{array}

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

Create a graph of the curve defined by the function \ds r=4+4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) .

Answer

 

The graph of r = 4 + 4 cosθ is given. It vaguely looks look a heart tipped on its side with a rounded bottom instead of a pointed one. Specifically, the graph starts at the origin, moves into the second quadrant and increases to a rounded circle-like figure. The graph is symmetric about the x axis, so it continues its rounded circle-like figure, goes into the third quadrant, and comes to a point at the origin.

The name of this shape is a cardioid, which we will study further later in this section.

Hint

Follow the problem-solving strategy for creating a graph in polar coordinates.

Because we already know how to sketch a variety of curves in Cartesian coordinates, transforming polar equations into rectangular coordinates may help sketching some polar curves as shown in the following example.

Transforming Polar Equations to Rectangular Coordinates

Rewrite each of the following equations in rectangular coordinates and identify the graph.

  1. \ds \theta =\frac{\pi }{3}
  2. \ds r=3
  3. \ds r=6\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) -8\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta)

Solution

  1. Take the tangent of both sides. This gives \ds \text{tan}\phantom{\rule{0.2em}{0ex}}(\theta) =\text{tan}\left(\frac{\pi} 3\right)=\sqrt{3}. Since \ds \text{tan}\phantom{\rule{0.2em}{0ex}}(\theta) =\frac yx we can replace the left-hand side of this equation by \ds \frac yx. This gives \ds \frac yx=\sqrt{3}, which can be rewritten as \ds y=x\sqrt{3}. This is the equation of a straight line passing through the origin with slope \ds \sqrt{3}. In general, any polar equation of the form \ds \theta =K represents a straight line through the pole with slope equal to \ds \text{tan}\phantom{\rule{0.2em}{0ex}}(K).
  2. First, square both sides of the equation. This gives \ds {r}^{2}=9. Next replace \ds {r}^{2} with \ds {x}^{2}+{y}^{2}. This gives the equation \ds {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 \ds r=k, where k is a 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, \ds \left(-3,\frac{\pi }{3}\right) is the same point as \ds \left(3,\frac{4\pi }{3}\right).)
  3. Multiply both sides of the equation by \ds r. This leads to \ds {r}^{2}=6r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) -8r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) . Next use the formulas

    \ds {r}^{2}={x}^{2}+{y}^{2},\phantom{\rule{1em}{0ex}}x=r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) ,\phantom{\rule{1em}{0ex}}y=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) .


     This gives

    \ds \begin{array}{ccc}\hfill {r}^{2}&\ds =\hfill &\ds 6\left(r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \right)-8\left(r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \right)\hfill \\[5mm]\ds \hfill {x}^{2}+{y}^{2}&\ds =\hfill &\ds 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.

    \ds \begin{array}{ccc}\hfill {x}^{2}+{y}^{2}&\ds =\hfill &\ds 6x-8y\hfill \\[5mm]\ds \hfill {x}^{2}-6x+{y}^{2}+8y&\ds =\hfill &\ds 0\hfill \\[5mm]\ds \hfill \left({x}^{2}-6x\right)+\left({y}^{2}+8y\right)&\ds =\hfill &\ds 0\hfill \\[5mm]\ds \hfill \left({x}^{2}-6x+9\right)+\left({y}^{2}+8y+16\right)&\ds =\hfill &\ds 9+16\hfill \\[5mm]\ds \hfill {\left(x-3\right)}^{2}+{\left(y+4\right)}^{2}&\ds =\hfill &\ds 25.\hfill \end{array}


     This is the equation of a circle with center at \ds \left(3,-4\right) and radius 5. Notice that the circle passes through the origin since the center is 5 units away.

Rewrite the equation \ds r=\text{sec}\phantom{\rule{0.2em}{0ex}}(\theta) \phantom{\rule{0.2em}{0ex}}\text{tan}\phantom{\rule{0.2em}{0ex}}(\theta) in rectangular coordinates and identify its graph.

