Graph each equation by plotting points.

1.r = cos

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r = cos 0 1

0.9

0.5

0

0.5

0.9 1

0.9

0.5

0

0.5

0.9 2 1

2.r = csc

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a line.

r = csc 0

2

1.2

1

1.2

2

2

1.2

2

1.2

2 2

3.r = cos

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r =

cos 0 0.5

0.4

0.3

0

0.3

0.4 0.5

0.4

0.3

0

0.3

0.4 2 0.5

4.r = 3 sin

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r = 3sin 0 0

1.5

2.6

3

2.6

1.5 0

1.5

2.6

3

2.6

1.5 2 0

5.r = sec

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a line.

r =

sec 0 1

1.2

2

2

1.2 1

1.2

2

2

1.2 2 1

6.r = sin

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r = sin

0 0

0.2

0.3

0.3

0.3

0.2 0

0.2

0.3

0.3

0.3

0.2 2 0

7.r = 4 cos

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r =

4cos 0 4

3.5

2

0

2

3.5 4

3.5

2

0

2

3.5 2 4

8.r = csc

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a line.

r = csc 0

2

1.2

1

1.2

2

2

1.2

1

1.2

2 2

Use symmetry to graph each equation.

9.r = 3 + 3 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 3 + 3 cos 0 6

5.6

5.1

4.5

3

1.5

0.9

0.4 0

10.r = 1 + 2 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 1 + 2 sin

1

0.7

0.4

0 0 1

2

2.4

2.7

3

11.r = 4 3 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 4 3

cos 0 1

1.4

1.9

2.5

4

5.5

6.12

6.6 7

12.r = 2 + 4 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 2 + 4

cos 0 6

5.5

4.8

4

2

0

0.8

1.5 2

13.r = 2 2 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 2 2

sin

4

3.7

3.4

3 0 2

1

0.6

0.3

0

14.r = 3 5 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 3 5 cos

0 2

1.3

0.5

0.5

3

5.5

6.5

7.3 8

15.r = 5 + 4 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 5 + 4

sin

1

1.5

2.2

3 0 5

7

7.8

8.5

9

16.r = 6 2 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 6 2

sin

8

7.7

7.4

7 0 6

5

4.6

4.3

4

Use symmetry, zeros, and maximum r-values tograph each function.

17.r = sin 4

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y = sin

4x on the interval . From the graph, you can

see that =1when and

y = 0 when

Interpreting these results in terms of the polar

equation r = sin 4 , we can say that hasamaximum value of 1 when

andr = 0 when

Since the function is symmetric with respect to the

line = , make a table and calculate the values of

r on .

Use these and a few additional points to sketch the graph of the function.

r = sin

4

0

0.9

0

0.9 0 0

0.9

0

0.9

0

18.r = 2 cos 2

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 2

cos 2x on the interval [0, ]. From the graph, you

can see that =2whenx = 0, , and and y = 0

when

Interpreting these results in terms of the polar

equation r = 2 cos 2 , we can say that hasa

maximum value of 2 when andr =

0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 2 cos

2 0 2

1

0

1

2

1

0

1 2

19.r = 5 cos 3

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 5

cos 3x on the interval [0, ]. From the graph, you

can see that =5when and y

= 0 when

Interpreting these results in terms of the polar

equation r = 5 cos 3 , we can say that hasa

maximum value of 5 when andr

= 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 5 cos 3 0 5

3.5

0

3.5

5

3.5

0

3.5

5

3.5

0

3.5 5

20.r = 3 sin 2

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y = 3

sin 2x on the interval . From the graph, you

can see that =3when andy = 0

when

Interpreting these results in terms of the polar

equation r = 3 sin 2 , we can say that hasa

maximum value of 3 when andr = 0

when

Since the function is symmetric with respect to the

line = , make a table and calculate the values of

r on .

Use these and a few additional points to sketch the graph of the function.

r = 3 sin

2

0

2.6

3

2.6 0 0

2.6

3

2.6

0

21.r = sin3

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y =

sin3x on the interval . From the graph,

you can see that = when

andy = 0 when

.

Interpreting these results in terms of the polar

equation r = sin3 , we can say that hasa

maximum value of when

andr = 0 when

Since the function is symmetric with respect to the

line = , make a table and calculate the values of

r on .

Use these and a few additional points to sketch the graph of the function.

r = sin

3

0.5

0.4

0

0.4

0.5

0.4 0 0

0.4

0.5

0.4

0

0.4

0.5

22.r = 4 cos 5

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 4

cos 5x on the interval [0, ]. From the graph, you can see that =4when

and y = 0 when

Interpreting these results in terms of the polar

equation r = 4 cos 5 , we can say that hasamaximum value of 4 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 4 cos

5 0 4

0

4

0

4

0

4

0

4

0 4

23.r = 2 sin 5

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y = 2

sin 5x on the interval . From the graph, you

can see that =2when

andy = 0 when

.

Interpreting these results in terms of the polar

equation r = 2 sin 5 , we can say that hasamaximum value of 2 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r = 2 sin

5

2

0

2

0

2 0 0

2

0

2

0

2

24.r = 3 cos 4

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 3

cos 4x on the interval [0, ]. From the graph, you

can see that =3when and

y = 0 when

Interpreting these results in terms of the polar

equation r = 3 cos 4 , we can say that hasa

maximum value of 3 when and

r = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 3 cos

4 0 3

1.5

1.5

3

1.5

1.5

3

1.5

1.5

3

1.5

1.5 3

25.MARINE BIOLOGY Rose curves can be observed in marine wildlife. Determine the symmetry, zeros, and maximum r-values of each

functionmodelingamarinespeciesfor0 . Then use the information to graph the function. a. The pores forming the petal pattern of a sand dollar (Refer to Figure 9.2.3 on Page 548) can be modeled by

r = 3 cos 5 .

b. The outline of the body of a crown-of-thorns sea star (Refer to Figure 9.2.4 on Page 548) can be modeled by

r = 20 cos 8 .

SOLUTION:a. Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 3

cos 5x on the interval . From the graph, you

can see that =3when

, and y = 0 when

Interpreting these results in terms of the polar

equation r = 3 cos 5 ,wecansaythat hasamaximum value of 3 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate a few

additional values of r on .

