10.3 day 2
Calculus of
Polar Curves
Lady Bird Johnson Grove,
Redwood National Park, California
Photo by Vickie Kelly, 2007
Greg Kelly, Hanford High School, Richland, Washington
r  2sin  2.15 
Try graphing this
on the TI-89.
0    16
To find the slope of a polar curve:
dy
dy
 d
dx
dx
d
d
r sin 
 d
d
r cos 
d
r  sin   r cos 

r  cos   r sin 
We use the product rule here.
To find the slope of a polar curve:
dy
dy
 d
dx
dx
d
d
r sin 
 d
d
r cos 
d
r  sin   r cos 

r  cos   r sin 
dy r  sin   r cos 

dx r  cos   r sin 

Example:
r  1  cos
r  sin
sin  sin   1  cos   cos 
Slope 
sin  cos   1  cos   sin 
sin   cos  cos 

sin  cos  sin   sin  cos
2
2
sin   cos   cos

2sin  cos  sin 
2
2
 cos 2  cos

sin 2  sin 

Area Inside a Polar Graph:
The length of an arc (in a circle) is given by r.  when  is
given in radians.
For a very small , the curve could be approximated by a
straight line and the area could be found using the triangle
1
formula:
A  bh
2
r
r  d
1
1 2
dA   rd   r  r d
2
2

1 2
dA  r d
2
We can use this to find the area inside a polar graph.
dA 
1 2
r d
2
A


1 2
r d
2

Example: Find the area enclosed by:
  2

2
1
0
-1
1
2
3
4

0
2
0
r  2 1  cos 
1 2
r d
2
1
2
 4 1  cos   d
2
-2
  2 1  2cos  cos   d
2
2
0

2
0
1  cos 2
2  4 cos   2 
d
2


2
0
1  cos 2
2  4 cos   2 
d
2
2
  3  4cos  cos 2 d
0
1
 3  4sin   sin 2
2
2
0
 6  0
 6

Notes:
To find the area between curves, subtract:
1  2 2
A   R  r d
2 
Just like finding the areas between Cartesian curves,
establish limits of integration where the curves cross.

When finding area, negative values of r cancel out:
r  2sin  2 
1
-1
0
1
-1
Area of one leaf times 4:
2
1 2
A  4    2sin  2   d
2 0
A  2
Area of four leaves:
2
1 2
A    2sin  2   d
2 0
A  2

To find the length of a curve:
Remember:
ds  dx 2  dy 2
For polar graphs:
x  r cos 
y  r sin 
If we find derivatives and plug them into the formula,
we (eventually) get:
2
 dr 
ds  r  
 d
 d 
2
So:
Length  


2
 dr 
r 
 d
 d 
2

Length  
2
 dr 
r 
 d
 d 

2

There is also a surface area equation similar to the
others we are already familiar with:
When rotated about the x-axis:
S


S


2
 dr 
2 y r  
 d
 d 
2
2
 dr 
2 r sin  r  
 d
 d 
2