Answer

\ds y={x}^{2}, which is the equation of a parabola opening upward.

Hint

Rewrite secant as the reciprocal of cosine and multiply both sides by cosine.

We have now seen several examples of curves defined by polar equations. A summary of some common curves is given in the table below. In each equation, a and b are arbitrary constants.

This table has three columns and 3 rows. The first row is a header row and is given from left to right as name, equation, and example. The second row is Line passing through the pole with slope tan K; θ = K; and a picture of a straight line on the polar coordinate plane with θ = π/3. The third row is Circle; r = a cosθ + b sinθ; and a picture of a circle on the polar coordinate plane with equation r = 2 cos(t) – 3 sin(t): the circle touches the origin but has center in the third quadrant.
This table has three columns and 3 rows. The first row is Spiral; r = a + bθ; and a picture of a spiral starting at the origin with equation r = θ/3. The second row is Cardioid; r = a(1 + cosθ), r = a(1 – cosθ), r = a(1 + sinθ), r = a(1 – sinθ); and a picture of a cardioid with equation r = 3(1 + cosθ): the cardioid looks like a heart turned on its side with a rounded bottom instead of a pointed one. The third row is Limaçon; r = a cosθ + b, r = a sinθ + b; and a picture of a limaçon with equation r = 2 + 4 sinθ: the figure looks like a deformed circle with a loop inside of it. The seventh row is Rose; r = a cos(bθ), r = a sin(bθ); and a picture of a rose with equation r = 3 sin(2θ): the rose looks like a flower with four petals, one petal in each quadrant, each with length 3 and reaching to the origin between each petal.

A cardioid is a special case of a limaçon (pronounced “lee-mah-son”), in which \ds a=b or \ds a= - b. The rose is a very interesting curve. Notice that the graph of \ds r=3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(2\theta) has four petals. However, the graph of \ds r=3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(3\theta) has three petals as shown below.

A rose with three petals, one in the first quadrant, another in the second quadrant, and the third in both the third and fourth quadrants, each with length 3. Each petal starts and ends at the origin.
Figure 5. Graph of \ds r=3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}3\theta .

If the coefficient of \ds \theta is even, the graph has twice as many petals as the coefficient. If the coefficient of \ds \theta is odd, then the number of petals equals the coefficient. You are encouraged to explore why this happens. Even more interesting graphs emerge when the coefficient of \ds \theta is not an integer. For example, if it is rational, then the curve is closed; that is, it eventually ends where it started, see Figure 6 (a) below. However, if the coefficient is irrational, then the curve never closes, see Figure 6 (b). Although it may appear that the curve is closed, a closer examination reveals that the petals just above the positive x axis are slightly thicker. This is because the petal does not quite match up with the starting point.

This figure has two figures. The first is a rose with so many overlapping petals that there are a few patterns that develop, starting with a sharp 10 pointed star in the center and moving out to an increasingly rounded set of petals. The second figure is a rose with even more overlapping petals, so many so that it is impossible to tell what is happening in the center, but on the outer edges are a number of sharply rounded petals.
Figure 6. Polar rose graphs of functions with (a) rational coefficient and (b) irrational coefficient. Note that the rose in part (b) would actually fill the entire circle if plotted in full.

Since the curve defined by the graph of \ds r=3\phantom{\rule{0.2em}{0ex}}\text{sin}\left(\pi \theta \right) never closes, the curve shown in Figure 6 (b) is only a partial depiction. In fact, this is an example of a space-filling curve. A space-filling curve is one that in fact occupies a two-dimensional subset of the real plane. In this case, the curve occupies the circle of radius 3 centered at the origin.

Chapter Opener: Describing a Spiral

Recall the chambered nautilus introduced in the chapter opener. This creature displays a spiral when half the outer shell is cut away. Figure 7 below shows a spiral in rectangular coordinates. How can we describe this curve analitically, that is, using formulas?

A spiral starting at the origin and continually increasing its radius to a point P(x, y).
Figure 7. How can we describe a spiral graph analitically?

