u
1
Table of Contents
Coordinate Conversion Analytical Example……………………………………………………………………………….3
Polar Derivative Multiple Choice Example…………………………………………………………………………………4
Conceptual Example………………………………………………………………………………………………………………….4
AP Free Response Example……………………………………………………………………………………………………….5
Classic Graphs Multiple Choice Example……………………………………………………………………………………8
Arclength Analytical Example…………………………………………………………………………………………………..9
Graphic Polar Area Example……………………………………………………………………………………………………11
Exercises……………………………………………………………………………………………………………………………12,13
Solution Guide……………………………………………………………………………………………………………………….14
Johannes Kepler was born in 1571
and died in 1630 after making vast
contributions to the scientific and
mathematical fields. Kepler’s main
interest with polar curves concerned
the orbits of the planets. He
constructed equations that could be
used to graph ellipses that described
planetary motion. He used polar
coordinate notation to write these
equations and published these
equations in his “First Law”.
Polar graphs have historically been
used to describe the paths of
planets and other celestial bodies.
Other naturally-occurring orbital
systems can be modeled with polar
equations as well. Polar is useful in
navigation, as location can be given
as an angle and a distance from a
set point. Air and nautical travel
rely on polar coordinates, and they
are used in air traffic control.
Generally, 360⁰ corresponds to
magnetic north, 180⁰ to magnetic
south, 90⁰ to magnetic east, and
270⁰ to magnetic west.
2
Gorgeous, isn’t he?
Example of Equation Conversion from Polar to Parametric Form
a) Convert r=2
to parametric form.
Explanation: Using the known polar conversion equations, plug the given
r into both x and y equations and simplify if necessary.
x=r
=
x=2
y=r
y=2
These two converted equations represent the polar equation in parametric form. Using each equation, we can plot x
and y coordinates in a Cartesian plane instead of on a polar plane. We can also use these formulas to find the x and y
coordinates of r in (x,y) coordinate form, as shown below.
b) Convert (2,2 ) to rectangular coordinates.
Explanation: Using the same equations, evaluate the expression by replacing r and
with the values given in the
coordinates. Remember, these coordinates are given in the form (r, ).
x =2
 replace
with 2
x=r
y=2
 replace
with 2
y=r
=
(x,y) coordinates = (2, 0)
Formula for Polar Derivative:
𝑑𝑦
𝑑𝑦 𝑑𝜃
=
,
𝑑𝑥 𝑑𝑥
𝑑𝜃
Note that when
𝑦 = 𝑟(𝜃) sin 𝜃,
𝑥 = 𝑟(𝜃) cos 𝜃
𝑑𝑦
𝑑𝜃
𝑑𝑥
= 0, 𝑑𝜃 ≠ 0, the particle
𝑑𝑥
movement is horizontal, and when 𝑑𝜃 =
𝑑𝑦
0, 𝑑𝜃 ≠ 0, the particle motion is vertical.
𝑑𝑦
When both 𝑑𝜃 = 0 and
𝑑𝑥
𝑑𝜃
= 0, the particle
is still and no conclusion can be drawn
about is motion.
Tangent Lines at the Pole:
Points at the pole have tangent lines that are particularly easy to calculate. When a curve passes
through the pole, that is, 𝑓(∅) = 0, and the function’s derivative is not equal to zero, the line
𝜃 = ∅ is tangent to the graph of 𝑓(𝜃) at that point. This can also be written as 𝑦 = 𝑥 tan ∅,
where tan ∅ is a constant.
3
Example AP Multiple Choice Problem
What is
for r =
?
A)
B)
C)
D)
E)
 These are ugly. But we know you like it!
Explanation: converting this equation in to parametric form enables us to find the y’ and x’. (Use equations given
above).
y=4
 use the chain rule and product rule y’ = 4 (
)
x=4
 use the chain rule and product rule  x’ = 4 (
)
=
/
=
The correct answer is choice (C).
(A) – The student forgot to use the chain rule on the
in
in the calculation of y’.
(B) – The student did not use the chain rule on
in the calculation of x’.
(C) – IF YOU GOT IT RIGHT YOU’RE A STUD. YES. YES YOU ARE.
(D) – The student did not use the product rule for 4
in y’.
(E) – The student forgot to use the chain rule on the
(outer function) of both y’ and x’.
