The document describes polar coordinates. Polar coordinates represent the location of a point P in a plane using two numbers: r, the distance from P to the origin O, and θ, the angle between the positive x-axis and a line from O to P. θ is positive for counter-clockwise angles and negative for clockwise angles. The polar coordinate (r, θ) uniquely identifies the point P. Conversions between polar coordinates (r, θ) and rectangular coordinates (x, y) are given by the equations x=r*cos(θ), y=r*sin(θ), and r=√(x2+y2).

1. Polar Coordinates

2. Polar Coordinates
The location of a point P in the plane may be given
by the following two numbers:
P
x
y
(r, )
O

3. Polar Coordinates
r = the distance between P and the origin O(0, 0)
The location of a point P in the plane may be given
by the following two numbers:
P
x
(r, )
r
O
y

4. Polar Coordinates
r = the distance between P and the origin O(0, 0)
 = a signed angle between the positive x–axis and
the direction to P,
The location of a point P in the plane may be given
by the following two numbers:
P
x
(r, )

r
O
y

5. Polar Coordinates
r = the distance between P and the origin O(0, 0)
 = a signed angle between the positive x–axis and
the direction to P, specifically,
 is + for counter clockwise measurements and
 is – for clockwise measurements.
The location of a point P in the plane may be given
by the following two numbers:
P
x
(r, )

r
O
y

6. Polar Coordinates
r = the distance between P and the origin O(0, 0)
 = a signed angle between the positive x–axis and
the direction to P, specifically,
 is + for counter clockwise measurements and
 is – for clockwise measurements.
The location of a point P in the plane may be given
by the following two numbers:
The ordered pair (r, ) is a polar coordinate of P.
P
x
(r, )

r
O
y

7. Polar Coordinates
r = the distance between P and the origin O(0, 0)
 = a signed angle between the positive x–axis and
the direction to P, specifically,
 is + for counter clockwise measurements and
 is – for clockwise measurements.
The location of a point P in the plane may be given
by the following two numbers:
The ordered pair (r, ) is a polar coordinate of P.
P
x
(r, )

r
The ordered pairs (r,  ±2nπ ) with
n = 0,1, 2, 3… give the same
geometric information hence lead to
the same location P(r, ).
O
y

8. Polar Coordinates
r = the distance between P and the origin O(0, 0)
 = a signed angle between the positive x–axis and
the direction to P, specifically,
 is + for counter clockwise measurements and
 is – for clockwise measurements.
The location of a point P in the plane may be given
by the following two numbers:
The ordered pair (r, ) is a polar coordinate of P.
P
x
(r, )

r
The ordered pairs (r,  ±2nπ ) with
n = 0,1, 2, 3… give the same
geometric information hence lead to
the same location P(r, ).
We also use signed distance,
so with negative values of r,
we are to step backward for
a distance of l r l. O
y

9. Polar Coordinates
If needed, we write (a, b)P for a polar coordinate
ordered pair, and (a, b)R for the rectangular
coordinate ordered pair.

10. Polar Coordinates
Conversion Rules
If needed, we write (a, b)P for a polar coordinate
ordered pair, and (a, b)R for the rectangular
coordinate ordered pair.

11. Polar Coordinates
Conversion Rules
Let (x, y)R and (r, )P be the rectangular and polar
coordinates of the same point P, then
P
x
y

r
O
The rectangular and polar
coordinates relations
x =
y =
r =
If needed, we write (a, b)P for a polar coordinate
ordered pair, and (a, b)R for the rectangular
coordinate ordered pair.
(r, )p ↔ (x, y)R

12. Polar Coordinates
Conversion Rules
Let (x, y)R and (r, )P be the rectangular and polar
coordinates of the same point P, then
P
x
y

r
O x = r*cos()
The rectangular and polar
coordinates relations
x = r*cos()
y =
r =
If needed, we write (a, b)P for a polar coordinate
ordered pair, and (a, b)R for the rectangular
coordinate ordered pair.
(r, )p ↔ (x, y)R

