Polar Coordinate
System
Hi, Hello it’s Avi
1 Corinthians 13:4-8
1
Learning Outcomes
At the end of the session, you should be able to:
1. Plot points in the polar coordinate system;
2. Find multiple sets of polar coordinates for a given
point;
3. Rewrite rectangular coordinates in polar form and vice
versa;
4. Identify some types of polar equations;
5. Graph polar equations by conversion to its Cartesian
form;
6. Graph a polar equation by plotting points;
7. Use symmetry in graphing polar equations
2
Polar Coordinate System
3
Polar Coordinate System
Plot the following points
𝜋 𝜋 𝜋 𝜋
A= 1, B= 2, − C= −2, D= −3, −
4 4 6 3
4
Polar Coordinate System
Remarks
𝜋 9𝜋 5𝜋
A= 1, = 1, = −1,
4 4 4
𝜋 7𝜋
B= 2, − 4 = 2, 4
𝜋 7𝜋
C= −2, 6 = 2, 6
𝜋 2𝜋
D= −3, − 3 = 3, 3
5
Polar Coordinate System
𝜋
Let 𝑃 3, 6 . Find another representation of this point in which
a. 𝑟 > 0 and 2𝜋 < 𝜃 < 4𝜋 c. 𝑟 > 0 and −2𝜋 < 𝜃 < 0
b. 𝑟 < 0 and 0 < 𝜃 < 2𝜋
Solution:
𝜋 𝜋 13𝜋
a. 3, 6 = 3, 6 + 2𝜋 = 3, 6
𝜋 𝜋 7𝜋
b. 3, 6 = −3, 6 + 𝜋 = −3, 6
𝜋 𝜋 11𝜋
c. 3, 6 = 3, 6 − 2𝜋 = 3, − 6
6
Polar Coordinate System
Polar to Cartesian Cartesian to Polar
• 𝑥 = 𝑟 cos 𝜃 𝑟2 = 𝑥 2 + 𝑦2
𝑦
• 𝑦 = 𝑟 sin 𝜃 tan 𝜃 = 𝑥
7
Polar Coordinate System
Determine the polar coordinates of the point having Cartesian
coordinates − 3, 1 .
𝑦
Solution: Recall that 𝑟 2 = 𝑥 2 + 𝑦 2 and tan 𝜃 = 𝑥. Thus,
2
𝑟2 = − 3 + 12 ⟹ 𝑟 = 2
1 5𝜋
tan 𝜃 = ⟹𝜃=
− 3 6
5𝜋 11𝜋
Hence, the polar coordinates are 2, 6 or −2, 6 .
Note: There are infinitely many polar points that coincides to
5𝜋
the point 2, .
6
8
Polar Coordinate System
Determine the cartesian coordinates of the point having polar
𝜋
coordinates −5, − .
3
Solution: Recall that 𝑥 = 𝑟 cos 𝜃 and 𝑦 = 𝑟 sin 𝜃. Thus,
𝜋 5
𝑥 = −5 cos − =−
3 2
𝜋 5 3
𝑦 = −5 sin − =
3 2
5 5 3
Hence, the Cartesian coordinates are −2, 2 .
9
Polar Coordinate System
Express the following Cartesian equations to polar equations.
a. 4𝑥𝑦 = 7 b. (𝑥 − 1)2 +𝑦 2 = 1
Solution
a. 4𝑥𝑦 = 7 ⟹ 4 𝑟 cos 𝜃 𝑟 sin 𝜃 = 7 ⟹ 4𝑟 2 cos 𝜃 sin 𝜃 = 7
⟹ 2𝑟 2 2 cos 𝜃 sin 𝜃 = 7 ⟹ 2𝑟 2 sin 2𝜃 = 7
2
7
⟹ 𝑟 = csc 2𝜃
2
b. (𝑥 − 1)2 +𝑦 2 = 1 ⟹ 𝑟 cos 𝜃 − 1 2 + 𝑟 sin 𝜃 2 = 1
⟹ 𝑟 2 cos2 𝜃 − 2𝑟 cos 𝜃 + 1 + 𝑟 2 sin2 𝜃 = 1
⟹ 𝑟 2 cos 2 𝜃 + sin2 𝜃 − 2𝑟 cos 𝜃 = 0
⟹ 𝑟 2 −2𝑟 cos 𝜃 = 0 ⟹ 𝑟 2 = 2𝑟 cos 𝜃 ⟹ 𝑟 = 2 cos 𝜃
10
Polar Coordinate System
Convert the following polar equations into Cartesian
equations.
5
a. 𝑟 = 6 b. 𝑟 = 2 cos 𝜃−3 sin 𝜃 c. 𝑟 = 2 sin 𝜃
Solution
a. 𝑟 = 6 ⟹ 𝑟 2 = 36 ⟹ 𝑥 2 + 𝑦 2 = 36 (circle)
5
b. 𝑟 = ⟹ 𝑟 2 cos 𝜃 − 3 sin 𝜃 = 5
2 cos 𝜃−3 sin 𝜃
⟹ 2𝑟 cos 𝜃 − 3𝑟 sin 𝜃 = 5 ⟹ 2𝑥 − 3𝑦 = 5 (line)
c. 𝑟 = 2 sin 𝜃 ⟹ 𝑟 2 = 2𝑟 sin 𝜃 ⟹ 𝑥 2 + 𝑦 2 − 2𝑦 = 0 (circle)
11
Polar Equations and Polar Curves
Sketch the graph of 𝑟 = 2.
