Parametric Equations & Polar Graphs - Unit 11
Plane curves:
This
Chap 11
□ Sample Problem (by Becca Bassett and Theo Kulczycki): This is an hard problem designed to give
students practice with parametric equations.
Problem:
d2y
Find 2 .
d x
y  4 5t 2  3t

x
12t
1 2t 3
Solution:

To find the second derivative we have to first find just dy/dx. We have been given parametric equations for x
dx
dy
and y so the first thing that you have to do is find
and
dt
dt
1/ 4
2
y  5t  3t 
3 / 4
dy 1 2
 5t  3t  (10t  3)
dt 4
dy
10t  3

dt 4(5t 2  3t) 3 / 4
12t
x
1 2t 3
3
2
dx 1 2t 12  12t 6t 

2
dt
1 2t 3 


dx 12  24t 3  72t 3

2
dt
1 2t 3 
dx 12  48t 3

dt 1 2t 3 2

now that we have dx/dt and dy/dt we can find dy/dx
dy dy dt
 
dx dt dx
so we are going to divide dy/dt by dx/dt


10t  3
45t 2  3t 

12  48t 3
1 2t 
10t  31 2t 

45t  3t  12  48t 
3/4
3 2
3 2
2
3/4
3
now that we have dy/dx we have to take the second derivative
use low* D High – high*D Low and the product rule to find the following IMMENSE derivative

45t  3t
2
3/4
12  48t  10t  321 2t6t   101 2t   10t  31 2t   45t
45t  3t 12  48t 
3
3 2
2
3 2
3/4
2
 3t 
3/4
144t  3
2
3

Chap 11
Sample Problem (by Becca Bassett and Theo Kulczycki): This is an intermediate problem designed to help
students graph polar conics and find points on the horizontal tangent line
Problem:
r  2  3cos find all points with a horizontal tangent line
Solution:
First, make a graph of the function.
To do this, look at the range of r. r  2  3cos , the largest cos can be is 1 and the smallest it can be is -1
So, 1  r  5 . Because the equation is the form r  a  bcos we know that the --- is a limaçon positioned
horizontally. Because r could be negative we also know that the limaçon has an inner loop.



Next we know that y  
rsin 

To find the horizontal tangent line, find y prime
dy
 r cos   r sin 
d

 2  3cos  cos   (3sin  )sin 


 2cos  3cos 2   3sin 2 
 2cos  3cos 2   sin 2  
2cos  3cos2  0



  0.4665,2.16996,4.1182,5.81669
r  0.6795,3.6795,3.6795,0.6795
20) Chap 11
□ Sample Problem (by Becca Bassett and Theo Kulczycki): This is a hard problem designed to help
students practice graphing polar conics and finding the area of a polar conic
Problem:
Find the area inside the circle but outside the cardiod (not including the area of the cardiod)
r  4 cos 
r  1 cos 

You should draw to draw your own picture to visualize where the circle and cardiod cross. We tried really hard
but could not get the program to draw a decent picture...
First you have to find where the cardiod and the circle cross. To do this, set the equations equal to each other.
4 cos   1 cos 
3cos   1
1
3
  0.9999830769, 5.28320223
cos   

now to find the area you have to do the area of the circle minus the area of the cardiod from the two angles of
intersection that you just found.
1 b 2
A   r d
2 a
A


2
2
1 0.999
4 cos   1 cos  d


2 5.283
use numerical integration and you should get, A=8.828

Chap 11: □ Sample problem (by Derek Caetano-Anolles & Ken Cebrian): Polar Bears
A bear is drawing a circle with radius 2 meters in the snow and wants to know how much area is under it from 0 to 
radians. Help him find out using polar integration.
r2
1
A
2

f
 A
r( ) 2 d
o
1
2


0
2 2 d  2  0 d


 2 0  2[  0]
 2 meters2

Chap 11: □ Sample problem (by Derek Caetano-Anolles & Ken Cebrian): Polar Areas Between Curves
Find the area outside the circle r=2acos , and inside the cardioid r=a(1+cos )
Since the object is symmetrical to the polar axis, we can use symmetry to find the area above the axis and double
it to find the total area. The area in the first quadrant can be found by subtracting the area of the cardioid from
the area of the circle. The area in the second quadrant can be found by integrating the cardioid alone.
And double it, of course.
A
1
2
 r d

