2. Polar Equations x = r*cos(θ)
We start with an example about lines y = r*sin(θ)
and we list the conversion rules for r = √ x2 + y2
easy references. tan(θ) = y/x
3. Polar Equations x = r*cos(θ)
We start with an example about lines y = r*sin(θ)
and we list the conversion rules for r = √ x2 + y2
easy references. tan(θ) = y/x
Example A.
a. Convert the linear equation y = 3x + 4 to the polar
form r = f(θ).
4. Polar Equations x = r*cos(θ)
We start with an example about lines y = r*sin(θ)
and we list the conversion rules for r = √ x2 + y2
easy references. tan(θ) = y/x
Example A.
a. Convert the linear equation y = 3x + 4 to the polar
form r = f(θ).
Replacing y with r*sin(θ) and x with r*cos(θ), we get
r*sin(θ) = 3r*cos(θ) + 4
5. Polar Equations x = r*cos(θ)
We start with an example about lines y = r*sin(θ)
and we list the conversion rules for r = √ x2 + y2
easy references. tan(θ) = y/x
Example A.
a. Convert the linear equation y = 3x + 4 to the polar
form r = f(θ).
Replacing y with r*sin(θ) and x with r*cos(θ), we get
r*sin(θ) = 3r*cos(θ) + 4
Solving the r in terms of θ we have that
r = 4
1*sin(θ) – 3*cos(θ)
6. Polar Equations x = r*cos(θ)
We start with an example about lines y = r*sin(θ)
and we list the conversion rules for r = √ x2 + y2
easy references. tan(θ) = y/x
Example A.
a. Convert the linear equation y = 3x + 4 to the polar
form r = f(θ).
Replacing y with r*sin(θ) and x with r*cos(θ), we get
r*sin(θ) = 3r*cos(θ) + 4
Solving the r in terms of θ we have that
r = 4
1*sin(θ) – 3*cos(θ)
Similarly, given a line Ax + By = C its polar form is
r = C
A*cos(θ) + B*sin(θ)
7. Polar Equations x = r*cos(θ)
b. Graph the polar equation y = r*sin(θ)
4 r = √ x2 + y2
r = tan(θ) = y/x
–3sin(θ) – 2cos(θ)
8. Polar Equations x = r*cos(θ)
b. Graph the polar equation y = r*sin(θ)
4 r = √ x2 + y2
r = tan(θ) = y/x
–3sin(θ) – 2cos(θ)
By its form, we know the graph is a straight line.
9. Polar Equations x = r*cos(θ)
b. Graph the polar equation y = r*sin(θ)
4 r = √ x2 + y2
r = tan(θ) = y/x
–3sin(θ) – 2cos(θ)
By its form, we know the graph is a straight line.
The x intercept is at θ = 0 with r(0) = –2
10. Polar Equations x = r*cos(θ)
b. Graph the polar equation y = r*sin(θ)
4 r = √ x2 + y2
r = tan(θ) = y/x
–3sin(θ) – 2cos(θ)
By its form, we know the graph is a straight line.
The x intercept is at θ = 0 with r(0) = –2 and at θ = π/2
the y intercept is r(π/2) = –4/3.
11. Polar Equations x = r*cos(θ)
b. Graph the polar equation y = r*sin(θ)
4 r = √ x2 + y2
r = tan(θ) = y/x
–3sin(θ) – 2cos(θ)
By its form, we know the graph is a straight line.
The x intercept is at θ = 0 with r(0) = –2 and at θ = π/2
the y intercept is r(π/2) = –4/3. Its graph is as shown.
(–2, 0)R
(0,–4/3,)R
12. Polar Equations x = r*cos(θ)
b. Graph the polar equation y = r*sin(θ)
4 r = √ x2 + y2
r = tan(θ) = y/x
–3sin(θ) – 2cos(θ)
What is the slope of this line?
By its form, we know the graph is a straight line.
The x intercept is at θ = 0 with r(0) = –2 and at θ = π/2
the y intercept is r(π/2) = –4/3. Its graph is as shown.
(–2, 0)R
(0,–4/3,)R
13. Polar Equations x = r*cos(θ)
b. Graph the polar equation y = r*sin(θ)
4 r = √ x2 + y2
r = tan(θ) = y/x
–3sin(θ) – 2cos(θ)
What is the slope of this line?
