1
~An
Pusheen.com
Introduction to
Polar
Coordinates, Graphs, and the
Calculus of ~
~as Presented by Pusheen with assistance
from Sara Dornblaser and Hannah
Schapiro~
2
Table of Contents
Introduction to Polar……………………………………………………………………………………………………………… 3
Important Mathematicians in Polar…………………………………………………………………….…………………. 4
Basic Analytical Examples………………………………………………………………………………………………………. 4
AP Level Multiple Choice……………………………………………………………………………………………………….. 6
Conceptual Examples…………………………………………………………………………………………………………….. 7
Real World Applications………………………………………………………………………………………………………… 8
Graphing Calculator and Graph Example……………………………………………………………………………….. 9
AP Level Free Response………………………………………………………………………………………………………… 10
Exercises……………………………………………………………………………………………………………………………….. 12
Solution Guide………………………………………………………………………………………………………………………. 16
Works Cited…………………………………………………………………………………………………………………………… 17
Pusheen.com
3
Important Polar Basics
2. Polar deals with circular motion!
4. Coordinate system: (𝑟, 𝜃)
a. r: directed distance from the pole
b. 𝜃: angle measured from the polar axis
5. x= r cos𝜃
6. y= r sin𝜃
7. r2=x2 + y2
𝑦
8. tan𝜃=
𝑥
Different types of common polar graphs
**it is important to know what each type of graph looks like so you can visualize the problem and not
waste time graphing out the equation unless specifically told to do so.
*all graphs are shown with cosine. For the sine graph, the image must be symmetrical over the y-axis




Circle
o r =a cos𝜃
o r = a sin𝜃
1
Radius = 2a
y

x





0≤𝜃 ≤𝜋

y




Rose
o r =a cos𝑛𝜃
o r = a sin𝑛𝜃
a= greatest r of the curve
Petals
o Odd n: n
o Even n: 2n

x






y


Limacon
o r = a + bcos𝜃
o r = a + bsin𝜃
a is the axis intercept
b-a= r of inner loop
b+a= r of outer loop
for cardioid graphs, b=a
loop becomes a dent, creating a heart-shaped graph


