How to Read Polar Graphs on a Calculator Table

The planets move through infinite in elliptical, periodic orbits almost the sunday, equally shown in (Figure). They are in constant motion, so fixing an exact position of any planet is valid only for a moment. In other words, we can gear up only a planet's instantaneous position. This is ane application of polar coordinates, represented every bit [latex]\,\left(r,\theta \correct).\,[/latex]
Nosotros interpret[latex]\,r\,[/latex]as the altitude from the sunday and[latex]\,\theta \,[/latex]every bit the planet'southward angular bearing, or its direction from a fixed signal on the sun. In this section, nosotros volition focus on the polar system and the graphs that are generated direct from polar coordinates.

Testing Polar Equations for Symmetry

Merely every bit a rectangular equation such as[latex]\,y={x}^{2}\,[/latex]describes the relationship between[latex]\,10\,[/latex]and[latex]\,y\,[/latex]on a Cartesian grid, a polar equation describes a relationship between[latex]\,r\,[/latex]and[latex]\,\theta \,[/latex]on a polar filigree. Recall that the coordinate pair[latex]\,\left(r,\theta \correct)\,[/latex]indicates that we motility counterclockwise from the polar axis (positive x-centrality) by an angle of[latex]\,\theta ,\,[/latex]and extend a ray from the pole (origin)[latex]\,r\,[/latex]units in the management of[latex]\,\theta .\,[/latex]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 i side would coincide with the portion on the other side. By performing three tests, we volition see how to employ the properties of symmetry to polar equations. Farther, we will use symmetry (in add-on to plotting key points, zeros, and maximums of[latex]\,r)\,[/latex] to determine the graph of a polar equation.

In the first exam, nosotros consider symmetry with respect to the line[latex]\,\theta =\frac{\pi }{ii}\,[/latex](y-centrality). We replace[latex]\,\left(r,\theta \right)\,[/latex]with[latex]\,\left(-r,-\theta \right)\,[/latex]to decide if the new equation is equivalent to the original equation. For case, suppose we are given the equation[latex]\,r=2\mathrm{sin}\,\theta ;[/latex]

[latex]\begin{array}{ll}\,\,\,\,r=2\mathrm{sin}\,\theta \hfill & \hfill \\ -r=2\mathrm{sin}\left(-\theta \right)\begin{array}{cccc}& & & \finish{array}\hfill & \text{Replace}\,\left(r,\theta \right)\,\text{with }\left(-r,-\theta \right).\hfill \\ -r=-2\mathrm{sin}\,\theta \hfill & \text{Identity: }\mathrm{sin}\left(-\theta \correct)=-\mathrm{sin}\,\theta .\hfill \\ \,\,\,\,\,r=2\mathrm{sin}\,\theta \hfill & \text{Multiply both sides by}-1.\hfill \terminate{array}[/latex]

This equation exhibits symmetry with respect to the line[latex]\,\theta =\frac{\pi }{2}.[/latex]

In the second test, we consider symmetry with respect to the polar axis ([latex]\,x[/latex]-axis). We replace[latex]\,\left(r,\theta \right)\,[/latex]with[latex]\,\left(r,-\theta \right)\,[/latex]or[latex]\,\left(-r,\pi -\theta \right)\,[/latex]to determine equivalency betwixt the tested equation and the original. For example, suppose we are given the equation[latex]\,r=ane-two\mathrm{cos}\,\theta .[/latex]

[latex]\brainstorm{array}{ll}r=i-2\mathrm{cos}\,\theta \hfill & \hfill \\ r=one-2\mathrm{cos}\left(-\theta \right)\begin{assortment}{cccc}& & & \end{array}\hfill & \text{Supercede }\left(r,\theta \right)\,\text{with}\,\left(r,-\theta \correct).\hfill \\ r=1-two\mathrm{cos}\,\theta \hfill & \text{Fifty-fifty/Odd identity}\hfill \end{assortment}[/latex]

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

In the 3rd exam, we consider symmetry with respect to the pole (origin). We supersede[latex]\,\left(r,\theta \correct)\,[/latex]with[latex]\,\left(-r,\theta \right)\,[/latex]to determine if the tested equation is equivalent to the original equation. For instance, suppose we are given the equation[latex]\,r=ii\mathrm{sin}\left(3\theta \right).[/latex]

[latex]\begin{assortment}{c}\,\,\,\,\,r=two\mathrm{sin}\left(iii\theta \right)\\ -r=2\mathrm{sin}\left(iii\theta \right)\terminate{array}[/latex]

The equation has failed the symmetry test, but that does non mean that it is not symmetric with respect to the pole. Passing i or more of the symmetry tests verifies that symmetry will be exhibited in a graph. Nevertheless, failing the symmetry tests does not necessarily indicate that a graph will non be symmetric nearly the line[latex]\,\theta =\frac{\pi }{2},\,[/latex]the polar axis, or the pole. In these instances, nosotros tin confirm that symmetry exists by plotting reflecting points beyond the apparent centrality of symmetry or the pole. Testing for symmetry is a technique that simplifies the graphing of polar equations, but its awarding is not perfect.

