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
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r = –cosθ 0 −1
−0.9
−0.5
0
0.5
0.9 π 1
0.9
0.5
0
−0.5
−0.9 2π −1
2. r = csc
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
line.
θ r = cscθ 0 −
2
1.2
1
1.2
2 π –
–2
–1.2
–2
−1.2
−2 2π −
3. r = cos
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ
r =
cosθ 0 0.5
0.4
0.3
0
−0.3
−0.4 π −0.5
−0.4
−0.3
0
0.3
0.4 2π 0.5
4. r = 3 sin
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r = 3sinθ 0 0
1.5
2.6
3
2.6
1.5 π 0
−1.5
−2.6
−3
−2.6
−1.5 2π 0
5. r = −sec
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
line.
θ r =
−secθ 0 −1
−1.2
−2
–
2
1.2 π 1
1.2
2
–
−2
−1.2 2π −1
6. r = sin
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r = sinθ
0 0
0.2
0.3
0.3
0.3
0.2 π 0
−0.2
−0.3
−0.3
−0.3
−0.2 2π 0
7. r = −4 cos
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r =
−4cosθ 0 −4
−3.5
−2
0
2
3.5 π 4
3.5
2
0
−2
−3.5 2π −4
8. r = −csc
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
line.
θ r = −cscθ 0 −
−2
−1.2
−1
−1.2
−2 π −
2
1.2
1
1.2
2 2π −
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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 3 + 3 cos θ 0 6
5.6
5.1
4.5
3
1.5
0.9
0.4 π 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.
θ r = 1 + 2 sin θ
−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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 4 − 3
cos θ 0 1
1.4
1.9
2.5
4
5.5
6.12
6.6 π 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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 2 + 4
cos θ 0 6
5.5
4.8
4
2
0
−0.8
−1.5 π −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.
θ r = 2 − 2
sin θ
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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 3 − 5 cos
0 –2
–1.3
–0.5
0.5
3
5.5
6.5
7.3 π 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.
θ r = 5 + 4
sin θ
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.
θ r = 6 − 2
sin θ
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 = 1 when and
y = 0 when
Interpreting these results in terms of the polar
equation r = sin 4 , we can say that
has a maximum value of 1 when
and r = 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.
θ r = sin
4θ
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, π]. From the graph, you
can see that = 2 when x = 0, , and π and y
= 0
when
Interpreting these results in terms of the polar
equation r = 2 cos 2 , we can say that
has a
maximum value of 2 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, π].
Use these and a few additional points to sketch the graph
of the function.
θ r = 2 cos
2θ 0 2
1
0
−1
−2
−1
0
1 π 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, π]. From the graph, you
can see that = 5 when and y
= 0 when
Interpreting these results in terms of the polar
equation r = 5 cos 3 , we can say that
has a
maximum value of 5 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, π].
Use these and a few additional points to sketch the graph
of the function.
θ r = 5 cos 3θ 0 5
3.5
0
−3.5
−5
−3.5
0
3.5
5
3.5
0
−3.5 π −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 = 3 when and y =
0
when
Interpreting these results in terms of the polar
equation r = 3 sin 2 , we can say that
has a
maximum value of 3 when and r = 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.
θ r = 3 sin
2θ
0
−2.6
−3
−2.6 0 0
2.6
3
2.6
0
21. r = sin 3
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 3x on the interval . From the graph,
you can see that = when
and y = 0 when
.
Interpreting these results in terms of the polar
equation r = sin 3 , we can say that
has a
maximum value of when
and r = 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.
θ
r = sin
3θ
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, π]. From the graph, you can see that
= 4 when
and y = 0 when
Interpreting these results in terms of the polar
equation r = 4 cos 5 , we can say that
has a maximum value of 4 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, π].
Use these and a few additional points to sketch the graph
of the function.
θ r = 4 cos
5θ 0 4
0
−4
0
4
0
−4
0
4
0 π −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 = 2 when
and y = 0 when
.
Interpreting these results in terms of the polar
equation r = 2 sin 5 , we can say that
has a maximum value of 2 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 .
Use these and a few additional points to sketch the graph
of the function.
θ r = 2 sin
5θ
−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, π]. From the graph, you
can see that = 3 when and
y = 0 when
Interpreting these results in terms of the polar
equation r = 3 cos 4 , we can say that
has a
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, π].