x















4
Important Mathematicians in Polar
One of the greatest contributors to the field of polar equations, as well as calculus in general, was Jakob
Bernoulli. Jakob Bernoulli, who lived from 1654 to 1705, was among the first in a long line of renowned
mathematicians in the Bernoulli family of Switzerland. He held the
mathematics chair at Basel University and is credited with the early
use of polar coordinates, the discovery of countless high level
applications of calculus, the use of mathematical probability (about
which he wrote the book Ars conjectandi, published 1713), and the
first use of the word “integral” (in reference to calculus) in his
solution of the isochrones curve. Named after him in honor of his
contributions to the field of mathematics are the Bernoulli
distribution, Bernoulli theorem of statistics and probability theory, Bernoulli equation, Bernoulli
numbers and Bernoulli polynomials of number theory interest, and the lemniscate of Bernoulli.
http://library.thinkquest.org/27694/Bernoulli%20Family.htm
**converting back and forth from polar to rectangular form is very important, as we are generally more
accustomed to rectangular coordinates and can more accurately graph the equations in this form. The
slope of a tangent line at any given point can also be determined once polar coordinates are transcribed.
This is useful when writing out the equation of any tangent line.
Basic Analytical Example I
𝜋
4
Find the rectangular coordinate of the point (5, )
1.
x = r cos𝜃
𝜋
x = 5 ∙ cos(4 )
x=5∙
y = r sin𝜃
𝜋
y = 5 ∙ sin(4 )
√2
2
y=5∙
√2
2
𝟓√𝟐 𝟓√𝟐
, 𝟐 )
𝟐
Rectangular coordinate: (
1. Use the basic formulas (found in Important Polar
Basics) and evaluate with given r and 𝜃
5
Basic Analytical Example II
𝜋
What is the line tangent to the graph r = 2sin3𝜃 at 3 ?
1.
x = r cos𝜃
x = (2sin3𝜃)(cos𝜃)
y = r sin𝜃
y = (2sin3𝜃)(sin𝜃)
𝑑𝑥
𝑑𝜃
𝑑𝑦
𝑑𝜃
= (2sin3𝜃)(-sin𝜃)+ (cos𝜃)(6cos3𝜃)
= (2sin3𝜃)(cos𝜃)+ (sin𝜃)(6cos3𝜃)
2.
𝜋
𝜋
𝜋
𝜋
𝜋
𝜋
x( 3 ) = (2sin33 )(cos3 )
y(3 ) = (2sin33 )(sin3 )
x( ) = (2sin𝜋) (cos )
y( ) = (2sin𝜋) (sin )
𝜋
3
𝜋
x( 3 )
𝜋
x( 3 )
𝜋
3
1
𝜋
3
𝜋
y(3 )
𝜋
y( )
3
= (2 ∙ 0)(2)
=0
𝜋
3
√3
= (2 ∙ 0)( 2 )
=0
3.
𝑑𝑥
𝑑𝜃
𝑑𝑥
𝑑𝜃
𝑑𝑥
𝑑𝜃
𝑑𝑥
𝑑𝜃
𝜋
𝜋
𝜋
𝜋
= (2sin33 )(-sin3 )+ (cos3 )(6cos33 )
𝜋
𝜋
= (2sin𝜋)(-sin3 )+ (cos3 )(6cos𝜋)
√3
1
= (2 ∙ 0)(- 2 )+ (2)(6 ∙ -1)
= -3
𝑑𝑦
𝑑𝜃
𝑑𝑦
𝑑𝜃
𝑑𝑦
𝑑𝜃
𝑑𝑦
𝑑𝜃
𝜋
=
𝜋
𝜋
1
√3
= (2 ∙ 0)(2)+ ( 2 )(6 ∙ -1)
= -3√3
−3√3
−3
y-0=√𝟑(x-0)
1. Create a formula bank utilizing the basic formulas to
obtain x, y, and their respective derivatives
𝜋
2. Evaluate position at 3
𝜋
𝜋
𝜋
= (2sin𝜋)(cos3 )+ (sin3 )(6cos𝜋)
4.
𝑑𝑦
𝑑𝜃
𝑑𝑥
𝑑𝜃
𝜋
= (2sin33 )(cos3 )+ (sin3 )(6cos3 3 )
3. Evaluate slope at 3
4. Combine slopes from part 3 to obtain the rectangular
slope for the tangent line
6
What is the ratio of
𝑠1
𝑠2
AP Level Multiple Choice Question
if 𝑠1 is the arc length of one petal of the polar curve 𝑟 = 2sin(3𝜃) and 𝑠2 is the
arc length of one petal of the polar curve 𝑟 = 3cos(2𝜃)?
a)
b)
c)
d)
e)
.613
.816
.889
1.226
1.631
http://chatstickers.com/pusheen-stickers
𝑠=
𝛽
∫𝛼 √(𝑟 2
+
𝑑𝑟 2
[𝑑𝜃] ) 𝑑𝜃
so…
𝜋
3
𝑠1 = ∫0 √[2sin(3𝜃)]2 + [6cos(3𝜃)]2 𝑑𝜃
𝑟 = 2sin(3𝜃)𝑟 = 𝑎sin(𝑛𝜃)
n=3; n is odd, so the rose curve has n petals, or 3 petals.
𝜋
1
Thus, the bounds of 1 petal are equivalent to [0, 3 ], or 3 of a circle.
𝜋
and…
𝑟 = 𝑎cos(𝑛𝜃)
𝑠2 = ∫02 √[3 cos(2𝜃)]2 + [−6 sin(2𝜃)]2 𝑑𝜃
𝑟 = 3cos(2𝜃)
n=2; n is even, so the rose curve has 2n petals, or 4 petals.
𝜋
2
1
4
Thus, the bounds of 1 petal are equivalent to [0, ], or of a circle.
𝜋
∫ 3 √[2sin(3𝜃)]2 + [6cos(3𝜃)]2 𝑑𝜃
𝑠1
= 𝜋0
≈ 𝟎. 𝟔𝟏𝟑
𝑠2
2
2
2
∫0 √[3 cos(2𝜃)] + [−6 sin(2𝜃)] 𝑑𝜃
Explanation of Answer Choices
a) .613
Correct solution.
b) .816
𝑠
𝑠
Solved for 𝑠2 rather than 𝑠1 .
1
2
c) .889
2𝜋
Left out radicals: solved
∫03 [2sin(3𝜃)]2 +[6cos(3𝜃)]2 𝑑𝜃
𝜋
instead.
∫02 [3 cos(2𝜃)]2 +[−6 sin(2𝜃)]2 𝑑𝜃
d) 1.226
2𝜋
𝜋
Solved with incorrect bounds for s1 using [0, 3 ] instead of [0, 3 ].
e) 1.631
𝑠
𝑠
𝜋
Solved for 2 rather than 1 , with 𝑠1’s bounds set [0, ] for a 6-petal rose or 𝑠2’s bounds set [0, 𝜋]
𝑠1
for a 2-petal rose.
𝑠2
3
7
Conceptual Example I
Find all the values of 𝜃 on the interval [0, 2𝜋) where the curve 𝑟 = 5 − 4sin(𝜃) has a vertical tangent.
1.
𝑥 = 𝑟cos(θ)
𝑥 = (5 − 4 sin(𝜃))cos(θ)
𝑦 = 𝑟sin(θ)
𝑦 = (5 − 4 sin(𝜃))sin(𝜃)
𝑥 = 5cos(θ) − 4sin(θ)cos(θ)
𝑑𝑥 = −5 sin(𝜃) − 4(cos2 (𝜃) − sin2 (𝜃))
𝑦 = 5 sin(θ) − 4 sin2(𝜃)
𝑑𝑦 = 5 cos(θ) − 8sin(θ)cos(θ)
2.
Vertical tangent occurs when
𝑑𝑥
𝑑𝑦
y
=0.