2
 Atotal  2 

2
0
 1
 1
1
[a(1 cos )]2 d  2  2 [2acos ]2 d  2   [a(1 cos  )]2 d
0 2
2
22


 a 2  0 2 (1 2cos  cos 2  )d  a 2  0 2 4 cos2  d  a 2   (1 2cos   cos2  )d

2
 a2 

0
2
(1 2cos  3cos 2  )d  a 2   (1 2cos  cos 2  )d

2



 a 2  0 2 d  2a 2  0 2 cos d  3a 2  0 2 cos 2  d  a 2   d  2a 2   cos  d  a 2   cos 2  d


2

2
2


 



 a 2 0 2  2sin  0 2  3  0 2 cos 2  d     2sin      cos 2  d


2
2
2
Substituting :
1 cos2
cos2  
2
 1 cos2

 1 cos2
 



 Atotal  a 2  0 2  2sin  0 2  3  2
d     2sin     
d
0
2
2


2
2
2



 

3 
3 


 a 2  0 2  2sin  0 2   2 d   2 cos2 d     2sin      d    cos2 d
0
0
2
2


2
2
2
2


 


1
 
3  2 3 1



2
2
2

 a  0  2sin  0   0   sin 2     2sin        sin 2 
2
2
2
0
2
 
2
2 2


2 

3

 
 a 2   2 
    2    
2
4
2
2 
3 
 a 2  
 4 


 Atotal 
3a 2
4
□ Sample problem (by Fan Huang & Fernanda Mendez):

Chap 11: Polar parab to rect.
a. Find the polar equation for a parabola whose focus is at the pole and whose vertex is at (4,0).
Solution: Because the vertex is to right of the pole,
the parabola is in the form of r 
ed
1  e cos( )
.
The eccentricity of a parabola is always 1.
d = twice the distance from the focus to the vertex =
8. Therefore, r 
b. r 
8
.
1  cos( )
25
Identify the conic and write the equation in rectangular form. Find a, b, c, d, and e.
9  4 sin 
Solution: Based on the equation r 
ed
, we know that e = 4, and d = 16/e = 4.
1  e sin 
25
25 25
4
25
9
The equation can be rewritten as r 
, and from that we can see e  , and d  9  9 
.
4
4
9
e
4
1  sin 
9
9
Plot the graph of the conic, which is an ellipse because e < 1.
P1 F
P2 F
a2
25
25
c  c2  a2 .
From the equation e 
, we eventually get d  c 
. Plug in
for d to get

c
4
4
P1Q1 P2 Q2
From the equation e 
Setting
c
c
c 2 81c 2
, we find a 2  2 
.
a
16
e
25
c 2 81c 2
81c 2 25
c  c 2  a 2 and a 2  2 

c  c 2 . Solve equation to get
equal to each other, we get
4
4
16
16
e
20
.
13
20
c
45
.
a   13 
4
e
13
9
b  a2  c2 
5 65
.
13
Therefore, a 
5 65
45
20
25
4
,b
, c
, d 
, and e  .
13
13
13
4
9
□ Sample problem (by Fan Huang & Fernanda Mendez):
parametric equations? Someone as smart as you, for sure,
knows the arc length formula, and how to do basic integration! Chap 11: A particle in a plane with positions:
x = sec(t) + tan(t)
y = sec(t) - tan(t)
at time t, where 0 ≤ t ≤ 1.5
Solution:
a. Find y in terms of x.
Solution: xy = [sec(t) + tan(t)] [sec(t) - tan(t)] = sec2(t) – tan2(t) = 1  y =
1
x
b. From t = 0 to t = 1.5, how far has the particle traveled?
Solution:
dx
 sec t tan t  sec 2 t
dt
tf
Arc length =

t0
dy
 sec t tan t  sec 2 t
dt
2
2
 dy   dx 
     dt
 dt   dt 
 sec t tan t  sec t   sec t tan t  sec t  dt = 27.391
1.5
=
2
2
2
0
□ Sample problem (by Fan Huang & Fernanda Mendez):
lines: r  7 cos( 4 )
Chap 11 Polar graphs, arc len, max r, horiz & vert tan
a. What polar graph is this?
Solution: A rose with 8 petals.
b. Find the area and the arc length of the graph.

