Math 37 Chapter 4

8/2/2019 Math 37 Chapter 4

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CHAPTER 4

POLAR

COORDINATES

-2

-1

0

1

2

-2 -1 1 2

-2

-1 0 1 2

1

2

3

4

2

Objectives:

At the end of the chapter, you shouldbe able to

1. plot polar points,

2. find the polar coordinates of acartesian point and vice-versa,

3. sketch polar curves and

4. find the area of a polar region.

3

OUTLINE

4.1 The polar coordinate system

4.2 Graphs of Polar Equations

4.3 Area of a polar region

4

A polar coordinate

system consists of a

horizontal ray called

thepolar axis(0-axis,

2 axis).The initial point of

the polar axis is called

thepole (O).

4.1 The polar coordinate system

O

pole

5

wherer is the distance of the point from

the pole,

A pointP on a polar plane has coordinates

(r,)

r

and

is the measure of the angle whichthe ray OP makes with the polar axis.

,rP

O

6

How do we plot a polar point (r,)?1. We first locate the -axis.2. a. Ifr > 0, the point is plotted

along the -axis.b. Ifr < 0, the point is plotted on

the opposite side of the -axis.


2/142

7

Example 4.1.1 Plot the following polar

points.

/43, .a /45,- .b /43,- .c /43.5,-- .d/4

a

5/4b

-/4c

3/4d

Solution:

/4/4

8

How are the cartesian coordinates

(x,y) and polar coordinates (r,) of apoint related?

r ,rO

x

y

(x,y)

r

xcos

r

ysin

cosrx sinry

222 ryx

0x,x

ytan

9

Example 4.1.2 Find the cartesian

coordinates of the given polar point.

/46, .aSolution:

cosrx /46, .a

64

cos 62

2 23sinry 6

4

sin 6

2

2 23 2323Answer. ,

/34,2 .b

10

cosrx 43

2cos 4

2

1 2

sinry 4 32

sin 4 2

3

32 322Answer. ,

/34,2 .b

11

While the cartesian coordinates of any

point are unique, the polar coordinates of any

point are not unique.

/4

5/4

/43,

/43,5-, etc./43,-3- , /43,-7

Consider the point with polar coordinates

The same point has

polar coordinates

-7/4

-3/4

12

Example 4.1.3 Find a set of polar coordinates

(r,) of the cartesian point (-3,3) such that-2 2 and

a. r > 0 and > 0 c. r < 0 and > 0b. r > 0 and < 0 d. r < 0 and < 0

Solution:222 ryx

22 yxr 2318

x

ytan

13

3

tan

4

3 (since (-3,3) QII)


3/143

13

3/4

a. r > 0 and > 0b. r > 0 and < 0c. r < 0 and > 0d. r < 0 and < 0

Answers:

4

323a.

,

4

523b.

,

-/4

4

723c.

,

423d.

,

-5/4

7/4

14

Example 4.1.4 Find a set of polar coordinates

(r,) of the cartesian point (-1,-2) such that-2

2

anda. r > 0 and > 0 c. r < 0 and > 0b. r > 0 and < 0 d. r < 0 and < 0

Solution:

22 yxr 5x

ytan 2

1

2

tan 2tanArc

(since (-1,-2) QIII)

222 ryx

15

a. r > 0 and > 0b. r > 0 and < 0c. r < 0 and > 0d. r < 0 and < 0

Answers:

25a. tanArc,

2tanArc

2tanArc

25b. tanArc,

25c. tanArc,

225d. tanArc,

16

Example 4.1.5 Find a polar equation of a curve

whose cartesian equation is given by

9a. 22 yx xy 3b. Solution:

9a. 22 yx 922 sinrcosr

92222 sinrcosr 9222 sincosr

92 r3r

17

922 yx

3r

The same graph is given by

.r 3or R,, 3 R,, 3

18

xy 3b. cosrsinr 3

03 cosrsinr 03 cossinr0r 03or cossin0r 3or tan0r

34or

3or


4/144

19

xy 3

The same graph is given by

.3

4or

3

/3

4/3

20

4.2 Graphs of polar equations

The graph of a cartesianequation consists of all points (x,y)

that satisfies the given equation.

The graph of a polar equation

consists of all points (r,) thatsatisfies the given equation.