Use these and a few additional points to sketch
the graph of the function.
θ r = 3 cos
4θ 0 3
1.5
−1.5
−3
−1.5
1.5
3
1.5
−1.5
−3
−1.5
1.5 π 3
25. MARINE BIOLOGY Rose curves can be observed in marine
wildlife. Determine the symmetry, zeros, and maximum r-values of
each
function modeling a marine species for 0 ≤
≤ π. 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 = 3 when
, and y = 0 when
Interpreting these results in terms of the
polar
equation r = 3 cos 5
, we can say that
has a 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 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
polar axis. Sketch the graph of the rectangular
function y = 20
cos 8x on the interval . From the graph, you
can see that = 20 when x = 0, , , , ,
, , , and π, and y = 0 when x = , ,
, , , , , and .
Interpreting these results in terms of the
polar
equation r = 20 cos 8
, we can say that
has a
maximum value of 20 when θ = 0, , , , ,
, , , and π, and r = 0 when θ = , ,
, , , , , 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 θ = , and pole to sketch the graph
of the function.
θ r = 3 cos
5θ 0 3
0.8
−2.6
−2.1
1.5
2.9
0
θ r = 20 cos 8θ
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 =
cos x on the interval . From the graph, you
can see that = when x = 0 and π and y = 0
when x = .
Interpreting these results in terms of the polar
equation r = cos , we can say that
has a
maximum value of when
= 0 and π 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 .
Use these and a few additional points to sketch the graph
of the function.
θ
r =
cos θ 0 0.33
0.29
0.24
0.17
0
−0.17
−0.24
−0.29 π −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, 4π] to sketch the graphof
the function.
θ r = 4θ + 1 0 1
4.1
7.3 π 13.6
19.9
2π 26.1 3π 38.7 4π 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 and y = 0 when
Interpreting these results in terms of the polar
equation r = 2 sin 4 , we can say that
has a maximum value of 2 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 .
Use these and a few additional points to sketch the graph
of the function.
θ r = 2 sin
4θ
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 and a = 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 = 12 when x = 0 and y = 0 when
x = π.
Interpreting these results in terms of the polar
equation r = 6 + 6 cos , we can say that
has a
maximum value of 12 when = 0 and r = 0 when
= π.
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.
θ r = 6 + 6
cos θ 0 12
11.2
9
6
3
0.8 π 0
30. r2 = 4 cos 2
SOLUTION:
The equation is of the form r2 = a
2 cos 2 , so its
graph is a lemniscate.
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 =
on the interval . From the graph,
you can see that = 2 when x = 0 and y = 0
when
.
Interpreting these results in terms of the polar
equation r2 = 4 cos 2 , we can say that
has a
maximum value of 2 when and r = 0 when
.
Use these points and the indicated symmetry to sketch the graph
of the function.
θ r2 = 4
cos 2θ
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, 4π] to sketch the graphof
the function.
θ r = 5θ
+ 2 0 2
5.9
9.9 π 17.7
25.6
2π 33.4 3π 49.2 4π 64.8
32. r = 3 − 2 sin
SOLUTION:
The equation is of the form r = a − b sin and b <
a <
2b, so its graph is a limaçon. More specifically, itis a dimpled limaçon. Because this polar equation is afunction
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 = 5 when . There are no
zeros.
Interpreting these results in terms of the polar
equation r = 3 − 2 sin , we can say that
has a
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 .
θ r = 3 − 2
sin θ
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
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r = –cosθ 0 −1
−0.9
−0.5
0
0.5
0.9 π 1
0.9
0.5
0
−0.5
−0.9 2π −1
2. r = csc
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
line.
θ r = cscθ 0 −
2
1.2
1
1.2
2 π –
–2
–1.2
–2
−1.2
−2 2π −
3. r = cos
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ
r =
cosθ 0 0.5
0.4
0.3
0
−0.3
−0.4 π −0.5
−0.4
−0.3
0
0.3
0.4 2π 0.5
4. r = 3 sin
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r = 3sinθ 0 0
1.5
2.6
3
2.6
1.5 π 0
−1.5
−2.6
−3
−2.6
−1.5 2π 0
5. r = −sec
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
line.