13. Polar Coordinates
Conversion Rules
Let (x, y)R and (r, )P be the rectangular and polar
coordinates of the same point P, then
P
x
y

r
O x = r*cos()
The rectangular and polar
coordinates relations
x = r*cos()
y = r*sin()
y = r*sin()
r =
If needed, we write (a, b)P for a polar coordinate
ordered pair, and (a, b)R for the rectangular
coordinate ordered pair.
(r, )p ↔ (x, y)R

14. Polar Coordinates
Conversion Rules
Let (x, y)R and (r, )P be the rectangular and polar
coordinates of the same point P, then
P
x
y

r
O x = r*cos()
y = r*sin()
The rectangular and polar
coordinates relations
x = r*cos()
y = r*sin()
r = √ x2 + y2
If needed, we write (a, b)P for a polar coordinate
ordered pair, and (a, b)R for the rectangular
coordinate ordered pair.
(r, )p ↔ (x, y)R

15. Polar Coordinates
Conversion Rules
Let (x, y)R and (r, )P be the rectangular and polar
coordinates of the same point P, then
P
x
y

r
O x = r*cos()
y = r*sin()
The rectangular and polar
coordinates relations
x = r*cos()
y = r*sin()
r = √ x2 + y2
For  we have that
tan() = y/x,
cos() = x/√x2 + y2
or if  is between 0 and π,
then  = cos–1 (x/√x2 + y2).
If needed, we write (a, b)P for a polar coordinate
ordered pair, and (a, b)R for the rectangular
coordinate ordered pair.
(r, )p ↔ (x, y)R

16. Polar Coordinates
Conversion Rules
Let (x, y)R and (r, )P be the rectangular and polar
coordinates of the same point P, then
P
x
y

r
O x = r*cos()
y = r*sin()
The rectangular and polar
coordinates relations
x = r*cos()
y = r*sin()
r = √ x2 + y2
For  we have that
tan() = y/x,
cos() = x/√x2 + y2
or if  is between 0 and π,
then  = cos–1 (x/√x2 + y2).
If needed, we write (a, b)P for a polar coordinate
ordered pair, and (a, b)R for the rectangular
coordinate ordered pair.
(r, )p ↔ (x, y)R

17. Polar Coordinates
Example A. a. Plot the following polar coordinates
A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P.
Find their corresponding rectangular coordinates.

18. Polar Coordinates
Example A. a. Plot the following polar coordinates
A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P.
Find their corresponding rectangular coordinates.
x
y
x = r*cos()
y = r*sin()
r2 = x2 + y2
tan() = y/x
For A(4, 60o)P

19. Polar Coordinates
Example A. a. Plot the following polar coordinates
A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P.
Find their corresponding rectangular coordinates.
x
y
60o
4
x = r*cos()
y = r*sin()
A(4, 60o)P
r2 = x2 + y2
tan() = y/x
For A(4, 60o)P

20. Polar Coordinates
Example A. a. Plot the following polar coordinates
A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P.
Find their corresponding rectangular coordinates.
For A(4, 60o)P
x = r*cos()
y = r*sin()
(x, y)R = (4*cos(60⁰), 4*sin(60⁰)),
= (2, 23)
r2 = x2 + y2
tan() = y/x
x
y
60o
4
A(4, 60o)P

21. Polar Coordinates
Example A. a. Plot the following polar coordinates
A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P.
Find their corresponding rectangular coordinates.
For A(4, 60o)P
x = r*cos()
y = r*sin()
for B(5, 0o)P, (x, y) = (5, 0),
r2 = x2 + y2
tan() = y/x
x
y
60o
4
A(4, 60o)P
B(5, 0)P
(x, y)R = (4*cos(60⁰), 4*sin(60⁰)),
= (2, 23)

22. Polar Coordinates
Example A. a. Plot the following polar coordinates
A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P.
Find their corresponding rectangular coordinates.
x
y
60o
4
For A(4, 60o)P
x = r*cos()
y = r*sin()
for B(5, 0o)P, (x, y) = (5, 0),
for C and D,
A(4, 60o)P
B(5, 0)P
4
r2 = x2 + y2
tan() = y/x
–45o
C
(x, y)R = (4*cos(60⁰), 4*sin(60⁰)),
= (2, 23)

