# 10.4: Polar Coordinates - Graphs - Mathematics

Learning Objectives

• Test polar equations for symmetry.
• Graph polar equations by plotting points.

Keplar's First Law of Planetary Motion argues that the planets move through space in elliptical, periodic orbits about the sun, as shown in Figure (PageIndex{1}). They are in constant motion, so fixing an exact position of any planet is valid only for a moment. In other words, we can fix only a planet’s instantaneous position. This is one application of polar coordinates, represented as ((r, heta)). We interpret (r) as the distance from the sun and ( heta) as the planet’s angular bearing, or its direction from a fixed point on the sun. In this section, we will focus on the polar system and the graphs that are generated directly from polar coordinates. #### Testing Polar Equations for Symmetry

Just as a rectangular equation such as (y=x^2) describes the relationship between (x) and (y) on a Cartesian grid, a polar equation describes a relationship between (r) and ( heta) on a polar grid. Recall that the coordinate pair ((r, heta)) indicates that we move counterclockwise from the polar axis (positive (x)-axis) by an angle of ( heta), and extend a ray from the pole (origin) (r) units in the direction of ( heta). All points that satisfy the polar equation are on the graph.

Symmetry is a property that helps us recognize and plot the graph of any equation. If an equation has a graph that is symmetric with respect to an axis, it means that if we folded the graph in half over that axis, the portion of the graph on one side would coincide with the portion on the other side. By performing three tests, we will see how to apply the properties of symmetry to polar equations. Further, we will use symmetry (in addition to plotting key points, zeros, and maximums of (r)) to determine the graph of a polar equation.

In the first test, we consider symmetry with respect to the line ( heta=dfrac{pi}{2}) ((y)-axis). We replace ((r, heta)) with ((−r,− heta)) to determine if the new equation is equivalent to the original equation. For example, suppose we are given the equation (r=2 sin heta);

[egin{align*} r&= 2 sin heta -r&= 2 sin - heta qquad ext{Replace } (r, heta) ext{ with }(-r,- heta). -r&= -2 sin heta qquad ext{Identity: }sin(- heta)=-sin heta. r&= 2 sin heta qquad ext{Multiply both sides by }-1 end{align*}]

This equation exhibits symmetry with respect to the line ( heta=dfrac{pi}{2}).

In the second test, we consider symmetry with respect to the polar axis ((x)-axis). We replace ((r, heta)) with ((r,− heta)) or ((−r,pi− heta)) to determine equivalency between the tested equation and the original. For example, suppose we are given the equation (r=1−2 cos heta).

[egin{align*} r&= 1-2 cos heta r&= 1-2 cos(- heta)qquad ext{Replace }(r, heta) ext{ with }(r,- heta). r&= 1-2 cos heta qquad ext{Even/Odd identity} end{align*}]

The graph of this equation exhibits symmetry with respect to the polar axis.

In the third test, we consider symmetry with respect to the pole (origin). We replace ((r, heta)) with ((−r, heta)) to determine if the tested equation is equivalent to the original equation. For example, suppose we are given the equation (r=2 sin(3 heta)).

(r=2 sin(3 heta))

(−r=2 sin(3 heta))

The equation has failed the symmetry test, but that does not mean that it is not symmetric with respect to the pole. Passing one or more of the symmetry tests verifies that symmetry will be exhibited in a graph. However, failing the symmetry tests does not necessarily indicate that a graph will not be symmetric about the line ( heta=dfrac{pi}{2}), the polar axis, or the pole. In these instances, we can confirm that symmetry exists by plotting reflecting points across the apparent axis of symmetry or the pole. Testing for symmetry is a technique that simplifies the graphing of polar equations, but its application is not perfect.

Note: SYMMETRY TESTS

A polar equation describes a curve on the polar grid. The graph of a polar equation can be evaluated for three types of symmetry, as shown in Figure (PageIndex{2}). How to: Given a polar equation, test for symmetry.

1. Substitute the appropriate combination of components for ((r, heta)): ((−r,− heta)) for ( heta=dfrac{pi}{2}) symmetry; ((r,− heta)) for polar axis symmetry; and ((−r, heta)) for symmetry with respect to the pole.
2. If the resulting equations are equivalent in one or more of the tests, the graph produces the expected symmetry.

Example (PageIndex{1}): Testing a Polar Equation for Symmetry

Test the equation (r=2 sin heta) for symmetry.

Solution

Test for each of the three types of symmetry.

 1) Replacing ((r, heta)) with ((−r,− heta)) yields the same result. Thus, the graph is symmetric with respect to the line ( heta=dfrac{pi}{2}). (−r=2 sin(− heta))(−r=−2 sin heta) Even-odd identity(r=2 sin heta) Multiply by (−1)Passed 2) Replacing ( heta) with (− heta) does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis. (r=2 sin(− heta))(r=−2 sin heta) Even-odd identity(r=−2 sin heta ≠ 2 sin heta)Failed 3) Replacing (r) with (–r) changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole. (−r=2 sin heta)(r=−2 sin heta ≠2 sin heta)Failed

Analysis

Using a graphing calculator, we can see that the equation (r=2 sin heta) is a circle centered at ((0,1)) with radius (r=1) and is indeed symmetric to the line ( heta=dfrac{pi}{2}). We can also see that the graph is not symmetric with the polar axis or the pole. See Figure (PageIndex{3}). Exercise (PageIndex{1})

Test the equation for symmetry: (r=−2 cos heta).

The equation fails the symmetry test with respect to the line ( heta=dfrac{pi}{2}) and with respect to the pole. It passes the polar axis symmetry test.

#### Graphing Polar Equations by Plotting Points

To graph in the rectangular coordinate system we construct a table of (x) and (y) values. To graph in the polar coordinate system we construct a table of ( heta) and (r) values. We enter values of ( heta) into a polar equation and calculate (r). However, using the properties of symmetry and finding key values of ( heta) and (r) means fewer calculations will be needed.

##### Finding Zeros and Maxima

To find the zeros of a polar equation, we solve for the values of ( heta) that result in (r=0). Recall that, to find the zeros of polynomial functions, we set the equation equal to zero and then solve for (x). We use the same process for polar equations. Set (r=0), and solve for ( heta).