By its form, we know the graph is a straight line.
The x intercept is at θ = 0 with r(0) = –2 and at θ = π/2
the y intercept is r(π/2) = –4/3. Its graph is as shown.
The slope in question is
(–2, 0)
–4/3 –2
R
m= 2 = 3
(0,–4/3,)R
14. Polar Equations x = r*cos(θ)
b. Graph the polar equation y = r*sin(θ)
4 r = √ x2 + y2
r = tan(θ) = y/x
–3sin(θ) – 2cos(θ)
What is the slope of this line?
Convert it to an x&y equation.
By its form, we know the graph is a straight line.
The x intercept is at θ = 0 with r(0) = –2 and at θ = π/2
the y intercept is r(π/2) = –4/3. Its graph is as shown.
The slope in question is
(–2, 0)
–4/3 –2
R
m= 2 = 3
(0,–4/3,)R
15. Polar Equations x = r*cos(θ)
b. Graph the polar equation y = r*sin(θ)
4 r = √ x2 + y2
r = tan(θ) = y/x
–3sin(θ) – 2cos(θ)
What is the slope of this line?
Convert it to an x&y equation.
By its form, we know the graph is a straight line.
The x intercept is at θ = 0 with r(0) = –2 and at θ = π/2
the y intercept is r(π/2) = –4/3. Its graph is as shown.
The slope in question is
(–2, 0)
–4/3 –2
R
m= 2 = 3
Clear the denominator of the (0,–4/3,)
R
polar equation, we have that
–3r*sin(θ) – 2r*cos(θ) = 4
16. Polar Equations x = r*cos(θ)
b. Graph the polar equation y = r*sin(θ)
4 r = √ x2 + y2
r = tan(θ) = y/x
–3sin(θ) – 2cos(θ)
What is the slope of this line?
Convert it to an x&y equation.
By its form, we know the graph is a straight line.
The x intercept is at θ = 0 with r(0) = –2 and at θ = π/2
the y intercept is r(π/2) = –4/3. Its graph is as shown.
The slope in question is
(–2, 0)
–4/3 –2
R
m= 2 = 3
Clear the denominator of the (0,–4/3,)
R
polar equation, we have that
–3r*sin(θ) – 2r*cos(θ) = 4 or –3y – 2x = 4.
18. Polar Equations
Polar Equations of Conic Sections
The cross–sectional boundaries of a
cone are known as the conic sections.
Conic Sections
(Wikipedia}
19. Polar Equations
Polar Equations of Conic Sections
The cross–sectional boundaries of a
cone are known as the conic sections.
They are circles, ellipses, parabolas
and hyperbolas.
Conic Sections
(Wikipedia}
20. Polar Equations
Polar Equations of Conic Sections
The cross–sectional boundaries of a
cone are known as the conic sections.
They are circles, ellipses, parabolas
and hyperbolas. In the standard
positions, i.e. not tilted, they are the
graphs of 2nd degree equations
Conic Sections
Ax2 + By2 + Cx + Dy = E. (Wikipedia}
21. Polar Equations
Polar Equations of Conic Sections
The cross–sectional boundaries of a
cone are known as the conic sections.
They are circles, ellipses, parabolas
and hyperbolas. In the standard
positions, i.e. not tilted, they are the
graphs of 2nd degree equations
Conic Sections
Ax2 + By2 + Cx + Dy = E. (Wikipedia}
Each non–circular conic section may
be defined geometrically by a constant
e > 0, its eccentricity.
22. Polar Equations
Polar Equations of Conic Sections
The cross–sectional boundaries of a
cone are known as the conic sections.
They are circles, ellipses, parabolas
and hyperbolas. In the standard
positions, i.e. not tilted, they are the
graphs of 2nd degree equations
Conic Sections
Ax2 + By2 + Cx + Dy = E. (Wikipedia}
Each non–circular conic section may D
be defined geometrically by a constant
e > 0, its eccentricity. Specifically,
a point F and a line D are given, F
23. Polar Equations
Polar Equations of Conic Sections
The cross–sectional boundaries of a
cone are known as the conic sections.