Symmetry Tests

A polar equation describes a bend on the polar grid. The graph of a polar equation tin be evaluated for 3 types of symmetry, every bit shown in (Figure).

Effigy 2. (a) A graph is symmetric with respect to the line[latex]\,\theta =\frac{\pi }{2}\,[/latex](y-axis) if replacing[latex]\,\left(r,\theta \correct)\,[/latex]with[latex]\,\left(-r,-\theta \right)\,[/latex]yields an equivalent equation. (b) A graph is symmetric with respect to the polar axis (10-axis) if replacing[latex]\,\left(r,\theta \right)\,[/latex]with[latex]\,\left(r,-\theta \right)\,[/latex]or[latex]\,\left(-r,\mathrm{\pi -}\theta \correct)\,[/latex]yields an equivalent equation. (c) A graph is symmetric with respect to the pole (origin) if replacing[latex]\,\left(r,\theta \right)\,[/latex]with[latex]\,\left(-r,\theta \correct)\,[/latex]yields an equivalent equation.

How To

Given a polar equation, examination for symmetry.

1. Substitute the appropriate combination of components for[latex]\,\left(r,\theta \right):[/latex][latex]\,\left(-r,-\theta \right)\,[/latex]for[latex]\,\theta =\frac{\pi }{2}\,[/latex]symmetry;[latex]\,\left(r,-\theta \right)\,[/latex]for polar centrality symmetry; and[latex]\,\left(-r,\theta \right)\,[/latex]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.

Testing a Polar Equation for Symmetry

Test the equation[latex]\,r=2\mathrm{sin}\,\theta \,[/latex] for symmetry.

Analysis

Using a graphing calculator, we can see that the equation[latex]\,r=two\mathrm{sin}\,\theta \,[/latex] is a circle centered at[latex]\,\left(0,1\correct)\,[/latex]with radius[latex]\,r=one\,[/latex]and is indeed symmetric to the line[latex]\,\theta =\frac{\pi }{2}.\,[/latex]We tin can besides see that the graph is not symmetric with the polar axis or the pole. See (Effigy).

Figure three.

Effort Information technology

Test the equation for symmetry:[latex]\,r=-2\mathrm{cos}\,\theta .[/latex]

Show Solution

The equation fails the symmetry test with respect to the line[latex]\,\theta =\frac{\pi }{ii}\,[/latex]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 organisation we construct a table of[latex]\,x\,[/latex]and[latex]\,y\,[/latex]values. To graph in the polar coordinate system we construct a table of[latex]\,\theta \,[/latex]and[latex]\,r\,[/latex]values. We enter values of[latex]\,\theta \,[/latex] into a polar equation and summate[latex]\,r.\,[/latex]However, using the properties of symmetry and finding central values of[latex]\,\theta \,[/latex]and[latex]\,r\,[/latex]means fewer calculations will exist needed.

Finding Zeros and Maxima

To find the zeros of a polar equation, we solve for the values of[latex]\,\theta \,[/latex]that event in[latex]\,r=0.\,[/latex] Call up that, to find the zeros of polynomial functions, we fix the equation equal to zilch and and then solve for[latex]\,x.\,[/latex]We apply the aforementioned procedure for polar equations. Set[latex]\,r=0,\,[/latex]and solve for[latex]\,\theta .[/latex]

For many of the forms we will see, the maximum value of a polar equation is found past substituting those values of[latex]\,\theta \,[/latex]into the equation that result in the maximum value of the trigonometric functions. Consider[latex]\,r=5\mathrm{cos}\,\theta ;\,[/latex]the maximum altitude betwixt the curve and the pole is 5 units. The maximum value of the cosine function is 1 when[latex]\,\theta =0,\,[/latex]so our polar equation is[latex]\,five\mathrm{cos}\,\theta ,\,[/latex]and the value[latex]\,\theta =0\,[/latex] will yield the maximum[latex]\,|r|.[/latex]

Similarly, the maximum value of the sine function is 1 when[latex]\,\theta =\frac{\pi }{ii},\,[/latex]and if our polar equation is[latex]\,r=5\mathrm{sin}\,\theta ,\,[/latex]the value[latex]\,\theta =\frac{\pi }{2}\,[/latex]volition yield the maximum[latex]\,|r|.\,[/latex]We may notice additional information by calculating values of[latex]\,r\,[/latex]when[latex]\,\theta =0.\,[/latex]These points would exist polar axis intercepts, which may be helpful in drawing the graph and identifying the curve of a polar equation.