AP Conceptual Example: A particle’s path in the polar plane is given by a function r(𝜃), such that
𝜃=t. At time t, the particle’s position is given by the polar coordinates (−𝜋, 𝜋/2). At this moment,
𝑑𝑟
𝑑𝜃
and
𝑑𝑥
𝑑𝑡
are positive. Describe the particle’s motion in terms of direction,
𝑑𝑦
,
𝑑𝑡
and distance from
the pole.
Solution: Remember that polar coordinates are oriented as (r,𝜃). Thus, the radial arm in the given
coordinate is of length 𝜋, and is negative. Thus, it is in the direction opposite of 𝜋/2, which is
𝑑𝑟
3𝜋/2. The particle is therefore below the pole, on the y-axis. Because 𝑑𝜃 is positive, the radial arm
is shrinking—it is becoming “less negative.” So, the distance from the pole is decreasing.
𝑑𝑦
Accordingly, 𝑑𝑡 is positive (a component of the particle’s velocity is in the upwards direction). It is
given that
𝑑𝑥
𝑑𝑡
is positive. Thus, the particle is moving northeast—upwards and to the right.
4
Example AP Free Response Question
Find the area of one petal of r = 3
1.
.
First, we need to investigate what this r really means.
COMMON POLAR GRAPHS:
According to the model r = a
the center of the graph at
, the graph of r will have four petals (petals = 2n) and will extend 3 beyond
0.
From the graphs and rules of rose curves to
the right, we can see that the graph of r must
fit a certain pattern. Firstly, because all
graphs involving
are oriented to the x
axis (and graphs of
are oriented in the
y direction) we know that the graph will be
symmetrical to the x axis. Secondly, we
know it will have 4 petals because n = 2 and
2(2) is 4. (That little calculation was for
those who, like the author, multiply 2 and 2
and arrive at 5 ¾. She feels your pain.)
Now that we have a general idea of what the
graph looks like, we can create a table and
graph at least one petal to know exactly the
bounds of one petal of the graph of r.
5
Example AP Free Response Question (continued)
2. Now, we need to create a table or a graph to determine the bounds of one petal of the graph. (The graph is
technically optional, but creating one will give you a visual representation of the graph itself. This helps you out
when determining the bounds, but either way is fine! Just know which one is right for you.)
0
r
3
0
-3
3
Create the table by
selecting values in the
(0, 2 ) interval and
evaluating them. This
will give you a
sufficient
representation of the
graph in question.
All right, now that we’ve got our graph and our table, we can examine the bounds of one petal of the graph. Let’s
take the horizontal petal on the right side, shall we? (We know this isn’t actually up to you, but you’d better get
over it. It’s a fresh looking petal and we’re enjoying the control!)
It is clear that the value of r
must begin at 0 and return
to 0 in order to complete
one petal. Now, we have to
find the values we can use
as bounds from our table.
From the table, we observe
that it the graph travels from
to = . Therefore, we
can double the area we
calculate from half the petal to
get the area of the whole petal.
THE AREA OF A SECTION OF A POLAR CURVE:
A=
d
6
Example AP Free Response Question (continued)
We have determined that the bounds of the petal are twice the area of the petal from (0,
So our equation is:
Insert r into the correct place in the area formula.
Remember to multiply the entire integral by 2
because we are doubling the area of half of the
petal.
Use the power reduction formula for
. Now,
this integral is much easier to evaluate!
Integrate the integral by finding the antiderivative
for each component.
Use the First Fundamental Theorem to evaluate
the new antiderivative.
Your answer is
!!!
Fantastic job. You get a cookie!
THIS POLAR
BEAR
DERIVES
DIRTY.
OH YEAH.
YOU KNOW
IT.
7
AP Multiple Choice Example: Which graph corresponds to the function 𝑟 = 2 sin(4𝜃)?
y
A.


