10. polar coordinates x

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

Polar Coordinates
r = the distance between P and the origin O(0, 0)
P
x
(r, )
r
O
y

Polar Coordinates
 = a signed angle between the positive x–axis and
the direction to P,
P
x
(r, )

r
O
y

Polar Coordinates
the direction to P, specifically,
 is + for counter clockwise measurements and
 is – for clockwise measurements.
P
x
(r, )

r
O
y

Polar Coordinates
The ordered pair (r, ) is a polar coordinate of P.
P
x
(r, )

r
O
y

Polar Coordinates
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

Polar Coordinates
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

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.

Polar Coordinates
Conversion Rules

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 =
(r, )p ↔ (x, y)R

Polar Coordinates
Conversion Rules
P
x
y

r
O x = r*cos()
x = r*cos()
y =
r =
(r, )p ↔ (x, y)R

Polar Coordinates
Conversion Rules
P
x
y

r
O x = r*cos()
x = r*cos()
y = r*sin()
y = r*sin()
r =
(r, )p ↔ (x, y)R

Polar Coordinates
Conversion Rules
P
x
y

r
O x = r*cos()
y = r*sin()
x = r*cos()
y = r*sin()
r = √ x2 + y2
(r, )p ↔ (x, y)R

Polar Coordinates
Conversion Rules
P
x
y

r
O x = r*cos()
y = r*sin()
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).
(r, )p ↔ (x, y)R

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.

Polar Coordinates
x
y
x = r*cos()
y = r*sin()
r2 = x2 + y2
tan() = y/x
For A(4, 60o)P

Polar 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

Polar Coordinates
For A(4, 60o)P
x = r*cos()
y = r*sin()
(x, y)R = (4*cos(60⁰), 4*sin(60⁰)),
r2 = x2 + y2
tan() = y/x
x
y
60o
4
A(4, 60o)P

Polar 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

Polar 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)

Polar 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)

Polar 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)

Polar 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)

Polar 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)

Polar 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)

Polar 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)

Polar 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)

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.

Polar Coordinates
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
x
y
E(–4, 3)

Polar Coordinates
(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)

Polar Coordinates
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
hence for E, r = 16 + 9 = 5.
x
y
E(–4, 3)
r=5

Polar Coordinates
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
There is no single formula that
would give .
hence for E, r = 16 + 9 = 5.
x
y
E(–4, 3)
r=5


Polar Coordinates
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
would give . This is because  has
to be expressed via the inverse
trig–functions hence the position of
E dictates which inverse function
hence for E, r = 16 + 9 = 5.
would be easier to use to extract .
x
y
E(–4, 3)
r=5


Polar Coordinates
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
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


Polar Coordinates
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
hence for E, r = 16 + 9 = 5.
x
y
E(–4, 3)

r=5
cosine inverse function (why?). So  = cos–1(–4/5) ≈
143o

Polar Coordinates
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
hence for E, r = 16 + 9 = 5.
x
y
E(–4, 3)

r=5
143o so that E(–4, 3)R ≈ (5,143o)P

Polar Coordinates
(with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
hence for E, r = 16 + 9 = 5.
x
y
E(–4, 3)

r=5
143o so that E(–4, 3)R ≈ (5,143o)P = (5,143o±n*360o)P

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

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

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

Polar Coordinates
For F(3, –2)R, r = 9 + 4 = √13.
x
y
F(3, –2,)
function. The tangent inverse has
the advantage of obtaining the answer directly from
the x and y coordinates.

r=√13

Polar Coordinates
For F(3, –2)R, r = 9 + 4 = √13.
x
y
F(3, –2,)

r=√13
the x and y coordinates. So  = tan–1(–2/3) ≈ –0.588rad

Polar Coordinates
For F(3, –2)R, r = 9 + 4 = √13.
x
y
F(3, –2,)

r=√13
= (√13, –0.588rad ± 2nπ)P
and that F(3, –2)R ≈ (√13, –0.588rad)P

Polar Coordinates
For F(3, –2)R, r = 9 + 4 = √13.
x
y
F(3, –2,)

r=√13
= (√13, –0.588rad ± 2nπ)P
For G(–3, –1)R, r = 9 + 1 = √10.
x
y
G(–3, –1)
r=√10
and that F(3, –2)R ≈ (√13, –0.588rad)P

Polar Coordinates
For F(3, –2)R, r = 9 + 4 = √13.
x
y
F(3, –2,)

r=√13
= (√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
and that F(3, –2)R ≈ (√13, –0.588rad)P

Polar Coordinates
For F(3, –2)R, r = 9 + 4 = √13.
x
y
F(3, –2,)

r=√13
= (√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
and that F(3, –2)R ≈ (√13, –0.588rad)P

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

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

Polar Coordinates
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


10. polar coordinates x

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