23. Polar Coordinates
Example A. a. Plot the following polar coordinates
A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P.
Find their corresponding rectangular coordinates.
x
y
60o
4
For A(4, 60o)P
x = r*cos()
y = r*sin()
for B(5, 0o)P, (x, y) = (5, 0),
for C and D,
A(4, 60o)P
B(5, 0)P
4
r2 = x2 + y2
tan() = y/x
–45o
3π/4
C
(x, y)R = (4*cos(60⁰), 4*sin(60⁰)),
= (2, 23)

24. Polar Coordinates
Example A. a. Plot the following polar coordinates
A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P.
Find their corresponding rectangular coordinates.
x
y
60o
4
For A(4, 60o)P
x = r*cos()
y = r*sin()
for B(5, 0o)P, (x, y) = (5, 0),
for C and D,
A(4, 60o)P
B(5, 0)P
4
r2 = x2 + y2
tan() = y/x
–45o
3π/4
C&D
(x, y)R = (4*cos(60⁰), 4*sin(60⁰)),
= (2, 23)

25. Polar Coordinates
Example A. a. Plot the following polar coordinates
A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P.
Find their corresponding rectangular coordinates.
x
y
60o
4
For A(4, 60o)P
x = r*cos()
y = r*sin()
for B(5, 0o)P, (x, y) = (5, 0),
for C and D,
A(4, 60o)P
B(5, 0)P
C(4, –45o)P
= D(–4, 3π/4 rad)P
4
r2 = x2 + y2
tan() = y/x
–45o
3π/4
C&D
(x, y)R = (4*cos(60⁰), 4*sin(60⁰)),
= (2, 23)

26. Polar Coordinates
Example A. a. Plot the following polar coordinates
A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P.
Find their corresponding rectangular coordinates.
x
y
60o
4
For A(4, 60o)P
x = r*cos()
y = r*sin()
for B(5, 0o)P, (x, y) = (5, 0),
for C and D,
(x, y)R = (4cos(–45⁰), 4sin(–45⁰))
= (–4cos(3π/4), –4sin(3π/4))
A(4, 60o)P
B(5, 0)P
C(4, –45o)P
= D(–4, 3π/4 rad)P
4
r2 = x2 + y2
tan() = y/x
–45o
3π/4
C&D
(x, y)R = (4*cos(60⁰), 4*sin(60⁰)),
= (2, 23)

27. Polar Coordinates
Example A. a. Plot the following polar coordinates
A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P.
Find their corresponding rectangular coordinates.
x
y
60o
4
For A(4, 60o)P
x = r*cos()
y = r*sin()
for B(5, 0o)P, (x, y) = (5, 0),
for C and D,
(x, y)R = (4cos(–45⁰), 4sin(–45⁰))
= (–4cos(3π/4), –4sin(3π/4))
= (22, –22)
A(4, 60o)P
B(5, 0)P
C(4, –45o)P
= D(–4, 3π/4 rad)P
4
r2 = x2 + y2
tan() = y/x
–45o
3π/4
C&D
(x, y)R = (4*cos(60⁰), 4*sin(60⁰)),
= (2, 23)

28. Polar Coordinates
Example A. a. Plot the following polar coordinates
A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P.
Find their corresponding rectangular coordinates.
x
y
60o
4
For A(4, 60o)P
x = r*cos()
y = r*sin()
for B(5, 0o)P, (x, y) = (5, 0),
for C and D,
(x, y)R = (4cos(–45⁰), 4sin(–45⁰))
= (–4cos(3π/4), –4sin(3π/4))
= (22, –22)
A(4, 60o)P
B(5, 0)P
C(4, –45o)P
= D(–4, 3π/4 rad)P
4
Converting rectangular positions
into polar coordinates requires
more care.
r2 = x2 + y2
tan() = y/x
–45o
3π/4
C&D
(x, y)R = (4*cos(60⁰), 4*sin(60⁰)),
= (2, 23)

29. Polar Coordinates
b. Find a polar coordinate then list all possible polar
coordinates for each of the following points
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.