For many of the forms we will encounter, the maximum value of a polar equation is found by substituting those values of ( heta) into the equation that result in the maximum value of the trigonometric functions. Consider (r=5 cos heta); the maximum distance between the curve and the pole is (5) units. The maximum value of the cosine function is (1) when ( heta=0), so our polar equation is (5 cos heta), and the value ( heta=0) will yield the maximum (| r |).

Similarly, the maximum value of the sine function is (1) when ( heta=dfrac{pi}{2}), and if our polar equation is (r=5 sin heta), the value ( heta=dfrac{pi}{2}) will yield the maximum (| r |). We may find additional information by calculating values of (r) when ( heta=0). These points would be polar axis intercepts, which may be helpful in drawing the graph and identifying the curve of a polar equation.

Example (PageIndex{2}): Finding Zeros and Maximum Values for a Polar Equation

Using the equation in Example (PageIndex{1}), find the zeros and maximum (| r |) and, if necessary, the polar axis intercepts of (r=2 sin heta).

Solution

To find the zeros, set (r) equal to zero and solve for ( heta).

[egin{align*} 2 sin heta &= 0 sin heta &= 0 heta &= {sin}^{-1} 0 heta &= npi qquad ext{where n is an integer} end{align*}]

Substitute any one of the ( heta) values into the equation. We will use (0).

[egin{align*} r&= 2 sin(0) r&= 0 end{align*}]

The points ((0,0)) and ((0,pm npi)) are the zeros of the equation. They all coincide, so only one point is visible on the graph. This point is also the only polar axis intercept.

To find the maximum value of the equation, look at the maximum value of the trigonometric function (sin heta), which occurs when ( heta=dfrac{pi}{2}pm 2kpi) resulting in (sinleft(dfrac{pi}{2} ight)=1). Substitute (dfrac{pi}{2}) for ( heta).

[egin{align*} r&= 2 sinleft(dfrac{pi}{2} ight) r&= 2(1) r&= 2 end{align*}]

Analysis

The point (left(2,dfrac{pi}{2} ight)) will be the maximum value on the graph. Let’s plot a few more points to verify the graph of a circle. See Table (PageIndex{1}) and Figure (PageIndex{4}).

Table (PageIndex{1})
( heta)(r=2 sin heta)(r)
(0)(r=2 sin(0)=0)(0)
(dfrac{pi}{6})(r=2 sinleft(dfrac{pi}{6} ight)=1)(1)
(dfrac{pi}{3})(r=2 sinleft(dfrac{pi}{3} ight)≈1.73)(1.73)
(dfrac{pi}{2})(r=2 sinleft(dfrac{pi}{2} ight)=2)(2)
(dfrac{2pi}{3})(r=2 sinleft(dfrac{2pi}{3} ight)≈1.73)(1.73)
(dfrac{5pi}{6})(r=2 sinleft(dfrac{5pi}{6} ight)=1)(1)
(pi)(r=2 sin(pi)=0)(0) Exercise (PageIndex{2})

Without converting to Cartesian coordinates, test the given equation for symmetry and find the zeros and maximum values of (| r |): (r=3 cos heta).

Tests will reveal symmetry about the polar axis. The zero is (left(0,dfrac{pi}{2} ight)), and the maximum value is ((3,0)).

##### Investigating Circles

Now we have seen the equation of a circle in the polar coordinate system. In the last two examples, the same equation was used to illustrate the properties of symmetry and demonstrate how to find the zeros, maximum values, and plotted points that produced the graphs. However, the circle is only one of many shapes in the set of polar curves.

There are five classic polar curves: cardioids, limaҫons, lemniscates, rose curves, and Archimedes’ spirals. We will briefly touch on the polar formulas for the circle before moving on to the classic curves and their variations.

FORMULAS FOR THE EQUATION OF A CIRCLE

Some of the formulas that produce the graph of a circle in polar coordinates are given by (r=a cos heta) and (r=a sin heta), wherea a is the diameter of the circle or the distance from the pole to the farthest point on the circumference. The radius is (dfrac{|a|}{2}), or one-half the diameter. For (r=a cos heta), the center is (left(dfrac{a}{2},0 ight)). For (r=a sin heta), the center is (left(dfrac{a}{2},pi ight)). Figure (PageIndex{5}) shows the graphs of these four circles.

0. The center is at (a/2,0). Second is r=acos(theta), a<0. Third is r=asin(theta), a>0. The center is at (a/2, pi). Fourth is r=asin(theta), a<0. The center is at (a/2, pi)." src="/@api/deki/files/7429/CNX_Precalc_Figure_08_04_005new.jpg">

Example (PageIndex{3}): Sketching the Graph of a Polar Equation for a Circle

Sketch the graph of (r=4 cos heta).

Solution

First, testing the equation for symmetry, we find that the graph is symmetric about the polar axis. Next, we find the zeros and maximum (| r |) for (r=4 cos heta). First, set (r=0), and solve for ( heta). Thus, a zero occurs at ( heta=dfrac{pi}{2}pm kpi). A key point to plot is (left(0,dfrac{pi}{2} ight)).

To find the maximum value of (r), note that the maximum value of the cosine function is (1) when ( heta=0pm 2kpi). Substitute ( heta=0) into the equation:

[egin{align*} r&= 4 cos heta r&= 4 cos(0) r&= 4(1) &= 4 end{align*}]

The maximum value of the equation is (4). A key point to plot is ((4, 0)).

As (r=4 cos heta) is symmetric with respect to the polar axis, we only need to calculate (r)-values for (θ) over the interval ([0, pi]). Points in the upper quadrant can then be reflected to the lower quadrant. Make a table of values similar to Table (PageIndex{2}). The graph is shown in Figure (PageIndex{6}).

