9-2 graphs of polar equations - montville township … the polar equation is a function of the sine...
TRANSCRIPT
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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