x






B.
y


x






C.
y


𝜋
x

Solution: The formula for a rose curve is
𝑟 = 𝑎 cos(𝑛𝜃). 𝑎 gives the length of the
petals, and 𝑛 determines the number of
petals. When 𝑛 is even, the curve has 2𝑛
petals, and when 𝑛 is odd, the curve has 𝑛
petals. Thus, the given curve has 8 petals,
each 2 units in length. This rules out
choices A and C. Choice E is not a rose
curve. Because the formula involves sine,
the function is an odd function, and is
therefore oriented so that it is symmetric
about the line 𝑦 = 𝑥. Answer D must be
the correct choice. To double check this



answer against answer B, plug in 4 for 𝜃.
The given equation yields 𝑟 = 0, not 𝑟 =
−1, and graph D agrees with this.


D.
y


x






E.
y


x






8
Formula for Polar Arclength:
𝜷
𝒔 = ∫ √[𝒓(𝜽)]𝟐 + [
𝜶
*If and only if the derivative of 𝑟(𝜃)is continuous on [𝛼, 𝛽]
𝒅𝒓 𝟐
] 𝒅𝜽
𝒅𝜽
Analytical Example: Find the arclength of the curve given by 𝑟 2 = sin(2𝜃).
Solution: First, graph the function.
y

x



We must make sure to only calculate the arc length for one revolution. To simplify things, we will
calculate the arc length for only one of the loops, and multiply be two. We will find the interval of
𝜃 over which one loop is graphed out. Examination of the graph shows that 𝑟 = 0 at both the start
and the finish of this loop. We solve for 𝜃. 0 = sin(2𝜃). 𝜃 = 0, 𝜋/2. Thus, the loop is graphed
out when 𝜃 is between these two values, and these are our bounds of integration when evaluating
2
𝛽
𝑑𝑟
the arc length. Remember the formula for polar arc length: 𝑠 = ∫𝛼 √[𝑟(𝜃)]2 + [𝑑𝜃] 𝑑𝜃.
The equation we are given is in terms of 𝑟 2 . For the upper loop, 𝑟 = √sin(2𝜃).
We differentiate to find
𝑑𝑟
𝑑𝜃
=
cos(2𝜃)
√sin(2𝜃)
𝜋/2
We now have the equation 𝑠 = 2 ∫0
𝑑𝑟 2
𝑑𝜃
and we square to obtain ( ) =
√sin(2𝜃) +
(cos(2𝜃))2
sin(2𝜃)
(cos(2𝜃)2
sin(2𝜃)
.
𝑑𝜃.
𝜋/2
(sin(2𝜃))2 + (cos(2𝜃))2
= 2∫ √
𝑑𝜃
sin(2𝜃)
0
Now use the trigonometric identity (cos 𝜃)2 + (sin 𝜃)2 = 1
𝜋/2
1
= 2∫ √
𝑑𝜃
sin(2𝜃)
0
Notice that by multiplying by two and adjusting the bounds of integration, we avoid an integral
with an infinite discontinuity between its bounds. To properly evaluate this integral, we would
take the limit as 𝛼 approached 0 and as 𝛽 approached 𝜋/2. However, we will use a calculator to
evaluate this integral.
𝑠 ≈ 5.244.
9
Polar Area
If the polar function 𝑟(𝜃) is continuous on the interval [𝛼, 𝛽], and this interval is no greater than
2𝜋 (that is, 𝛽 ≤ 2𝜋 + 𝛼 ), then the area bounded by the graph of 𝑟(𝜃) and the radial lines on
either side of the interval is given by the following integral:
𝛽
1
𝐴 = ∫ 𝑟 2 𝑑𝜃
2
𝛼
A good tactic for finding polar area, especially when dealing with more than one curve, is to
consider Sauron’s eye. In The Lord of the Rings, Sauron’s eye sees all things, missing NOTHING.
With polar area, the radial arm sweeps from the pole (just like the sweeping motion of radar)
and includes all parts of the polar curve. This means that inner loops, for example (as found in
certain limaçons), are counted TWICE if the boundaries 0 and 2𝜋 are used for area evaluation. In
this case, one must adjust the bounds of integration so that the inner loop is excluded the
second time; that is, so that only the values of theta that do not trace out the inner loop are
included between the bounds of integration. If ever in doubt, draw on the analogy to Sauron’s
eye.
10
Graphic Example: Find the area common to the curves given by 𝑟1 (𝜃) = 3 − 3 sin 𝜃 and
𝑟2 (𝜃) = 2 sin 𝜃, as shown in the graph.
y