30. Polar Coordinates
b. Find a polar coordinate then list all possible polar
coordinates for each of the following points
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
x
y
E(–4, 3)

31. Polar Coordinates
b. Find a polar coordinate then list all possible polar
coordinates for each of the following points
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
We have the distance formula r = x2 + y2,
x
y
E(–4, 3)

32. Polar Coordinates
b. Find a polar coordinate then list all possible polar
coordinates for each of the following points
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
We have the distance formula r = x2 + y2,
hence for E, r = 16 + 9 = 5.
x
y
E(–4, 3)
r=5

33. Polar Coordinates
b. Find a polar coordinate then list all possible polar
coordinates for each of the following points
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
There is no single formula that
would give .
We have the distance formula r = x2 + y2,
hence for E, r = 16 + 9 = 5.
x
y
E(–4, 3)
r=5


34. Polar Coordinates
b. Find a polar coordinate then list all possible polar
coordinates for each of the following points
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
There is no single formula that
would give . This is because  has
to be expressed via the inverse
trig–functions hence the position of
E dictates which inverse function
We have the distance formula r = x2 + y2,
hence for E, r = 16 + 9 = 5.
would be easier to use to extract .
x
y
E(–4, 3)
r=5


35. Polar Coordinates
b. Find a polar coordinate then list all possible polar
coordinates for each of the following points
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
We have the distance formula r = x2 + y2,
hence for E, r = 16 + 9 = 5.
would be easier to use to extract . Since E is in the
2nd quadrant, the angle  may be recovered by the
cosine inverse function (why?).
x
y
E(–4, 3)
r=5
There is no single formula that
would give . This is because  has
to be expressed via the inverse
trig–functions hence the position of
E dictates which inverse function


36. Polar Coordinates
b. Find a polar coordinate then list all possible polar
coordinates for each of the following points
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
We have the distance formula r = x2 + y2,
hence for E, r = 16 + 9 = 5.
x
y
E(–4, 3)

r=5
would be easier to use to extract . Since E is in the
2nd quadrant, the angle  may be recovered by the
cosine inverse function (why?). So  = cos–1(–4/5) ≈
143o
There is no single formula that
would give . This is because  has
to be expressed via the inverse
trig–functions hence the position of
E dictates which inverse function

37. Polar Coordinates
b. Find a polar coordinate then list all possible polar
coordinates for each of the following points
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
We have the distance formula r = x2 + y2,
hence for E, r = 16 + 9 = 5.
x
y
E(–4, 3)

r=5
would be easier to use to extract . Since E is in the
2nd quadrant, the angle  may be recovered by the
cosine inverse function (why?). So  = cos–1(–4/5) ≈
143o so that E(–4, 3)R ≈ (5,143o)P
There is no single formula that
would give . This is because  has
to be expressed via the inverse
trig–functions hence the position of
E dictates which inverse function

38. Polar Coordinates
b. Find a polar coordinate then list all possible polar
coordinates for each of the following points
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
We have the distance formula r = x2 + y2,
hence for E, r = 16 + 9 = 5.
x
y
E(–4, 3)

r=5
would be easier to use to extract . Since E is in the
2nd quadrant, the angle  may be recovered by the
cosine inverse function (why?). So  = cos–1(–4/5) ≈
143o so that E(–4, 3)R ≈ (5,143o)P = (5,143o±n*360o)P
There is no single formula that
would give . This is because  has
to be expressed via the inverse
trig–functions hence the position of
E dictates which inverse function

39. Polar Coordinates
For F(3, –2)R,
x
y
F(3, –2,)

40. Polar Coordinates
For F(3, –2)R, r = 9 + 4 = √13.
x
y
F(3, –2,)
r=√13

41. Polar Coordinates
For F(3, –2)R, r = 9 + 4 = √13.
x
y
F(3, –2,)
Since F is in the 4th quadrant, the
angle  may be recovered by the
sine inverse or the tangent inverse
function.