θ r =
−secθ 0 −1
−1.2
−2
–
2
1.2 π 1
1.2
2
–
−2
−1.2 2π −1
6. r = sin
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r = sinθ
0 0
0.2
0.3
0.3
0.3
0.2 π 0
−0.2
−0.3
−0.3
−0.3
−0.2 2π 0
7. r = −4 cos
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r =
−4cosθ 0 −4
−3.5
−2
0
2
3.5 π 4
3.5
2
0
−2
−3.5 2π −4
8. r = −csc
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
line.
θ r = −cscθ 0 −
−2
−1.2
−1
−1.2
−2 π −
2
1.2
1
1.2
2 2π −
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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 3 + 3 cos θ 0 6
5.6
5.1
4.5
3
1.5
0.9
0.4 π 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.
θ r = 1 + 2 sin θ
−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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 4 − 3
cos θ 0 1
1.4
1.9
2.5
4
5.5
6.12
6.6 π 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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 2 + 4
cos θ 0 6
5.5
4.8
4
2
0
−0.8
−1.5 π −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.
θ r = 2 − 2
sin θ
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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 3 − 5 cos
0 –2
–1.3
–0.5
0.5
3
5.5
6.5
7.3 π 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.
θ r = 5 + 4
sin θ
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.
θ r = 6 − 2
sin θ
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 = 1 when and
y = 0 when
Interpreting these results in terms of the polar
equation r = sin 4 , we can say that
has a maximum value of 1 when
and r = 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.
θ r = sin
4θ
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, π]. From the graph, you
can see that = 2 when x = 0, , and π and y
= 0
when
Interpreting these results in terms of the polar
equation r = 2 cos 2 , we can say that
has a
maximum value of 2 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, π].
Use these and a few additional points to sketch the graph
of the function.
θ r = 2 cos
2θ 0 2
1
0
−1
−2
−1
0
1 π 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, π]. From the graph, you
can see that = 5 when and y
= 0 when
Interpreting these results in terms of the polar
equation r = 5 cos 3 , we can say that
has a
maximum value of 5 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, π].
Use these and a few additional points to sketch the graph
of the function.
θ r = 5 cos 3θ 0 5
3.5
0
−3.5
−5
−3.5
0
3.5
5
3.5
0
−3.5 π −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 = 3 when and y =
0
when
Interpreting these results in terms of the polar
equation r = 3 sin 2 , we can say that
has a
maximum value of 3 when and r = 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.
θ r = 3 sin
2θ
0
−2.6
−3
−2.6 0 0
2.6
3
2.6
0
21. r = sin 3
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 3x on the interval . From the graph,
you can see that = when
and y = 0 when
.
Interpreting these results in terms of the polar
equation r = sin 3 , we can say that
has a
maximum value of when
and r = 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.
θ
r = sin
3θ
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, π]. From the graph, you can see that
= 4 when
and y = 0 when
Interpreting these results in terms of the polar
equation r = 4 cos 5 , we can say that
has a maximum value of 4 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, π].
Use these and a few additional points to sketch the graph
of the function.
θ r = 4 cos
5θ 0 4
0
−4
0
4
0
−4
0
4
0 π −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 = 2 when
and y = 0 when
.
Interpreting these results in terms of the polar
equation r = 2 sin 5 , we can say that
has a maximum value of 2 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 .
Use these and a few additional points to sketch the graph
of the function.
θ r = 2 sin
5θ
−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, π]. From the graph, you
can see that = 3 when and
y = 0 when
Interpreting these results in terms of the polar
equation r = 3 cos 4 , we can say that
has a
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, π].
Use these and a few additional points to sketch
the graph of the function.
θ r = 3 cos
4θ 0 3
1.5
−1.5
−3
−1.5
1.5
3
1.5
−1.5
−3
−1.5
1.5 π 3
25. MARINE BIOLOGY Rose curves can be observed in marine
wildlife. Determine the symmetry, zeros, and maximum r-values of
each
function modeling a marine species for 0 ≤
≤ π. 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 = 3 when
, and y = 0 when
Interpreting these results in terms of the
polar
equation r = 3 cos 5
, we can say that
has a 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 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
polar axis. Sketch the graph of the rectangular
function y = 20
cos 8x on the interval . From the graph, you
can see that = 20 when x = 0, , , , ,
, , , and π, and y = 0 when x = , ,
, , , , , and .