They are circles, ellipses, parabolas
and hyperbolas. In the standard
positions, i.e. not tilted, they are the
graphs of 2nd degree equations
Conic Sections
Ax2 + By2 + Cx + Dy = E. (Wikipedia}
Each non–circular conic section may D
be defined geometrically by a constant P
e > 0, its eccentricity. Specifically,
a point F and a line D are given, let P F
be a point
24. Polar Equations
Polar Equations of Conic Sections
The cross–sectional boundaries of a
cone are known as the conic sections.
They are circles, ellipses, parabolas
and hyperbolas. In the standard
positions, i.e. not tilted, they are the
graphs of 2nd degree equations
Conic Sections
Ax2 + By2 + Cx + Dy = E. (Wikipedia}
Each non–circular conic section may D
be defined geometrically by a constant P
e > 0, its eccentricity. Specifically,
a point F and a line D are given, let P F
be a point and that PF, PD be the
distances from to F and P to D
respectively.
25. Polar Equations
Polar Equations of Conic Sections
The cross–sectional boundaries of a
cone are known as the conic sections.
They are circles, ellipses, parabolas
and hyperbolas. In the standard
positions, i.e. not tilted, they are the
graphs of 2nd degree equations
Conic Sections
Ax2 + By2 + Cx + Dy = E. (Wikipedia}
Each non–circular conic section may D
be defined geometrically by a constant P
e > 0, its eccentricity. Specifically, PF
a point F and a line D are given, let P F
be a point and that PF, PD be the
distances from to F and P to D
respectively.
26. Polar Equations
Polar Equations of Conic Sections
The cross–sectional boundaries of a
cone are known as the conic sections.
They are circles, ellipses, parabolas
and hyperbolas. In the standard
positions, i.e. not tilted, they are the
graphs of 2nd degree equations
Conic Sections
Ax2 + By2 + Cx + Dy = E. (Wikipedia}
Each non–circular conic section may D
be defined geometrically by a constant P PD
e > 0, its eccentricity. Specifically, PF
a point F and a line D are given, let P F
be a point and that PF, PD be the
distances from to F and P to D
respectively. A non–circular conic A conic section
with PF = e*PD
section consists of all points P such that where e ≈ ½
27. Polar Equations Directrix
P
The point F is called the focus and the PD
line D is called the directrix. PF
Focus
A conic section
where PF = e*PD
28. Polar Equations Directrix
P
The point F is called the focus and the PD
line D is called the directrix. PF
Focus
For 0 < e < 1, we have ellipses,
for e = 1, we have parabolas,
A conic section
and for 1 < e, we have hyperbolas. where PF = e*PD
29. Polar Equations Directrix
P PD
The point F is called the focus and the
line D is called the directrix. PF
Focus
For 0 < e < 1, we have ellipses,
for e = 1, we have parabolas,
A conic section
and for 1 < e, we have hyperbolas. where PF = e*PD
The entire graph for e = ½ is shown below.
30. Polar Equations Directrix
P PD
The point F is called the focus and the
line D is called the directrix. PF
Focus
For 0 < e < 1, we have ellipses,
for e = 1, we have parabolas,
A conic section
and for 1 < e, we have hyperbolas. where PF = e*PD
The entire graph for e = ½ is shown below.
Note that it has two foci and two directrices.
D2 D1
P
F1 An ellipse
F2 with e=½
i.e. PF = ½*PD
31. Polar Equations Directrix
P PD
The point F is called the focus and the
line D is called the directrix. PF
Focus
For 0 < e < 1, we have ellipses,
for e = 1, we have parabolas,
A conic section
and for 1 < e, we have hyperbolas. where PF = e*PD
The entire graph for e = ½ is shown below.
Note that it has two foci and two directrices.
Ellipses and hyperbolas have two focus–directrix pairs,
i.e. each focus is paired with a directrix.
D2 D1
P
F1 An ellipse
F2 with e=½
i.e. PF = ½*PD
32. Polar Equations Directrix
P PD
The point F is called the focus and the
line D is called the directrix. PF
Focus
For 0 < e < 1, we have ellipses,
for e = 1, we have parabolas,
A conic section
and for 1 < e, we have hyperbolas. where PF = e*PD
The entire graph for e = ½ is shown below.
