Download Section 10.3 Polar Coordinates and Functions

Section 10.3 Polar Coordinates and Functions
Ruipeng Shen
March 10, 12
1
Polar Coordinates
Pole and Polar Axis We choose a point in the plane that is called the pole (or origin) and
is labeled O. Then we draw a ray (half-line) starting at O called the polar axis. This axis
is usually drawn horizontally to the right and corresponds to the positive x-axis in Cartesian
coordinates.
Polar Coordinates If P is any other point in the plane. Let r be the distance from O to P
and let θ be the angle (usually measured in radians) from the polar axis to the line OP as in
Figure 1. Then the point P is represented by the ordered pair (r, θ).
P(r,θ)
P(r,θ+2𝛑)
r
r
θ+2𝛑
θ
O
r
O
O
-θ
r
P(-r,θ)
P(r,-θ)
Figure 1: Polar Coordinates
1
θ
Negative Values of θ or r The angle is considered positive if the rotation from the polar axis
to OP is in the counter-clockwise direction, negative if the rotation is in the clockwise direction.
The polar coordinates (−r, θ) and (r, θ) represent two points that are symmetric about the origin.
Remark 1. A single point may have a lot of different representations of polar coordinates. For
example, the polar coordinates (r, θ + 2π), (−r, θ + π) represent the same point as (r, θ). We
usually use positive radius and angle between 0 and 2π.
Example 2. Plot the points where the polar coordinates are given.
(2, 5π/4)
Solution
(2, −π/4)
(3, π/2)
See the figure 2.
P2(3,𝛑/2)
O
P1(2,5𝛑/4)
P3(2,-𝛑/4)
Figure 2: Example 2 - Polar Coordinates
Switch between two coordinate systems By basic definition of trigonometric functions,
we can convert polar coordinates into Cartesian coordinates
x = r cos θ
y = r sin θ
We can also convert Cartesian coordinates into polar coordinates
y
r 2 = x2 + y 2
tan θ =
x
Example 3. Convert the point (2, π/3) from polar coordinates to Cartesian coordinates
Solution
We have
x = 2 cos
π
= 1;
3
y = 2 sin
√
Thus the Cartesian coordinates are (1, 3).
π √
= 3.
3
Example 4. Represent the point with Cartesian coordinates (1, −1) in term of polar coordinates.
2
Solution
Let us choose r to be positive; we can calculate
r=
p
12 + (−1)2 =
√
2;
tan θ =
−1
= −1
1
Since√the point is in the fourth quadrant,
we can choose θ = 7π/4. Thus the polar coordinates
√
are ( 2, 7π/4). One can also use ( 2, −π/4).
2
Polar Curves
The graph of a polar equation r = f (θ) or more generally F (r, θ) = 0, consists of all points
P that have at least one polar representation (r, θ) whose coordinates satisfy the equation.
Example 5. What curve is represented by the polar equation r = 2? How about θ = 1?
Solution This can be understood by considering the geometrical meaning of polar coordinates. The curves are the circle centered at the origin with radius 2 and the line y = (tan 1)x,
respectively.
Example 6. Let us consider the curves with a polar equation r = 2 cos θ. Find a Cartesian
equation and sketch the curve.
Solution
Using the identities x = r cos θ and r2 = x2 + y 2 , we have
r = 2 cos θ ⇐⇒ r2 = 2r cos θ ⇐⇒ x2 + y 2 = 2x ⇐⇒ (x − 1)2 + y 2 = 1
This is a circle centered at (1, 0) with a radius 1.
r=2cosθ
O
θ
2
Figure 3: Circle: Example 6
Example 7 (Cardioid). Sketch the curve r = 1 − sin θ.
Solution Let us first sketch the graph of r = 1 − sin θ in the Cartesian coordinates (on a r-θ
plane). This enables us to read at a glance the values of r that correspond to increasing values
of θ.
3
2
1
-1
0
1
2
3
4
5
6
7
8
Figure 4: Graph of r = 1 − sin θ in the r-θ coordinate system
0.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
-0.5
r = 1- sin θ
-1
-1.5
-2
-2.5
Figure 5: Cardioid: polar equation r = 1 − sin θ
Example 8. Sketch the curve r = cos 2θ.
Solution Again we first sketch the graph of r = cos 2θ in the Cartesian coordinates, then
sketch the curve.
1
-1
0
1
2
3
4
5
6
7
8
-1
Figure 6: Graph of r = cos 2θ in the r-θ coordinate system
4
1.5
θ=3π/4
θ=π/4
1
0.5
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
r = cos 2θ
-0.5
θ=5π/4
θ=7π/4
-1
-1.5
Figure 7: Graph of the polar equation r = cos 2θ
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Tangents to Polar Curves
To find a tangent line to a polar curve r = f (θ), we regard θ as a parameter and write its
parametric equation as
x = r cos θ = f (θ) cos θ;
y = r sin θ = f (θ) sin θ.
Thus the slope of the parametric curve is given by
dy
dr
sin θ + r cos θ
dy
f 0 (θ) sin θ + f (θ) cos θ
dθ
dθ
=
.
=
= 0
dx
dr
dx
f (θ) cos θ − f (θ) sin θ
cos θ − r sin θ
dθ
dθ
Remark 9. In particular, if f (θ0 ) = 0, then we have
dy
= tan θ0 .
dx
Example 10. Consider the cardioid r = 1 − sin θ in the Figure 5.
(a) Find the formula for the slope of the curve in term of θ.
(b) Find the points where the tangent line is horizontal or vertical.
5
(1)
2.5
Solution
(a) By the formula (1), we have
dr
sin θ + r cos θ
dy
(− cos θ) sin θ + (1 − sin θ) cos θ
= dθ
=
dr
dx
(− cos θ) cos θ − (1 − sin θ) sin θ
cos θ − r sin θ
dθ
cos θ(1 − 2 sin θ)
=
−(1 − sin θ)(1 + 2 sin θ)
(b) In order to seek horizontal or vertical tangents, we solve the equation
1
π 3π π 5π
dy
= cos θ(1 − 2 sin θ) = 0 =⇒ cos θ = 0 or sin θ =
=⇒ θ = ,
, ,
dθ
2
2 2 6 6
dx
1
π 7π 11π
= − (1 − sin θ)(1 + 2 sin θ) = 0 =⇒ sin θ = 1 or sin θ = − =⇒ θ = ,
,
dθ
2
2 6
6
Thus we have
θ
π
6
π
2
5π
6
7π
6
3π
2
11π
6
y 0 (t) = 0?
Yes
Yes
Yes
No
Yes
No
x0 (t) = 0?
No
Yes
No
Yes
No
Yes
Tangent
Horizontal
Indeterminate
Horizontal
Vertical
Horizontal
Vertical
Finally when θ = π/2, the radius r = 0. Thus the tangent line is vertical, according to Remark
9. In summary, the tangent is horizontal at the points (1/2, π/6), (1/2, 5π/6) and (2, 3π/2); the
tangent is vertical at the points (3/2, 7π/6), (3/2, 11π/6) and the pole.
6
0.5π
0.5
-2
-1.5
π
-1
-0.5
0
0.5
-0.5
-11.5π
-1.5
-2
-2.5
Figure 8: Cardioid in Polar Coordinates
7
1
1.5
2
2.5