Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
9.3
Polar Coordinates
Up to this point we have dealt exclusively with the Cartesian coordinate
system. However, as we will see, this is not always the easiest coordinate
system to work in. In this section we will start looking at the polar coordinate system.
The Cartesian system consists of two rectangular axes. A point P in this
system is uniquely determined by two points x and y as shown in Figure
9.3.1(a). The polar coordinate system consists of a point O, called the
pole, and a half-axis starting at O and pointing to the right, known as the
polar axis. A point P in this system is determined by two numbers: the
distance r between P and O and an angle θ between the ray OP and the
polar axis as shown in Figure 9.3.1(b).
Figure 9.3.1
The Cartesian and polar coordinates can be combined into one figure as
shown in Figure 9.3.2.
Figure 9.3.2
Figure 9.3.2 reveals the relationship between the Cartesian and polar coordinates:
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r=
p
x2 + y 2
x = r cos θ
y = r sin θ.
In our definition above, r is positive. However, graphs of curves in polar coordinates are traditionally drawn using negative values of r as well, because
this makes the graphs symmetric. If an equation r = f (θ) gives a negative
r−value, it is plotted in the quadrant on the opposite side of the pole. The
point (−r, θ) represents the same point as (r, θ + π). For example, the point
7π
(−3, 3π
4 ) is located at (3, 4 ).
The following example illustrates the process of converting between polar
and Cartesian coordinates.
Example 9.3.1 (a) Convert 4, 2π
to Cartesian coordinates.
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(b) Convert (−1, −1) into polar coordinates.
Solution.
(a) Plugging into the above formulas we find
x = 4 cos 2π
− 12 = −2
3 = 4 √
√
3
y = 4 sin 2π
=2 3
3 =4
2
√
So, in Cartesian coordinates this point
is
(−2,
2
3).
√
(b) Choosing r > 0, we find r = 2 and tan θ = xy = 1. Since (−1, −1) is
3π
in the third quadrant, we can choose θ = 5π
4 or θ = − 4 . Thus one possible
√ 5π √
answer is
2, 4 ; another is
2, − 3π
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Plotting Curves in Polar Coordinates
Several important types of graphs have equations that are simpler in polar
form than in Cartesian form. For example, the polar equation of a circle having radius a and centered at the origin is simply r = a. In other words, polar
coordinate system is useful in describing two dimensional regions that may
be difficult to describe using Cartesian coordinates. For example, graphing
the circle x2 + y 2 = a2 in Cartesian coordinates requires two functions, one
for the upper half and one for the lower half. In polar coordinate system,
the same circle has the very simple representation r = a.
The equation of a curve expressed in polar coordinates is known as a polar equation, and is usually written in the form r = f (θ). To construct
graphs within the polar coordinate system you need to find how the value
of r changes as θ changes. It is best to construct a table with certain values
of θ and work out the respective value of r using the equation r = f (θ).
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Example 9.3.2 (Polar Roses)
A polar rose is a famous mathematical curve which looks like a petalled
flower, and which can only be expressed as a polar equation. It is given by
the equations r = a sin kθ or r = a cos kθ where a > 0 and k is an integer.
These equations will produce a k−petalled rose if k is odd, or a 2k−petalled
rose if k is even. Graph the roses r = 4 sin 2θ and r = 4 sin 3θ.
Solution.
The graph of r = 4 sin 2θ is given in Figure 9.3.3(a). Petals in quadrants II
and IV have negative r−values. The graph of r = 4 sin 3θ is given in Figure
9.3.3(b)
Figure 9.3.3
Example 9.3.3 (Archemidean Spiral)
The Archimedean spiral is a famous spiral that was discovered by Archimedes,
which also can be expressed only as a polar equation. It is represented by
the equation r = a + bθ. Graph r = θ.
Solution.
The graph of r = θ is given in Figure 9.3.4
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Figure 9.3.4
Example 9.3.4 (Limaçons)
Limaçons are curves with polar equations r = b+a cos θ (horizontal limaçons)
or r = b + a sin θ (vertical limaçons) with a, b > 0. Graph r = 1 + 2 cos θ and
r = 3 + 2 cos θ.
Solution.
The graph of r = 1 + 2 cos θ is given in Figure 9.3.5(a) and that of r =
3 + 2 cos θ is given in Figure 9.3.5(b)
Figure 9.3.5
Example 9.3.5 (Cardioids)
Cardioids are curves with polar equations r = a(1+cos θ) or r = a(1+sin θ).
Graph r = 2(1 + cos θ) and r = 2(1 + sin θ).
Solution.
The graph of r = 2(1 + cos θ) is given in Figure 9.3.6(a) and that of r =
2(1 + sin θ) is given in Figure 9.3.6 (b)
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Figure 9.3.6
Slope in Polar Coordinates
Since x = r cos θ = f (θ) cos θ and y = f (θ) sin θ, by the chain rule we have
dy
=
dx
dy
dθ
.
dx
dθ
Example 9.3.6
Find the slope of the curve r = θ at the point
π π
2, 2
.
Solution.
We are given that x = θ cos θ and y = θ sin θ. Thus, dx
dθ = cos θ − θ sin θ and
dy
dθ = sin θ + θ cos θ. Thus,
dy sin θ + θ cos θ 2
π
π = −
=
dx θ= 2
cos θ − θ sin θ θ= 2
π
Graphing in Polar Coordinates Using TI-83
Consider graphing the function r = 4 sin 2θ with a TI-83 calculator.
1. Hit the MODE key.
2. Arrow down to where it says Func (short for “function”).
3. Now, use the right arrow to choose Pol (short for “polar”).
4. Hit ENTER.(The calculator is now in polar coordinates mode).
5. Hit the Y = key. Note that, instead of Y 1 =, Y 2 =, and so on, you now
have r1 = and so on.
6. In the r1 = slot, type 4 sin (.
7. Now hit the familiar X, T, θ, n key.
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8. Finally, close off the parentheses and hit GRAPH.
If you did everything right, you just asked the calculator to graph the polar
equation r = 4 sin 2θ as shown in Figure 9.3.3(a).
The WINDOW options are a little different in this mode too. You can still
specify X and Y ranges, which define the viewing screen. But you can also
specify the θ values that the calculator begins and ends with; for instance,
you may limit the graph to 0 < θ < π2 . This would not change the viewing
window, but it would only draw part of the graph.
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