x














Solution: First, identify the shared area and realize that it is equal on both sides of the yaxis. Thus, we need only calculate the shared area on one side of the y-axis, and then
multiply by two.
Find the two points of intersection that bound the shared area in the first quadrant:
𝑟1 (𝜃) = 𝑟2 (𝜃) , 2 sin 𝜃 = 3 − 3 sin 𝜃, 3 = 5 sin 𝜃, 𝜃 = sin−1(35)
Both curves also go through the pole, but at different values of 𝜃. Solve to obtain that
𝑟1 (0) = 0, and 𝑟2 (𝜋2) = 0.
Inspection of the graph shows that two separate integrals must be used to calculate the
area: one for each curve. This shared area cannot be found through the subtraction of one
equation from the other. We will work with 𝑟2 first. 𝑟1 intersects 𝑟2 at 𝜃 = 0 for 𝛼, and
𝜃 = sin−1 (35) for 𝛽. This will sweep out the lower segment of the shared area. The
𝛽
formula for polar area is 𝐴 = 12 ∫𝛼 [𝑟(𝜃)]2 𝑑𝜃 .
sin−1 (.6)
sin−1(.6)
sin−1(.6)
𝐴1 = ∫
(2 sin 𝜃)2 𝑑𝜃 = ∫
(4[sin 𝜃]2 ) 𝑑𝜃 = ∫
0
0
0
(2 − 2 cos(2𝜃)) 𝑑𝜃
1
2
where we have multiplied the in the formula by 2 to account for both areas. Now
integrate, using the formulas sin(2𝜃) = 2 sin 𝜃 cos 𝜃 and (cos 𝜃)2 + (sin 𝜃)2 = 1, which
3
4
shows that cos (sin−1 (5)) = 5.
sin−1(.6)
𝐴1 = [2𝜃 − sin(2𝜃)]0
3 4
24
= 2 sin−1(. 6) − (2) ( ) ( ) = 2 sin−1 (. 6) −
5 5
25
3
5
𝜋
2
We now turn to 𝑟1 , which has the bounds 𝛼 = sin−1 ( ) and 𝛽 = , as found above. This
will give the area of the upper segment of the shared area, and once again we multiply be 2
to account for both sides of the y-axis, canceling out the formula’s
𝜋
2
sin−1 (.6)
=
27𝜋
4
27
∫sin−1(.6)( 2 −
27
3
sin−1 (5)
2
𝐴 = 2 sin−1(. 6) −
sin−1 (.6)
𝜋
𝜋
2
−
as before.
(9 − 18 sin 𝜃 + 9(sin 𝜃)2 ) 𝑑𝜃
(3 − 3 sin 𝜃) 𝑑𝜃 = ∫
𝐴2 = ∫
𝐴2 =
𝜋
2
2
1
2
18 sin 𝜃 −
−
24
25
72
5
+
9
cos(2𝜃)) 𝑑𝜃
2
=
27𝜃
[ 2
+ 18 cos 𝜃
9
2
− 4 sin(2𝜃)] −1
sin (.6)
54
+ 25 (Area of the upper sector) Now, let’s combine areas.
27𝜋
4
−
27
72
sin−1(. 6) −
2
5
+
54
25
=
27𝜋
4
−
66
23
3
− sin−1 ( )
5
2
5
11
≈ .605
Works Cited
Dana, Peter H. "Coordinate Systems Overview." University of Colorado Boulder. 15 Dec. 1999. Web.
24 May 2011. <http://www.colorado.edu/geography/gcraft/notes/coordsys/coordsys.html>.
Larson, Ron, Robert P. Hostetler, and Bruce H. Edwards. Calculus with Analytic Geometry. 8th ed.
Boston: Houghton Mifflin, 2006. Print.
Weisstein, Eric W. "Kepler's Equation -- from Wolfram MathWorld." Wolfram MathWorld: The Web's