Finding Zeros and Maximum Values for a Polar Equation

Using the equation in (Figure), observe the zeros and maximum[latex]\,|r|\,[/latex]and, if necessary, the polar axis intercepts of[latex]\,r=2\mathrm{sin}\,\theta .[/latex]

Analysis

The point[latex]\,\left(ii,\frac{\pi }{2}\right)\,[/latex]volition be the maximum value on the graph. Let's plot a few more points to verify the graph of a circle. See (Figure) and (Figure).

[latex]\theta [/latex]	[latex]r=2\mathrm{sin}\,\theta [/latex]	[latex]r[/latex]
0	[latex]r=two\mathrm{sin}\left(0\right)=0[/latex]	[latex]0[/latex]
[latex]\frac{\pi }{6}[/latex]	[latex]r=2\mathrm{sin}\left(\frac{\pi }{6}\correct)=1[/latex]	[latex]1[/latex]
[latex]\frac{\pi }{3}[/latex]	[latex]r=2\mathrm{sin}\left(\frac{\pi }{3}\right)\approx one.73[/latex]	[latex]1.73[/latex]
[latex]\frac{\pi }{2}[/latex]	[latex]r=ii\mathrm{sin}\left(\frac{\pi }{2}\right)=2[/latex]	[latex]2[/latex]
[latex]\frac{2\pi }{3}[/latex]	[latex]r=2\mathrm{sin}\left(\frac{2\pi }{iii}\right)\approx 1.73[/latex]	[latex]one.73[/latex]
[latex]\frac{5\pi }{half-dozen}[/latex]	[latex]r=2\mathrm{sin}\left(\frac{v\pi }{6}\correct)=1[/latex]	[latex]1[/latex]
[latex]\pi [/latex]	[latex]r=2\mathrm{sin}\left(\pi \correct)=0[/latex]	[latex]0[/latex]

Figure 4.

Try Information technology

Without converting to Cartesian coordinates, test the given equation for symmetry and find the zeros and maximum values of[latex]\,|r|:\,[/latex][latex]\,r=3\mathrm{cos}\,\theta .[/latex]

Prove Solution

Tests will reveal symmetry about the polar axis. The zero is[latex]\,\left(0,\frac{\pi }{ii}\correct),\,[/latex]and the maximum value is[latex]\,\left(three,0\correct).[/latex]

Investigating Circles

Now we have seen the equation of a circle in the polar coordinate system. In the final ii 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 circumvolve is simply i of many shapes in the prepare of polar curves.

In that location are five classic polar curves: cardioids, limaҫons, lemniscates, rose curves, and Archimedes' spirals. We will briefly bear upon on the polar formulas for the circumvolve 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 circumvolve in polar coordinates are given by[latex]\,r=a\mathrm{cos}\,\theta \,[/latex]and[latex]\,r=a\mathrm{sin}\,\theta ,[/latex] where[latex]\,a\,[/latex]is the diameter of the circumvolve or the distance from the pole to the farthest point on the circumference. The radius is[latex]\,\frac{|a|}{2},[/latex] or one-half the bore. For[latex]\,r=a\mathrm{cos}\,\theta , [/latex] the center is[latex]\,\left(\frac{a}{2},0\right).\,[/latex]For[latex]\,r=a\mathrm{sin}\,\theta ,[/latex] the center is[latex]\,\left(\frac{a}{2},\frac{\pi }{2}\right).\,[/latex](Figure) shows the graphs of these four circles.

Effigy 5.

Sketching the Graph of a Polar Equation for a Circumvolve

Sketch the graph of[latex]\,r=4\mathrm{cos}\,\theta .[/latex]

Investigating Cardioids

While translating from polar coordinates to Cartesian coordinates may seem simpler in some instances, graphing the archetype curves is really 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 unit 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[latex]\,r=a±b\mathrm{cos}\,\theta \,[/latex]and[latex]\,r=a±b\mathrm{sin}\,\theta \,[/latex]where[latex]\,a>0,\,\,b>0,\,[/latex]and[latex]\,\frac{a}{b}=1.\,[/latex]The cardioid graph passes through the pole, as we can run across in (Figure).

Figure 7.

How To

Given the polar equation of a cardioid, sketch its graph.

1. Cheque equation for the 3 types of symmetry.
2. Find the zeros. Set[latex]\,r=0.[/latex]
3. Notice the maximum value of the equation according to the maximum value of the trigonometric expression.
4. Make a tabular array of values for[latex]\,r\,[/latex]and[latex]\,\theta .[/latex]
5. Plot the points and sketch the graph.