Interpreting these results in terms of the
polar
equation r = 20 cos 8
, we can say that
has a
maximum value of 20 when θ = 0, , , , ,
, , , and π, and r = 0 when θ = , ,
, , , , , 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 θ = , and pole to sketch the graph
of the function.
θ r = 3 cos
5θ 0 3
0.8
−2.6
−2.1
1.5
2.9
0
θ r = 20 cos 8θ
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 =
cos x on the interval . From the graph, you
can see that = when x = 0 and π and y = 0
when x = .
Interpreting these results in terms of the polar
equation r = cos , we can say that
has a
maximum value of when
= 0 and π 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 .
Use these and a few additional points to sketch the graph
of the function.
θ
r =
cos θ 0 0.33
0.29
0.24
0.17
0
−0.17
−0.24
−0.29 π −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, 4π] to sketch the graphof
the function.
θ r = 4θ + 1 0 1
4.1
7.3 π 13.6
19.9
2π 26.1 3π 38.7 4π 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 and y = 0 when
Interpreting these results in terms of the polar
equation r = 2 sin 4 , we can say that
has a maximum value of 2 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 .
Use these and a few additional points to sketch the graph
of the function.
θ r = 2 sin
4θ
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 and a = 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 = 12 when x = 0 and y = 0 when
x = π.
Interpreting these results in terms of the polar
equation r = 6 + 6 cos , we can say that
has a
maximum value of 12 when = 0 and r = 0 when
= π.
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.
θ r = 6 + 6
cos θ 0 12
11.2
9
6
3
0.8 π 0
30. r2 = 4 cos 2
SOLUTION:
The equation is of the form r2 = a
2 cos 2 , so its
graph is a lemniscate.
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 =
on the interval . From the graph,
you can see that = 2 when x = 0 and y = 0
when
.
Interpreting these results in terms of the polar
equation r2 = 4 cos 2 , we can say that
has a
maximum value of 2 when and r = 0 when
.
Use these points and the indicated symmetry to sketch the graph
of the function.
θ r2 = 4
cos 2θ
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, 4π] to sketch the graphof
the function.
θ r = 5θ
+ 2 0 2
5.9
9.9 π 17.7
25.6
2π 33.4 3π 49.2 4π 64.8
32. r = 3 − 2 sin
SOLUTION:
The equation is of the form r = a − b sin and b <
a <
2b, so its graph is a limaçon. More specifically, itis a dimpled limaçon. Because this polar equation is afunction
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 = 5 when . There are no
zeros.
Interpreting these results in terms of the polar
equation r = 3 − 2 sin , we can say that
has a
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 .
θ r = 3 − 2
sin θ
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
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r = –cosθ 0 −1
−0.9
−0.5
0
0.5
0.9 π 1
0.9
0.5
0
−0.5
−0.9 2π −1
2. r = csc
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
line.
θ r = cscθ 0 −
2
1.2
1
1.2
2 π –
–2
–1.2
–2
−1.2
−2 2π −
3. r = cos
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ
r =
cosθ 0 0.5
0.4
0.3
0
−0.3
−0.4 π −0.5
−0.4
−0.3
0
0.3
0.4 2π 0.5
4. r = 3 sin
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r = 3sinθ 0 0
1.5
2.6
3
2.6
1.5 π 0
−1.5
−2.6
−3
−2.6
−1.5 2π 0
5. r = −sec
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
line.
θ r =
−secθ 0 −1
−1.2
−2
–
2
1.2 π 1
1.2
2
–
−2
−1.2 2π −1
6. r = sin
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r = sinθ
0 0
0.2
0.3
0.3
0.3
0.2 π 0
−0.2
−0.3
−0.3
−0.3
−0.2 2π 0
7. r = −4 cos
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r =
−4cosθ 0 −4
−3.5
−2
0
2
3.5 π 4
3.5
2
0
−2
−3.5 2π −4
8. r = −csc
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
line.
θ r = −cscθ 0 −
−2
−1.2
−1
−1.2
−2 π −
2
1.2
1
1.2
2 2π −
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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 3 + 3 cos θ 0 6
5.6
5.1
4.5
3
1.5
0.9
0.4 π 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.