Note that it has two foci and two directrices.
Ellipses and hyperbolas have two focus–directrix pairs,
i.e. each focus is paired with a directrix.
D2 D1
P
PD1
PF1
F1 An ellipse
F2 with e=½
i.e. PF = ½*PD
33. Polar Equations Directrix
P PD
The point F is called the focus and the
line D is called the directrix. PF
Focus
For 0 < e < 1, we have ellipses,
for e = 1, we have parabolas,
A conic section
and for 1 < e, we have hyperbolas. where PF = e*PD
The entire graph for e = ½ is shown below.
Note that it has two foci and two directrices.
Ellipses and hyperbolas have two focus–directrix pairs,
i.e. each focus is paired with a directrix.
D2 D1
PD2 P
PF2 PD1
PF1
F1 An ellipse
F2 with e=½
i.e. PF = ½*PD
34. Polar Equations Directrix
P PD
The point F is called the focus and the
line D is called the directrix. PF
Focus
For 0 < e < 1, we have ellipses,
for e = 1, we have parabolas,
A conic section
and for 1 < e, we have hyperbolas. where PF = e*PD
The entire graph for e = ½ is shown below.
Note that it has two foci and two directrices.
Ellipses and hyperbolas have two focus–directrix pairs,
i.e. each focus is paired with a directrix.
Hence, if P is any point on an ellipse
then PF1 = e*PD1, D 2
P
D 1
PD 2
PF2 = e*PD2, PF PF
2
PD 1
1
but PF1 ≠ e*PD2 F
F 1
An ellipse
2 with e=½
and PF2 ≠ e*PD1. i.e. PF = ½*PD
35. Polar Equations
(0,0) (0,0)
(0,0) (0,0)
e = 0.5 e = 0.5, 0.8 e = 0.5, 0.8, 1 e = 0.5, 0.8, 1, 2
We give the
polar forms of
the conic
sections below
without proofs.
37. Polar Equations
The Intersection of Polar Equations
As noted before, unlike the
rectangular coordinate system where each point is
addressed by a unique ordered pair (x, y), the polar
coordinates (r, θ+2nπ)P give the same location as (r,
θ)P where n = ±1, ±2..
38. Polar Equations
The Intersection of Polar Equations
As noted before, unlike the
rectangular coordinate system where each point is
addressed by a unique ordered pair (x, y), the polar
coordinates (r, θ+2nπ)P give the same location as (r,
θ)P where n = ±1, ±2.. One consequence of this is that
a polar graph may be represented by infinitely many
“different” equations.
39. Polar Equations
The Intersection of Polar Equations
As noted before, unlike the
rectangular coordinate system where each point is
addressed by a unique ordered pair (x, y), the polar
coordinates (r, θ+2nπ)P give the same location as (r,
θ)P where n = ±1, ±2.. One consequence of this is that
a polar graph may be represented by infinitely many
“different” equations. For example, the polar
equations θ = π/4 + nπ give the same graph as θ = π/4,
the diagonal line y = x.
40. Polar Equations
The Intersection of Polar Equations
As noted before, unlike the
rectangular coordinate system where each point is
addressed by a unique ordered pair (x, y), the polar
coordinates (r, θ+2nπ)P give the same location as (r,
θ)P where n = ±1, ±2.. One consequence of this is that
a polar graph may be represented by infinitely many
“different” equations. For example, the polar
equations θ = π/4 + nπ give the same graph as θ = π/4,
Fact: The graph of = x.θ + 2nπ) = 0 for n = ±1, ±2..
the diagonal line y f(r,
is the same as the graph of polar equation f(r, θ) = 0
41. Polar Equations
The Intersection of Polar Equations
As noted before, unlike the
rectangular coordinate system where each point is
addressed by a unique ordered pair (x, y), the polar
coordinates (r, θ+2nπ)P give the same location as (r,
θ)P where n = ±1, ±2.. One consequence of this is that
a polar graph may be represented by infinitely many
“different” equations. For example, the polar
equations θ = π/4 + nπ give the same graph as θ = π/4,
Fact: The graph of = x.θ + 2nπ) = 0 for n = ±1, ±2..
the diagonal line y f(r,
is the same as the graph of polar equation f(r, θ) = 0
Proof: Let (a, b)P be a point on the graph of f(r, θ) = 0
so that f(a, b) = 0.