Sketching the Graph of a Cardioid

Sketch the graph of[latex]\,r=two+2\mathrm{cos}\,\theta .[/latex]

Investigating Limaçons

The give-and-take 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 unit, 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[latex]\,i<\frac{a}{b}<2\,[/latex]and convex limaçons when[latex]\,\frac{a}{b}\ge two.\,[/latex]

Formulas for One-Loop Limaçons

The formulas that produce the graph of a dimpled 1-loop limaçon are given by[latex]\,r=a±b\mathrm{cos}\,\theta \,[/latex]and[latex]\,r=a±b\mathrm{sin}\,\theta \,[/latex]where[latex]\,a>0,\,b>0,\,\,\text{and 1<}\frac{a}{b}<two.\,[/latex]All four graphs are shown in (Effigy).

Effigy ix. Dimpled limaçons

How To

Given a polar equation for a i-loop limaçon, sketch the graph.

1. Test the equation for symmetry. Recollect that failing a symmetry test does non mean that the shape volition not exhibit symmetry. Ofttimes the symmetry may reveal itself when the points are plotted.
2. Detect the zeros.
3. Find the maximum values co-ordinate to the trigonometric expression.
4. Make a table.
5. Plot the points and sketch the graph.

Sketching the Graph of a One-Loop Limaçon

Graph the equation[latex]\,r=iv-3\mathrm{sin}\,\theta .[/latex]

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 neglect; the zero is undefined. While it may be apparent that an equation involving[latex]\,\mathrm{sin}\,\theta \,[/latex] is probable symmetric with respect to the line[latex]\,\theta =\frac{\pi }{2},[/latex] evaluating more points helps to verify that the graph is right.

Endeavor It

Sketch the graph of[latex]\,r=3-2\mathrm{cos}\,\theta .[/latex]

Show Solution

Another type of limaçon, the inner-loop limaçon, is named for the loop formed within the full general limaçon shape. It was discovered by the German language 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 on, 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 past[latex]\,r=a±b\mathrm{cos}\,\theta \,[/latex]and[latex]\,r=a±b\mathrm{sin}\,\theta \,[/latex]where[latex]\,a>0,\,\,b>0,\,[/latex]and[latex]\,\,a<b.\,[/latex]The graph of the inner-loop limaçon passes through the pole twice: once for the outer loop, and once for the inner loop. Come across (Figure) for the graphs.

Figure eleven.

Sketching the Graph of an Inner-Loop Limaçon

Sketch the graph of[latex]\,r=2+5\text{cos}\,\theta .[/latex]

Investigating Lemniscates

The lemniscate is a polar curve resembling the infinity symbol[latex]\,\infty \,[/latex]or a figure eight. Centered at the pole, a lemniscate is symmetrical past definition.

Formulas for Lemniscates

The formulas that generate the graph of a lemniscate are given by[latex]\,{r}^{2}={a}^{2}\mathrm{cos}\,two\theta \,[/latex]and[latex]\,{r}^{2}={a}^{2}\mathrm{sin}\,ii\theta \,[/latex]where[latex]\,a\ne 0.\,[/latex]The formula[latex]\,{r}^{ii}={a}^{2}\mathrm{sin}\,two\theta \,[/latex]is symmetric with respect to the pole. The formula[latex]\,{r}^{2}={a}^{2}\mathrm{cos}\,2\theta \,[/latex]is symmetric with respect to the pole, the line[latex]\,\theta =\frac{\pi }{2},\,[/latex]and the polar axis. Run into (Figure) for the graphs.

Effigy 13.

Sketching the Graph of a Lemniscate

Sketch the graph of[latex]\,{r}^{2}=4\mathrm{cos}\,2\theta .[/latex]

Analysis

Making a commutation such equally[latex]\,u=2\theta \,[/latex]is a mutual do in mathematics because information technology can make calculations simpler. Yet, nosotros must not forget to replace the exchange term with the original term at the end, and so solve for the unknown.

Some of the points on this graph may non show up using the Trace function on the TI-84 graphing reckoner, and the computer tabular array may show an fault for these same points of[latex]\,r.\,[/latex]This is because there are no real square roots for these values of[latex]\,\theta .\,[/latex]In other words, the corresponding r-values of[latex]\,\sqrt{4\mathrm{cos}\left(2\theta \right)}\,[/latex]
are circuitous numbers because there is a negative number under the radical.

Investigating Rose Curves

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

Rose Curves

The formulas that generate the graph of a rose curve are given by[latex]\,r=a\mathrm{cos}\,n\theta \,[/latex]and[latex]\,r=a\mathrm{sin}\,n\theta \,[/latex]where[latex]\,a\ne 0.\,[/latex]If[latex]\,north\,[/latex]is fifty-fifty, the bend has[latex]\,2n\,[/latex]petals. If[latex]\,northward\,[/latex]is odd, the bend has[latex]\,n\,[/latex]petals. See (Effigy).

Effigy fifteen.

Sketching the Graph of a Rose Curve (due north Even)

Sketch the graph of[latex]\,r=two\mathrm{cos}\,4\theta .[/latex]

Analysis

When these curves are drawn, information technology is best to plot the points in guild, as in the (Figure). This allows usa to meet 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 activity is continuous until all the petals are drawn.