Solution: Area =
2
Area =
1
 r d
 2
2
 2 7 cos(4 ) d 
1
2
0
49
2

 r 

' 2
Arc Length =
 r 2 d
2
=
2
  28sin( 4 
2
 7 cos(4 ) d = 120.098
2
0
 
?
 4 
c. Where, if anywhere, does the graph have vertical or horizontal tangents over the interval 0,
Solution: There are horizontal tangents at points where
dy
=0.
d
r  7 cos( 4 )
y  r sin(  )  7 cos( 4 ) sin(  )
dy
 7 cos( ) cos( 4 )  28 sin(  ) sin( 4 )
d
7 cos( ) cos( 4 )  28 sin(  ) sin( 4 )  0 when  = 0.213874
dx
There are vertical tangents at points where
= 0.
d
r  7 cos( 4 )
x  r cos( )  7 cos( 4 ) cos( )
dx
 7 sin(  ) cos( 4 )  28 cos( ) sin( 4 )
d
 7 sin(  ) cos( 4 )  28 cos( ) sin( 4 )  0 when  = 0 and  = 0.730336 .
 
?
 4 
dr
 0.
Solution: r is a relative extremum, where
d
dr
 28 sin( 4 )
d
d. Where is r a maximum, over the interval 0,
 28 sin( 4 )  0 when  = 0,

4
20 □ Sample problem (by Fan Huang & Fernanda Mendez):
r1 ( )  4  4 sin 
r2 ( )  8 sin 
Chap 11 Polar graphs, arc len, area
a. Graph these equations and identify what types of polar graphs they are.
Solution:
r1 ( ) is a cardioid
r2 ( ) is a circle
b. Find the arc length of r1 ( ) .

Solution: Arc Length =
 r 
' 2

 r 2 d =
2
 4 cos( )
2
 4  4 sin(  ) d = 32
2
0
c. Find the area inside of r1 ( ) , but outside of r2 ( ) .

Solution: Area =
1
 r d
 2
2
2

1
1
2
2
0 2 r1 ( ) d  0 2 r2 ( ) d
(The limits for r2(  ) are 0 and  because the circle
Area
of green region =
is
complete from 0 to  )
1
=
2
2
 4  4 sin(  )
0
2

2
1
d   8 sin(  ) d = 8 
20
Chap 11: □ Sample problem (by Frances Ha, Lusiana Hadi, & Amity Xu): advanced] Find the area of the region that
lies outside the circle r  1 and inside the first-quadrant
petal of the curve r  2 sin 3 .
  /3
Solution: Since the region consists of same three
pieces, first let us find the area of one piece of whole
region. In the 1st quadrant, the graph is as the above
for 0   

3
r  2 sin 3
.
-1
1
r 1

r

0
18
0
2

12

9

6

3
2 2
2 3
2
0
So the whole area is

3
1
A  3 {( 2 sin 3 ) 2  12 } d
2
0

3
1
 3 {4 sin 2 3  12 } d
2
0


33
1  cos 6
{4
 1} d

20
2

3
sin 6
3
 2( 
)  
2
6
0


3  0 
2(  )  
2  3 6
3

2
Chap 11: □ Sample problem (by David Mesri & Jake Mathis): You are given the parametric equation: xt  
y t   3t 2  4 . You are then asked to find the arc length from 0 to 30.
3t ,
Solution: In order to find the parametric arc length, you must remember how to derive the equation. First, here is a
review to deriving the original arc length formula.
xt  
dx
3

dt
2 3t
3t
y t   3t 2  4
ds 2
dx 2
 dx   dy 
2
b
s 
dy
 6t
dt
 dx 
2
2
 dy 
2
 dy 
2
2

 dy  
1    
 dx  


dx 

2

 dy  
1

   dx

 dx  

2
a
b
s 

a
2
 dy 
1    dx
 dx 
Now it’s time to derive the parametric arc length equation.
2
ds 
 dy 
1  
 dx 
ds 
 dy dt 