21

A. Circles centered at the pole

0where a,arIllustration:

4r

.r 4bygiveniscirclesameThe 22

B. Circles tangent to the pole

0where a,cosar 0where a,sinar

23

cosar If

22

22

yx

xayx

axyx 22022 yaxx

2

2

2

22

ayax

20

2

ar,,

aC

an equation of

a circle

24

cosr 4a = 4

Illustrations:

202 r,,C

cosr 4a = -4 202 r,,C


5/145

25

sinar If

2222

yxyayx

ayyx 22022 ayyx

22

2

22

aayx

220

ar,

a,C

equation of a

circle

26

sinr 4a = 4

Illustrations:

220 r,,C

sinr 4a = -4 220 r,,C

27

C.i. Lines through the pole

Ra,a Illustrations:

6

6

7 28

C.ii. Vertical lines

Ra,acosr Ra,ax

Illustrations:

4cosr 2cosr

29

C.iii. horizontal lines

Ra,asinr Ra,ay

Illustrations:

4sinr

2sinr

30

Symmetry of a polar curve

a. The graph of a polar equation is

symmetric with respect to thex-axis

when an equivalent polar equation is

obtained when (r,) is replaced byeither

(r,-) or (-r, - ).


6/146

31

Illustration:

a. The graph ofis symmetric with respect to thex-axis.

cosr 55

cosr 55.cosr 55

When (r,) is replaced by (r,-), theequation becomes

32

cosr 55

(r,)

(r,-)

3333

b. The graph of a polar equation is

symmetric with respect to they-axis

when an equivalent polar equation is

obtained when (r,) is replaced byeither

(-r,-) or (r, - ).

34

Illustration:

b. The graph of

is symmetric with respect to they-axis.

sinr 33

sinr 33When (r,) is replaced by (r, - ), theequation becomes

sincoscossin33 r 0 -1sinr 33

35

sinr 33

(r,)(r, -)

36

D. Limacons

,sinbarorcosbar where a 0 andb 0.

Types of Limacons

1b

a

1b

a

21 ba

2b

a3. Limacon with a dent

4. Convex Limacon

1. Limacon with a loop

2. Cardioid


7/147

37

32 b,a.cosr 32

Illustration. Sketch the graph of

Solution:

13

20

3

2 b

a

The graph is a limacon with a loop which

is symmetric with respect to thex-axis.

38

0 /6 /4 /3 /2 2/3 3/4 5/6 cos

3cos

2+ 3cos

cosr 321 23 / 22 / 21/ 0 21/ 22 / 23 / -1

3 233 / 223 / 23/ 0 23/ 223 / 233 / -3

5 4.6 4.1 3.5 2 .5 .12 -.6 -1

cosr 32 sind

dr3

0,when0 ddr

39

cosr 32

032 cos

3

2

cos

3

2cosArc

3

2cosArc

40

33 b,a

.sinr 33Illustration. Sketch the graph of

Solution:

133

ba

The graph is a cardioid which is symmetric

with respect to they-axis.

41

23 b,a

.sinr 23Illustration. Sketch the graph of

Solution:

212

3 b

a

b

a

The graph is a limacon with a dent whichis symmetric with respect to they-axis.

42

48 b,a.sinr 48


Solution:

24

8 b

a

The graph is a convex limacon which is

symmetric with respect to they-axis.


8/148

43

E. Roses

1. Ifn is odd, then the rose has

n congruent leaves.

2. Ifn is even, then the rose has

2n congruent leaves.

nsinarorncosar .n 2where

44

22 n,aSolution:The graph is a

rose with 4 leaves

and which is

symmetric with

respect to thex-

axis andy-axis.


.cosr 22

45

12 cos...,, 202

...,, 0

22cosr 02 cos

...,, 2

3

22

...,,

4

3

4

90

24

2

46

4

Half a leaf can be generated by considering

the interval .,

40

47

33 n,aSolution:

The graph is a

rose with 3 leaves

and which is

symmetric with

respect to thex-axis.


.cosr 33

48

13 cos...,, 203 ...,,

3

20

33cosr 03 cos

...,, 2

3

23

...,,

26

1203

2


9/149

49

33cosr

6

Half a leaf can be generated by considering

the interval .,

60

50

F. Lemniscates

22 22 cosarorsinar .a 0where

The graph of a lemniscate is a figure

8.

51

22 sinar 2sinar

0If a 02 sin

20

20

0If a 02 sin

22

2

52

22 cosar 2cosar

0If a 02 cos2

22

44

0If a 02 cos2

3

22

4

3

4

53

Illustrations:

242 sinr 242 cosr

54

232 cosr 2162 sinr

4

4

5


10/1410

55

43

4

7

222 sinr

56

G. Spirals

Spiral of Archimedes

Logarithmic spiral

Exponential spiral

0 ,r

0 ,logr

0 ,er

57

Illustrations:01 ,r.