Most Extensive Mathematics Resource. Web. 23 May 2011.
<http://mathworld.wolfram.com/KeplersEquation.html>.
Analytical Exercises:
*Sauron’s eye indicates greater difficulty.
1. Convert (8, 3𝜋) to rectangular form.
𝜋
4
2. Given tan 6 = 𝑥, find the rectangular coordinate (x).
3. Given (6, 3) in rectangular form, convert to polar coordinates.
4. Convert (9,4) to polar form.
5. Find
𝑑𝑦
𝑑𝑥
of r = 2 + 3cos 𝜃.
6. Find the vertical tangents of r = 2 + 3cos 𝜃.
7.
8.
9.
10.
11.
Find the arclength of the curve given by 𝑟(𝜃) = 5 sin 𝜃, and identify the curve by name.
Find the arclength of the curve given by 𝑟(𝜃) = 2 + 2 cos 𝜃, and identify the curve by name.
Find the arclength of the curve given by 𝑟(𝜃) = 3 cos(2𝜃), and identify the curve by name.
Find the area of the lemniscate described by the equation 𝑟 2 = 9 cos(2𝜃).
Find the area of one petal of the rose curve given by 𝑟(𝜃) = 5 cos(4𝜃).
12.
Find the area of the limaçon given by 𝑟(𝜃) = 1 + 2 sin 𝜃.
13.
Find the area common to the two curves given by 𝑟1 (𝜃) = 4 sin(2𝜃) and 𝑟2 = 2.
AP Free Response:
The equation 𝑟(𝜃) = 5 − 2 sin 𝜃 describes the path of a particle traveling in the polar coordinate plane.
a) Describe the graph of the above function (i.e. the path of the particle).
b) Find
c) Find
𝑑𝑟
𝑑𝜃
𝑑𝑦
𝑑𝑥
𝜋
3
when 𝜃 = , and describe its meaning.
𝜋
when = 3 , and describe its meaning.
d) Calculate the area under the curve in the first quadrant.
12
AP Multiple Choice:
1. What is
a) d) -
𝒅𝒚
𝒅𝒙
for r = 𝟔 𝐜𝐨𝐬 𝟒𝜽?
cos 4𝜃 cos 𝜃− sin 4𝜃 sin 𝜃
cos 4𝜃 sin 𝜃+ sin 4𝜃 cos 𝜃
cos 4𝜃 cos 𝜃− 4sin 4𝜃 sin 𝜃
cos 4𝜃 sin 𝜃
b)
cos 4𝜃 cos 𝜃− 4sin 4𝜃 sin 𝜃
cos 4𝜃 sin 𝜃+ 4 sin 4𝜃 cos 𝜃
e) -
c)-
cos 4𝜃 cos 𝜃
cos 4𝜃 sin 𝜃+ 4 sin 4𝜃 cos 𝜃
cos 4𝜃 cos 𝜃− 4 sin 4𝜃 sin 𝜃
cos 4𝜃 sin 𝜃+ 4sin 4𝜃 cos 𝜃
2. Find the area of one petal of 𝒓 = 𝟑 𝐜𝐨𝐬 𝟑𝜽.
𝜋
a) 6
b)
3𝜋
c)
2
3𝜋
4
3
-2
d)
3𝜋
4
e)
3𝜋
4
9
-2
𝒅𝒚
3. Find 𝒅𝒙 at the pole for r = 𝟐 𝐬𝐢𝐧 𝟒𝜽.
𝜋 𝜋
a) 𝜃 = 0, 4 , 2
𝜋 𝜋
b) 𝜃 = 4 , 2
c) 𝜃 = 0
𝜋
d) 𝜃 = 0, 2
e) 𝜃 =
𝜋
4
𝒅𝒓
4. What is 𝒅𝜽 for r = 𝐬𝐢𝐧 𝟒𝜽 (𝐜𝐨𝐬 𝜽)𝟐 ?
a) 18 (2cos 4𝜃(cos 𝜃)2 - sin 𝜃 sin 4𝜃 cos 𝜃)
b) 9 (cos 4𝜃 (cos 𝜃)2 + 2cos 𝜃 sin 4𝜃
c) 2cos 4𝜃(cos 𝜃)2 - sin 𝜃 sin 4𝜃 cos 𝜃
d) 9 (4cos 4𝜃 (cos 𝜃)2 + 2sin 4𝜃 sin 𝜃 cos 𝜃
e) 9sin 4𝜃2cos 𝜃
5. At which values of 𝜽 does r = 𝟒𝐜𝐨𝐬 𝟓𝜽 have horizontal tangent lines on the
𝜋
interval (0, 2 )?
a) 𝜃 = 0.171, 0.895, 0.3234, 1.431, 6.789, 4.736, ∞
b) 𝜃 = 0.677, 1.269,
c) 𝜃 = 0.568, 1.243
d) 𝜃 = 0.601, 1.169, 1.571
e) 𝜃 = 0.171, 0.677, 1.269
13