Effort It

Sketch the graph of[latex]\,r=4\mathrm{sin}\left(two\theta \correct).[/latex]

Sketching the Graph of a Rose Curve (n Odd)

Sketch the graph of[latex]\,r=2\mathrm{sin}\left(v\theta \correct).[/latex]

Attempt Information technology

Sketch the graph of[latex]r=3\mathrm{cos}\left(3\theta \correct).[/latex]

Investigating the Archimedes' Screw

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' screw is given by[latex]\,r=\theta \,[/latex]
for[latex]\,\theta \ge 0.\,[/latex]As[latex]\,\theta \,[/latex]increases,[latex]\,r\,[/latex]
increases at a constant rate in an ever-widening, never-catastrophe, spiraling path. See (Figure).

Effigy 18.

How To

Given an Archimedes' spiral over[latex]\,\left[0,two\pi \right],[/latex]sketch the graph.

1. Brand a tabular array of values for[latex]\,r\,[/latex]and[latex]\,\theta \,[/latex]over the given domain.
2. Plot the points and sketch the graph.

Sketching the Graph of an Archimedes' Spiral

Sketch the graph of[latex]\,r=\theta \,[/latex]over[latex]\,\left[0,ii\pi \right].[/latex]

Assay

The domain of this polar curve is[latex]\,\left[0,ii\pi \right].\,[/latex]In full general, however, the domain of this function is[latex]\,\left(-\infty ,\infty \correct).\,[/latex]Graphing the equation of the Archimedes' spiral is rather simple, although the paradigm makes it seem like it would be circuitous.

Attempt It

Sketch the graph of[latex]\,r=-\theta \,[/latex]over the interval[latex]\,\left[0,iv\pi \right].[/latex]

Show Solution

Summary of Curves

Figure twenty.

We accept explored a number of seemingly circuitous polar curves in this section. (Figure) and (Figure) summarize the graphs and equations for each of these curves.

Figure 21.

Key Concepts

• It is easier to graph polar equations if we can exam the equations for symmetry with respect to the line[latex]\,\theta =\frac{\pi }{two},\,[/latex]the polar axis, or the pole.
• There are 3 symmetry tests that indicate whether the graph of a polar equation volition exhibit symmetry. If an equation fails a symmetry test, the graph may or may not exhibit symmetry. See (Effigy).
• Polar equations may be graphed by making a table of values for[latex]\,\theta \,[/latex]and[latex]\,r.[/latex]
• The maximum value of a polar equation is institute by substituting the value[latex]\,\theta \,[/latex]that leads to the maximum value of the trigonometric expression.
• The zeros of a polar equation are found by setting[latex]\,r=0\,[/latex]and solving for[latex]\,\theta .\,[/latex]See (Figure).
• Some formulas that produce the graph of a circle in polar coordinates are given by[latex]\,r=a\mathrm{cos}\,\theta \,[/latex]and[latex]\,r=a\mathrm{sin}\,\theta .\,[/latex]See (Figure).
• The formulas that produce the graphs of a cardioid are given by[latex]\,r=a±b\mathrm{cos}\,\theta \,[/latex]and[latex]\,r=a±b\mathrm{sin}\,\theta ,\,[/latex]for[latex]\,a>0,\,\,b>0,\,[/latex]and[latex]\,\frac{a}{b}=1.\,[/latex]Run into (Figure).
• The formulas that produce the graphs of a i-loop limaçon are given by[latex]\,r=a±b\mathrm{cos}\,\theta \,[/latex]and[latex]\,r=a±b\mathrm{sin}\,\theta \,[/latex]for[latex]\,1<\frac{a}{b}<2.\,[/latex]Meet (Effigy).
• The formulas that produce the graphs of an inner-loop limaçon are given by[latex]\,r=a±b\mathrm{cos}\,\theta \,[/latex]and[latex]\,r=a±b\mathrm{sin}\,\theta \,[/latex]for[latex]\,a>0,\,\,b>0,\,[/latex] and[latex]\,a<b.\,[/latex] See (Figure).
• The formulas that produce the graphs of a lemniscates are given past[latex]\,{r}^{two}={a}^{2}\mathrm{cos}\,ii\theta \,[/latex]and[latex]\,{r}^{ii}={a}^{2}\mathrm{sin}\,2\theta ,\,[/latex]where[latex]\,a\ne 0.[/latex]See (Figure).
• The formulas that produce the graphs of rose curves are given by[latex]\,r=a\mathrm{cos}\,north\theta \,[/latex]and[latex]\,r=a\mathrm{sin}\,north\theta ,\,[/latex]where[latex]\,a\ne 0;\,[/latex]if[latex]\,n\,[/latex]is even, at that place are[latex]\,2n\,[/latex]petals, and if[latex]\,n\,[/latex]is odd, there are[latex]\,n\,[/latex]petals. Encounter (Figure) and (Effigy).
• The formula that produces the graph of an Archimedes' spiral is given by[latex]\,r=\theta ,\,\,\theta \ge 0.\,[/latex]See (Figure).

