Post on 22-Dec-2015
極座標
第十七單元
Polar Coordinate System
Cartesian Coordinate System Polar Coordinate System
To plot points in Polar Coordinates
Relation to Cartesian Coordinates
cosx r q=
siny r q={
tany
xq=
2 2 2x y r+ ={
Example
Graph 1 sinr q= -
2
p6
p3
p
2 3
2
-1
2
0
1
q
r 0
5
6
pp
12 3
2
-
2
3
p
1
2
q
r
3
2
p
2
ExampleFind the Cartesian coordinates corresponding t
o (4 ,π/6 )Solution:
2 3=
( 4 , ) ( 2 3 , 2 )6
p®
y =
34
2= ×4 cos
6
p×x=
4 sin6
p×
14
2= × 2=
( , ) ( 4 , )6
rp
q =
Graphs of Polar Equations
cos
sin
r a b
r a b
q
q
= ±
= ±
Cardioids and Limaçons蚶
The Graph of Roses
Polar equation of the form
cos
sin
r a n
r a n
q
q
=
=
The Graph of Spirals
Spiral of Archimedes
Logarithmic Spiral
r aq=
br ae q=
Conchoid de Nicomedes2 2 2 2 2Equation : ( )( ) 0x y y b l y+ - - =
2 2 2 2Polar eq.: ( )r y b l y- =2 2 2( sin ) sinr b lq q- =
( sin ) sinr b lq q- =±
sin sinr l bq q=± +
cscr l b q=± +
Nicomedes280BC- 210BC
Conchoid de Nicomedes (graph)
l a<
l a=
l a>
5 , 1 10a l= £ £
Area of a Sector
21
2A r q=
A
Calculus in Polar Coordinates
Area in Polar Coordinates
21[ ( )]
2A f d
b
aq q= ò
21[ ( )]
2dA f dq q=
21
2dA r dq=
Example
Find the area of the region inside the limaçon
2 cosr q= +
Solution
0
9
2
p
q=
0
9
2d
p
q=ò0
4dp
q=ò
2
0(2 cos ) d
p
q q= +ò 2
0(4 4cos cos )d
p
q q q= + +ò
04 cos d
p
q q+ ò 0
1(1 cos 2 )
2d
p
q q+ +ò
04 cos d
p
q q+ ò 0
1(cos 2 ) 2
4d
p
q q+ ×ò
04sin
pq+
0
1sin 2
4
p
q+9
2p=
22
0
1(2 cos )
2A d
p
q q= +ò
Arc Length
( ) ( )2 2ds dx dy= +
cosx r q=Qcos
cosdx dr d
rd d d
q q q= +
cos sindr dx rd q q q= -
siny r q=Q sinsin
dy dr dr
d d d
q q q= +
sin cosdrd r dy q q+=
Arc Length(continued)cos sindr dx rd q q q= -
sin cosdrd r dy q q+=
( ) ( )2 2dx dy+
2cos in( s )dr r dq q q= - 2sin os( c )dr r dq q q+ +
( ) ( )2 22 2 2cos 2 sin cos sindr r drd r dq q q q q q= - +
( ) ( )2 22 2 2sin 2 sin cos cosdr r drd r dq q q q q q+ + +
( ) ( )2 22dr r dq= +
( ) ( )2 22dr dds r q+\ =2
2 drr
ddq
q
æ ö÷ç ÷ç ÷ç ø= +
è
Arc Length Formula
22 dr
s r dd
b
aq
q
æ ö÷ç= + ÷ç ÷çè øò
Example
2
Find the length of the arc of the function
from 0 to 2r e q q q p= = =Solution :
1
s=ò
2 2dr
ed
q
q= ×Q
2r + 2( / )dr dq dq0
2p
24 4
04e e d
pq q q= +ò
22
05 e d
pq q=ò
5=2e q1
20
2p
5
2= ( 4e p 1- )
2p
單元結語
還有很多以極座標方式表達的函數在此無法一一列舉，同學可參考本校數學網站
http://www.chit.edu.tw/mathmet
再進入函數圖形。
以極座標方式有時比直角座標表示方便簡捷很多。