Solution

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 \ds \theta that the line segment OP makes with the positive x-axis. Therefore \ds d\left(P,O\right)=k\theta , where \ds O is the origin. Now use the distance formula and some trigonometry:

\ds \begin{array}{ccc}\hfill d\left(P,O\right)&\ds =\hfill &\ds k\theta \hfill \\[5mm]\ds \hfill \sqrt{{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}}&\ds =\hfill &\ds k\phantom{\rule{0.2em}{0ex}}\text{arctan}\left(\frac{y}{x}\right)\hfill \\[5mm]\ds \hfill \sqrt{{x}^{2}+{y}^{2}}&\ds =\hfill &\ds k\phantom{\rule{0.2em}{0ex}}\text{arctan}\left(\frac{y}{x}\right)\hfill \\[5mm]\ds \hfill \text{arctan}\left(\frac{y}{x}\right)&\ds =\hfill &\ds \frac{\sqrt{{x}^{2}+{y}^{2}}}{k}\hfill \\[5mm]\ds \hfill y&\ds =\hfill &\ds x\phantom{\rule{0.2em}{0ex}}\text{tan}\left(\frac{\sqrt{{x}^{2}+{y}^{2}}}{k}\right).\hfill \end{array}

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, \ds d\left(P,O\right)=r, and \ds \theta is the second coordinate. Therefore the equation for the spiral becomes \ds r=k\theta . Note that when \ds \theta =0 we also have \ds 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 \ds r=a+k\theta for arbitrary constants \ds a and \ds 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 \ds r=a\cdot {b}^{\theta }. A graph of the function \ds r=1.2\left({1.25}^{\theta }\right) is given in Figure 8 below. This spiral describes the shell shape of the chambered nautilus.

This figure has two figures. The first is a shell with many chambers that increase in size from the center out. The second is a spiral with equation r = 1.2(1.25θ).
Figure 8. A logarithmic spiral is similar to the shape of the chambered nautilus shell. (credit: modification of work by Jitze Couperus, Flickr)

Suppose a curve is described in the polar coordinate system via the function \ds r=f\left(\theta \right). Since we have conversion formulas from polar to rectangular coordinates given by \ds \begin{array}{c}x=r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta),\ y=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) ,\hfill \end{array} we can rewrite the polar equation in Cartesian coordinates:

\ds \begin{array}{c}x=f\left(\theta \right)\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) ,\ y=f\left(\theta \right)\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) .\hfill \end{array}

This step gives a parameterization of the curve in rectangular coordinates using \ds \theta as the parameter and allows to perform calculus methods we developed in the previous section to parametric curves.

Calculus with Polar Curves

Find the slope of the tangent line to the spiral with polar equation r=\pi-\theta at the point corresponding to \ds\theta=\frac{2\pi}3.

Solution

We first find the parametrization of the spiral in rectangular coordinates by using conversion formulas (*) and replacing r with f(\theta) as was shown above:

\ds \begin{array}{l} &x=r\cos(\theta)=(\pi-\theta)\cos(\theta)\hfill\\[3mm] &y=r\sin(\theta)=(\pi-\theta)\sin(\theta)\hfill\end{array}

Next we find \ds\frac{dy}{dx} as a function of \theta using the methods of the previous section.

\ds\begin{array}{ll}\ds\frac{dy}{dx}&\ds=\frac{dy\text{/}d\theta}{dx\text{/}d\theta}=\frac{\frac d{d\theta}\big((\pi-\theta)\sin(\theta)\big)}{\frac d{d\theta}\big((\pi-\theta)\cos(\theta)\big)}=\frac{-\sin(\theta)+(\pi-\theta)\cos(\theta)}{-\cos(\theta)+(\pi-\theta)(-\sin(\theta))}\hfill\end{array}

The slope m of the tangent line to the spiral at the point corresponding to \ds\theta=\frac{2\pi}3 can be found by substituting \ds\theta=\frac{2\pi}3 into the above formula for \frac{dy}{dx}:

\ds \begin{array}{ll} \ds m&\ds=\frac{dy}{dx}\left(\frac{2\pi}3\right)=\frac{-\sin\left(\frac{2\pi}3\right)+\left(\pi-\frac{2\pi}3\right)\cos\left(\frac{2\pi}3\right)}{-\cos\left(\frac{2\pi}3\right)+\left(\pi-\frac{2\pi}3\right)\left(-\sin\left(\frac{2\pi}3\right)\right)}\hfill\\[5mm] &\ds=\frac{-\frac{\sqrt3}2+\frac{\pi}3\cdot\left(-\frac12\right)}{-\left(-\frac12\right)+\frac{\pi}3\cdot\left(-\frac{\sqrt3}2\right)} =\frac{3\sqrt3+\pi}{-3+\sqrt3\pi}.\hfill\end{array}

Find the slope of the tangent line to the polar curve r=1+\sin(\theta) at the point corresponding to \ds\theta=-\frac{\pi}4.

Answer

The slope is \ds 1-\sqrt 2.

Symmetry in Polar Coordinates

When studying symmetry of functions in rectangular coordinates (i.e., in the form \ds y=f\left(x\right),) we talk about symmetry with respect to the y-axis and symmetry with respect to the origin. In particular, if \ds f\left( - x\right)=f\left(x\right) for all \ds x in the domain of \ds f, then \ds f is an even function and its graph is symmetric with respect to the y-axis. If \ds f\left( - x\right)= - f\left(x\right) for all \ds x in the domain of \ds f, then \ds f is an odd function and its graph is symmetric with respect to the origin. By determining which types of symmetry a graph exhibits, we can learn more about the shape and appearance of the graph. Symmetry can also reveal other properties of the function that generates the graph. Symmetry in polar curves works in a similar fashion.

Symmetry in Polar Curves and Equations

Consider a polar curve with equation \ds r=f\left(\theta \right).

  1. The curve is symmetric about the polar axis if for every point \ds \left(r,\theta \right) on the graph, the point \ds \left(r, - \theta \right) is also on the graph. This happens if f(-\theta)=f(\theta) or f(\pi-\theta)=-f(\theta).
  2. The curve is symmetric about the pole if for every point \ds \left(r,\theta \right) on the graph, the point \ds \left(r,\pi +\theta \right) is also on the graph. This happens if f(\pi+\theta)=f(\theta).
  3. The curve is symmetric about the vertical line \ds \theta =\frac{\pi }{2} if for every point \ds \left(r,\theta \right) on the graph, the point \ds \left(r,\pi -\theta \right) is also on the graph. This happens if f(\pi-\theta)=f(\theta) or f(-\theta)=-f(\theta).

The following table shows examples of each type of symmetry.

This table has three rows and two columns. The first row reads “Symmetry with respect to the polar axis: For every point (r, θ) on the graph, there is also a point reflected directly across the horizontal (polar) axis” and it has a picture of a cardioid with equation r = 2 – 2 cosθ: this cardioid has points marked (r, θ) and (r, −θ), which are symmetric about the x axis, and the entire cardioid is symmetric about the x axis. The second row reads “Symmetry with respect to the pole: For every point (r, θ) on the graph, there is also a point on the graph that is reflected through the pole as well” and it has a picture of a skewed infinity symbol with equation r2 = 9 cos(2θ – π/2): this figure has points marked (r, θ) and (−r, θ), which are symmetric about the pole, and the entire figure is symmetric about the pole. The third row reads “Symmetry with respect to the vertical line θ = π/2: For every point (r, θ) on the graph, there is also a point reflected directly across the vertical axis” and there is a picture of a cardioid with equation r = 2 – 2 sinθ: this figure has points marked (r, θ) and (r, π − θ), which are symmetric about the vertical line θ = π/2, and the entire cardioid is symmetric about the vertical line θ = π/2.