Section Exercises

Verbal

Describe the three types of symmetry in polar graphs, and compare them to the symmetry of the Cartesian plane.

Evidence Solution

Symmetry with respect to the polar centrality is similar to symmetry nearly the[latex]\,x[/latex]-axis, symmetry with respect to the pole is similar to symmetry nigh the origin, and symmetric with respect to the line[latex]\,\theta =\frac{\pi }{2}\,[/latex]is similar to symmetry nearly the[latex]\,y[/latex]-axis.

Which of the 3 types of symmetries for polar graphs stand for to the symmetries with respect to the ten-axis, y-axis, and origin?

What are the steps to follow when graphing polar equations?

Show Solution

Exam for symmetry; find zeros, intercepts, and maxima; make a table of values. Decide the general type of graph, cardioid, limaçon, lemniscate, etc., and then plot points at [latex]\,\theta =0,\,\frac{\pi }{2},\,\,\pi \,\,\text{and }\frac{three\pi }{two},\,[/latex]and sketch the graph.

Draw the shapes of the graphs of cardioids, limaçons, and lemniscates.

What part of the equation determines the shape of the graph of a polar equation?

Show Solution

The shape of the polar graph is determined past whether or non it includes a sine, a cosine, and constants in the equation.

Graphical

For the post-obit exercises, test the equation for symmetry.

[latex]r=5\mathrm{cos}\,3\theta [/latex]

[latex]r=3-3\mathrm{cos}\,\theta [/latex]

Show Solution

symmetric with respect to the polar centrality

[latex]r=3+two\mathrm{sin}\,\theta [/latex]

[latex]r=3\mathrm{sin}\,2\theta [/latex]

Show Solution

symmetric with respect to the polar centrality, symmetric with respect to the line [latex]\theta =\frac{\pi }{2},[/latex] symmetric with respect to the pole

[latex]r=ii\theta [/latex]

Show Solution

no symmetry

[latex]r=iv\mathrm{cos}\,\frac{\theta }{2}[/latex]

[latex]r=\frac{ii}{\theta }[/latex]

Show Solution

no symmetry

[latex]r=3\sqrt{1-{\mathrm{cos}}^{two}\theta }[/latex]

[latex]r=\sqrt{5\mathrm{sin}\,2\theta }[/latex]

Testify Solution

symmetric with respect to the pole

For the following exercises, graph the polar equation. Place the name of the shape.

[latex]r=iii\mathrm{cos}\,\theta [/latex]

[latex]r=4\mathrm{sin}\,\theta [/latex]

Bear witness Solution

circumvolve

[latex]r=2+2\mathrm{cos}\,\theta [/latex]

[latex]r=two-2\mathrm{cos}\,\theta [/latex]

[latex]r=5-5\mathrm{sin}\,\theta [/latex]

[latex]r=3+3\mathrm{sin}\,\theta [/latex]

Show Solution

cardioid

[latex]r=3+ii\mathrm{sin}\,\theta [/latex]

[latex]r=7+four\mathrm{sin}\,\theta [/latex]

Show Solution

ane-loop/dimpled limaçon

[latex]r=4+3\mathrm{cos}\,\theta [/latex]

[latex]r=5+iv\mathrm{cos}\,\theta [/latex]

[latex]r=ten+9\mathrm{cos}\,\theta [/latex]

[latex]r=1+3\mathrm{sin}\,\theta [/latex]

[latex]r=2+5\mathrm{sin}\,\theta [/latex]

[latex]r=v+seven\mathrm{sin}\,\theta [/latex]

[latex]r=2+four\mathrm{cos}\,\theta [/latex]

[latex]r=5+half dozen\mathrm{cos}\,\theta [/latex]

[latex]{r}^{2}=36\mathrm{cos}\left(ii\theta \correct)[/latex]

[latex]{r}^{2}=x\mathrm{cos}\left(two\theta \right)[/latex]

[latex]{r}^{2}=four\mathrm{sin}\left(two\theta \right)[/latex]

[latex]{r}^{2}=ten\mathrm{sin}\left(2\theta \right)[/latex]

[latex]r=iii\text{sin}\left(2\theta \right)[/latex]

[latex]r=3\text{cos}\left(2\theta \right)[/latex]

[latex]r=5\text{sin}\left(iii\theta \right)[/latex]

[latex]r=4\text{sin}\left(4\theta \right)[/latex]

[latex]r=4\text{sin}\left(five\theta \right)[/latex]

[latex]r=-\theta [/latex]

[latex]r=ii\theta [/latex]

[latex]r=-3\theta [/latex]

Technology

For the following exercises, use a graphing calculator to sketch the graph of the polar equation.