42. Polar Equations
The Intersection of Polar Equations
As noted before, unlike the
rectangular coordinate system where each point is
addressed by a unique ordered pair (x, y), the polar
coordinates (r, θ+2nπ)P give the same location as (r,
θ)P where n = ±1, ±2.. One consequence of this is that
a polar graph may be represented by infinitely many
“different” equations. For example, the polar
equations θ = π/4 + nπ give the same graph as θ = π/4,
Fact: The graph of = x.θ + 2nπ) = 0 for n = ±1, ±2..
the diagonal line y f(r,
is the same as the graph of polar equation f(r, θ) = 0
Proof: Let (a, b)P be a point on the graph of f(r, θ) = 0
so that f(a, b) = 0. Let g(r, θ) = f(r, θ + 2nπ) for some n
43. Polar Equations
The Intersection of Polar Equations
As noted before, unlike the
rectangular coordinate system where each point is
addressed by a unique ordered pair (x, y), the polar
coordinates (r, θ+2nπ)P give the same location as (r,
θ)P where n = ±1, ±2.. One consequence of this is that
a polar graph may be represented by infinitely many
“different” equations. For example, the polar
equations θ = π/4 + nπ give the same graph as θ = π/4,
Fact: The graph of = x.θ + 2nπ) = 0 for n = ±1, ±2..
the diagonal line y f(r,
is the same as the graph of polar equation f(r, θ) = 0
Proof: Let (a, b)P be a point on the graph of f(r, θ) = 0
so that f(a, b) = 0. Let g(r, θ) = f(r, θ + 2nπ) for some n
then g(a, b – 2nπ) = f(a, b – 2nπ + 2nπ) = f(a, b) = 0.
44. Polar Equations
The Intersection of Polar Equations
As noted before, unlike the
rectangular coordinate system where each point is
addressed by a unique ordered pair (x, y), the polar
coordinates (r, θ+2nπ)P give the same location as (r,
θ)P where n = ±1, ±2.. One consequence of this is that
a polar graph may be represented by infinitely many
“different” equations. For example, the polar
equations θ = π/4 + nπ give the same graph as θ = π/4,
Fact: The graph of = x.θ + 2nπ) = 0 for n = ±1, ±2..
the diagonal line y f(r,
is the same as the graph of polar equation f(r, θ) = 0
Proof: Let (a, b)P be a point on the graph of f(r, θ) = 0
so that f(a, b) = 0. Let g(r, θ) = f(r, θ + 2nπ) for some n
then g(a, b – 2nπ) = f(a, b – 2nπ + 2nπ) = f(a, b) = 0.
Hence (a, b – 2nπ)P is on the graph of g(r, θ) = 0
which is the same as (a, b) .
45. Polar Equations
When we solve algebraically for the intersection points
of graphs of rectangular equations, we solve the
system of the given equations.
46. Polar Equations
When we solve algebraically for the intersection points
of graphs of rectangular equations, we solve the
system of the given equations. But to solve for the
intersection points of polar equations, we must
consider other possible polar equations that give rise
the same graphs.
47. Polar Equations
When we solve algebraically for the intersection points
of graphs of rectangular equations, we solve the
system of the given equations. But to solve for the
intersection points of polar equations, we must
consider other possible polar equations that give rise
the same graphs.
Example B.
a. Find the intersection of the polar graphs r = sin(2θ)
and θ = π/4. Graph.
48. Polar Equations
When we solve algebraically for the intersection points
of graphs of rectangular equations, we solve the
system of the given equations. But to solve for the
intersection points of polar equations, we must
consider other possible polar equations that give rise
the same graphs.
Example B.
a. Find the intersection of the polar graphs r = sin(2θ)
and θ = π/4. Graph.
49. Polar Equations
When we solve algebraically for the intersection points
of graphs of rectangular equations, we solve the
system of the given equations. But to solve for the
intersection points of polar equations, we must
consider other possible polar equations that give rise
the same graphs.