θ r = 1 + 2 sin θ
−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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 4 − 3
cos θ 0 1
1.4
1.9
2.5
4
5.5
6.12
6.6 π 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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 2 + 4
cos θ 0 6
5.5
4.8
4
2
0
−0.8
−1.5 π −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.
θ r = 2 − 2
sin θ
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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 3 − 5 cos
0 –2
–1.3
–0.5
0.5
3
5.5
6.5
7.3 π 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.
θ r = 5 + 4
sin θ
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.
θ r = 6 − 2
sin θ
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 = 1 when and
y = 0 when
Interpreting these results in terms of the polar
equation r = sin 4 , we can say that
has a maximum value of 1 when
and r = 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.
θ r = sin
4θ
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, π]. From the graph, you
can see that = 2 when x = 0, , and π and y
= 0
when
Interpreting these results in terms of the polar
equation r = 2 cos 2 , we can say that
has a
maximum value of 2 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, π].
Use these and a few additional points to sketch the graph
of the function.
θ r = 2 cos
2θ 0 2
1
0
−1
−2
−1
0
1 π 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, π]. From the graph, you
can see that = 5 when and y
= 0 when
Interpreting these results in terms of the polar
equation r = 5 cos 3 , we can say that
has a
maximum value of 5 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, π].
Use these and a few additional points to sketch the graph
of the function.
θ r = 5 cos 3θ 0 5
3.5
0
−3.5
−5
−3.5
0
3.5
5
3.5
0
−3.5 π −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 = 3 when and y =
0
when
Interpreting these results in terms of the polar
equation r = 3 sin 2 , we can say that
has a
maximum value of 3 when and r = 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.
θ r = 3 sin
2θ
0
−2.6
−3
−2.6 0 0
2.6
3
2.6
0
21. r = sin 3
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 3x on the interval . From the graph,
you can see that = when
and y = 0 when
.
Interpreting these results in terms of the polar
equation r = sin 3 , we can say that
has a
maximum value of when
and r = 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.
θ
r = sin
3θ
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, π]. From the graph, you can see that
= 4 when
and y = 0 when
Interpreting these results in terms of the polar
equation r = 4 cos 5 , we can say that
has a maximum value of 4 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, π].
Use these and a few additional points to sketch the graph
of the function.
θ r = 4 cos
5θ 0 4
0
−4
0
4
0
−4
0
4
0 π −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 = 2 when
and y = 0 when
.
Interpreting these results in terms of the polar
equation r = 2 sin 5 , we can say that
has a maximum value of 2 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 .
Use these and a few additional points to sketch the graph
of the function.
θ r = 2 sin
5θ
−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, π]. From the graph, you
can see that = 3 when and
y = 0 when
Interpreting these results in terms of the polar
equation r = 3 cos 4 , we can say that
has a
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, π].
Use these and a few additional points to sketch
the graph of the function.
θ r = 3 cos
4θ 0 3
1.5
−1.5
−3
−1.5
1.5
3
1.5
−1.5
−3
−1.5
1.5 π 3
25. MARINE BIOLOGY Rose curves can be observed in marine
wildlife. Determine the symmetry, zeros, and maximum r-values of
each
function modeling a marine species for 0 ≤
≤ π. 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 = 3 when
, and y = 0 when
Interpreting these results in terms of the
polar
equation r = 3 cos 5
, we can say that
has a 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 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
polar axis. Sketch the graph of the rectangular
function y = 20
cos 8x on the interval . From the graph, you
can see that = 20 when x = 0, , , , ,
, , , and π, and y = 0 when x = , ,
, , , , , and .
Interpreting these results in terms of the
polar
equation r = 20 cos 8
, we can say that
has a
maximum value of 20 when θ = 0, , , , ,
, , , and π, and r = 0 when θ = , ,
, , , , , 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 θ = , and pole to sketch the graph
of the function.
θ r = 3 cos
5θ 0 3
0.8
−2.6
−2.1
1.5
2.9
0
θ r = 20 cos 8θ
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 =
cos x on the interval . From the graph, you
can see that = when x = 0 and π and y = 0
when x = .
Interpreting these results in terms of the polar
equation r = cos , we can say that
has a
maximum value of when
= 0 and π 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 .
Use these and a few additional points to sketch the graph
of the function.