Using Symmetry to Graph a Polar Equation

Determine all symmetries of the rose defined by the equation \ds r=3\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2\theta \right) and create a graph.

Solution

Suppose the point \ds \left(r,\theta \right) is on the graph of \ds r=3\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2\theta \right).

Let f(\theta)=3\sin(2\theta). We first substitute -\theta instead of \theta into f:

f(-\theta)=3\sin(-2\theta)=-3\sin(2\theta)=-f(\theta),

since sine is an odd function. According to iii in the statement above, this implies symmetry with respect to the vertical line \ds\thet\frac{\pi}2.

To test for symmetry with respect to the polar axis, we consider f(\pi-\theta):

f(\pi-\theta)=3\sin(2\pi-2\theta)=3\sin(-2\theta)=-3\sin(2\theta),

since sine function is 2\pi-periodic and odd. Hence, by i, we have that the curve is symmetric with respect to the polar axis as well.

Geometrically, the above two symmetries automatically imply symmetry with respect to the pole, but this can also be verified analytically by checking that f(\pi+\theta)=f(\theta).

So this graph has symmetry with respect to the vertical line going through the pole, the polar axis and the origin. To sketch the curve, tabulate values of \ds \theta between 0 and \ds \frac{\pi}2 and then reflect the resulting graph about the polar axis and the line \ds\theta=\frac{\pi}2.

\ds \theta \ds r
\ds 0 \ds 0
\ds \frac{\pi }{6} \ds \frac{3\sqrt{3}}{2}\approx 2.6
\ds \frac{\pi }{4} \ds 3
\ds \frac{\pi }{3} \ds \frac{3\sqrt{3}}{2}\approx 2.6
\ds \frac{\pi }{2} \ds 0

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

A single petal is graphed with equation r = 3 sin(2θ) for 0 ≤ θ ≤ π/2. It starts at the origin and reaches a maximum distance from the origin of 3.
Figure 9. The graph of the equation between \ds \theta =0 and \ds \theta =\pi \text{/}2.

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

A four-petaled rose is graphed with equation r = 3 sin(2θ). Each petal starts at the origin and reaches a maximum distance from the origin of 3.
Figure 10. The entire graph of the equation is called a four-petaled rose.

Determine all symmetries of the polar curve \ds r=2\phantom{\rule{0.2em}{0ex}}\text{cos}\left(3\theta \right) and create a graph.

Answer

Symmetric with respect to the polar axis.

A three-petaled rose is graphed with equation r = 2 cos(3θ). Each petal starts at the origin and reaches a maximum distance from the origin of 2.

Key Concepts

  • The polar coordinate system provides an alternative way to locate points in the plane.
  • Convert points between rectangular and polar coordinates using the formulas

    \ds x=r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta);

    \ds r=\sqrt{{x}^{2}+{y}^{2}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{tan}\phantom{\rule{0.2em}{0ex}}(\theta) =\frac{y}{x}.

  • To sketch a polar curve, make a table of values and take advantage of periodic properties.
  • Use the conversion formulas to convert equations between rectangular and polar coordinates.
  • Identify symmetry in polar curves, which can occur through the pole, the horizontal axis, or the vertical axis.

Exercises

In the following exercises, plot the point whose polar coordinates are given by first constructing the angle \ds \theta and then marking off the distance r along the ray.

1. \ds \left(3,\frac{\pi }{6}\right)

Answer

 

On the polar coordinate plane, a ray is drawn from the origin marking π/6 and a point is drawn when this line crosses the circle with radius 3.

2. \ds \left(-2,\frac{5\pi }{3}\right)

3. \ds \left(0,\frac{7\pi }{6}\right)

Answer

 

On the polar coordinate plane, a ray is drawn from the origin marking 7π/6 and a point is drawn when this line crosses the circle with radius 0, that is, it marks the origin.