Example B.
a. Find the intersection of the polar graphs r = sin(2θ)
and θ = π/4. Graph.
If we just set θ = π/4, then r = sin(π/2) = 1 we would
obtain only one solution (1, π/4).
50. Polar Equations
When we solve algebraically for the intersection points
of graphs of rectangular equations, we solve the
system of the given equations. But to solve for the
intersection points of polar equations, we must
consider other possible polar equations that give rise
the same graphs.
Example B.
a. Find the intersection of the polar graphs r = sin(2θ)
and θ = π/4. Graph.
If we just set θ = π/4, then r = sin(π/2) = 1 we would
obtain only one solution (1, π/4).
(0, π/4)
51. Polar Equations
When we solve algebraically for the intersection points
of graphs of rectangular equations, we solve the
system of the given equations. But to solve for the
intersection points of polar equations, we must
consider other possible polar equations that give rise
the same graphs.
Example B.
a. Find the intersection of the polar graphs r = sin(2θ)
and θ = π/4. Graph.
If we just set θ = π/4, then r = sin(π/2) = 1 we would
obtain only one solution (1, π/4). However, there are
two other points. (0, π/4)
52. Polar Equations
When we solve algebraically for the intersection points
of graphs of rectangular equations, we solve the
system of the given equations. But to solve for the
intersection points of polar equations, we must
consider other possible polar equations that give rise
the same graphs.
Example B.
a. Find the intersection of the polar graphs r = sin(2θ)
and θ = π/4. Graph.
If we just set θ = π/4, then r = sin(π/2) = 1 we would
obtain only one solution (1, π/4). However, there are
two other points. At θ = –3π/4, we obtain (0, π/4)
the point (1, –3π/4) and the pole (0, #).
which always have to be checked
separately.
53. Polar Equations
b. Find the intersection points of the polar graphs r = θ
and θ = π/4 where 0 ≤ r. Graph.
54. Polar Equations
b. Find the intersection points of the polar graphs r = θ
and θ = π/4 where 0 ≤ r. Graph.
For n = 0, 1, 2,.. the equations θ = π/4 + nπ give the
same graph as θ = π/4.
55. Polar Equations
b. Find the intersection points of the polar graphs r = θ
and θ = π/4 where 0 ≤ r. Graph.
For n = 0, 1, 2,.. the equations θ = π/4 + nπ give the
same graph as θ = π/4. Hence the intersection points
where 0 ≤ r are (π/4 + nπ, π/4 + nπ).
56. Polar Equations
b. Find the intersection points of the polar graphs r = θ
and θ = π/4 where 0 ≤ r. Graph.
For n = 0, 1, 2,.. the equations θ = π/4 + nπ give the
same graph as θ = π/4. Hence the intersection points
where 0 ≤ r are (π/4 + nπ, π/4 + nπ).
θ = π/4
(9π/4,9π/4)
(π/4, π/4)
o
(5π/4, 5π/4)
(13π/4, 13π/4)
r=θ
57. Polar Equations
b. Find the intersection points of the polar graphs r = θ
and θ = π/4 where 0 ≤ r. Graph.
For n = 0, 1, 2,.. the equations θ = π/4 + nπ give the
same graph as θ = π/4. Hence the intersection points
where 0 ≤ r are (π/4 + nπ, π/4 + nπ). We check that
the point (0,0) is also θ = π/4
a solution.
The graphs are
shown here. (9π/4,9π/4)
(π/4, π/4)
o
(5π/4, 5π/4)
(13π/4, 13π/4)
r=θ
58. Polar Equations
b. Find the intersection points of the polar graphs r = θ
and θ = π/4 where 0 ≤ r. Graph.
For n = 0, 1, 2,.. the equations θ = π/4 + nπ give the
same graph as θ = π/4. Hence the intersection points
where 0 ≤ r are (π/4 + nπ, π/4 + nπ). We cheek that
the point (0,0) is also θ = π/4
a solution.
The graphs are
shown here. (9π/4,9π/4)
(π/4, π/4)
In general, we need to o
(5π/4, 5π/4)
graph when solving for (13π/4, 13π/4)
intersection points of
polar equations to make
sure that we have all the r=θ
solutions.