1  
 dx dt 
ds 
 dx   dy 
   
 dt   dt 
2
 dx 
 
 dt 
dx
2
tf
S 
 dx 
 
 dt 

to
30
S 

0
2
 3

 2 3t
2
dx
2
 dy 
  
 dt 



2
dx
 dt
dt
2
dt
 6t  dt
2
S  2700.95
Chap 11: □ Sample problem (by David Mesri & Jake Mathis): Find all the relative extrema of r  5  cos2  .
Solution: Relative extrema occur where the derivative is zero.
dr
 2 sin 2   0
d
Chap 11: □ Sample problem (by David Mesri & Jake Mathis): You are given the polar equation: r  3  4 cos  .
a. Determine what the graph is and how it is orientated.
Solution: From the equation r  3  4 cos , it can be determined that you are dealing with a limaçon with an
inner loop (because of minus) with symmetry over the x-axis (because of cos).
b) Find the area.
Solution: In order to figure out the area, we must derive a new method for finding the area since we will be in polar
coordinates.
d
dA
r 2 d


dA

 A 
2
2
 r2
A  53.41

r 2 d
 2 So, A 
2

0
3
 4 cos  
d
2
2
*Found using numerical integration
c. Find the arc length.
Solution: To find the arc length, another formula must be derived in order to apply polar coordinates.
x  r cos
y  r sin 
dx
 r sin   r ' cos
d
dy
 r cos   r ' sin 
d

2
2
 dx 
 dy 
s 
  
 d
d

d






r '2 cos 2 
=
r2
 r 2 sin 2   2rr ' sin  cos  r 2 cos 2   r ' sin 2   2rr ' sin  cos
2
 r '
2
*Through ugly algebra this is found

So a simplified equation is: s 
2
s
 3
*The radican is as follows
 4 cos


r 2  r ' d . Now we can solve the problem
2
  4 sin  
2
2
d  s  28.81 .
0
2
2
Chap 11: □ Sample problem (by Morgan Holbrook & Justin Parks): Given 5x  5y  50x  25y  0 .
to polar.
a. Convert
2
2
2
2
Solution: 5r cos   5r sin   50r cos  25r sin   0
5r 2 (cos2   sin 2  )  25r sin   50r cos
r  5sin  10 cos
b. Find r when θ=(п/2)


r  5 sin( )  10 cos( )
2
2 r = 5.
Solution:
chap 11: □ Sample problem (by Merla Hübler & Lisa Portis): polar graphs: You are sitting and doing your calculus
homework when you see an ant walking along a polar graph on your paper. The ant’s radial velocity is given by the
equation 2 cos5 2 and the ant has an initial polar coordinate of (5, 4 ). Find ant’s position in terms of x and y coordinates
when  
. At what values of  is the ant farthest away from the pole? If the ant started to walk along a straight

6
horizontal line at the previous y coordinate (when  

6
), write the polar equation for its new path.
dr
dt

2 cos 2
5
Solution: You can solve the differential equation
the initial polar coordinate given. Then, solve for r when  
for r and then find C by using

6
and use the conversion equations
x  r cos  , y  r sin  to get the ant’s x and y coordinates. To find out when the ant is farthest away from the pole,
you have to find the values of  for which ddr  0 . The conversion equations can be used again to get the polar
equation for the ant’s horizontal path.
 ddr   2 cos5 2
 5dr   2 cos 2
u  2
du  2d
5r   cos udu  sin u  C  sin 2  C
r  15 sin 2  C
5  15 sin 2( 4 )  C
25  sin

2
 5C
25  1  5C
C
24
5
When  

6
:
r  15 sin 2( 6 ) 
r  15 sin( 3 ) 
24
5
24
5
 15 ( 23 ) 
24
5

3
10

24
5
)
24
5
x  r cos 
x  ( 103  245 )(cos 6 ) 
3
2
( 103 
3
20
 125 3
y  r sin 
y  ( 103  245 )(sin 6 )  12 ( 103 
24
5
So, the ant’s position when  

6
)
3
20
 125
3
is ( 20
 125 3 ,
3
20
 125 ).
dr
d

2 cos 2
5
0
 cos 2  0
   4 , 34 , 54 ,...
5π
The ant is farthest away from the pole when θ  π4 , 3π
4 , 4 ,...
When y 
r sin  
r
3
3
20
3
20
12
20
5
sin 