r=√13

42. Polar Coordinates
For F(3, –2)R, r = 9 + 4 = √13.
x
y
F(3, –2,)
Since F is in the 4th quadrant, the
angle  may be recovered by the
sine inverse or the tangent inverse
function. The tangent inverse has
the advantage of obtaining the answer directly from
the x and y coordinates.

r=√13

43. Polar Coordinates
For F(3, –2)R, r = 9 + 4 = √13.
x
y
F(3, –2,)

r=√13
Since F is in the 4th quadrant, the
angle  may be recovered by the
sine inverse or the tangent inverse
function. The tangent inverse has
the advantage of obtaining the answer directly from
the x and y coordinates. So  = tan–1(–2/3) ≈ –0.588rad

44. Polar Coordinates
For F(3, –2)R, r = 9 + 4 = √13.
x
y
F(3, –2,)

r=√13
Since F is in the 4th quadrant, the
angle  may be recovered by the
sine inverse or the tangent inverse
function. The tangent inverse has
= (√13, –0.588rad ± 2nπ)P
the advantage of obtaining the answer directly from
the x and y coordinates. So  = tan–1(–2/3) ≈ –0.588rad
and that F(3, –2)R ≈ (√13, –0.588rad)P

45. Polar Coordinates
For F(3, –2)R, r = 9 + 4 = √13.
x
y
F(3, –2,)

r=√13
Since F is in the 4th quadrant, the
angle  may be recovered by the
sine inverse or the tangent inverse
function. The tangent inverse has
= (√13, –0.588rad ± 2nπ)P
For G(–3, –1)R, r = 9 + 1 = √10.
x
y
G(–3, –1)
r=√10
the advantage of obtaining the answer directly from
the x and y coordinates. So  = tan–1(–2/3) ≈ –0.588rad
and that F(3, –2)R ≈ (√13, –0.588rad)P

46. Polar Coordinates
For F(3, –2)R, r = 9 + 4 = √13.
x
y
F(3, –2,)

r=√13
Since F is in the 4th quadrant, the
angle  may be recovered by the
sine inverse or the tangent inverse
function. The tangent inverse has
= (√13, –0.588rad ± 2nπ)P
For G(–3, –1)R, r = 9 + 1 = √10.
G is the 3rd quadrant. Hence  can’t
be obtained directly via the inverse–
trig functions.
x
y
G(–3, –1)
r=√10
the advantage of obtaining the answer directly from
the x and y coordinates. So  = tan–1(–2/3) ≈ –0.588rad
and that F(3, –2)R ≈ (√13, –0.588rad)P

47. Polar Coordinates
For F(3, –2)R, r = 9 + 4 = √13.
x
y
F(3, –2,)

r=√13
Since F is in the 4th quadrant, the
angle  may be recovered by the
sine inverse or the tangent inverse
function. The tangent inverse has
= (√13, –0.588rad ± 2nπ)P
For G(–3, –1)R, r = 9 + 1 = √10.
G is the 3rd quadrant. Hence  can’t
be obtained directly via the inverse–
trig functions. We will find the angle
A as shown first, then  = A + 180⁰.
x
y
G(–3, –1)
r=√10
A
the advantage of obtaining the answer directly from
the x and y coordinates. So  = tan–1(–2/3) ≈ –0.588rad
and that F(3, –2)R ≈ (√13, –0.588rad)P

48. Polar Coordinates
Again, using tangent inverse
A = tan–1(1/3) ≈ 18.3o
x
y
G(–3, –1)
r=√10
A

49. Polar Coordinates
Again, using tangent inverse
A = tan–1(1/3) ≈ 18.3o so
 = 180 + 18.3o = 198.3o
x
y
G(–3, –1)
r=√10
A


50. Polar Coordinates
Again, using tangent inverse
A = tan–1(1/3) ≈ 18.3o so
 = 180 + 18.3o = 198.3o or
G ≈ (√10, 198.3o ± n x 360o)P
x
y
G(–3, –1)
r=√10
A