 ( heta) (r) (0) (dfrac{pi}{6}) (dfrac{pi}{4}) (dfrac{pi}{3}) (dfrac{pi}{2}) (dfrac{2pi}{3}) (dfrac{3pi}{4}) (dfrac{5pi}{6}) (pi) (4) (3.46) (2.83) (2) (0) (−2) (−2.83) (−3.46) (4) ##### Investigating Cardioids

While translating from polar coordinates to Cartesian coordinates may seem simpler in some instances, graphing the classic curves is actually less complicated in the polar system. The next curve is called a cardioid, as it resembles a heart. This shape is often included with the family of curves called limaçons, but here we will discuss the cardioid on its own.

FORMULAS FOR A CARDIOID

The formulas that produce the graphs of a cardioid are given by (r=apm b cos heta) and (r=apm b sin heta) where (a>0), (b>0), and (dfrac{a}{b}=1). The cardioid graph passes through the pole, as we can see in Figure (PageIndex{7}). How to: Given the polar equation of a cardioid, sketch its graph

1. Check equation for the three types of symmetry.
2. Find the zeros. Set (r=0).
3. Find the maximum value of the equation according to the maximum value of the trigonometric expression.
4. Make a table of values for (r) and ( heta).
5. Plot the points and sketch the graph.

Example (PageIndex{4}): Sketching the Graph of a Cardioid

Sketch the graph of (r=2+2 cos heta).

Solution

First, testing the equation for symmetry, we find that the graph of this equation will be symmetric about the polar axis. Next, we find the zeros and maximums. Setting (r=0), we have ( heta=pi+2kpi). The zero of the equation is located at ((0,pi)). The graph passes through this point.

The maximum value of (r=2+2 cos heta) occurs when (cos heta) is a maximum, which is when (cos heta=1) or when ( heta=0). Substitute ( heta=0) into the equation, and solve for (r).

[egin{align*} r&= 2+2 cos(0) r&= 2+2(1) &= 4 end{align*}]

The point ((4,0)) is the maximum value on the graph.

We found that the polar equation is symmetric with respect to the polar axis, but as it extends to all four quadrants, we need to plot values over the interval ([0, pi]). The upper portion of the graph is then reflected over the polar axis. Next, we make a table of values, as in Table (PageIndex{3}), and then we plot the points and draw the graph. See Figure (PageIndex{8}).

 θ r 0 (dfrac{π}{4}) (dfrac{π}{2}) (dfrac{2π}{3}) (π) 4 3.41 2 1 0 ##### Investigating Limaçons

The word limaçon is Old French for “snail,” a name that describes the shape of the graph. As mentioned earlier, the cardioid is a member of the limaçon family, and we can see the similarities in the graphs. The other images in this category include the one-loop limaçon and the two-loop (or inner-loop) limaçon. One-loop limaçons are sometimes referred to as dimpled limaçons when (1

FORMULAS FOR ONE-LOOP LIMAÇONS

The formulas that produce the graph of a dimpled one-loop limaçon are given by (r=apm b cos heta) and (r=apm b sin heta) where (a>0), (b>0), and (1 How to: Given a polar equation for a one-loop limaçon, sketch the graph

1. Test the equation for symmetry. Remember that failing a symmetry test does not mean that the shape will not exhibit symmetry. Often the symmetry may reveal itself when the points are plotted.
2. Find the zeros.
3. Find the maximum values according to the trigonometric expression.
4. Make a table.
5. Plot the points and sketch the graph.

Example (PageIndex{5}): Sketching the Graph of a One-Loop Limaçon

Graph the equation (r=4−3 sin heta).

Solution

First, testing the equation for symmetry, we find that it fails all three symmetry tests, meaning that the graph may or may not exhibit symmetry, so we cannot use the symmetry to help us graph it. However, this equation has a graph that clearly displays symmetry with respect to the line ( heta=dfrac{pi}{2}), yet it fails all the three symmetry tests. A graphing calculator will immediately illustrate the graph’s reflective quality.

Next, we find the zeros and maximum, and plot the reflecting points to verify any symmetry. Setting (r=0) results in ( heta) being undefined. What does this mean? How could ( heta) be undefined? The angle ( heta) is undefined for any value of (sin heta>1). Therefore, ( heta) is undefined because there is no value of ( heta) for which (sin heta>1). Consequently, the graph does not pass through the pole. Perhaps the graph does cross the polar axis, but not at the pole. We can investigate other intercepts by calculating (r) when ( heta=0).

[egin{align*} r(0)&= 4-3 sin(0) r&= 4-3cdot 0 &= 4 end{align*}]

So, there is at least one polar axis intercept at ((4,0)).

Next, as the maximum value of the sine function is (1) when ( heta=dfrac{pi}{2}), we will substitute ( heta=dfrac{pi}{2}) into the equation and solve for (r). Thus, (r=1).

Make a table of the coordinates similar to Table (PageIndex{4}).

 ( heta) (r) (0) (dfrac{pi}{6}) (dfrac{pi}{3}) (dfrac{pi}{2}) (dfrac{2pi}{3}) (dfrac{5pi}{6}) (pi) (dfrac{7pi}{6}) (dfrac{4pi}{3}) (dfrac{3pi}{2}) (dfrac{5pi}{3}) (dfrac{11pi}{6}) (2pi) (4) (2.5) (1.4) (1) (1.4) (2.5) (4) (5.5) (6.6) (7) (6.6) (5.5) (4)

The graph is shown in Figure (PageIndex{10}). Analysis

This is an example of a curve for which making a table of values is critical to producing an accurate graph. The symmetry tests fail; the zero is undefined. While it may be apparent that an equation involving (sin heta) is likely symmetric with respect to the line ( heta=dfrac{pi}{2}), evaluating more points helps to verify that the graph is correct.

Exercise (PageIndex{3})

Sketch the graph of (r=3−2 cos heta). Another type of limaçon, the inner-loop limaçon, is named for the loop formed inside the general limaçon shape. It was discovered by the German artist Albrecht Dürer(1471-1528), who revealed a method for drawing the inner-loop limaçon in his 1525 book Underweysung der Messing. A century later, the father of mathematician Blaise Pascal, Étienne Pascal(1588-1651), rediscovered it.