x




3.






𝑑𝑥
𝑑𝑦
=
−5 sin(𝜃)−4(cos2(𝜃)−sin2(𝜃))
5 cos(θ)−8sin(θ)cos(θ)
𝑑𝑥 = 0, 𝑑𝑦 ≠ 0



𝑑𝑥 = −5 sin(𝜃) − 4[cos 2(𝜃) − sin2 (𝜃)] = 0

−5 sin(𝜃) = 4[cos
2 (𝜃)
2

− sin (𝜃)]

5
−4 =
cos2(𝜃)−sin2(𝜃)
sin(θ)
𝛉 = 𝟑. 𝟔𝟐𝟎
𝛉 = 𝟓. 𝟖𝟎𝟓
1. Create a formula bank
2. Find the derivative
𝑑𝑥
3. Set
= 0 and simplify
𝑑𝑦
*Check that 𝑑𝑦 ≠ 0.
Winplot.com







8
Conceptual Example II
Convert the rectangular equation 𝑥 2 + 𝑦 2 + 3𝑥𝑦 − 4 = 0 to polar form.
1.
𝑥2 + 𝑦2 = 𝑟2
𝑥 = 𝑟cos(𝜃)
𝑦 = 𝑟sin(𝜃)
2.
𝑟 2 + 3𝑟cos(𝜃)𝑟sin(𝜃) − 4 = 0
𝑟 2 + 3𝑟 2 cos(𝜃)sin(𝜃) = 4
𝑟 2 [1 + 3 cos(𝜃)sin(𝜃)] = 4
𝑟2 =
4
1 + 3 cos(𝜃)sin(𝜃)
𝑟=±
2
√1 + 3 cos(𝜃)sin(𝜃)
1. Create a formula bank
2. Rearrange and simplify the
equation to arrive at the polar form
http://www.teacherschoice.com.au/maths_library/coordinates/polar_-_rectangular_conversion.htm
Real World Applications
One of the first applications of polar was to understand the circular motion
of planets as ancient astronomers tried to chart the heavens. Later on,
scientists used this system to decipher the path of an electron traveling
around a nucleus and developed radar to aid in nautical travel and aviation.
The newest popular function of these polar graphs is their usage in
microphones, as the pick-up pattern to most are cardioid-shaped.
http://en.wikipedia.org/wiki/Polar_coordinate_system
http://www.wmich.edu/math/kaap/modules/polar-complex/polar-complex.swf
9
**polar integration is also very important and a concept often tested on the AP exam. Make sure to take
note of the bounds, as most graphs are between 0 and 2𝜋. An inacurate bound can result in an
exaggerated answer as the same area will be accounted for multiple times.
Graphing Calculator and Graph Example
Find the area of the inner loop of r = 1 + 2sin𝜃.
3.0
1.
r = 1 + 2sin𝜃
2.5
0 = 1 + 2sin𝜃
2.0
-1 = 2sin𝜃
1.5
−1
2
1.0
= sin𝜃
𝜃=
7𝜋 11𝜋
,
6
6
0.5
2.
2
1
2
1
1
2
http://demonstrations.wolfram.com/PolarAreaSweep/
0.5
∫ 𝑟 2 𝑑𝜃
11𝜋
1
∫7𝜋6
2
(1 + 2sin𝜃)2 𝑑𝜃
6
.5 (fnInt (r12, 𝜃,
7𝜋 11𝜋
,
))
6
6
0.544
1. Set r equal to zero to find the bounds
of the inner loop
2. Integrate with the bounds found in
part 1 using the basic formula to
determine the area of the inner loop
http://chatstickers.com/pusheen-stickers
10
AP Level Free Response