Use these points and symmetry with respect to the polar axis to sketch the graph of the function.

b. Because the polar equation is a function of the cosine function, it is symmetric with respect to the polaraxis. Sketch the graph of the rectangular function y = 20

cos 8x on the interval . From the graph, you

can see that =20whenx = 0, , , , ,

, , , and , and y = 0 when x = , ,

, , , , , and .

Interpreting these results in terms of the polar

equation r = 20 cos 8 ,wecansaythat hasa

maximum value of 20 when = 0, , , , ,

, , , and , and r = 0 when = , ,

, , , , , and

Make a table and calculate a few additional values

of r on .

Use these points and symmetry with respect to the

polar axis, line = , and pole to sketch the graph

of the function.

r = 3 cos

5 0 3

0.8

2.6

2.1

1.5

2.9

0

r = 20 cos 8

0 20

10

10

20

10

10

20

Identify the type of curve given by each equation. Then use symmetry, zeros, and maximum r-values to graph the function.

26.r = cos

SOLUTION:

The equation is of the form r = a cos , so its graph is a circle. Because this polar equation is a function of the cosine function, it is symmetric with respect tothe polar axis. Sketch the graph of the rectangular function y =

cosx on the interval . From the graph, you

can see that = when x = 0 and and y = 0

when x = .

Interpreting these results in terms of the polar

equation r = cos , we can say that hasa

maximum value of when =0andand r = 0

when =

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r =

cos 0 0.33

0.29

0.24

0.17

0

0.17

0.24

0.29 0.3

27.r = 4 + 1; >0

SOLUTION:

The equation is of the form r = a + b, so its graph is a spiral of Archimedes. The function has no symmetry. Spirals are unbounded. Therefore, the function has no maximum r-values and only one zero. To find the zero, substitute r = 0 into the function and solve for

.

So, r = 0 when , which is not in the given

domain. Use points on the interval [0, 4] to sketch the graphof the function.

r = 4 + 1 0 1

4.1

7.3 13.6

19.9

2 26.1 3 38.7 4 51.3

28.r = 2 sin 4

SOLUTION:

The equation is of the form r = a sin n , so its graph is a rose. Because this polar equation is a function of the sine function, it is symmetric with

respect to the line = . Sketch the graph of the

rectangular function y = 2 sin 4x on the interval

. From the graph, you can see that =2

when andy = 0 when

Interpreting these results in terms of the polar

equation r = 2 sin 4 , we can say that hasamaximum value of 2 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r = 2 sin

4

0

1.7

1.7

0

1.7

1.7 0 0

1.7

1.7

0

1.7

1.7

0

29.r = 6 + 6 cos

SOLUTION:

The equation is of the form r = a + b cos anda = b, so its graph is a cardioid. Because this polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 6 +

6 cos x on the interval . From the graph, you

can see that =12whenx = 0 and y = 0 when x = .

Interpreting these results in terms of the polar

equation r = 6 + 6 cos , we can say that hasa

maximum value of 12 when =0andr = 0 when =.

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r = 6 + 6

cos 0 12

11.2

9

6

3

0.8 0

30.r2 = 4 cos 2

SOLUTION:

The equation is of the form r2 = a

2 cos 2 , so its

graphisalemniscate.

Replacing (r, ) with (r, ) yields r2 = 4 cos 2

( ). Since cosine is an even function, r2 = 4 cos 2

( ) can be written as r2 = 4 cos 2 , and the

function has symmetry with respect to the polar axis.

Replacing (r, ) with (r, ) yields (r)2 = 4 cos

2( ) or r2 = 4 cos 2( ). Again, since cosine is an

even function, r2 = 4 cos 2( ) can be written as r2

= 4 cos 2 , and the function has symmetry with

respect to the line = .

Finally, replacing (r, ) with (r, ) yields (r)2 = 4

cos 2 or r2 = 4 cos 2 . Therefore, the function

has symmetry with respect to the pole.

The equation r2 = 4 cos 2 is equivalent to r =

, which is undefined when cos 2 < 0. Therefore, the domain of the function is restricted to

the intervals and .

Sketch the graph of the rectangular function y =

ontheinterval . From the graph,

you can see that =2whenx = 0 and y = 0 when

.

Interpreting these results in terms of the polar

equation r2 = 4 cos 2 , we can say that hasa

maximum value of 2 when andr = 0 when

.

Use these points and the indicated symmetry to sketch the graph of the function.

r2 = 4

cos 2

0

1.4

1.7 0 2

1.7

1.4

0

31.r = 5 + 2; >0

SOLUTION:

The equation is of the form r = a + b, so its graph is a spiral of Archimedes. The function has no symmetry. Spirals are unbounded. Therefore, the function has no maximum r-values and only one zero. To find the zero, substitute r = 0 into the function and solve for

.

So, r = 0 when , which is not in the given

domain. Use points on the interval [0, 4] to sketch the graphof the function.

r = 5

+ 2 0 2

5.9

9.9 17.7

25.6

2 33.4 3 49.2 4 64.8

32.r = 3 2 sin

SOLUTION:

The equation is of the form r = a b sin andb < a < 2b,soitsgraphisalimaon.Morespecifically,itisadimpledlimaon.Becausethispolarequationisafunction of the sine function, it is symmetric with

respect to the line = .

Sketch the graph of the rectangular function y = 3

2 sin x on the interval . From the graph, you

can see that =5when . There are no

zeros.

Interpreting these results in terms of the polar

equation r = 3 2 sin , we can say that hasa

maximum value of 5 when .

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

r = 3 2

sin

5

4.9

4.7

4.4

Graph each equation by plotting points.