FORMULAS FOR INNER-LOOP LIMAÇONS

The formulas that generate the inner-loop limaçons are given by (r=apm bcos heta) and (r=apm b sin heta) where (a>0), (b>0), and (a Example (PageIndex{6}): Sketching the Graph of an Inner-Loop Limaçon

Sketch the graph of (r=2+5 cos heta).

Solution

Testing for symmetry, we find that the graph of the equation is symmetric about the polar axis. Next, finding the zeros reveals that when (r=0), ( heta=1.98). The maximum (| r |) is found when (cos heta=1) or when ( heta=0). Thus, the maximum is found at the point ((7, 0)).

Even though we have found symmetry, the zero, and the maximum, plotting more points will help to define the shape, and then a pattern will emerge. See Table (PageIndex{5}).

 ( heta) (r) (0) (dfrac{pi}{6}) (dfrac{pi}{3}) (dfrac{pi}{2}) (dfrac{2pi}{3}) (dfrac{5pi}{6}) (pi) (dfrac{7pi}{6}) (dfrac{4pi}{3}) (dfrac{3pi}{2}) (dfrac{5pi}{3}) (dfrac{11pi}{6}) (2pi) (7) (6.3) (4.5) (2) (−0.5) (−2.3) (−3) (−2.3) (−0.5) (2) (4.5) (6.3) (7)

As expected, the values begin to repeat after ( heta=pi). The graph is shown in Figure (PageIndex{13}). ##### Investigating Lemniscates

The lemniscate is a polar curve resembling the infinity symbol (infty) or a figure (8). Centered at the pole, a lemniscate is symmetrical by definition.

FORMULAS FOR LEMNISCATES

The formulas that generate the graph of a lemniscate are given by (r^2=a^2 cos 2 heta) and (r^2=a^2 sin 2 heta) where (a≠0). The formula (r^2=a^2 sin 2 heta) is symmetric with respect to the pole. The formula (r^2=a^2 cos 2 heta) is symmetric with respect to the pole, the line ( heta=dfrac{pi}{2}), and the polar axis. See Figure (PageIndex{14}) for the graphs. Example (PageIndex{7}): Sketching the Graph of a Lemniscate

Sketch the graph of (r^2=4 cos 2 heta).

Solution

The equation exhibits symmetry with respect to the line ( heta=dfrac{pi}{2}), the polar axis, and the pole.

Let’s find the zeros. It should be routine by now, but we will approach this equation a little differently by making the substitution (u=2 heta).

[egin{align*} 0 &= 4 cos 2 heta 0 &= 4 cos u 0 &= cos u {cos}^{-1} 0 &= dfrac{pi}{2} u &= dfrac{pi}{2} qquad ext{Substitute } 2 heta ext{ back in for } u. 2 heta &= dfrac{pi}{2} heta &= dfrac{pi}{4} end{align*}]

So, the point (left(0,dfrac{pi}{4} ight))is a zero of the equation.

Now let’s find the maximum value. Since the maximum of (cos u=1) when (u=0), the maximum (cos 2 heta=1) when (2 heta=0). Thus,

[egin{align*} r^2 &= 4 cos(0) r^2 &= 4(1) r^2&= 4 r&= pm 4 &=2 end{align*}]

We have a maximum at ((2, 0)). Since this graph is symmetric with respect to the pole, the line ( heta=dfrac{pi}{2}), and the polar axis, we only need to plot points in the first quadrant.

Make a table similar to Table (PageIndex{6}).

 ( heta) (r) (0) (dfrac{pi}{6}) (dfrac{pi}{4}) (dfrac{pi}{3}) (dfrac{pi}{2}) (2) (sqrt{2}) (0) (sqrt{2}) (0)

Plot the points on the graph, such as the one shown in Figure (PageIndex{15}). Analysis

Making a substitution such as (u=2 heta) is a common practice in mathematics because it can make calculations simpler. However, we must not forget to replace the substitution term with the original term at the end, and then solve for the unknown.

Some of the points on this graph may not show up using the Trace function on the TI-84 graphing calculator, and the calculator table may show an error for these same points of (r). This is because there are no real square roots for these values of (θ). In other words, the corresponding (r)-values of (sqrt{4 cos(2 heta)}) are complex numbers because there is a negative number under the radical.

##### Investigating Rose Curves

The next type of polar equation produces a petal-like shape called a rose curve. Although the graphs look complex, a simple polar equation generates the pattern.

ROSE CURVES

The formulas that generate the graph of a rose curve are given by (r=a cos n heta) and (r=a sin n heta) where (a≠0). If (n) is even, the curve has (2n) petals. If (n) is odd, the curve has (n) petals. See Figure (PageIndex{16}). Example (PageIndex{8}): Sketching the Graph of a Rose Curve ((n) Even)

Sketch the graph of (r=2 cos 4 heta).

Solution

Testing for symmetry, we find again that the symmetry tests do not tell the whole story. The graph is not only symmetric with respect to the polar axis, but also with respect to the line ( heta=dfrac{pi}{2}) and the pole.

Now we will find the zeros. First make the substitution (u=4 heta).

[egin{align*} 0 &= 2 cos 4 heta 0 &= cos 4 heta 0 &= cos u {cos}^{-1} 0 &=u u &= dfrac{pi}{2} 4 heta &= dfrac{pi}{2} heta &=dfrac{pi}{8} end{align*}]

The zero is ( heta=dfrac{pi}{8}). The point (left(0,dfrac{pi}{8} ight)) is on the curve.

Next, we find the maximum (| r |). We know that the maximum value of (cos u=1) when ( heta=0). Thus,

[egin{align*} r &=2 cos(4cdot 0) r &=2 cos(0) r &=2(1) &= 2 end{align*}]

The point ((2,0)) is on the curve.

The graph of the rose curve has unique properties, which are revealed in Table (PageIndex{7}).