θ
r =
cos θ 0 0.33
0.29
0.24
0.17
0
−0.17
−0.24
−0.29 π −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, 4π] to sketch the graphof
the function.
θ r = 4θ + 1 0 1
4.1
7.3 π 13.6
19.9
2π 26.1 3π 38.7 4π 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 and y = 0 when
Interpreting these results in terms of the polar
equation r = 2 sin 4 , we can say that
has a maximum value of 2 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 .
Use these and a few additional points to sketch the graph
of the function.
θ r = 2 sin
4θ
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 and a = 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 = 12 when x = 0 and y = 0 when
x = π.
Interpreting these results in terms of the polar
equation r = 6 + 6 cos , we can say that
has a
maximum value of 12 when = 0 and r = 0 when
= π.
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.
θ r = 6 + 6
cos θ 0 12
11.2
9
6
3
0.8 π 0
30. r2 = 4 cos 2
SOLUTION:
The equation is of the form r2 = a
2 cos 2 , so its
graph is a lemniscate.
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 =
on the interval . From the graph,
you can see that = 2 when x = 0 and y = 0
when
.
Interpreting these results in terms of the polar
equation r2 = 4 cos 2 , we can say that
has a
maximum value of 2 when and r = 0 when
.
Use these points and the indicated symmetry to sketch the graph
of the function.
θ r2 = 4
cos 2θ
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, 4π] to sketch the graphof
the function.
θ r = 5θ
+ 2 0 2
5.9
9.9 π 17.7
25.6
2π 33.4 3π 49.2 4π 64.8
32. r = 3 − 2 sin
SOLUTION:
The equation is of the form r = a − b sin and b <
a <
2b, so its graph is a limaçon. More specifically, itis a dimpled limaçon. Because this polar equation is afunction
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 = 5 when . There are no
zeros.
Interpreting these results in terms of the polar
equation r = 3 − 2 sin , we can say that
has a
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 .
θ r = 3 − 2
sin θ
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
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r = –cosθ 0 −1
−0.9
−0.5
0
0.5
0.9 π 1
0.9
0.5
0
−0.5
−0.9 2π −1
2. r = csc
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
line.
θ r = cscθ 0 −
2
1.2
1
1.2
2 π –
–2
–1.2
–2
−1.2
−2 2π −
3. r = cos
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ
r =
cosθ 0 0.5
0.4
0.3
0
−0.3
−0.4 π −0.5
−0.4
−0.3
0
0.3
0.4 2π 0.5
4. r = 3 sin
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r = 3sinθ 0 0
1.5
2.6
3
2.6
1.5 π 0
−1.5
−2.6
−3
−2.6
−1.5 2π 0
5. r = −sec
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
line.
θ r =
−secθ 0 −1
−1.2
−2
–
2
1.2 π 1
1.2
2
–
−2
−1.2 2π −1
6. r = sin
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r = sinθ
0 0
0.2
0.3
0.3
0.3
0.2 π 0
−0.2
−0.3
−0.3
−0.3
−0.2 2π 0
7. r = −4 cos
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
smooth curve.
θ r =
−4cosθ 0 −4
−3.5
−2
0
2
3.5 π 4
3.5
2
0
−2
−3.5 2π −4
8. r = −csc
SOLUTION: Make a table of values to find the
r-values
corresponding to various values of
on the interval [0, 2π]. Round each r-value to
the nearest tenth.
Graph the ordered pairs (r, ) and connect them with a
line.
θ r = −cscθ 0 −
−2
−1.2
−1
−1.2
−2 π −
2
1.2
1
1.2
2 2π −
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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 3 + 3 cos θ 0 6
5.6
5.1
4.5
3
1.5
0.9
0.4 π 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.
θ r = 1 + 2 sin θ
−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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 4 − 3
cos θ 0 1
1.4
1.9
2.5
4
5.5
6.12
6.6 π 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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 2 + 4
cos θ 0 6
5.5
4.8
4
2
0
−0.8
−1.5 π −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.
θ r = 2 − 2
sin θ
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,
π].
Use these points and polar axis symmetry to graph the
function.
θ r = 3 − 5 cos
0 –2
–1.3
–0.5
0.5
3
5.5
6.5
7.3 π 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.
θ r = 5 + 4
sin θ
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.