[latex]r=\frac{one}{\theta }[/latex]

[latex]r=\frac{1}{\sqrt{\theta }}[/latex]

Evidence Solution

[latex]r=2\mathrm{sin}\,\theta \mathrm{tan}\,\theta ,[/latex] a cissoid

[latex]r=2\sqrt{1-{\mathrm{sin}}^{2}\theta }[/latex], a hippopede

Show Solution

[latex]r=5+\mathrm{cos}\left(four\theta \right)[/latex]

[latex]r=ii-\mathrm{sin}\left(2\theta \right)[/latex]

Show Solution

[latex]r={\theta }^{two}[/latex]

[latex]r=\theta +1[/latex]

Prove Solution

[latex]r=\theta \mathrm{sin}\,\theta [/latex]

[latex]r=\theta \mathrm{cos}\,\theta [/latex]

Prove Solution

For the post-obit exercises, employ a graphing utility to graph each pair of polar equations on a domain of [latex]\,\left[0,4\pi \right]\,[/latex] and then explain the differences shown in the graphs.

[latex]r=\theta ,r=-\theta [/latex]

[latex]r=\theta ,r=\theta +\mathrm{sin}\,\theta [/latex]

Show Solution

They are both spirals, merely not quite the same.

[latex]r=\mathrm{sin}\,\theta +\theta ,r=\mathrm{sin}\,\theta -\theta [/latex]

[latex]r=2\mathrm{sin}\left(\frac{\theta }{two}\right),r=\theta \mathrm{sin}\left(\frac{\theta }{2}\right)[/latex]

Bear witness Solution

Both graphs are curves with 2 loops. The equation with a coefficient of[latex]\,\theta \,[/latex] has 2 loops on the left, the equation with a coefficient of ii has two loops side by side. Graph these from 0 to[latex]\,4\pi \,[/latex]to become a better picture.

[latex]r=\mathrm{sin}\left(\mathrm{cos}\left(3\theta \right)\right)\,\,r=\mathrm{sin}\left(iii\theta \right)[/latex]

On a graphing utility, graph[latex]\,r=\mathrm{sin}\left(\frac{xvi}{5}\theta \right)\,[/latex]on[latex]\,\left[0,4\pi \right],\left[0,8\pi \right],\left[0,12\pi \right],\,[/latex]and[latex]\,\left[0,xvi\pi \right].\,[/latex]Describe the effect of increasing the width of the domain.

Show Solution

When the width of the domain is increased, more petals of the flower are visible.

On a graphing utility, graph and sketch[latex]\,r=\mathrm{sin}\,\theta +{\left(\mathrm{sin}\left(\frac{v}{ii}\theta \right)\right)}^{3}\,[/latex]on[latex]\,\left[0,four\pi \right].[/latex]

On a graphing utility, graph each polar equation. Explicate the similarities and differences yous observe in the graphs.

[latex]\begin{array}{l}\begin{array}{50}\\ {r}_{1}=3\mathrm{sin}\left(3\theta \right)\end{array}\hfill \\ {r}_{two}=2\mathrm{sin}\left(iii\theta \right)\hfill \\ {r}_{3}=\mathrm{sin}\left(3\theta \right)\hfill \end{assortment}[/latex]

Evidence Solution

The graphs are three-petal, rose curves. The larger the coefficient, the greater the bend's distance from the pole.

On a graphing utility, graph each polar equation. Explain the similarities and differences you lot observe in the graphs.

[latex]\brainstorm{assortment}{l}\brainstorm{assortment}{l}\\ {r}_{1}=3+3\mathrm{cos}\,\theta \end{array}\hfill \\ {r}_{2}=ii+2\mathrm{cos}\,\theta \hfill \\ {r}_{3}=1+\mathrm{cos}\,\theta \hfill \finish{assortment}[/latex]

On a graphing utility, graph each polar equation. Explain the similarities and differences you lot observe in the graphs.

[latex]\begin{array}{50}\begin{assortment}{50}\\ {r}_{ane}=3\theta \end{assortment}\hfill \\ {r}_{2}=ii\theta \hfill \\ {r}_{three}=\theta \hfill \terminate{array}[/latex]

Evidence Solution

The graphs are spirals. The smaller the coefficient, the tighter the screw.

Extensions

For the post-obit exercises, draw each polar equation on the aforementioned set of polar axes, and detect the points of intersection.