y



Find the area of the shaded parts of the graph.



x
𝑟 = 4 + 8 cos(𝜃)




















𝑟=3



1.
Where the limaçon has a
positive radius, the
curves intersect at the
same value 𝜃; here we
set the two curves equal
to each other and solve
for 𝜃:
4 + 8 cos(𝜃) = 3
8 cos(𝜃) = −1
1
cos(𝜃) = −
8
𝜃 = 1.696
𝜃 = 4.587


Where the limaçon has a
negative radius, the
curves do not intersect at
the same value 𝜃; here
we solve for the value
𝜃 𝑜𝑓 4 + 8 cos (𝜃):
The values 𝜃 for 𝑟 = 3 at
these same points of
intersection are exactly 𝜋
radians away (opposite
the circle because 𝑟 = 3
has a positive radius):
4 + 8 cos(𝜃) = −3
8 cos(𝜃) = −7
7
cos(𝜃) = −
8
𝜃 = 2.636
𝜃 = 3.647
𝜃 = 2.636 − 𝜋
𝜃 = 3.647 − 𝜋
Intersection of 𝑟 =
4 + 8 cos(𝜃) with
the pole:
4 + 8 cos(𝜃) = 0
8 cos(𝜃) = −4
1
cos(𝜃) = −
2
2𝜋 4𝜋
𝜃=
,
3 3
2.
Little part:
[Area within limaçon curve]-[Area within circle]
1 3.647
1 3.647−π 2
∫
(4 + 8 cos(θ))2 dθ − ∫
3 dθ
2 2.636
2 2.636−π
1
Big part x2 (2 cancels 2):
[Area within limaçon curve]-[Area within circle]
1.696
∫
2
4π
3
3 dθ − ∫
3.647−π
(4 + 8 cos(θ))2 dθ
3.647
3.
4𝜋
1.696
1 3.647
1 3.647−𝜋 2
3
(4 + 8 cos(𝜃))2 𝑑𝜃 − ∫
[ ∫
3 𝑑𝜃] + [∫
32 𝑑𝜃 − ∫ (4 + 8 𝑐𝑜𝑠(𝜃))2 ] = 𝟏𝟏. 𝟏𝟐𝟗
2 2.636
2 2.636−𝜋
3.647−𝜋
3.647
~remember to take this problem in steps, solving the parts instead of being overwhelmed by the whole
1. Find the points of intersection to set bounds
2. Use integrals to calculate area
3. Add together all of the areas and arrive at the final answer
http://chatstickers.com/pusheen-stickers
11
~and here concludes Pusheen’s desire to assist you with polar equations; he is now scheduled to sleep on
a marshmallow~
pusheen.com
12
Exercises
Follow the specific directions given for each question to solve using your knowledge
of polar equations. *denotes calculator problems.
Pusheen.com
*Analytical Exercises
1. Convert the rectangular coordinate (1, −√3) to a polar coordinate.
2. Sketch the curves of 𝑟 = 3sin(2𝜃) and 𝑟 = 3. Label all points of intersection and indicate direction.
3. Convert the rectangular equation 3𝑥 2 + 𝑦 2 − 𝑥𝑦 − 4 = 0 to a polar equation.
4. Convert the polar equation 𝑟 =
−2 sin(θ)−3
1−cos2(𝜃)
to a rectangular equation.
𝜋
5. Find the bounds of 𝜃 for a single tracing of the curve 𝑟 = 3sin(𝜃 − 4 ).
𝑑𝑦
𝜋
6. Find 𝑑𝑥 of 𝑟 = −7cos(𝜃) at 𝜃 = 3 .