1.r = cos

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r = cos 0 1

0.9

0.5

0

0.5

0.9 1

0.9

0.5

0

0.5

0.9 2 1

2.r = csc

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a line.

r = csc 0

2

1.2

1

1.2

2

2

1.2

2

1.2

2 2

3.r = cos

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r =

cos 0 0.5

0.4

0.3

0

0.3

0.4 0.5

0.4

0.3

0

0.3

0.4 2 0.5

4.r = 3 sin

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r = 3sin 0 0

1.5

2.6

3

2.6

1.5 0

1.5

2.6

3

2.6

1.5 2 0

5.r = sec

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a line.

r =

sec 0 1

1.2

2

2

1.2 1

1.2

2

2

1.2 2 1

6.r = sin

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r = sin

0 0

0.2

0.3

0.3

0.3

0.2 0

0.2

0.3

0.3

0.3

0.2 2 0

7.r = 4 cos

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r =

4cos 0 4

3.5

2

0

2

3.5 4

3.5

2

0

2

3.5 2 4

8.r = csc

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a line.

r = csc 0

2

1.2

1

1.2

2

2

1.2

1

1.2

2 2

Use symmetry to graph each equation.

9.r = 3 + 3 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 3 + 3 cos 0 6

5.6

5.1

4.5

3

1.5

0.9

0.4 0

10.r = 1 + 2 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 1 + 2 sin

1

0.7

0.4

0 0 1

2

2.4

2.7

3

11.r = 4 3 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 4 3

cos 0 1

1.4

1.9

2.5

4

5.5

6.12

6.6 7

12.r = 2 + 4 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 2 + 4

cos 0 6

5.5

4.8

4

2

0

0.8

1.5 2

13.r = 2 2 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 2 2

sin

4

3.7

3.4

3 0 2

1

0.6

0.3

0

14.r = 3 5 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 3 5 cos

0 2

1.3

0.5

0.5

3

5.5

6.5

7.3 8

15.r = 5 + 4 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 5 + 4

sin

1

1.5

2.2

3 0 5

7

7.8

8.5

9

16.r = 6 2 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 6 2

sin

8

7.7

7.4

7 0 6

5

4.6

4.3

4

Use symmetry, zeros, and maximum r-values tograph each function.

17.r = sin 4

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y = sin

4x on the interval . From the graph, you can

see that =1when and

y = 0 when

Interpreting these results in terms of the polar

equation r = sin 4 , we can say that hasamaximum value of 1 when

andr = 0 when

Since the function is symmetric with respect to the

line = , make a table and calculate the values of

r on .

Use these and a few additional points to sketch the graph of the function.

r = sin

4

0

0.9

0

0.9 0 0

0.9

0

0.9

0

18.r = 2 cos 2

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 2

cos 2x on the interval [0, ]. From the graph, you

can see that =2whenx = 0, , and and y = 0

when

Interpreting these results in terms of the polar

equation r = 2 cos 2 , we can say that hasa

maximum value of 2 when andr =

0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 2 cos

2 0 2

1

0

1

2

1

0

1 2

19.r = 5 cos 3

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 5

cos 3x on the interval [0, ]. From the graph, you

can see that =5when and y

= 0 when

Interpreting these results in terms of the polar

equation r = 5 cos 3 , we can say that hasa

maximum value of 5 when andr

= 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 5 cos 3 0 5

3.5

0

3.5

5

3.5

0

3.5

5

3.5

0

3.5 5

20.r = 3 sin 2

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y = 3

sin 2x on the interval . From the graph, you

can see that =3when andy = 0

when

Interpreting these results in terms of the polar

equation r = 3 sin 2 , we can say that hasa

maximum value of 3 when andr = 0

when

Since the function is symmetric with respect to the

line = , make a table and calculate the values of

r on .

Use these and a few additional points to sketch the graph of the function.

r = 3 sin

2

0

2.6

3

2.6 0 0

2.6

3

2.6

0

21.r = sin3

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y =

sin3x on the interval . From the graph,

you can see that = when

andy = 0 when

.

Interpreting these results in terms of the polar

equation r = sin3 , we can say that hasa

maximum value of when

andr = 0 when

Since the function is symmetric with respect to the

line = , make a table and calculate the values of

r on .

Use these and a few additional points to sketch the graph of the function.

r = sin

3

0.5

0.4

0

0.4

0.5

0.4 0 0

0.4

0.5

0.4

0

0.4

0.5

22.r = 4 cos 5

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 4

cos 5x on the interval [0, ]. From the graph, you can see that =4when

and y = 0 when

Interpreting these results in terms of the polar

equation r = 4 cos 5 , we can say that hasamaximum value of 4 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 4 cos

5 0 4

0

4

0

4

0

4

0

4

0 4

23.r = 2 sin 5

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y = 2

sin 5x on the interval . From the graph, you

can see that =2when

andy = 0 when

.

Interpreting these results in terms of the polar

equation r = 2 sin 5 , we can say that hasamaximum value of 2 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r = 2 sin

5

2

0

2

0

2 0 0

2

0

2

0

2

24.r = 3 cos 4

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 3

cos 4x on the interval [0, ]. From the graph, you

can see that =3when and

y = 0 when

Interpreting these results in terms of the polar

equation r = 3 cos 4 , we can say that hasa

maximum value of 3 when and

r = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 3 cos

4 0 3

1.5

1.5

3

1.5

1.5

3

1.5

1.5

3

1.5

1.5 3

25.MARINE BIOLOGY Rose curves can be observed in marine wildlife. Determine the symmetry, zeros, and maximum r-values of each

functionmodelingamarinespeciesfor0 . Then use the information to graph the function. a. The pores forming the petal pattern of a sand dollar (Refer to Figure 9.2.3 on Page 548) can be modeled by

r = 3 cos 5 .

b. The outline of the body of a crown-of-thorns sea star (Refer to Figure 9.2.4 on Page 548) can be modeled by

r = 20 cos 8 .

SOLUTION:a. Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 3

cos 5x on the interval . From the graph, you

can see that =3when

, and y = 0 when

Interpreting these results in terms of the polar

equation r = 3 cos 5 ,wecansaythat hasamaximum value of 3 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate a few

additional values of r on .

Use these points and symmetry with respect to the polar axis to sketch the graph of the function.

b. Because the polar equation is a function of the cosine function, it is symmetric with respect to the polaraxis. Sketch the graph of the rectangular function y = 20

cos 8x on the interval . From the graph, you

can see that =20whenx = 0, , , , ,

, , , and , and y = 0 when x = , ,

, , , , , and .

Interpreting these results in terms of the polar

equation r = 20 cos 8 ,wecansaythat hasa

maximum value of 20 when = 0, , , , ,

, , , and , and r = 0 when = , ,

, , , , , and

Make a table and calculate a few additional values

of r on .