 ( heta) (r) (0) (dfrac{pi}{8}) (dfrac{pi}{4}) (dfrac{3pi}{8}) (dfrac{pi}{2}) (5π8) (3π4) (2) (0) (−2) (0) (2) (0) (−2)

As (r=0) when ( heta=dfrac{pi}{8}), it makes sense to divide values in the table by (dfrac{pi}{8}) units. A definite pattern emerges. Look at the range of (r)-values: (2, 0, −2, 0) and so on. This represents the development of the curve one petal at a time. Starting at (r=0), each petal extends out a distance of (r=2), and then turns back to zero (2n) times for a total of eight petals. See the graph in Figure (PageIndex{17}). Analysis

When these curves are drawn, it is best to plot the points in order, as in the Table (PageIndex{7}). This allows us to see how the graph hits a maximum (the tip of a petal), loops back crossing the pole, hits the opposite maximum, and loops back to the pole. The action is continuous until all the petals are drawn.

Exercise (PageIndex{4})

Sketch the graph of (r=4 sin(2 heta)).

The graph is a rose curve, (n) even Example (PageIndex{9}): Sketching the Graph of a Rose Curve ((n) Odd)

Sketch the graph of (r=2 sin(5 heta)).

Solution

The graph of the equation shows symmetry with respect to the line ( heta=dfrac{pi}{2}). Next, find the zeros and maximum. We will want to make the substitution (u=5 heta).

[egin{align*} 0 &=2 sin(5 heta) 0 &=sin u {sin}^{-1} 0 &=0 u &=0 5 heta &=0 heta &=0 end{align*}]

The maximum value is calculated at the angle where (sin heta) is a maximum. Therefore,

[egin{align*} r&= 2 sinleft(5cdot dfrac{pi}{2} ight) r&= 2(1) &= 2 end{align*}]

Thus, the maximum value of the polar equation is (2). This is the length of each petal. As the curve for (n) odd yields the same number of petals as (n), there will be five petals on the graph. See Figure (PageIndex{19}).

Create a table of values similar to Table (PageIndex{8}).

 ( heta) (r) (0) (dfrac{pi}{6}) (dfrac{pi}{3}) (dfrac{pi}{2}) (dfrac{2pi}{3}) (dfrac{5pi}{6}) (pi) (0 (1) (−1.73) (2) (−1.73) (1) (0) Exercise (PageIndex{5})

Sketch the graph of (r=3 cos(3 heta)). Rose curve, (n) odd

##### Investigating the Archimedes’ Spiral

The final polar equation we will discuss is the Archimedes’ spiral, named for its discoverer, the Greek mathematician Archimedes (c. 287 BCE - c. 212 BCE), who is credited with numerous discoveries in the fields of geometry and mechanics.

ARCHIMEDES’ SPIRAL

The formula that generates the graph of the Archimedes’ spiral is given by (r= heta) for ( heta≥0). As ( heta) increases, (r) increases at a constant rate in an ever-widening, never-ending, spiraling path. See Figure (PageIndex{21}). How to: Given an Archimedes’ spiral over ([ 0,2pi ]),sketch the graph

1. Make a table of values for (r) and ( heta) over the given domain.
2. Plot the points and sketch the graph.

Example (PageIndex{10}): Sketching the Graph of an Archimedes’ Spiral

Sketch the graph of (r= heta) over ([0,2pi]).

Solution

As (r) is equal to ( heta), the plot of the Archimedes’ spiral begins at the pole at the point ((0, 0)). While the graph hints of symmetry, there is no formal symmetry with regard to passing the symmetry tests. Further, there is no maximum value, unless the domain is restricted.

Create a table such as Table (PageIndex{9}).

 ( heta) (r) (dfrac{pi}{4}) (dfrac{pi}{2}) (pi) (dfrac{3pi}{2}) (dfrac{7pi}{4}) (2pi) (0.785) (1.57) (3.14) (4.71) (5.50) (6.28)

Notice that the r-values are just the decimal form of the angle measured in radians. We can see them on a graph in Figure (PageIndex{22}). Analysis

The domain of this polar curve is ([ 0,2pi ]). In general, however, the domain of this function is ((−infty,infty)). Graphing the equation of the Archimedes’ spiral is rather simple, although the image makes it seem like it would be complex.

Exercise (PageIndex{6})

Sketch the graph of (r=− heta) over the interval ([ 0,4pi ]). #### Summary of Curves

We have explored a number of seemingly complex polar curves in this section. Figure (PageIndex{24}) and Figure (PageIndex{25}) summarize the graphs and equations for each of these curves.

0, b>0, a/b=1. (C) is one-loop limaçons. r= a + or - bcos(theta), or r= a + or - bsin(theta). a>0, b>0, 10, b>0, a=0." src="/@api/deki/files/7449/CNX_Precalc_Figure_08_04_024n.jpg">

Media

Access these online resources for additional instruction and practice with graphs of polar coordinates.

• Graphing Polar Equations Part 1
• Graphing Polar Equations Part 2
• Animation: The Graphs of Polar Equations
• Graphing Polar Equations on the TI-84

## Key Concepts

• It is easier to graph polar equations if we can test the equations for symmetry with respect to the line ( heta=dfrac{pi}{2}), the polar axis, or the pole.
• There are three symmetry tests that indicate whether the graph of a polar equation will exhibit symmetry. If an equation fails a symmetry test, the graph may or may not exhibit symmetry. See Example (PageIndex{1}).
• Polar equations may be graphed by making a table of values for ( heta) and (r).
• The maximum value of a polar equation is found by substituting the value ( heta) that leads to the maximum value of the trigonometric expression.
• The zeros of a polar equation are found by setting (r=0) and solving for ( heta). See Example (PageIndex{2}).
• Some formulas that produce the graph of a circle in polar coordinates are given by (r=a cos heta) and (r=a sin heta). See Example (PageIndex{3}).
• The formulas that produce the graphs of a cardioid are given by (r=apm b cos heta) and (r=apm b sin heta), for (a>0), (b>0), and (ab=1). See Example (PageIndex{4}).
• The formulas that produce the graphs of a one-loop limaçon are given by (r=apm b cos heta) and (r=apm b sin heta) for (1
• The formulas that produce the graphs of an inner-loop limaçon are given by (r=apm b cos heta) and (r=apm b sin heta) for (a>0), (b>0), and (a
• The formulas that produce the graphs of a lemniscates are given by (r^2=a^2 cos 2 heta) and (r^2=a^2 sin 2 heta), where (a≠0).See Example (PageIndex{7}).
• The formulas that produce the graphs of rose curves are given by (r=a cos n heta) and (r=a sin n heta), where (a≠0); if (n) is even, there are (2n) petals, and if (n) is odd, there aren n petals. See Example (PageIndex{8}) and Example (PageIndex{9}).
• The formula that produces the graph of an Archimedes’ spiral is given by (r= heta), ( heta≥0). See Example (PageIndex{10}).