12
5
 125

3  48
20 sin 
The ant’s horizontal path is given by the polar equation r=
3  48
20sinθ
.
chap 11: □ Sample problem (by Merla Hübler & Lisa Portis): area of a polar graph: Find area contained by r =
10  cos .
Solution: A 
1 f
1 2
1 2
[r ( ) 2 ] d =  [10  cos  ] 2 d =  [100  20 cos   cos 2  ] d

2 0
2 0
2 0
(use double angle formula for cos2θ)
1
1 2
1 2
1
1  cos 2
100

20
cos


]
d

[
100


20
sin



]

cos 2 d [u = 2θ, du = 2 dθ]
=
0
2 0
2
2
2
2 0
201
1
201
1
201
201π
1
1
A  [( 200   )  (0)]   cos u du =
 [sin u ] =
 [sin 2 ]02 =
0 
.
2
4
2
4
2
4
2
2
A
2
chap 11: basic Patrick McCall & Nathan Dornfeld: polar area Find the area of the polar curve r = 3 f rom  = 0 to .

Solution: A 
9 
9
1 2
.
3 dx  x 0 

2
2
20
Chap 11: (by Shaofeng Sun & Artem Rogachev): basic physics and parametric equations. A particle moves in an xy coordinated system, its speed vector is (2t+1, 5-t). What is the particle’s position after 5 seconds if the particle started
out from the origin with initial velocity = 0?
Solution: We can find x and y displacements separately using the fact that displacement is the integral of velocity.
X direction displacement:
5
5
0
0
 (2t  1)dt =30. Y direction displacement:  (5  t )dt =12.5. So final position is (30,12.5).
Chap 11: (by Shaofeng Sun & Artem Rogachev): advanced parametric equations. There is a system of 3 bodies in
space, green moon is going around yellow planet, and yellow planet orbits the black matter. Period of green moon is P 1
and its radius of curvature is r1. The period of yellow planet is P2 and radius is r2. Assume at t=0 all three bodies are lined
up on X-axis as shown in the figure above. Describe the position of the green planet with a function of time.
Solution: This problem may seem hard to you at first but if you think about it’s not bad. All we need to do is express
the motion of the green moon relative to the yellow planet, and express the motion of yellow planet relative to the dark
matter. Later we can combine these two equations. First let’s write the position of green planet relative to the yellow
planet:
x  r1 cos y  r1 sin  . Now express θ in terms
2
2
of t:  
t , now substitute: x  r1 cos( t
) ,
P1
P1
2
) . now express the position of yellow
P1
planet relative to the dark matter: x  r2 cos
2
2
y  r2 sin    t
x  r2 cos(t
),
P2
P2
2
y  r2 sin(t
) express the position of green planet
P2
y  r1 sin(t
relative to the dark matter:
x  r1 cos(t
2
2
)  r2 cos(t )
P1
P2
y  r1 sin(t
2
2
)  r2 sin(t )
P1
P2
Chap 11: (by Shaofeng Sun & Artem Rogachev): basic polar-rectangular coordinate conversions. Change the
following rectangular equation into polar form:
y2 x2

 1.
25 36
r 2 sin 2  r 2 cos2 
Solution: Remember that rcosθ = x, and rsinθ = y. so after substituting those in you get

 1.
25
36
Then multiply both sides by 900. you get: 36r 2 sin 2   25r 2 cos2   900 . Distribute r2 and divide 900 by what is
900
900
left, you get: r 2 
, hence r 
.
2
2
2
(36 sin   25 cos2  )
(36 sin   25 cos  )
Chap 11: (by Shaofeng Sun & Artem Rogachev): intermediate areas in polar form. Find the outer area of the limacone
(not counting the loop) with equation: r = 2-3cosθ.
Solution: This isn’t too bad of a problem, just find the whole area and then subtract the area of the inner loop.
1 2
.
Total area of limo is  (2  3 cos ) 2 d  3104
. Inner loop area: during this time the r is negative. To find out the
2 0
interval set r =0 and that will give you the limits. 0=2-3cosθ, cosθ=2/3, θ=±0.84
Inner area =
1 0.841
(2  3 cos ) 2 d  0.841


0
.
841
2
Outer – Inner = 2.263.