In general, the graph of the equation 𝑟 = 𝑘 is a circle centered at the pole
of radius |𝑘|. Note that 𝑟 = 𝑘 and 𝑟 = −𝑘 represent the same curve.
12
Polar Equations and Polar Curves
𝜋
Sketch the graph of 𝜃 = .
4
In general, the graph of the equation 𝜃 = 𝑘 is a line passing through the
pole making an angle 𝑘 with the polar axis. Also, its Cartesian form is
𝑦 = tan 𝑘 𝑥, when non-vertical, or 𝑥 = 0, when vertical.
13
Polar Equations and Polar Curves
Graph of 𝑟 = 4 cos 𝜃 Graph of 𝑟 = −5 sin 𝜃.
In general, the graph of the equation 𝑟 = 2𝑎 cos 𝜃 is a circle centered at
(𝑎, 0) with radius |𝑎|. The graph of the equation 𝑟 = 2𝑏 sin 𝜃 is a circle
centered at (0, 𝑏) with radius |𝑏|.
Note: Both circles are traced by 𝜃 ∈ 0, 𝜋 and contains the pole.
14
Your Turn
A. Sketch the graph of the following polar equations.
𝜋
1. 𝜃 = 3. 𝑟 = −4 cos 𝜃 5. 𝑟 = −2
3
2. 𝑟 = 3 4. 𝑟 = 6 sin 𝜃 6. 𝑟 = −5 sin 𝜃
B. Find the polar equivalent of the following.
1. 𝑥 = 2 3. 𝑥 2 + (𝑦 − 3)2 = 9
2. 𝑥𝑦 = 1 4. 𝑥 = 2 cos 𝑡 , 𝑦 = 2 sin 𝑡 , 𝑡 ∈ ℝ
C. Find the Cartesian equation of the following.
4
1. 𝑟 2 = 4𝑟 cos 𝜃 2. 2 cos 𝜃−sin 𝜃
15
Symmetry in the Polar Plane
Symmetry about the polar axis 𝜽 = 𝟎
A polar curve is symmetric about the polar axis (or x-axis)
whenever 𝑟, 𝜃 , in its equation, is replaced by 𝑟, −𝜃 or by
−𝑟, 𝜋 − 𝜃 , and an equivalent equation is obtained.
16
Symmetry in the Polar Plane
Symmetry about the pole
A polar curve is symmetric about the pole whenever 𝑟, 𝜃 , in its
equation, is replaced by −𝑟, 𝜃 or by 𝑟, 𝜋 + 𝜃 , and an equivalent
equation is obtained.
17
Symmetry in the Polar Plane
𝝅
Symmetry about 𝜽 =
𝟐
𝜋
A polar curve is symmetric about the line 𝜃 = 2 (or y-axis)
whenever 𝑟, 𝜃 , in its equation, is replaced by −𝑟, −𝜃 or by
𝑟, 𝜋 − 𝜃 , and an equivalent equation is obtained.
18
Special Curves
Limacons
Limacons are curves whose equations are of the form
• 𝑟 = 𝑎 ± 𝑏 cos 𝜃
• 𝑟 = 𝑎 ± 𝑏 sin 𝜃
where 𝑎, 𝑏 > 0
Testing for symmetry:
• 𝑟 = 𝑎 ± 𝑏 cos 𝜃
• 𝑟 = 𝑎 ± 𝑏 cos −𝜃 ⟹ 𝑟 = 𝑎 ± 𝑏 cos 𝜃
• Thus, symmetric with respect to the polar axis
• 𝑟 = 𝑎 ± 𝑏 sin 𝜃
• 𝑟 = 𝑎 ± 𝑏 sin 𝜋 − 𝜃 ⟹ 𝑟 = 𝑎 ± 𝑏 sin 𝜃
𝜋
• Thus, symmetric with respect to the line 𝜃 = 2
19
Limacons
Sketch the graph of 𝑟 = 1 + 2 cos 𝜃
The graph is called a limacon with a loop.
𝑎 𝑎 1
The type of limacon depends on the ratio 𝑏
. Here, 𝑏
= 2
.
20
Limacons
Sketch the graph of 𝑟 = 1 + cos 𝜃
𝑎
The graph is called a cardiod. Note that = 1.
𝑏 21
Limacons
Sketch the graph of 𝑟 = 3 + 2 cos 𝜃
𝑎 3
The graph is called a limacon with a dent. Note that = .
𝑏 2 22
Limacons
Sketch the graph of 𝑟 = 2 + cos 𝜃
𝑎
The graph is called a convex limacon. Note that = 2.