## Polar Coordinates and Graphs

I remember being frightened by the polar coordinate system and graphing equations using it when I was in my students’ seats many years ago. With graphing calculators, this doesn’t have to be intimidating.

I started the unit with an introduction to the polar coordinate system and reminded my students about how intimately familiar they are with the unit circle and how closely this resembles it. I passed out the Vizual Notes and we learned how to plot points, name a point in three additional ways, and how to convert between the two systems. Then my students practiced using their new knowledge. Next we discussed how to convert rectangular equations to polar and vice versa.

Then it’s on to an investigation to see how much my students can learn by graphing polar equations using a graphing calculator.

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## A Collection of Problems in Differential Calculus: Problems from Calculus I Final Examinations, 2000-2020 Department of Mathematics, Simon Fraser University

Express the polar equation (r=cos 2 heta) in rectangular coordinates.

For the each of the following circles find a polar equation, i.e. an equation in (r) and ( heta ext<:>)

Find the maximum height above the (x)-axis of the cardioid (r=1+cos heta ext<.>)

Sketch the graph of the curve whose equation in polar coordinates is (r=1-2cos heta ext<,>) (0leq heta lt 2pi ext<.>)

Sketch the graph of the curve whose equation in polar coordinates is (r=3cos 3 heta ext<.>)

Sketch the curve whose polar equation is (r=-1+cos heta ext<,>) indicating any symmetries. Mark on your sketch the polar coordinates of all points where the curve intersects the polar axis.

Sketch a polar coordinate plot of:

(displaystyle ds r=frac<1><2>+sin heta)

(displaystyle r=2cos 3 heta)

(displaystyle r^2=-4sin 2 heta)

(displaystyle r=4+7cos heta)

Consider the curve given by the polar equation (r=1-cos heta ext<,>) for (0leq heta lt 2pi ext<.>)

Given a point (P) on this curve with polar coordinates ((r, heta) ext<,>) represent its Cartesian coordinates ((x,y)) in terms of ( heta ext<.>)

Find the slope of the tangent line to the curve where (ds heta = frac<2> ext<.>)

Find the points on this curve where the tangent line is horizontal or vertical.

Consider the curve given by the polar equation (r=cos (2 heta) ext<,>) for (0leq heta lt 2pi ext<.>)

Find (ds frac) in terms of ( heta ext<.>)

Find the Cartesian coordinates for the point on the curve corresponding to (ds heta = frac<8> ext<.>)

Find the tangent line to the curve at the point corresponding to (ds heta = frac<8> ext<.>)

Sketch this curve for (displaystyle 0leq heta leq frac<4>) and label the point from part (b) on your curve.

Consider the curve given by the polar equation (r=4cos (3 heta) ext<,>) for (0leq heta lt 2pi ext<.>)

Find the Cartesian coordinates for the point on the curve corresponding to (ds heta = frac<3> ext<.>)

One of graphs in Figure 4.3.1 is the graph of (r=4cos(3 heta) ext<.>) Indicate which one by circling it.

Find the slope of the tangent line to the curve where (ds heta = frac<3> ext<.>)

Consider the curve given by the polar equation (r=4sin (3 heta) ext<,>) for (0leq heta lt 2pi ext<.>)

Find the Cartesian coordinates for the point on the curve corresponding to (ds heta = frac<6> ext<.>)

One of graphs in Figure 4.3.1 is the graph of (r=4sin(3 heta) ext<.>) Indicate which one by circling it.

Find the slope of the tangent line to the curve where (ds heta = frac<3> ext<.>)

Consider the curve given by the polar equation (r=1+3cos(2 heta) ext<,>) for (0leq heta lt 2pi ext<.>)

Find the Cartesian coordinates for the point on the curve corresponding to (ds heta = frac<6> ext<.>)

One of graphs in Figure 4.3.2 is the graph of (r=1+3cos(2 heta) ext<.>) Indicate which one by putting a checkmark in the box below the graph you chose.

Find the slope of the tangent line to the curve where (ds heta = frac<6> ext<.>)

Consider the curve given by the polar equation (r=1-2sin heta ext<,>) for (0leq heta lt 2pi ext<.>)

Find the Cartesian coordinates for the point on the curve corresponding to (ds heta = frac<3pi ><2> ext<.>)

The curve intersects the (x)-axis at two points other than the pole. Find polar coordinates for these other points.

On Figure 4.3.3 identify the graphs that correspond to the following two polar curves.

Consider the curve (C) given by the polar equation (r=1+2cos heta ext<,>) for (0leq heta lt 2pi ext<.>)

Find the Cartesian coordinates for the point on the curve corresponding to (ds heta = frac<3> ext<.>)

Find the slope of the tangent line where (ds heta = frac<3> ext<.>)

On Figure 4.3.4 identify the graph of (C ext<.>)

Sketch a polar coordinate plot of

How many points lie in the intersection of the two polar graphs

Algebraically find all values of ( heta) that

Explain in a sentence or two why the answer to part (b) differs from (or is the same as) the number of solutions you found in part (c).

Consider the following curve (C) given in polar coordinates as

Calculate the value of (r( heta )) for (ds heta =0, frac<2>, frac<3pi ><2> ext<.>)

What is the minimum distance from a point on the curve (C) to the origin? (i.e. determine the minimum of (|r( heta )|=r( heta )=1+sin heta +e^) for ( heta in [0,2pi ]))

Give polar coordinates for each of the points (A ext<,>) (B ext<,>) (C) and (D) on Figure 4.3.5.

On Figure 4.3.6 identify the graphs that correspond to the following three polar curves.