θ r = 6 − 2
sin θ
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 = 1 when and
y = 0 when
Interpreting these results in terms of the polar
equation r = sin 4 , we can say that
has a maximum value of 1 when
and r = 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.
θ r = sin
4θ
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, π]. From the graph, you
can see that = 2 when x = 0, , and π and y
= 0
when
Interpreting these results in terms of the polar
equation r = 2 cos 2 , we can say that
has a
maximum value of 2 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, π].
Use these and a few additional points to sketch the graph
of the function.
θ r = 2 cos
2θ 0 2
1
0
−1
−2
−1
0
1 π 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, π]. From the graph, you
can see that = 5 when and y
= 0 when
Interpreting these results in terms of the polar
equation r = 5 cos 3 , we can say that
has a
maximum value of 5 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, π].
Use these and a few additional points to sketch the graph
of the function.
θ r = 5 cos 3θ 0 5
3.5
0
−3.5
−5
−3.5
0
3.5
5
3.5
0
−3.5 π −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 = 3 when and y =
0
when
Interpreting these results in terms of the polar
equation r = 3 sin 2 , we can say that
has a
maximum value of 3 when and r = 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.
θ r = 3 sin
2θ
0
−2.6
−3
−2.6 0 0
2.6
3
2.6
0
21. r = sin 3
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 3x on the interval . From the graph,
you can see that = when
and y = 0 when
.
Interpreting these results in terms of the polar
equation r = sin 3 , we can say that
has a
maximum value of when
and r = 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.
θ
r = sin
3θ
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, π]. From the graph, you can see that
= 4 when
and y = 0 when
Interpreting these results in terms of the polar
equation r = 4 cos 5 , we can say that
has a maximum value of 4 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, π].
Use these and a few additional points to sketch the graph
of the function.
θ r = 4 cos
5θ 0 4
0
−4
0
4
0
−4
0
4
0 π −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 = 2 when
and y = 0 when
.
Interpreting these results in terms of the polar
equation r = 2 sin 5 , we can say that
has a maximum value of 2 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 .
Use these and a few additional points to sketch the graph
of the function.
θ r = 2 sin
5θ
−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, π]. From the graph, you
can see that = 3 when and
y = 0 when
Interpreting these results in terms of the polar
equation r = 3 cos 4 , we can say that
has a
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, π].
Use these and a few additional points to sketch
the graph of the function.
θ r = 3 cos
4θ 0 3
1.5
−1.5
−3
−1.5
1.5
3
1.5
−1.5
−3
−1.5
1.5 π 3
25. MARINE BIOLOGY Rose curves can be observed in marine
wildlife. Determine the symmetry, zeros, and maximum r-values of
each
function modeling a marine species for 0 ≤
≤ π. 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 = 3 when
, and y = 0 when
Interpreting these results in terms of the
polar
equation r = 3 cos 5
, we can say that
has a 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 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
polar axis. Sketch the graph of the rectangular
function y = 20
cos 8x on the interval . From the graph, you
can see that = 20 when x = 0, , , , ,
, , , and π, and y = 0 when x = , ,
, , , , , and .
Interpreting these results in terms of the
polar
equation r = 20 cos 8
, we can say that
has a
maximum value of 20 when θ = 0, , , , ,
, , , and π, and r = 0 when θ = , ,
, , , , , 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 θ = , and pole to sketch the graph
of the function.
θ r = 3 cos
5θ 0 3
0.8
−2.6
−2.1
1.5
2.9
0
θ r = 20 cos 8θ
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 =
cos x on the interval . From the graph, you
can see that = when x = 0 and π and y = 0
when x = .
Interpreting these results in terms of the polar
equation r = cos , we can say that
has a
maximum value of when
= 0 and π 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 .
Use these and a few additional points to sketch the graph
of the function.
θ
r =
cos θ 0 0.33
0.29
0.24
0.17
0
−0.17
−0.24
−0.29 π −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, 4π] to sketch the graphof
the function.
θ r = 4θ + 1 0 1
4.1
7.3 π 13.6
19.9
2π 26.1 3π 38.7 4π 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 and y = 0 when
Interpreting these results in terms of the polar
equation r = 2 sin 4 , we can say that
has a maximum value of 2 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 .
Use these and a few additional points to sketch the graph
of the function.
θ r = 2 sin
4θ
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 and a = b,
so its graph is a cardioid.
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