Use these points and symmetry with respect to the

polar axis, line = , and pole to sketch the graph

of the function.

r = 3 cos

5 0 3

0.8

2.6

2.1

1.5

2.9

0

r = 20 cos 8

0 20

10

10

20

10

10

20

Identify the type of curve given by each equation. Then use symmetry, zeros, and maximum r-values to graph the function.

26.r = cos

SOLUTION:

The equation is of the form r = a cos , so its graph is a circle. Because this polar equation is a function of the cosine function, it is symmetric with respect tothe polar axis. Sketch the graph of the rectangular function y =

cosx on the interval . From the graph, you

can see that = when x = 0 and and y = 0

when x = .

Interpreting these results in terms of the polar

equation r = cos , we can say that hasa

maximum value of when =0andand r = 0

when =

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r =

cos 0 0.33

0.29

0.24

0.17

0

0.17

0.24

0.29 0.3

27.r = 4 + 1; >0

SOLUTION:

The equation is of the form r = a + b, so its graph is a spiral of Archimedes. The function has no symmetry. Spirals are unbounded. Therefore, the function has no maximum r-values and only one zero. To find the zero, substitute r = 0 into the function and solve for

.

So, r = 0 when , which is not in the given

domain. Use points on the interval [0, 4] to sketch the graphof the function.

r = 4 + 1 0 1

4.1

7.3 13.6

19.9

2 26.1 3 38.7 4 51.3

28.r = 2 sin 4

SOLUTION:

The equation is of the form r = a sin n , so its graph is a rose. Because this polar equation is a function of the sine function, it is symmetric with

respect to the line = . Sketch the graph of the

rectangular function y = 2 sin 4x on the interval

. From the graph, you can see that =2

when andy = 0 when

Interpreting these results in terms of the polar

equation r = 2 sin 4 , we can say that hasamaximum value of 2 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r = 2 sin

4

0

1.7

1.7

0

1.7

1.7 0 0

1.7

1.7

0

1.7

1.7

0

29.r = 6 + 6 cos

SOLUTION:

The equation is of the form r = a + b cos anda = b, so its graph is a cardioid. Because this polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 6 +

6 cos x on the interval . From the graph, you

can see that =12whenx = 0 and y = 0 when x = .

Interpreting these results in terms of the polar

equation r = 6 + 6 cos , we can say that hasa

maximum value of 12 when =0andr = 0 when =.

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r = 6 + 6

cos 0 12

11.2

9

6

3

0.8 0

30.r2 = 4 cos 2

SOLUTION:

The equation is of the form r2 = a

2 cos 2 , so its

graphisalemniscate.

Replacing (r, ) with (r, ) yields r2 = 4 cos 2

( ). Since cosine is an even function, r2 = 4 cos 2

( ) can be written as r2 = 4 cos 2 , and the

function has symmetry with respect to the polar axis.

Replacing (r, ) with (r, ) yields (r)2 = 4 cos

2( ) or r2 = 4 cos 2( ). Again, since cosine is an

even function, r2 = 4 cos 2( ) can be written as r2

= 4 cos 2 , and the function has symmetry with

respect to the line = .

Finally, replacing (r, ) with (r, ) yields (r)2 = 4

cos 2 or r2 = 4 cos 2 . Therefore, the function

has symmetry with respect to the pole.

The equation r2 = 4 cos 2 is equivalent to r =

, which is undefined when cos 2 < 0. Therefore, the domain of the function is restricted to

the intervals and .

Sketch the graph of the rectangular function y =

ontheinterval . From the graph,

you can see that =2whenx = 0 and y = 0 when

.

Interpreting these results in terms of the polar

equation r2 = 4 cos 2 , we can say that hasa

maximum value of 2 when andr = 0 when

.

Use these points and the indicated symmetry to sketch the graph of the function.

r2 = 4

cos 2

0

1.4

1.7 0 2

1.7

1.4

0

31.r = 5 + 2; >0

SOLUTION:

The equation is of the form r = a + b, so its graph is a spiral of Archimedes. The function has no symmetry. Spirals are unbounded. Therefore, the function has no maximum r-values and only one zero. To find the zero, substitute r = 0 into the function and solve for

.

So, r = 0 when , which is not in the given

domain. Use points on the interval [0, 4] to sketch the graphof the function.

r = 5

+ 2 0 2

5.9

9.9 17.7

25.6

2 33.4 3 49.2 4 64.8

32.r = 3 2 sin

SOLUTION:

The equation is of the form r = a b sin andb < a < 2b,soitsgraphisalimaon.Morespecifically,itisadimpledlimaon.Becausethispolarequationisafunction of the sine function, it is symmetric with

respect to the line = .

Sketch the graph of the rectangular function y = 3

2 sin x on the interval . From the graph, you

can see that =5when . There are no

zeros.

Interpreting these results in terms of the polar

equation r = 3 2 sin , we can say that hasa

maximum value of 5 when .

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

r = 3 2

sin

5

4.9

4.7

4.4

eSolutions Manual - Powered by Cognero Page 1

9-2 Graphs of Polar Equations

Graph each equation by plotting points.

1.r = cos

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r = cos 0 1

0.9

0.5

0

0.5

0.9 1

0.9

0.5

0

0.5

0.9 2 1

2.r = csc

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a line.

r = csc 0

2

1.2

1

1.2

2

2

1.2

2

1.2

2 2

3.r = cos

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r =

cos 0 0.5

0.4

0.3

0

0.3

0.4 0.5

0.4

0.3

0

0.3

0.4 2 0.5

4.r = 3 sin

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r = 3sin 0 0

1.5

2.6

3

2.6

1.5 0

1.5

2.6

3

2.6

1.5 2 0

5.r = sec

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a line.

r =

sec 0 1

1.2

2

2

1.2 1

1.2

2

2

1.2 2 1

6.r = sin

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r = sin

0 0

0.2

0.3

0.3

0.3

0.2 0

0.2

0.3

0.3

0.3

0.2 2 0

7.r = 4 cos

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r =

4cos 0 4

3.5

2

0

2

3.5 4

3.5

2

0

2

3.5 2 4

8.r = csc

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a line.

r = csc 0

2

1.2

1

1.2

2

2

1.2

1

1.2

2 2

Use symmetry to graph each equation.

