section 6.3: polar forms & area. (x,y) 3 polar coordinates
Post on 27-Dec-2015
227 Views
Preview:
TRANSCRIPT
Section 6.3: Polar Forms & Area
Rectangular coordinates were horizontal/vertical
directions for reaching the point
Eg: 3, in polar4
Polar coordinates are "as the crow flies"directions
(x,y)
4
3cos ,
3
xso
2 2
1
x = r cos
y = r sin tan
x y r
y
x
Polar Coordinates
Notation
43, 3 3cis 3e4 4 4
i
, r rcis reir
Warnings 5 93, and 3, and 3,
4 4 4
Same point
0, is always the origin
Graph r = 3 Graph =4
Graph r =
Graph r = 4 sin
r θ
sin
Graph r = 4 sin
r θ
0
sin
Graph r = 4 sin
r θ
0 0
sin
Graph r = 4 sin
r θ
0 0
π/2
sin
Graph r = 4 sin
r θ
0 0
4 π/2
sin
Graph r = 4 sin
r θ
0 0
4 π/2
π
sin
Graph r = 4 sin
r θ
0 0
4 π/2
0 π
sin
Graph r = 4 sin
r θ
0 0
4 π/2
0 π
3π/2
sin
Graph r = 4 sin
r θ
0 0
4 π/2
0 π
-4 3π/2
sin
Graph r = 4 sin
r θ
0 0
4 π/2
0 π
-4 3π/2
2π
sin
Graph r = 4 sin
r θ
0 0
4 π/2
0 π
-4 3π/2
0 2π
sin
r = 2(1 - cos )Cardiod
r θ
r = 2(1 - cos )Cardiod
r θ
0
r = 2(1 - cos )Cardiod
r θ
0 0
r = 2(1 - cos )Cardiod
r θ
0 0
π/2
r = 2(1 - cos )Cardiod
r θ
0 0
2 π/2
r = 2(1 - cos )Cardiod
r θ
0 0
2 π/2
π
r = 2(1 - cos )Cardiod
r θ
0 0
2 π/2
4 π
r = 2(1 - cos )Cardiod
r θ
0 0
2 π/2
4 π
3π/2
r = 2(1 - cos )Cardiod
r θ
0 0
2 π/2
4 π
2 3π/2
r = 2(1 - cos )Cardiod
r θ
0 0
2 π/2
4 π
2 3π/2
2π
r = 2(1 - cos )Cardiod
r θ
0 0
2 π/2
4 π
2 3π/2
0 2π
4
2
2
r θ
r = 1 2 cosLimacon
r θ
-1 0
r = 1 2 cosLimacon
r θ
-1 0
1 π/2
r = 1 2 cosLimacon
r θ
-1 0
1 π/2
3 π
r = 1 2 cosLimacon
r θ
-1 0
1 π/2
3 π
1 3π/2
r = 1 2 cosLimacon
3 -1
1
1
r θ
-1 0
1 π/2
3 π
1 3π/2
r = 1 2 cosLimacon
3 -1
1
1
r θ-1 0
1 π/2
r θ
-1 0
1 π/2
3 π
1 3π/2
r = 1 2 cosLimacon
3 -1
1
1
r θ-1 0
1 - √ 2 π/4
1 π/2
r θ
-1 0
1 π/2
3 π
1 3π/2
r = 1 2 cosLimacon
3 -1
1
1
r θ-1 0
1 - √ 2 π/4
0 π/3
1 π/2
r = 4 cos(2 )Rose
r θ
0
π/4
π/2
3π/4
π
5pi/4
3π/2
7π/4
r = 4 cos(2 )Rose
r θ
4 0
π/4
π/2
3π/4
π
5pi/4
3π/2
7π/4
r = 4 cos(2 )Rose
r θ
4 0
0 π/4
π/2
3π/4
π
5pi/4
3π/2
7π/4
r = 4 cos(2 )Rose
r θ
4 0
0 π/4
-4 π/2
3π/4
π
5pi/4
3π/2
7π/4
r = 4 cos(2 )Rose
r θ
4 0
0 π/4
-4 π/2
0 3π/4
π
5pi/4
3π/2
7π/4
r = 4 cos(2 )Rose
r θ
4 0
0 π/4
-4 π/2
0 3π/4
4 π
5pi/4
3π/2
7π/4
r = 4 cos(2 )Rose
r θ
4 0
0 π/4
-4 π/2
0 3π/4
4 π
0 5pi/4
3π/2
7π/4
r = 4 cos(2 )Rose
r θ
4 0
0 π/4
-4 π/2
0 3π/4
4 π
0 5pi/4
-4 3π/2
7π/4
r = 4 cos(2 )Rose
r θ
4 0
0 π/4
