極座標 第十七單元 polar coordinate system to plot points in polar coordinates
TRANSCRIPT
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。