9.r = 3 + 3 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 3 + 3 cos 0 6

5.6

5.1

4.5

3

1.5

0.9

0.4 0

10.r = 1 + 2 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 1 + 2 sin

1

0.7

0.4

0 0 1

2

2.4

2.7

3

11.r = 4 3 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 4 3

cos 0 1

1.4

1.9

2.5

4

5.5

6.12

6.6 7

12.r = 2 + 4 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 2 + 4

cos 0 6

5.5

4.8

4

2

0

0.8

1.5 2

13.r = 2 2 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 2 2

sin

4

3.7

3.4

3 0 2

1

0.6

0.3

0

14.r = 3 5 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 3 5 cos

0 2

1.3

0.5

0.5

3

5.5

6.5

7.3 8

15.r = 5 + 4 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 5 + 4

sin

1

1.5

2.2

3 0 5

7

7.8

8.5

9

16.r = 6 2 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 6 2

sin

8

7.7

7.4

7 0 6

5

4.6

4.3

4

Use symmetry, zeros, and maximum r-values tograph each function.

17.r = sin 4

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y = sin

4x on the interval . From the graph, you can

see that =1when and

y = 0 when

Interpreting these results in terms of the polar

equation r = sin 4 , we can say that hasamaximum value of 1 when

andr = 0 when

Since the function is symmetric with respect to the

line = , make a table and calculate the values of

r on .

Use these and a few additional points to sketch the graph of the function.

r = sin

4

0

0.9

0

0.9 0 0

0.9

0

0.9

0

18.r = 2 cos 2

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 2

cos 2x on the interval [0, ]. From the graph, you

can see that =2whenx = 0, , and and y = 0

when

Interpreting these results in terms of the polar

equation r = 2 cos 2 , we can say that hasa

maximum value of 2 when andr =

0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 2 cos

2 0 2

1

0

1

2

1

0

1 2

19.r = 5 cos 3

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 5

cos 3x on the interval [0, ]. From the graph, you

can see that =5when and y

= 0 when

Interpreting these results in terms of the polar

equation r = 5 cos 3 , we can say that hasa

maximum value of 5 when andr

= 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 5 cos 3 0 5

3.5

0

3.5

5

3.5

0

3.5

5

3.5

0

3.5 5

20.r = 3 sin 2

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y = 3

sin 2x on the interval . From the graph, you

can see that =3when andy = 0

when

Interpreting these results in terms of the polar

equation r = 3 sin 2 , we can say that hasa

maximum value of 3 when andr = 0

when

Since the function is symmetric with respect to the

line = , make a table and calculate the values of

r on .

Use these and a few additional points to sketch the graph of the function.

r = 3 sin

2

0

2.6

3

2.6 0 0

2.6

3

2.6

0

21.r = sin3

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y =

sin3x on the interval . From the graph,

you can see that = when

andy = 0 when

.

Interpreting these results in terms of the polar

equation r = sin3 , we can say that hasa

maximum value of when

andr = 0 when

Since the function is symmetric with respect to the

line = , make a table and calculate the values of

r on .

Use these and a few additional points to sketch the graph of the function.

r = sin

3

0.5

0.4

0

0.4

0.5

0.4 0 0

0.4

0.5

0.4

0

0.4

0.5

22.r = 4 cos 5

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 4

cos 5x on the interval [0, ]. From the graph, you can see that =4when

and y = 0 when

Interpreting these results in terms of the polar

equation r = 4 cos 5 , we can say that hasamaximum value of 4 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 4 cos

5 0 4

0

4

0

4

0

4

0

4

0 4

23.r = 2 sin 5

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y = 2

sin 5x on the interval . From the graph, you

can see that =2when

andy = 0 when

.

Interpreting these results in terms of the polar

equation r = 2 sin 5 , we can say that hasamaximum value of 2 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r = 2 sin

5

2

0

2

0

2 0 0

2

0

2

0

2

24.r = 3 cos 4

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 3

cos 4x on the interval [0, ]. From the graph, you

can see that =3when and

y = 0 when

Interpreting these results in terms of the polar

equation r = 3 cos 4 , we can say that hasa

maximum value of 3 when and

r = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 3 cos

4 0 3

1.5

1.5

3

1.5

1.5

3

1.5

1.5

3

1.5

1.5 3

25.MARINE BIOLOGY Rose curves can be observed in marine wildlife. Determine the symmetry, zeros, and maximum r-values of each

functionmodelingamarinespeciesfor0 . Then use the information to graph the function. a. The pores forming the petal pattern of a sand dollar (Refer to Figure 9.2.3 on Page 548) can be modeled by

r = 3 cos 5 .

b. The outline of the body of a crown-of-thorns sea star (Refer to Figure 9.2.4 on Page 548) can be modeled by

r = 20 cos 8 .

SOLUTION:a. Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 3

cos 5x on the interval . From the graph, you

can see that =3when

, and y = 0 when

Interpreting these results in terms of the polar

equation r = 3 cos 5 ,wecansaythat hasamaximum value of 3 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate a few

additional values of r on .

Use these points and symmetry with respect to the polar axis to sketch the graph of the function.

b. Because the polar equation is a function of the cosine function, it is symmetric with respect to the polaraxis. Sketch the graph of the rectangular function y = 20

cos 8x on the interval . From the graph, you

can see that =20whenx = 0, , , , ,

, , , and , and y = 0 when x = , ,

, , , , , and .

Interpreting these results in terms of the polar

equation r = 20 cos 8 ,wecansaythat hasa

maximum value of 20 when = 0, , , , ,

, , , and , and r = 0 when = , ,

, , , , , and

Make a table and calculate a few additional values

of r on .

Use these points and symmetry with respect to the

polar axis, line = , and pole to sketch the graph

of the function.

r = 3 cos

5 0 3

0.8

2.6

2.1

1.5

2.9

0

r = 20 cos 8

0 20

10

10

20

10

10

20

Identify the type of curve given by each equation. Then use symmetry, zeros, and maximum r-values to graph the function.

