1.
Find the equation (in rectangular form) of the line tangent to the graph of r  3  2sin  at  = .
y  r sin    3  2sin   sin  
dy
 3cos   4sin  cos 
d
x  r cos    3  2sin   cos  
dx
  3  sin   ( sin  )  2cos  cos 
d
 3sin   sin 2   2cos 2 
dy
dy d
3cos   4sin  cos 


So,
dx dx 3sin   sin 2   2 cos 2 
d
dy 3 3
3
 . (r , )  (3,  )  ( x, y)  ( 3, 0). Equation of tangent line: y  ( x  3)
When    , 
dx 2 2
2
2.
For r = 3 – 2 sin , find all points (r, ), 0 ≤  < 2for which the tangent line to the graph is:
a.
horizontal
b.
dx
 0  3sin   sin 2   2 cos 2   0
d
 3sin   sin 2   2(1  sin 2  )  0
dy
 0  3cos   4sin  cos   0
dx
 cos   3  4sin    0
 3sin 2   3sin   2  0  (from calculator)
3
 cos   0 or sin  
4
 3
3
3
  , ,sin 1   , or  -sin 1  
2 2
4
4
So,
 3
 , , 0.848, or 2.29
2 2
2.29
vertical
 =3.48 or 5.94
0.848
3.48
0.848
3.
Given r = 2a sin , show that the formula for the area inside this circle found by evaluating the
appropriate polar integral agrees with the standard A = r2 formula for this circle.
Picture to right is graph with a = 2.
In order to find the area inside the circle, it is
important to note that the circle is completely
trace out as  goes from 0 to  . So,
 1

2
A    2a sin   d 2a 2  sin 2  d 
0 2
0
 1

2a 2  1  cos 2  d a 2  1  cos 2  d 
0 2
0
1


a 2    sin 2   a 2   0   (0  0)    a 2
2

0
4.
Find the maximum value (exact) of all the y-coordinates on the graph of r = 1 + cos .
dy
dy d
Remember that

, where y  r sin   (1  cos  ) sin   sin   sin  cos  , an
dx dx
d
x  r cos   (1  cos  ) cos   cos   cos 2  . Since we are looking for a maximum, we want
dy
dy
 0 . That is
  cos   cos 2   sin 2   0 . So
dx
d
 cos   cos 2   sin 2   0   cos   cos 2   (1  cos 2  )  0
 2 cos 2   cos   1  0
 (2 cos   1)(cos   1)  0
1
 cos    or cos   1
2
2 4
 
,
,or 2
3 3
2
From the graph, we can see that  
gives the maximum value for y and
3
3
 2 
 2 
 2  1 3
y  r sin 

  (1  cos 
)sin 
 
4
 3 
 3 
 3  2 2
5.
Find the exact area inside the graph of:
r  2sin(3 )
a.
r  2 cos(4 )
b.
The first petal is graphed for 0   
r  2sin(3 )  0 when  = 0 and


3
since
. Therefore,
3
the area inside one petal is given by:
The first petal is graphed for 
r  2 cos(4 )  0 when  = 

3
8


3
3
1
2
 2sin(3 )  d  6  sin 2 (3 )d
2
0
0
3

1  cos(6 )
 6
d
2
0
3



3
sin(6

)


3  


6
0 

 


 3   0    0  0  

 3


8
 
1
 2  2 cos(4 ) 

2
and


8
since
.
8
8
Therefore, the area inside one petal is given by:

1
2
0 2  2sin(3 )  d . So the area inside the
graph is given by:


d . So the area inside the
8
graph is given by:

8
8



8
1
2
 2 cos(4 )  d  16  cos 2 (4 )d 
2


8
8

1  cos(8 )
d
2

8
 16 

8

sin(8 )

 8  
8






 
8 

8
 
  

 8   0      0  
  8

 8
 2
6.
Find the exact area inside the graphs of both r = 2 and r = 2 – 2 sin . (area shared by these graphs)
2  2  sin( )  sin( )  0    0 or 
So,

A
1
1
2
 Area of circle w/ radius 2     2  2sin( )  d
2
2
0

1
2
 2  2sin( )  d
2
0
 2  

 2    2  4sin( )  2sin 2 ( ) d
0

 2   2  4 cos( ) 

 
   1  cos(2 ) d
0
 0


sin(2 )  
 2   2  0   (4  4)     
 
2  0 

 5
7.
Set up the integral(s) necessary to determine the exact area shared by
the graphs of r = 2a cos  and r = 2a sin . (Assume a > 0)

8.
Set up the integral(s) necessary to determine the exact area shared by
the graphs of r = a(1 + cos ) and r = a(1 – cos ).(Assume a > 0)
9.
Find the exact area inside the large loop but outside the small loop of r = 2 cos  + 1.
10.
Show that the width of a petal of r = cos(2) is
11.
Set up the integral(s) necessary to determine the exact area inside the graph of r2 = 2a2cos(2.
2 6
.
9