4. \ds \left(-4,\frac{3\pi }{4}\right)

5. \ds \left(1,\frac{\pi }{4}\right)

Answer

 

On the polar coordinate plane, a ray is drawn from the origin marking π/4 and a point is drawn when this line crosses the circle with radius 1.

6. \ds \left(2,\frac{5\pi }{6}\right)

7. \ds \left(1,\frac{\pi }{2}\right)

Answer

 

On the polar coordinate plane, a ray is drawn from the origin marking π/2 and a point is drawn when this line crosses the circle with radius 1.

For the following exercises, consider the polar graph below. Give two sets of polar coordinates for each point.

The polar coordinate plane is divided into 12 pies. Point A is drawn on the first circle on the first spoke above the θ = 0 line in the first quadrant. Point B is drawn in the fourth quadrant on the third circle and the second spoke below the θ = 0 line. Point C is drawn on the θ = π line on the third circle. Point D is drawn on the fourth circle on the first spoke below the θ = π line.

8. Point A.

9. Point B.

Answer

\ds \begin{array}{cc}\ds \left(3,\frac{ - \pi }{3}\right),\hfill &\ds \left(-3,\frac{2\pi }{3}\right)\hfill \end{array}

10. Point C.

11. Point D.

Answer

\ds \left(5,\frac{7\pi }{6}\right),\quad\left(-5,\frac{\pi }{6}\right)

For the following exercises, the rectangular coordinates of a point are given. Find two sets of polar coordinates for the point with the angular coordinate \theta in \ds \left(-\pi,\pi \right].

12. \ds \left(2,2\right)

13. \ds \left(3,-4\right)

Answer

\ds \begin{array}{cc}\ds \left(5,-\tan^{-1}\left(\frac43\right)\right),\hfill &\ds \left(-5,\pi-\tan^{-1}\left(\frac43\right)\right)\hfill \end{array}

14. \ds \left(8,15\right)

15. \ds \left(-2,5\right)

Answer

\ds \left(\sqrt{29},\pi-\tan^{-1}\left(\frac52\right)\right),\ \left(-\sqrt{29},-\tan^{-1}\left(\frac52\right) \right)

16. \ds \left(-4,-3\right)

17. \ds \left(3, - \sqrt{3}\right)

Answer

\ds \left(2\sqrt{3},-\frac{\pi}6\right),\ \left(-2\sqrt{3},\frac{5\pi}6 \right)

For the following exercises, find rectangular coordinates for the given point in polar coordinates.

18. \ds \left(2,\frac{5\pi }{4}\right)

19. \ds \left(-2,\frac{\pi }{6}\right)

Answer

\ds \left(-\sqrt3,-1\right)

20. \ds \left(-3,\frac{3\pi }{4}\right)

21. \ds \left(1,\frac{17\pi }{3}\right)

Answer

\ds \left(\frac12,-\frac{\sqrt 3}2\right)

22. \ds \left(5,-\frac{5\pi }{2}\right)

23. \ds \left(0,\frac{\pi }{8}\right)

Answer

\ds (0,0)

24. \ds \left(\pi,-1\right)

(There is no typo in this question.)

For the following exercises, determine the shape of each polar curve.

25. \ds r=-3

Answer

Circle.

26. \ds \theta =\frac{7\pi }{4}

27. \ds r=e^{\theta}

Answer

Spiral.

For the following exercises, convert the polar equation to rectangular form and sketch its graph.

28. \ds r=\text{csc}\phantom{\rule{0.2em}{0ex}}(\theta)

29. \ds r=4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta)

Answer

\ds {x}^{2}+(y-2)^2=4

A circle of radius 2 with center at (2, π/2).

30. \ds r=6\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta)

31. Consider polar curve r=f(\theta).  Then \ds x=r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) =f\left(\theta \right)\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) and \ds y=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) =f\left(\theta \right)\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) are parametric equations of the curve in rectangular coordinates. Using calculus for parametric curves, derive the formula for the derivative \ds\frac{dy}{dx} of a polar equation.