Sketch the curve defined by (r=1+2sin heta ext<.>)

For what values of ( heta ext<,>) ( heta in [-pi ,pi ) ext<,>) is the radius (r) positive?

For what values of ( heta ext<,>) ( heta in [-pi ,pi ) ext<,>) is the radius (r) maximum and for what values is it minimum?

Sketch the graph described in polar coordinates by the equation (r= heta) where (-pi leq heta leq 3pi ext<.>)

Find the slope of this curve when (ds heta =frac<5pi ><2> ext<.>) Simplify your answer for full credit.

Express the polar equation (r= heta) in cartesian coordinates, as an equation in (x) and (y ext<.>)

Let (C) denote the graph of the polar equation (r=5sin heta ext<.>) Find the rectangular coordinates of the point on (C) corresponding to (ds heta =frac<3pi ><2> ext<.>)

Write a rectangular equation (i.e. using the variables (x) and (y)) for (C ext<.>) (in other words, convert the equation for (C) into rectangular coordinates.)

Rewrite the equation of (C) in parametric form, i.e. express both (x) and (y) as functions of ( heta ext<.>)

Find an expression for (ds frac) in terms of ( heta ext<.>)

Find the equation of the tangent line to (C) at the point corresponding to (ds heta =frac<6> ext<.>)

Find the slope of the tangent line to the polar curve (r=2) at the points where it intersects the polar curve (r=4cos heta ext<.>) (Hint: After you find the intersection points, convert one of the curves to a pair of parametric equations with ( heta) as the perimeter.

A bee goes out from its hive in a spiral path given in polar coordinates by (r=be^) and ( heta =ct ext<,>) where (b ext<,>) (k ext<,>) and (c) are positive constants. Show that the angle between the bee's velocity and acceleration remains constant as the bee moves outward.

## Relationships Between Polar and Cartesian Coordinates

The same point ( P ) may be located using Cartesian or polar coordinates. The relationship between the two sets of coordinates may be found using trigonometry in a right triangle.

( x = r cos heta ) and ( y = r sin heta )
( r^2 = x^2 + y^2 ) and ( heta = arctan (dfrac) )   Fig.5 - Relationship Between Cartesian and Polar Coordinates More on converting polar to rectangular coordinates and vice versa are included.

## 10.4: Polar Coordinates - Graphs - Mathematics  When we describe a curve using polar coordinates, it is still a curve in the $x$-$y$ plane. We would like to be able to compute slopes and areas for these curves using polar coordinates.

We have seen that $x=rcos heta$ and $y=rsin heta$ describe the relationship between polar and rectangular coordinates. If in turn we are interested in a curve given by $r=f( heta)$, then we can write $x=f( heta)cos heta$ and $y=f( heta)sin heta$, describing $x$ and $y$ in terms of $heta$ alone. The first of these equations describes $heta$ implicitly in terms of $x$, so using the chain rule we may compute $=.$ Since $d heta/dx=1/(dx/d heta)$, we can instead compute $== .$

Example 10.2.1 Find the points at which the curve given by $r=1+cos heta$ has a vertical or horizontal tangent line. Since this function has period $2pi$, we may restrict our attention to the interval $[0,2pi)$ or $(-pi,pi]$, as convenience dictates. First, we compute the slope: $=<(1+cos heta)cos heta-sin hetasin hetaover -(1+cos heta)sin heta-sin hetacos heta>= .$ This fraction is zero when the numerator is zero (and the denominator is not zero). The numerator is $ds ds 2cos^2 heta+cos heta-1$ so by the quadratic formula $cos heta=<-1pmsqrt<1+4cdot2>over 4> = -1 quadhboxquad <1over 2>.$ This means $heta$ is $pi$ or $pm pi/3$. However, when $heta=pi$, the denominator is also $, so we cannot conclude that the tangent line is horizontal. Setting the denominator to zero we get$eqalign < -sin heta-2sin hetacos heta &= 0cr sin heta(1+2cos heta)&=0,cr>$so either$sin heta=0$or$cos heta=-1/2$. The first is true when$ heta$is$ or $pi$, the second when $heta$ is $2pi/3$ or $4pi/3$. However, as above, when $heta=pi$, the numerator is also $, so we cannot conclude that the tangent line is vertical. Figure 10.2.1 shows points corresponding to$ heta$equal to$, $pmpi/3$, $2pi/3$ and $4pi/3$ on the graph of the function. Note that when $heta=pi$ the curve hits the origin and does not have a tangent line.

We know that the second derivative $f''(x)$ is useful in describing functions, namely, in describing concavity. We can compute $f''(x)$ in terms of polar coordinates as well. We already know how to write $dy/dx=y'$ in terms of $heta$, then $= ==.$

## 10.4: Polar Coordinates - Graphs - Mathematics

By now you're used to the Cartesian coordinate system, the x-y grid that we generally use to graph functions and plot points. Each point has a set of coordinates (x, y), a recipe for finding the point starting from the origin.

Aside from its convenience—axes at 90˚ angles, evenly-spaced grid points—there's nothing unique about the Cartesian plane. In fact, there are an infinite number of different coordinate systems, mostly silly ones, that we could use. Though less convenient, any of them will work to represent any mathematical problem or function in a plane.

For example, if you're reading this section, you've probably dealt with balls rolling down ramps in physics. Recall that in those problems you redefined your coordinate system. You tilted it to make the x-axis parallel to the ramp and the y-axis perpendicular or normal to the ramp.

This helped you to resolve vectors into x- and y-components, and made the problem much easier to deal with.

But the fact is, nature couldn't care less about your coordinate system. Whatever system you choose, the ball will still roll down the ramp just the same. #### Examples of coordinate systems

Drawn below are a Cartesian grid (left) and a couple of goofy coordinate systems no one uses, but they can locate a point just the same. The same point on each square is represented with an ordered pair of coordinates in each system.