𝑏 23
Limacons
Type of Limacons
In general, for 𝑟 = 𝑎 ± 𝑏 cos 𝜃 or 𝑟 = 𝑎 ± 𝑏 sin 𝜃, with 𝑎, 𝑏 > 0, the
polar curve will be
𝑎
i. 0 < 𝑏 < 1 (limacon with a loop)
𝑎
ii. = 1 (cardiod)
𝑏
𝑎
iii. 1 < 𝑏 < 2 (limacon with a dent)
𝑎
iv. 𝑏
> 2 (convex limacon)
Remarks
• The graph of 𝑟 = −𝑎 ± 𝑏 cos 𝜃 is the same as that of 𝑟 = −𝑎 ± 𝑏 cos 𝜃.
• The graph of 𝑟 = −𝑎 ± 𝑏 sin 𝜃 is the same as that of 𝑟 = −𝑎 ± 𝑏 sin 𝜃.
24
Limacons
The graph of 𝑟 = 𝑎 ± 𝑏 cos 𝜃 is a limacon oriented horizontally
(symmetric with respect to the polar axis)
25
Limacons
The graph of 𝑟 = 𝑎 ± 𝑏 sin 𝜃 is a limacon oriented vertically
𝜋
(symmetric with respect to the line 𝜃 = 2 )
26
Special Curves
Roses
Roses are curves whose equations are of the form
• 𝑟 = 𝑎 cos 𝑛𝜃
• 𝑟 = 𝑎 sin 𝑛𝜃
where 𝑎 > 0, 𝑛 ∈ ℕ
Testing for symmetry:
• 𝑟 = 𝑎 cos 𝑛𝜃
• 𝑟 = 𝑎 cos −𝑛𝜃 ⟹ 𝑟 = 𝑎 cos 𝑛𝜃
• Thus, symmetric with respect to the polar axis
𝜋
• Additionally, symmetric with respect to 𝜃 = 2 if 𝑛 is even
• 𝑟 = 𝑎 sin 𝑛𝜃
• −𝑟 = 𝑎 sin −𝑛𝜃 ⟹ −𝑟 = −𝑎 sin 𝑛𝜃 ⟹ 𝑟 = 𝑎 sin 𝑛𝜃
𝜋
• Thus, symmetric with respect to the line 𝜃 = 2
• Additionally, symmetric with respect to the polar axis if 𝑛 is even
27
Roses
Sketch the graph of 𝑟 = 2 cos 2𝜃.
The graph is a rose with 4 petals and length 2. In fact, the
number of petals is 2𝑛 if 𝑛 is even. And it’s 𝑛 if 𝑛 is odd. 28
Roses
Sketch the graph of 𝑟 = 2 sin 3𝜃.
The graph is a rose with 3 petals and length 2. Here, 𝑛 = 3 is
odd. Hence, 𝑛 = 3 is the number of petals. 29
Roses
30
Roses
31
Interesting curves
The graph of 𝑟 2 = 6 cos 2𝜃 is a lemniscate.
32
Interesting curves
The graph of 𝑟 = 𝜃, 𝜃 > 0 is the Archimedean spiral.
33
Interesting curves
8𝜃
𝑟 = 1 + 4 cos 5𝜃 𝑟 = sin 5
34
Interesting curves
𝑟 = 𝑒 sin 𝜃 − 2 cos 4𝜃 𝑟 = sin2 2.4𝜃 + cos 4 2𝜃
35
Interesting curves
𝜃
𝑟 = 𝑒 cos 𝜃 − 2 cos 4𝜃 + sin3 𝑟 = sin2 1.2𝜃 + cos 6 6𝜃
3
36
Interesting curves
The “cannabis” curve
9 1 9 1
𝑟 = 1 + cos 8𝜃 1 + cos 24𝜃 + cos 200𝜃 1 + sin 𝜃
10 10 10 10
37
Your Turn
A. Determine the type of limacon the following equations
pertain to.
1. 𝑟 = 3 − 2 cos 𝜃 4. 𝑟 = 3 sin 𝜃
2. 𝑟 = 1 + 4 sin 𝜃 5. 𝑟 = 3 + 4 cos 𝜃
3. 𝑟 = 2 − 2 cos 𝜃 6. 𝑟 = −1 + cos 𝜃
B. Sketch the graph of the following curves.
1. 𝑟 = 4 cos 𝜃 5. 𝑟 = 3 sin 3𝜃
𝜋
2. 𝜃 = 6. 𝑟 = 1 − 2 sin 𝜃
3
3. 𝑟 = 1 − sin 𝜃 7. 𝑟 = −3 cos 2𝜃
4. 𝑟 = 2 cos 2𝜃 8. 𝑟 = sin 4𝜃
38
References
◼ Stewart J. (2012). Calculus Early Transcendentals, 7th
Edition, Brooks/Cole Cengage Learning
◼ Leithold, L. (1996). The Calculus 7. Addison-Wesley
Publishing Company
◼ PSHS System Math 6 Student Learning Guides
◼ UP Diliman, Math 54 Lecture Slides
(https://sites.google.com/site/updmathsite/math-54-
summer-2014/lecture-slides)
39
40
41