26.r = cos

SOLUTION:

The equation is of the form r = a cos , so its graph is a circle. Because this polar equation is a function of the cosine function, it is symmetric with respect tothe polar axis. Sketch the graph of the rectangular function y =

cosx on the interval . From the graph, you

can see that = when x = 0 and and y = 0

when x = .

Interpreting these results in terms of the polar

equation r = cos , we can say that hasa

maximum value of when =0andand r = 0

when =

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r =

cos 0 0.33

0.29

0.24

0.17

0

0.17

0.24

0.29 0.3

27.r = 4 + 1; >0

SOLUTION:

The equation is of the form r = a + b, so its graph is a spiral of Archimedes. The function has no symmetry. Spirals are unbounded. Therefore, the function has no maximum r-values and only one zero. To find the zero, substitute r = 0 into the function and solve for

.

So, r = 0 when , which is not in the given

domain. Use points on the interval [0, 4] to sketch the graphof the function.

r = 4 + 1 0 1

4.1

7.3 13.6

19.9

2 26.1 3 38.7 4 51.3

28.r = 2 sin 4

SOLUTION:

The equation is of the form r = a sin n , so its graph is a rose. Because this polar equation is a function of the sine function, it is symmetric with

respect to the line = . Sketch the graph of the

rectangular function y = 2 sin 4x on the interval

. From the graph, you can see that =2

when andy = 0 when

Interpreting these results in terms of the polar

equation r = 2 sin 4 , we can say that hasamaximum value of 2 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r = 2 sin

4

0

1.7

1.7

0

1.7

1.7 0 0

1.7

1.7

0

1.7

1.7

0

29.r = 6 + 6 cos

SOLUTION:

The equation is of the form r = a + b cos anda = b, so its graph is a cardioid. Because this polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 6 +

6 cos x on the interval . From the graph, you

can see that =12whenx = 0 and y = 0 when x = .

Interpreting these results in terms of the polar

equation r = 6 + 6 cos , we can say that hasa

maximum value of 12 when =0andr = 0 when =.

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r = 6 + 6

cos 0 12

11.2

9

6

3

0.8 0

30.r2 = 4 cos 2

SOLUTION:

The equation is of the form r2 = a

2 cos 2 , so its

graphisalemniscate.

Replacing (r, ) with (r, ) yields r2 = 4 cos 2

( ). Since cosine is an even function, r2 = 4 cos 2

( ) can be written as r2 = 4 cos 2 , and the

function has symmetry with respect to the polar axis.

Replacing (r, ) with (r, ) yields (r)2 = 4 cos

2( ) or r2 = 4 cos 2( ). Again, since cosine is an

even function, r2 = 4 cos 2( ) can be written as r2

= 4 cos 2 , and the function has symmetry with

respect to the line = .

Finally, replacing (r, ) with (r, ) yields (r)2 = 4

cos 2 or r2 = 4 cos 2 . Therefore, the function

has symmetry with respect to the pole.

The equation r2 = 4 cos 2 is equivalent to r =

, which is undefined when cos 2 < 0. Therefore, the domain of the function is restricted to

the intervals and .

Sketch the graph of the rectangular function y =

ontheinterval . From the graph,

you can see that =2whenx = 0 and y = 0 when

.

Interpreting these results in terms of the polar

equation r2 = 4 cos 2 , we can say that hasa

maximum value of 2 when andr = 0 when

.

Use these points and the indicated symmetry to sketch the graph of the function.

r2 = 4

cos 2

0

1.4

1.7 0 2

1.7

1.4

0

31.r = 5 + 2; >0

SOLUTION:

The equation is of the form r = a + b, so its graph is a spiral of Archimedes. The function has no symmetry. Spirals are unbounded. Therefore, the function has no maximum r-values and only one zero. To find the zero, substitute r = 0 into the function and solve for

.

So, r = 0 when , which is not in the given

domain. Use points on the interval [0, 4] to sketch the graphof the function.

r = 5

+ 2 0 2

5.9

9.9 17.7

25.6

2 33.4 3 49.2 4 64.8

32.r = 3 2 sin

SOLUTION:

The equation is of the form r = a b sin andb < a < 2b,soitsgraphisalimaon.Morespecifically,itisadimpledlimaon.Becausethispolarequationisafunction of the sine function, it is symmetric with

respect to the line = .

Sketch the graph of the rectangular function y = 3

2 sin x on the interval . From the graph, you

can see that =5when . There are no

zeros.

Interpreting these results in terms of the polar

equation r = 3 2 sin , we can say that hasa

maximum value of 5 when .

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

r = 3 2

sin

5

4.9

4.7

4.4

Graph each equation by plotting points.

1.r = cos

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r = cos 0 1

0.9

0.5

0

0.5

0.9 1

0.9

0.5

0

0.5

0.9 2 1

2.r = csc

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a line.

r = csc 0

2

1.2

1

1.2

2

2

1.2

2

1.2

2 2

3.r = cos

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r =

cos 0 0.5

0.4

0.3

0

0.3

0.4 0.5

0.4

0.3

0

0.3

0.4 2 0.5

4.r = 3 sin

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r = 3sin 0 0

1.5

2.6

3

2.6

1.5 0

1.5

2.6

3

2.6

1.5 2 0

5.r = sec

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a line.

r =

sec 0 1

1.2

2

2

1.2 1

1.2

2

2

1.2 2 1

6.r = sin

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r = sin

0 0

0.2

0.3

0.3

0.3

0.2 0

0.2

0.3

0.3

0.3

0.2 2 0

7.r = 4 cos

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r =

4cos 0 4

3.5

2

0

2

3.5 4

3.5

2

0

2

3.5 2 4

8.r = csc

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a line.

r = csc 0

2

1.2

1

1.2

2

2

1.2

1

1.2

2 2

Use symmetry to graph each equation.

9.r = 3 + 3 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 3 + 3 cos 0 6

5.6

5.1

4.5

3

1.5

0.9

0.4 0

10.r = 1 + 2 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 1 + 2 sin

1

0.7

0.4

0 0 1

2

2.4

2.7

3

11.r = 4 3 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 4 3

cos 0 1

1.4

1.9

2.5

4

5.5

6.12

6.6 7

12.r = 2 + 4 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 2 + 4

cos 0 6

5.5

4.8

4

2

0

0.8

1.5 2

13.r = 2 2 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 2 2

sin

4

3.7

3.4

3 0 2

1

0.6

0.3

0

14.r = 3 5 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 3 5 cos

0 2

1.3

0.5

0.5

3

5.5

6.5

7.3 8

15.r = 5 + 4 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 5 + 4

sin

1

1.5

2.2

3 0 5

7

7.8

8.5

9

16.r = 6 2 sin

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =. Therefore, make a table and calculate the values

of r on .