## Math Insight

In two dimensions, the Cartesian coordinates $(x,y)$ specify the location of a point $P$ in the plane. Another two-dimensional coordinate system is polar coordinates. Instead of using the signed distances along the two coordinate axes, polar coordinates specifies the location of a point $P$ in the plane by its distance $r$ from the origin and the angle $heta$ made between the line segment from the origin to $P$ and the positive $x$-axis. The polar coordinates $(r, heta)$ of a point $P$ are illustrated in the below figure. As $r$ ranges from 0 to infinity and $heta$ ranges from 0 to $2pi$, the point $P$ specified by the polar coordinates $(r, heta)$ covers every point in the plane. Adding $2pi$ to $heta$ brings us back to the same point, so if we allowed $heta$ to range over an interval larger than $2pi$, each point would have multiple polar coordinates. Hence, we typically restrict $heta$ to be in the interal le heta

Polar coordinates. When you change the values of the polar coordinates $r$ and $heta$ by dragging the red points on the sliders, the blue point moves to the corresponding position $(x,y)$ in Cartesian coordinates. Alternatively, you can move the blue point in the Cartesian plane directly with the mouse and observe how the polar coordinates on the sliders change. The coordinate $r$ is the length of the line segment from the point $(x,y)$ to the origin and the coordinate $heta$ is the angle between the line segment and the positive $x$-axis.

#### Conversion formulas

We can calculate the Cartesian coordinates of a point with polar coordinates $(r, heta)$ by forming the right triangle illustrated in the below figure. The hypotenuse is the line segment from the origin to the point, and its length is $r$. The projection of this line segment on the $x$-axis is the leg of the triangle adjacent to the angle $heta$, so $x=rcos heta$. The $y$-component is determined by the other leg, so $y=rsin heta$. Our conversion formula is egin x &= rcos heta otag y &= rsin heta. label ag <1>end To go the other direction, one can use the same right triangle. Since $r$ is the distance from the origin to $(x,y)$, it is the magnitude $r=sqrt$. Alternatively, from the equation eqref, one can calculate directly that egin x^2+y^2 &=r^2cos^2 heta+r^2sin^2 heta &= r^2(cos^2 heta+sin^2 heta)=r^2. end

Taking the ratio of $y$ and $x$ from equation eqref, one can obtain a formula for $heta$, egin frac &= frac = an heta. end One can also see this relationshp from the above right triangle. We can set $heta= arctan frac$, but a problem is that $arctan$ gives a value between $-pi/2$ and $pi/2$. One might neet to add $pi$ or $2pi$ to get the correct angle. With this caveat (and also mapping points where $x=0$ to $heta=pi/2$ or $-pi/2$), one obtains the following formula to convert from Cartesian to polar coordinates egin r &= sqrt otag heta &= arctan frac. label ag <2>end

What about the point where $(x,y)=(0,0)$? In this case, the angle $heta$ isn't well defined. You could just take it to be $heta=0$ to be concrete.

#### Another perspective on polar coordinates

In the above applet, we represented polar coordinates by two separate points on a $r$-axis and a $heta$-axis (the two sliders). We could also put those two axes together into a single $(r, heta)$ plane (just like the $(x,y)$ plane of Cartesian coordinates). This simple change of representing the polar coordinates as a single point on the polar $(r, heta)$ plane is reflected in the following applet. Such a slight shift in perspective has more consequences than you might imagine, as you can read about in a page about polar coordinates as a mapping from the $(r, heta)$ plane to the $(x,y)$ plane.

Polar coordinates with polar axes. The red point in the inset polar $(r, heta)$ axes represent the polar coordinates of the blue point on the main Cartesian $(x,y)$ axes. When you drag the red point, you change the polar coordinates $(r, heta)$, and the blue point moves to the corresponding position $(x,y)$ in Cartesian coordinates. Alternatively, you can move the blue point in the Cartesian plane directly with the mouse and observe how the polar coordinates represented by the red point change. The coordinate $r$ is the length of the line segment from the point $(x,y)$ to the origin and the coordinate $heta$ is the angle between the line segment and the positive $x$-axis.

## Introduction to Polar Coordinates - Concept Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Another form of plotting positions in a plane is using polar coordinates. We are used to using rectangular coordinates, or xy-coordinates. Polar coordinates use a graphing system based on circles, and we specify positions using the radius and angle of a point on a circle centered at the origin. We must also know how to convert from rectangular to polar coordinates and from polar coordinates to rectangular.

I want to talk about the polar coordinate system. First, just a review of the rectangular coordinate system. Remember that the rectangular coordinate system has 2 axes, an x axis and a y axis and every point in the plane can be described by a pair of coordinates that gives the x and y coordinates of the point. So every point can be described with a sort of an address. And the polar coordinates system does something similar, only instead of using the x and y axis, it uses the polar axis. This is the polar axis. And it calls the origin the pole.
Now, a point let's call it p would be described by r and theta where r is its distance from the origin or the pole. This is r and theta is an angle measured from the polar axis. So this is theta. So every point in the plane can be described this way too. Now one of the interesting things about polar coordinates is that the polar coordinates of a particular point aren't necessarily unique. For example this point 3 coma pi over 2, right. It's 3 units away from the pole and it's an angle of pi over 2 from the polar axis, this point can also be located by 3 coma 5 pi over 2, right? It's 3 units away from the pole but you go round pi over 2 and then another 2 pi and get to the same point. You could also describe this point as -3, negative pi over 2 right? You go negative pi over 2 and then go backwards 3 to get here. So what's important to know is that the same point in the polar coordinate system can be described multiple ways. Right? So that's something to be aware of when you're dealing with polar coordinates.
Now another thing to be aware of is how to convert from polar coordinates to rectangular and to do that, we need a little Trigonometry. So let's call this this distance x and this distance y. And then the point p can also be described by the coordinates x and y if this is the origin. Well, power x, y, r and theta are related. Well, for one thing they have the Pythagorean Theorem x squared plus y squared equals r squared. For another, tangent of theta equals y over x. And you could also say that x over r equals cosine theta so x equals r cosine theta and y over r equals sine theta. So y equals r sine theta.
These equations are going to become really important in the next couple of lessons where we convert back and forth between rectangular and polar coordinates.
So once again, the polar coordinate system, the origin's called the pole. We have one axis called the polar axis and we describe points by listing their distance from the pole and their angle from the polar axis.