7. Find all the values of 𝜃 (to 3 decimal places) for when the curve 𝑟 = 1 + 4sin(θ) has a horizontal or
vertical tangent.
𝑑2 𝑦
8. Find 𝑑𝑥 2 of 𝑟 = 3sin(𝜃) at 𝜃 =
𝑑𝑟
5𝜋
.
6
𝜋
9. Find 𝑑𝜃 of 𝑟 = 4 sin(𝜃) cos(2𝜃) at 𝜃 = 4 .
10. Find the area of 3 petals of the rose curve 𝑟 = 2 sin(4𝜃).
11. Find the area of the shaded sections of the graphs 𝑟 = 3 − 4 cos(θ) and 𝑟 =
−5cos(𝜃).
12. Find the arc length of 2 petals of the rose curve 𝑟 = 4sin(3𝜃).
13. Find the rectangular equation for all the tangent lines at the pole of the curve 𝑟 = 1 − 4sin(𝜃).
14. Find all the values 𝜃 on the interval [0, 𝜋) where the curve 𝑟 = 6sin(5𝜃) intersects the pole.
15. Find the rectangular coordinates of all the points of intersection between the curves 𝑟 = −4sin(3𝜃)
and 𝑟 = 3.
16. Find the surface area of the shape formed by revolving the curve 𝑟 =
𝜋
sin(θ) + 3sin(2θ) about the line 𝜃 = 2 .
17. Find the area of the shaded sections of the graph 𝑟 = 2 −
3 sin2(𝜃) − 3cos(𝜃).
13
18. Given the triangle pictured, derive a polar function 𝑟1 with the hypotenuse as
the radius and a polar function 𝑟2 with the leg 𝑥 as the radius. Position 𝜃 at
the pole. Calculate [area enclosed by 𝑟1 ] – [area enclosed by 𝑟2 ].
AP Multiple Choice
1. Find the surface area of the graph 𝑟 = 6𝑠𝑖𝑛𝜃 rotated about the 𝑥 axis.
a) 36𝜋
b) 0
c) 72𝜋 2
d) 36𝜋 2
e) 18𝜋
Use the graphs 𝑟 = 3 + 2𝑠𝑖𝑛2 𝜃 and 𝑟 = 6 − 6𝑐𝑜𝑠 2 𝜃 for questions 2 and 3.
2. At what point(s) do the above graphs intersect?
a) No points of intersection
b)
c)
d)
e)
𝜋 2𝜋
,
3 3
𝜋 3𝜋
,
2 2
𝜋 2𝜋 4𝜋 5𝜋
, , ,
3 3 3 3
𝜋 5𝜋 7𝜋 11𝜋
, , ,
6 6 6
6
3. *Which integral represents the common area of the above graphs?
1
𝜋
𝜋
1
a) 4 ∙ ∫03 (6 − 6𝑐𝑜𝑠 2 𝜃)2 𝑑𝜃 + 4 ∙ ∫𝜋2 (3 + 2𝑠𝑖𝑛2 𝜃)2 𝑑𝜃
2
2
3
b)
𝜋
3
1
∫ (6 −
2 0
1
1
𝜋
2
𝜋
3
6𝑐𝑜𝑠 2 𝜃)2 𝑑𝜃 + ∫ (3 + 2𝑠𝑖𝑛2 𝜃)2 𝑑𝜃
2
𝜋
1
𝜋
c) 4 ∙ 2 ∫03 (3 + 2𝑠𝑖𝑛2 𝜃)2 + 4 ∙ 2 ∫𝜋2 (6 − 6𝑐𝑜𝑠 2 𝜃)2 𝑑𝜃
3
𝜋
d) 4 ∙
1 2
∫ (6 −
2 0
1
𝜋
6𝑐𝑜𝑠 2 𝜃)2 𝑑𝜃
1
𝜋
e) 4 ∙ 2 ∫03 (6 − 6𝑐𝑜𝑠 2 𝜃)𝑑𝜃 + 4 ∙ 2 ∫𝜋2 (3 + 2𝑠𝑖𝑛2 𝜃)𝑑𝜃
3
14
4. Using 𝑟 = 𝑐𝑜𝑠𝜃 to represent the litter box and 𝑟 = 2 − 𝑐𝑜𝑠𝑥 to represent the area wherein the
litter is spread on the floor by Pusheen, find the total
contaminated area. The area A is bounded by the y-axis, x-axis,
shown graph, and the line 𝑥 = 𝜋.
𝜋
a) 𝐴 − ∫0 𝑐𝑜𝑠 2 𝜃𝑑𝜃
2𝜋
b) ∫0 (2 − 𝑐𝑜𝑠𝑥)2 𝑑𝜃 − 2𝐴
𝜋
c) 2𝐴 − ∫0 𝑐𝑜𝑠 2 𝜃𝑑𝜃
2𝜋
d) 2𝐴 − ∫0 𝑐𝑜𝑠 2 𝜃𝑑𝜃
2𝜋
e) ∫0 (2 − 𝑐𝑜𝑠𝑥)2 𝑑𝜃 − 𝐴
y