[latex]{r}_{one}=3+two\mathrm{sin}\,\theta ,\,{r}_{ii}=2[/latex]

[latex]{r}_{1}=vi-4\mathrm{cos}\,\theta ,\,{r}_{2}=four[/latex]

Show Solution

[latex]\left(four,\frac{\pi }{3}\right),\left(4,\frac{five\pi }{3}\correct)[/latex]

[latex]{r}_{i}=ane+\mathrm{sin}\,\theta ,\,{r}_{2}=iii\mathrm{sin}\,\theta [/latex]

[latex]{r}_{1}=ane+\mathrm{cos}\,\theta ,\,{r}_{two}=3\mathrm{cos}\,\theta [/latex]

Show Solution

[latex]\left(\frac{3}{2},\frac{\pi }{3}\right),\left(\frac{three}{2},\frac{5\pi }{three}\right)[/latex]

[latex]{r}_{1}=\mathrm{cos}\left(2\theta \right),\,{r}_{2}=\mathrm{sin}\left(2\theta \right)[/latex]

[latex]{r}_{1}={\mathrm{sin}}^{2}\left(2\theta \right),\,{r}_{two}=1-\mathrm{cos}\left(4\theta \right)[/latex]

Show Solution

[latex]\left(0,\frac{\pi }{two}\right),\,\left(0,\pi \correct),\,\left(0,\frac{three\pi }{two}\right),\,\left(0,2\pi \right)[/latex]

[latex]{r}_{i}=\sqrt{3},\,{r}_{2}=two\mathrm{sin}\left(\theta \right)[/latex]

[latex]{r}_{ane}{}^{2}=\mathrm{sin}\,\theta ,{r}_{2}{}^{ii}=\mathrm{cos}\,\theta [/latex]

Show Solution

[latex]\left(\frac{\sqrt[4]{8}}{ii},\frac{\pi }{4}\right),\,\left(\frac{\sqrt[iv]{8}}{2},\frac{5\pi }{four}\right)\,[/latex] and at[latex]\,\theta =\frac{iii\pi }{4},\,\,\frac{7\pi }{4}\,\,[/latex] since[latex]\,r\,[/latex]is squared

[latex]{r}_{1}=one+\mathrm{cos}\,\theta ,\,{r}_{two}=1-\mathrm{sin}\,\theta [/latex]

Glossary

Archimedes' spiral

a polar bend given by[latex]\,r=\theta .\,[/latex]When multiplied by a constant, the equation appears equally[latex]\,r=a\theta .\,[/latex]As[latex]\,r=\theta ,\,[/latex]the curve continues to widen in a spiral path over the domain.

cardioid

a member of the limaçon family of curves, named for its resemblance to a centre; its equation is given as[latex]\,r=a±b\mathrm{cos}\,\theta \,[/latex] and[latex]\,r=a±b\mathrm{sin}\,\theta ,\,[/latex]where[latex]\,\frac{a}{b}=1[/latex]

convex limaҫon

a type of one-loop limaçon represented by[latex]\,r=a±b\mathrm{cos}\,\theta \,[/latex]and[latex]\,r=a±b\mathrm{sin}\,\theta \,[/latex]such that[latex]\,\frac{a}{b}\ge 2[/latex]

dimpled limaҫon

a type of i-loop limaçon represented by[latex]\,r=a±b\mathrm{cos}\,\theta \,[/latex] and[latex]\,r=a±b\mathrm{sin}\,\theta \,[/latex] such that[latex]\,1<\frac{a}{b}<2[/latex]

inner-loop limaçon

a polar curve like to the cardioid, but with an inner loop; passes through the pole twice; represented by[latex]\,r=a±b\mathrm{cos}\,\theta \,[/latex] and[latex]\text{ }r=a±b\mathrm{sin}\,\theta \text{ }[/latex]where[latex]\,a<b[/latex]

lemniscate

a polar bend resembling a effigy viii and given past the equation[latex]\,{r}^{2}={a}^{ii}\mathrm{cos}\,2\theta \,[/latex]and[latex]\,{r}^{two}={a}^{2}\mathrm{sin}\,ii\theta ,\,\,a\ne 0[/latex]

ane-loop limaҫon

a polar curve represented by[latex]\,r=a±b\mathrm{cos}\,\theta \,[/latex] and[latex]\,r=a±b\mathrm{sin}\,\theta \,[/latex] such that [latex]a>0,b>0,[/latex]and[latex]\,\frac{a}{b}>1;[/latex] may be dimpled or convex; does not pass through the pole

polar equation

an equation describing a curve on the polar grid.

rose curve

a polar equation resembling a flower, given by the equations[latex]\,r=a\mathrm{cos}\,due north\theta \,[/latex] and [latex]\,r=a\mathrm{sin}\,due north\theta ;\,[/latex]when [latex]\,n\,[/latex] is even there are [latex]\,2n\,[/latex] petals, and the curve is highly symmetrical; when[latex]\,n\,[/latex]is odd there are [latex]n[/latex] petals.

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Source: https://courses.lumenlearning.com/suny-osalgebratrig/chapter/polar-coordinates-graphs/

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