Use these points and symmetry with respect to the

line = to graph the function.

r = 6 2

sin

8

7.7

7.4

7 0 6

5

4.6

4.3

4

Use symmetry, zeros, and maximum r-values tograph each function.

17.r = sin 4

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y = sin

4x on the interval . From the graph, you can

see that =1when and

y = 0 when

Interpreting these results in terms of the polar

equation r = sin 4 , we can say that hasamaximum value of 1 when

andr = 0 when

Since the function is symmetric with respect to the

line = , make a table and calculate the values of

r on .

Use these and a few additional points to sketch the graph of the function.

r = sin

4

0

0.9

0

0.9 0 0

0.9

0

0.9

0

18.r = 2 cos 2

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 2

cos 2x on the interval [0, ]. From the graph, you

can see that =2whenx = 0, , and and y = 0

when

Interpreting these results in terms of the polar

equation r = 2 cos 2 , we can say that hasa

maximum value of 2 when andr =

0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 2 cos

2 0 2

1

0

1

2

1

0

1 2

19.r = 5 cos 3

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 5

cos 3x on the interval [0, ]. From the graph, you

can see that =5when and y

= 0 when

Interpreting these results in terms of the polar

equation r = 5 cos 3 , we can say that hasa

maximum value of 5 when andr

= 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 5 cos 3 0 5

3.5

0

3.5

5

3.5

0

3.5

5

3.5

0

3.5 5

20.r = 3 sin 2

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y = 3

sin 2x on the interval . From the graph, you

can see that =3when andy = 0

when

Interpreting these results in terms of the polar

equation r = 3 sin 2 , we can say that hasa

maximum value of 3 when andr = 0

when

Since the function is symmetric with respect to the

line = , make a table and calculate the values of

r on .

Use these and a few additional points to sketch the graph of the function.

r = 3 sin

2

0

2.6

3

2.6 0 0

2.6

3

2.6

0

21.r = sin3

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y =

sin3x on the interval . From the graph,

you can see that = when

andy = 0 when

.

Interpreting these results in terms of the polar

equation r = sin3 , we can say that hasa

maximum value of when

andr = 0 when

Since the function is symmetric with respect to the

line = , make a table and calculate the values of

r on .

Use these and a few additional points to sketch the graph of the function.

r = sin

3

0.5

0.4

0

0.4

0.5

0.4 0 0

0.4

0.5

0.4

0

0.4

0.5

22.r = 4 cos 5

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 4

cos 5x on the interval [0, ]. From the graph, you can see that =4when

and y = 0 when

Interpreting these results in terms of the polar

equation r = 4 cos 5 , we can say that hasamaximum value of 4 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 4 cos

5 0 4

0

4

0

4

0

4

0

4

0 4

23.r = 2 sin 5

SOLUTION:Because the polar equation is a function of the sine

function, it is symmetric with respect to the line =.

Sketch the graph of the rectangular function y = 2

sin 5x on the interval . From the graph, you

can see that =2when

andy = 0 when

.

Interpreting these results in terms of the polar

equation r = 2 sin 5 , we can say that hasamaximum value of 2 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r = 2 sin

5

2

0

2

0

2 0 0

2

0

2

0

2

24.r = 3 cos 4

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 3

cos 4x on the interval [0, ]. From the graph, you

can see that =3when and

y = 0 when

Interpreting these results in terms of the polar

equation r = 3 cos 4 , we can say that hasa

maximum value of 3 when and

r = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on [0, ].

Use these and a few additional points to sketch the graph of the function.

r = 3 cos

4 0 3

1.5

1.5

3

1.5

1.5

3

1.5

1.5

3

1.5

1.5 3

25.MARINE BIOLOGY Rose curves can be observed in marine wildlife. Determine the symmetry, zeros, and maximum r-values of each

functionmodelingamarinespeciesfor0 . Then use the information to graph the function. a. The pores forming the petal pattern of a sand dollar (Refer to Figure 9.2.3 on Page 548) can be modeled by

r = 3 cos 5 .

b. The outline of the body of a crown-of-thorns sea star (Refer to Figure 9.2.4 on Page 548) can be modeled by

r = 20 cos 8 .

SOLUTION:a. Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 3

cos 5x on the interval . From the graph, you

can see that =3when

, and y = 0 when

Interpreting these results in terms of the polar

equation r = 3 cos 5 ,wecansaythat hasamaximum value of 3 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate a few

additional values of r on .

Use these points and symmetry with respect to the polar axis to sketch the graph of the function.

b. Because the polar equation is a function of the cosine function, it is symmetric with respect to the polaraxis. Sketch the graph of the rectangular function y = 20

cos 8x on the interval . From the graph, you

can see that =20whenx = 0, , , , ,

, , , and , and y = 0 when x = , ,

, , , , , and .

Interpreting these results in terms of the polar

equation r = 20 cos 8 ,wecansaythat hasa

maximum value of 20 when = 0, , , , ,

, , , and , and r = 0 when = , ,

, , , , , and

Make a table and calculate a few additional values

of r on .

Use these points and symmetry with respect to the

polar axis, line = , and pole to sketch the graph

of the function.

r = 3 cos

5 0 3

0.8

2.6

2.1

1.5

2.9

0

r = 20 cos 8

0 20

10

10

20

10

10

20

Identify the type of curve given by each equation. Then use symmetry, zeros, and maximum r-values to graph the function.

26.r = cos

SOLUTION:

The equation is of the form r = a cos , so its graph is a circle. Because this polar equation is a function of the cosine function, it is symmetric with respect tothe polar axis. Sketch the graph of the rectangular function y =

cosx on the interval . From the graph, you

can see that = when x = 0 and and y = 0

when x = .