A

x













5. Find Pusheen’s rectangular (cartesian) velocity vector at 𝑡 = 𝜃 =
11𝜋 11𝜋√3
, 6 ⟩
6
11𝜋 3
11𝜋
⟨ √ − 1,
+ √3, ⟩
6
6
−11𝜋√3
11𝜋
⟨
− 1, 6 + √3, ⟩
6
−11𝜋√3
−11𝜋
⟨
− 1,
+ √3, ⟩
6
6
11𝜋
11𝜋√3
⟨
+ √3, 6 − 1⟩
6
a) ⟨
b)
c)
d)
e)
11𝜋
6
as he travels along 𝑟 = 2𝜃.
15
*AP Free Response
y


Pusheen’s living room is represented by
the graph 𝑟 = 3 + 𝑐𝑜𝑠4𝜃. A
portion of that room is
indicated by 𝑟 = 5𝑐𝑜𝑠2𝜃.


Note that the shaded
area is bounded by 𝑟 =
3 + 𝑐𝑜𝑠4𝜃!


x






a) If Pusheen travels around the perimeter of the room, exploring his domain, what are the
rectangular coordinates of his position at 𝑡 = 3.25 if 𝜃 = 𝑡?
b) Find the total distance he has walked to reach this point.
c) Pusheen is allowed only in designated areas of the living room after committing the “crimes”
seen below. The shaded region of the graph indicates the section of the room he does not have
access to. Please find the area of his new habitable quarters.
𝑑𝑟
𝑑𝜃
at 𝜃 =
11𝜋
6
and describe Pusheen’s movement at that point in relation to the food in the
center of the room.