Answer

\ds \frac{dy}{\,dx }=\frac{{f}^{\prime }\left(\theta \right)\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) +f\left(\theta \right)\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) }{{f}^{\prime }\left(\theta \right)\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) -f\left(\theta \right)\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) }

For the following exercises, find the slope of the tangent line to the given polar curve at the point that corresponds to the specified value of angular coordinate \theta.

32. \ds r=1-\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta);\ \ds \theta=\frac{5\pi }{6}

33. \ds r=4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta);\ \theta=\frac{\pi }{3}

Answer

The slope is \ds \frac{1}{\sqrt{3}}.

34. \ds r=\sin(3\theta);\ \theta=\frac{\pi}2

35. \ds r=1-\cos\left(\frac{\theta}2\right);\ \theta=\frac{2\pi}3

Answer

The slope is \ds-\frac{\sqrt3}{9}.

36. \ds r=\sqrt3\tan(\theta);\ \theta=-\frac{\pi}{6}

37. \ds r=2\theta;\ \theta=\frac{\pi }{4}

Answer

The slope is \ds\frac{4+\pi}{4-\pi}.

For the following exercises, find the Cartesian coordinates of the points at which the following polar curves have horizontal or vertical tangent line.

38. \ds r=2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta)

39.* The cardioid \ds r=1+\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta).

(Hint: if for some \theta_0 both \ds\frac{dx}{d\theta}(\theta_0)=0 and \ds\frac{dy}{d\theta}(\theta_0)=0, one needs to consider \ds\lim_{\theta\to\theta_0} \frac{dy}{dx} to determine if the tangent line at the corresponding point is horizontal, vertical, or neither.)

Answer

Horizontal tangents at \ds (0,0), \left(\frac34,\pm\frac{3\sqrt3}4\right). Vertical tangents at \ds (0,2), \left(-\frac14,\pm\frac{\sqrt3}4\right).

40. Find all points on the polar curve \ds r=1+\cot(\theta), where the tangent line is horizontal.

For the following exercises, determine whether the polar curves are symmetric with respect to the polar axis, the pole, or the vertical line \ds\theta=\frac{\pi}2 through the pole.

41. \ds r=\phantom{\rule{0.2em}{0ex}}\text{cos}\left(2\theta \right)

Answer

Symmetric with respect to the polar axis, the pole, and the vertical line \ds\theta=\frac{\pi}2.

42. \ds r=1-2\sin(\theta)

43. \ds r=\text{cos}\left(\frac{\theta }{5}\right)

Answer

Symmetric with respect to the polar axis only.

44. \ds r=2\phantom{\rule{0.2em}{0ex}}\text{sec}\phantom{\rule{0.2em}{0ex}}(\theta)

45. \ds r=\theta^3

Answer

Symmetry with respect to the vertical line \ds\theta=\frac{\pi}2 only.

For the following exercises, identify any symmetries of a given polar curve and then sketch it.

46. \ds r=-4\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta)

47. \ds r=1+\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta)

Answer

 Symmetric with respect to the vertical line \ds\theta=\frac{\pi}2.

A cardioid with the upper heart part at the origin and the rest of the cardioid oriented up.

48. \ds r=3-2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta)

49. \ds r=2-2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta)

Answer

 Symmetric with respect to the vertical line \ds\theta=\frac{\pi}2.

A cardioid with the upper heart part at the origin and the rest of the cardioid oriented down.

50. \ds r=1-2\cos\left(\theta\right)

51. \ds r=3\phantom{\rule{0.2em}{0ex}}\text{cos}\left(2\theta \right)

Answer

Symmetric with respect to the pole, the polar axis and the vertical line \ds\theta=\frac{\pi}2.

 

A rose with four petals that reach their furthest extent from the origin at θ = 0, π/2, π, and 3π/2.

52. \ds r=2\phantom{\rule{0.2em}{0ex}}\text{sin}\left(\frac{\theta}2 \right)

53. \ds r=2\phantom{\rule{0.2em}{0ex}}\text{cos}\left(3\theta \right)

Answer

 Symmetric with respect to the polar axis.