-4 π/2
0 3π/4
4 π
0 5pi/4
-4 3π/2
0 7π/4
4
2 r 4cos(2 )Lemniscate 4cos r (2 )Or
r θ
0
π/4
π/2
3π/4
π
5pi/4
3π/2
7π/4
2 r 4cos(2 )Lemniscate 4cos r (2 )Or
r θ
2 0
π/4
π/2
3π/4
π
5pi/4
3π/2
7π/4
2 r 4cos(2 )Lemniscate 4cos r (2 )Or
r θ
2 0
0 π/4
π/2
3π/4
π
5pi/4
3π/2
7π/4
2 r 4cos(2 )Lemniscate 4cos r (2 )Or
r θ
2 0
0 π/4
Undefined π/2
3π/4
π
5pi/4
3π/2
7π/4
2 r 4cos(2 )Lemniscate 4cos r (2 )Or
r θ
2 0
0 π/4
Undefined π/2
0 3π/4
π
5pi/4
3π/2
7π/4
2 r 4cos(2 )Lemniscate 4cos r (2 )Or
r θ
2 0
0 π/4
Undefined π/2
0 3π/4
2 π
5pi/4
3π/2
7π/4
2 r 4cos(2 )Lemniscate 4cos r (2 )Or
r θ
2 0
0 π/4
Undefined π/2
0 3π/4
2 π
0 5pi/4
3π/2
7π/4
2 r 4cos(2 )Lemniscate 4cos r (2 )Or
r θ
2 0
0 π/4
Undefined π/2
0 3π/4
2 π
0 5pi/4
Undefined 3π/2
7π/4
2 r 4cos(2 )Lemniscate 4cos r (2 )Or
r θ
2 0
0 π/4
Undefined π/2
0 3π/4
2 π
0 5pi/4
Undefined 3π/2
0 7π/4
2
1. Something is changing, so we can’t use the old algebra formulas.
2. Break the problem into pieces.
3. Pretend everything is constant on each piece and use the old formulas.
4. Add up the pieces. (This is called a Riemann Sum)
5. If we use more and more pieces, the limit is the right answer! (This limit is a definite integral.)
Big Idea
Insted of breaking the area into little rectangles,
we use little sectors...
Polar Area
That's not surprising if we think of graphing by running
The area of seFact: ctor 2 1is
2r
r
2 1That's easy if you remember the area of circle is 2
2r
Example: Find the area of o rn e = leaf 4 co of s(2 )
Always grImportant: aph first
2
*1Riemann sum
24cos 2 kk
4
2
4
1Integral
24cos 2 d
4 4
4
4
2cos 28 d
4
2
4
1Integral 4cos 2
2d
4
4
11 c8 os 4
2d
2
2
1cos
Two impor
1 cos 22
tant identi
1sin
tie s
1 c s2
:
o 2
x x
x x
4
4
4 1 cos (4 ) d
4
4
14 sin( )4
4
2
Example: Find the area enclosed by and rr = 2s = in 2cos
Find Intersections
= 2 2 s
in cos
But that misses the interaction at origin
because we can only
divide by sin if it's not zero!
!!!Always Graph
tan = 1
5=
4 4
2 - 2
or
r or
4 2
2 2
04
1 12cos2sin
2 2d d
Example: Find the area enclosed by and rr = 2s = in 2cos
r = 2 cosExample: Find the area outside the lemniscate
and inside the circl r = 3 c
2
e os
2
: This is not the same as the area
between curves in rectangular coordinates!
You can't do 3cos co1
22 s 2
Warn
d
ing
2 4 22
2 4
1 12 cos 23cos
2 2d d
2 3
top related