d) Find

winplot.com
16
Solution Guide
Analytical Exercises Solutions
5𝜋
)
3
𝜋
3𝜋
5𝜋
7𝜋
(3, 4 ) (3, 4 ) (3, 4 ) (3, 4 )
2
1. (2,
2.
3. ±
√2𝑐𝑜𝑠2 𝜃+1−𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃
2
2
4. 𝑦 = −2𝑦 − 3√𝑥 + 𝑦 2
𝜋 5𝜋
5. [ 4 , 4 )
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
𝑑𝑦
𝑑𝑥
=
−7
)
4
−7√3
2(
)
4
−7−2(
𝜃 = 0.704, 2.437, 4.024, 5.401 (vertical) 𝜃 = 1.571, 3.267, 4.712, 6.158 (horizontal)
16
3
−4√2
2.356
33.773
17.820
𝑦 = −0.258𝑥 𝑦 = 0.258
𝜋 2𝜋 3𝜋 4𝜋
0, , , ,
5 5 5 5
(−2.883, −0.838)(0.716, 2.914)(2.165, −2.076)(−2.161, −2.075)(−0.716, 2.911)(2.880, −0.836)
107.565
9.228
78.540
Multiple Choice Solutions
1. D
𝑏
𝑑𝑟
a) No multiplication by 2𝜋, incorrect equation ∫𝑎 𝑟𝑠𝑖𝑛𝜃√(𝑟)2 + (𝑑𝜃)2 𝑑𝜃
b) Rotation around the wrong axis, incorrect equation
𝜋
2𝜋 ∫0 6𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃√(6𝑠𝑖𝑛𝜃)2 + (6𝑐𝑜𝑠𝜃)2 𝑑𝜃
c) Incorrect bounds [0, 2𝜋] used
d) Correct answer
e) U-substitution mistake when integrating 1-cos2a, basic mathematical error when evaluating
𝜋
cos2a]0
2. D
a) dropped negative produces non-real answer 𝑠𝑖𝑛𝜃 = √
b)
c)
d)
e)
3. A
a)
b)
c)
−3
8
Both positive and negative answers not considered evaluating √𝑠𝑖𝑛2 𝜃
Incorrect application of sin2+cos2=1, basic math errors
Correct answer
1
𝜋 5𝜋 7𝜋 11𝜋
Incorrect evaluation of trig function 𝑐𝑜𝑠 ≠ , , ,
2
6
6
6
6
Correct answer
Only 1 quadrant taken into account, no multiplication by 4
Incorrect equations for specified integrals and bounds
17
d)
e)
4. C
a)
b)
Area of only 1 graph, included area not shared by both graphs
Incorrect application of basic formula, 𝑟 2 replaced by 𝑟
Incorrect area for 𝑟 = 2 − 𝑐𝑜𝑠𝑥, evaluated [0, 𝜋] rather than [0, 2𝜋]
Used 2𝐴 to represent area of the litter box, incorrect equations matched with incorrect
bounds for each equation
c) Correct answer
d) Incorrect bounds, should be [0, 𝜋] rather than [0,2𝜋]
e) Incorrect equation paired with graph, bounds match the equation but are again in the
wrong spot
5. E
a) Incorrect velocity formula, wrong derivation of 2𝜃
b) Incorrect position functions, 𝑥 = 𝑟𝑠𝑖𝑛𝜃 and 𝑦 = 𝑟𝑐𝑜𝑠𝜃 used
c) Dropped negative
d) Incorrect evaluation of trig functions, 𝑐𝑜𝑠
e) Correct answer
11𝜋
6
≠
−1
2
and 𝑠𝑖𝑛
11𝜋
6
≠
√3
2
Free Response Solutions
a) correct answer (1): (-3.885, -0.423)
3.25
b) correct equation (1): ∫0 √(3 + 𝑐𝑜𝑠4𝜃)2 + (−4𝑠𝑖𝑛4𝜃)2 𝑑𝜃
correct answer (1): 13.258
1 2𝜋
1 𝐿
c) correct equations, half credit may be given(2): ∫0 (3 + 𝑐𝑜𝑠4𝜃)2 𝑑𝜃 − [ ∫𝐾 (3 + 𝑐𝑜𝑠4𝜃)2 𝑑𝜃 +
𝐾
∫𝐽 (5𝑐𝑜𝑠2𝜃)2 𝑑𝜃]
2
2
correct bounds (1): 3 + 𝑐𝑜𝑠4𝜃 = 5𝑐𝑜𝑠2𝜃 yields 2.618 (K) and 3.665 (L)
0 = 5𝑐𝑜𝑠2𝜃 yields 2.356 (J)
correct answer (1): 23.343
𝑑𝑟
d) correct derivative (1): 𝑑𝜃 = −4𝑠𝑖𝑛4𝜃
correct answer with interpretation (1): −2√3 Pusheen is moving towards the food (origin)
𝑑𝑟
because 𝑑𝜃 is negative
Works Cited
http://en.wikipedia.org/wiki/Polar_coordinate_system
http://www.wmich.edu/math/kaap/modules/polar-complex/polar-complex.swf
http://demonstrations.wolfram.com/PolarAreaSweep/
www.pusheen.com
www.winplot.com
http://library.thinkquest.org/27694/Bernoulli%20Family.htm
http://www.teacherschoice.com.au/maths_library/coordinates/polar_-_rectangular_conversion.htm
http://apcentral.collegeboard.com/