A rose with three petals that reach their furthest extent from the origin at θ = 0, 2π/3, and 4π/3.

54. \ds r=\phantom{\rule{0.2em}{0ex}}\text{cos}\left(\frac{\theta }{3}\right)

55. \ds {r}^{2}=4\phantom{\rule{0.2em}{0ex}}\text{cos}\left(2\theta \right)

Answer

Symmetric with respect to the pole, the polar axis and the vertical line \ds\theta=\frac{\pi}2.

The infinity symbol with the crossing point at the origin and with the furthest extent of the two petals being at θ = 0 and π.

56. \ds {r}^{2}=4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta)

57. \ds r=2\theta, \theta\ge0

Answer

 No symmetries.

A spiral that starts at the origin crossing the line θ = π/2 between 3 and 4, θ = π between 6 and 7, θ = 3π/2 between 9 and 10, θ = 0 between 12 and 13, θ = π/2 between 15 and 16, and θ = π between 18 and 19.

58. [T] The graph of \ds r=2\phantom{\rule{0.2em}{0ex}}\text{cos}\left(2\theta \right)\text{sec}\left(\theta \right) is called a strophoid. Use a graphing utility to sketch the graph, and, from the graph, determine the asymptote.

59. [T] Use a graphing utility and sketch the graph of \ds r=\frac{6}{2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) -3\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) }.

Answer

A line that crosses the y axis at roughly 3 and has slope roughly 3/2.

60. [T] Use a graphing utility to graph \ds r=\frac{1}{1-\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) }.

61. [T] Use technology to graph \ds r={e}^{\text{sin}\left(\theta \right)}-2\phantom{\rule{0.2em}{0ex}}\text{cos}\left(4\theta \right).

Answer

 

A geometric shape that resembles a butterfly with larger wings in the first and second quadrants, smaller wings in the third and fourth quadrants, a body along the θ = π/2 line and legs along the θ = 0 and π lines.

62. [T] Use technology to plot \ds r=\text{sin}\left(\frac{3\theta }{7}\right) (use the interval \ds 0\le \theta \le 14\pi \right).

63. [T] Use technology to plot \ds r={e}^{-0.1\theta } for \ds -10\le \theta \le 10.

Answer

 

A spiral that starts in the third quadrant.

64. [T] There is a curve known as the “Black Hole.” Use technology to plot \ds r={e}^{-0.01\theta } for \ds -100\le \theta \le 100.

Glossary

angular coordinate
\ds \theta the angle formed by a line segment connecting the origin to a point in the polar coordinate system with the positive radial (x) axis, measured counterclockwise
cardioid
a plane curve traced by a point on the perimeter of a circle that is rolling around a fixed circle of the same radius; the equation of a cardioid is \ds r=a\left(1+\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \right) or \ds r=a\left(1+\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \right)
limaçon
the graph of the equation \ds r=a+b\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) or \ds r=a+b\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) . If \ds a=b then the graph is a cardioid
polar axis
the horizontal axis in the polar coordinate system corresponding to \ds r\ge 0
polar coordinate system
a system for locating points in the plane. The coordinates are \ds r, the radial coordinate, and \ds \theta , the angular coordinate
polar equation
an equation or function relating the radial coordinate to the angular coordinate in the polar coordinate system
pole
the central point of the polar coordinate system, equivalent to the origin of a Cartesian system
radial coordinate
\ds r the coordinate in the polar coordinate system that measures the distance from a point in the plane to the pole
rose
graph of the polar equation \ds r=a\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}2\theta or \ds r=a\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}2\theta for a positive constant a
space-filling curve
a curve that completely occupies a two-dimensional subset of the real plane

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Calculus: Volume 2 (Second University of Manitoba Edition) Copyright © 2021 by Gilbert Strang and Edward 'Jed' Herman, modified by Varvara Shepelska is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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