Interpreting these results in terms of the polar

equation r = cos , we can say that hasa

maximum value of when =0andand r = 0

when =

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r =

cos 0 0.33

0.29

0.24

0.17

0

0.17

0.24

0.29 0.3

27.r = 4 + 1; >0

SOLUTION:

The equation is of the form r = a + b, so its graph is a spiral of Archimedes. The function has no symmetry. Spirals are unbounded. Therefore, the function has no maximum r-values and only one zero. To find the zero, substitute r = 0 into the function and solve for

.

So, r = 0 when , which is not in the given

domain. Use points on the interval [0, 4] to sketch the graphof the function.

r = 4 + 1 0 1

4.1

7.3 13.6

19.9

2 26.1 3 38.7 4 51.3

28.r = 2 sin 4

SOLUTION:

The equation is of the form r = a sin n , so its graph is a rose. Because this polar equation is a function of the sine function, it is symmetric with

respect to the line = . Sketch the graph of the

rectangular function y = 2 sin 4x on the interval

. From the graph, you can see that =2

when andy = 0 when

Interpreting these results in terms of the polar

equation r = 2 sin 4 , we can say that hasamaximum value of 2 when

andr = 0 when

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r = 2 sin

4

0

1.7

1.7

0

1.7

1.7 0 0

1.7

1.7

0

1.7

1.7

0

29.r = 6 + 6 cos

SOLUTION:

The equation is of the form r = a + b cos anda = b, so its graph is a cardioid. Because this polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Sketch the graph of the rectangular function y = 6 +

6 cos x on the interval . From the graph, you

can see that =12whenx = 0 and y = 0 when x = .

Interpreting these results in terms of the polar

equation r = 6 + 6 cos , we can say that hasa

maximum value of 12 when =0andr = 0 when =.

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

Use these and a few additional points to sketch the graph of the function.

r = 6 + 6

cos 0 12

11.2

9

6

3

0.8 0

30.r2 = 4 cos 2

SOLUTION:

The equation is of the form r2 = a

2 cos 2 , so its

graphisalemniscate.

Replacing (r, ) with (r, ) yields r2 = 4 cos 2

( ). Since cosine is an even function, r2 = 4 cos 2

( ) can be written as r2 = 4 cos 2 , and the

function has symmetry with respect to the polar axis.

Replacing (r, ) with (r, ) yields (r)2 = 4 cos

2( ) or r2 = 4 cos 2( ). Again, since cosine is an

even function, r2 = 4 cos 2( ) can be written as r2

= 4 cos 2 , and the function has symmetry with

respect to the line = .

Finally, replacing (r, ) with (r, ) yields (r)2 = 4

cos 2 or r2 = 4 cos 2 . Therefore, the function

has symmetry with respect to the pole.

The equation r2 = 4 cos 2 is equivalent to r =

, which is undefined when cos 2 < 0. Therefore, the domain of the function is restricted to

the intervals and .

Sketch the graph of the rectangular function y =

ontheinterval . From the graph,

you can see that =2whenx = 0 and y = 0 when

.

Interpreting these results in terms of the polar

equation r2 = 4 cos 2 , we can say that hasa

maximum value of 2 when andr = 0 when

.

Use these points and the indicated symmetry to sketch the graph of the function.

r2 = 4

cos 2

0

1.4

1.7 0 2

1.7

1.4

0

31.r = 5 + 2; >0

SOLUTION:

The equation is of the form r = a + b, so its graph is a spiral of Archimedes. The function has no symmetry. Spirals are unbounded. Therefore, the function has no maximum r-values and only one zero. To find the zero, substitute r = 0 into the function and solve for

.

So, r = 0 when , which is not in the given

domain. Use points on the interval [0, 4] to sketch the graphof the function.

r = 5

+ 2 0 2

5.9

9.9 17.7

25.6

2 33.4 3 49.2 4 64.8

32.r = 3 2 sin

SOLUTION:

The equation is of the form r = a b sin andb < a < 2b,soitsgraphisalimaon.Morespecifically,itisadimpledlimaon.Becausethispolarequationisafunction of the sine function, it is symmetric with

respect to the line = .

Sketch the graph of the rectangular function y = 3

2 sin x on the interval . From the graph, you

can see that =5when . There are no

zeros.

Interpreting these results in terms of the polar

equation r = 3 2 sin , we can say that hasa

maximum value of 5 when .

Since the function is symmetric with respect to the polar axis, make a table and calculate the values of r

on .

r = 3 2

sin

5

4.9

4.7

4.4

eSolutions Manual - Powered by Cognero Page 2

9-2 Graphs of Polar Equations

Graph each equation by plotting points.

1.r = cos

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r = cos 0 1

0.9

0.5

0

0.5

0.9 1

0.9

0.5

0

0.5

0.9 2 1

2.r = csc

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a line.

r = csc 0

2

1.2

1

1.2

2

2

1.2

2

1.2

2 2

3.r = cos

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r =

cos 0 0.5

0.4

0.3

0

0.3

0.4 0.5

0.4

0.3

0

0.3

0.4 2 0.5

4.r = 3 sin

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r = 3sin 0 0

1.5

2.6

3

2.6

1.5 0

1.5

2.6

3

2.6

1.5 2 0

5.r = sec

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a line.

r =

sec 0 1

1.2

2

2

1.2 1

1.2

2

2

1.2 2 1

6.r = sin

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r = sin

0 0

0.2

0.3

0.3

0.3

0.2 0

0.2

0.3

0.3

0.3

0.2 2 0

7.r = 4 cos

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a smooth curve.

r =

4cos 0 4

3.5

2

0

2

3.5 4

3.5

2

0

2

3.5 2 4

8.r = csc

SOLUTION:Make a table of values to find the r-values

corresponding to various values of ontheinterval[0, 2]. Round each r-value to the nearest tenth.

Graph the ordered pairs (r, ) and connect them with a line.

r = csc 0

2

1.2

1

1.2

2

2

1.2

1

1.2

2 2

Use symmetry to graph each equation.

9.r = 3 + 3 cos

SOLUTION:Because the polar equation is a function of the cosine function, it is symmetric with respect to the polar axis. Therefore, make a table and calculate thevalues of r on [0, ].

Use these points and polar axis symmetry to graph the function.

r = 3 + 3 cos 0 6

5.6

5.1

4.5