0 /6 /4 /3 /2 2/3 3/4 5/6

r 0 .52 .79 1.05 1.57 2.09 2.35 2.62 3.1

01ddr

As increases,r increases.

58

0 ,r

59 60

01

2 ,r.


11/1411

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4.3 Area of polar regionsWe recall

the area of a sector of a circle of radiusr

and which subtends a central angle ofradians is

r

.r 22

1

62

LetR be the

region enclosed by the

graph ofr =f()

and the lines given by

= and = ,wherefis continuous

and non-negative on

the interval [ , ].

=

=

r =f()

63

=

= r =f()

Subdivide the closed intervalinton sub-intervals by choosing

intermediate numbers ,

where

1n ,121 n,...,,

.... nn 121064

Denote the ith sub-

interval byIi so that

101 ,I

212 ,I

323 ,I iii ,I 1

nnn ,I 1For each ,

choose a number

n,...,,i 21

.Iii

=

= r =f()

i

Construct a sector of

radius .f i

i

65

The area of the ith sector is

22

1r 2

2

1if i

The sum of areas of then sectors is

iini

f 2

1 2

1

The area of the region is

iini

f 2

1 2

1

iin

in

flim 2

1 2

1

df

2

2

1 66

Illustration:Find the area of the region

enclosed by the graph of

Solution:

The graph is symmetric

with respect to thex-

axis.

We may consider the

area of the region above

or below the x-axis.

.cosr 2


12/1412

67

dcosA / 220 2212

dcos

/ 22

0 2

dcos

/2

2

04 dcos/

2

214

2

0

dcos/ 212 20 2

0

22

12

/

sin

022

102

22

2

1

22 sinsin

0 0 The area of the region is graph is square

units.

68


enclosed by the graph of

.sinr 33Solution:

The graph is symmetric

with respect to they-axis.

We may consider the

area of the region to the

right or to the left of the

y-axis.

69

dsinA 222

332

12

dsin 222

33

dsinsin 222

189

d

cossin

2 211892

2

dcossin

221182192

2

70

dcossin

221182192

2

2

2

2

4

118

2

19

sincos

2

24

1

218

22

19 sincos

2

24

1

218

22

19 sincos

2

19

4

19

4

19 The area of the region is graph is square units.

2

19

0 0

0 0

71

Solution:The graph is symmetric

with respect to thex-

axis.

We may consider the

area of the region to theabove below thex-axis.


enclosed by the loop of the graph of

.cosr 42

72

042 cosr24 cos2

1cos

3

2

242 cosr44 cos

1cos


13/1413

73

dcosA / 232 42212 dcos/ 232 42 dcoscos/ 232 16164

dcos

cos/

2 2116164 32

dcoscos/ 218164 32 dcoscos/ 281612 32

74

dcoscos/ 281612 32 32241612 /sinsin

241612 sinsin

3

224

3

216

3

212

sinsin

0 0

2

34

2

316812

344 The area of the region enclosed by the loop

is square units. 344

75

Area between two polar curves

IfR is the region enclosed by thegraphs of

on [,], wherefandg are continuous andnon-negative and

for each in [,], then the area ofR isgiven by

grfr and

gf

.dgfA 21 22 76

=

=

r =g()

r =f()

R

dfA 21 2 dg 21 2 .dgfA 21 22

77

Illustration. Find the area of the region

inside the graph of but

outside the graph of .

cosr 11r

Solution:

11 cos0cos

2

3

2

,

78

dcosA / 2220 11212 .dgfA 21 22

dcos/ 2220

11 dcoscos/ 22

02

dcoscos/

2 21220 dcoscos/

22121220


14/14

79

2

0

24

1

2

12

/

sinsin

224122122 sinsin02

4

10

2

102 sinsin

0

0 0

42

The area of the region is

square units.

4

2

80

Illustration. Find the area of the region

common to the regions enclosed by the

graphs of .sinr 2and 1rSolution:

12 sin2

1sin

6

5

6

,6

6

5

81

6

6

5

21 AAA

dA

26/

01 sin2

2

12

dA

22/

6/2 1

2

12

82

2sin 2.15

0 16

r

The end !!!

Math 37 Chapter 4

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