Sec 5

Symmetry in polar coordinate

Definitions

Symmetry about the Polar Axis

The curve is symmetric about the polar axis ( the x-axis)if replacing the point (r , θ) by either the point (r , - θ) or an equivalent polar representation of it results into an equation equivalent to the original equation

Symmetry about the Polar Axis

The curve is symmetric about the line θ=π/2 ( the y-axis) if replacing the point

(r , θ) by either the point (r , π - θ) or an equivalent polar representation of it, such as (-r , - θ) results into an equation equivalent to the original equation.

Symmetry about the Pole

. The curve is symmetric about the pole (the origin) if replacing the point(r , θ) by either the point (- r , θ) or an equivalent polar representation of it, such as ( r, π+θ) results into an equation equivalent to the original equation.

Examples

Test all type of symmetry for each of the following functions

r = f(θ) = cos4θ r = f(θ) = cos 2θ + 5 r = f(θ) = sin2θ

Solutions

1. r = f(θ) = cos4θ

a. f(-θ) = cos(-4θ) = cos4θ = f(θ) , which means that, the curve is symmetric about the polar axis b. f(π - θ) = cos( (4(π - θ) ) = cos( (4π - 4θ ) = cos(-4θ) = cos4θ = f(θ) which means that, the curve is symmetric about the y-axis

c. Replacing the point (r , θ) by the point ( -r , θ) does not result into an equation equivalent to the original one. However, replacing the point (r , θ) by the equivalent representation ( r , π + θ) in the original equation, we get: r = cos[4(π + θ)] = cos4θ

Which is equivalent to the original equation r = cos4θ. This show that the curve is symmetric about the pole.

The rose curve r = cos4θ exhibits all of the three types of polar symmetry

Example 2.

r = f(θ) = cos 2θ + 5

Do it!

What types of polar symmetry does the curve r = cos 2θ + 5 have?

3. r = f(θ) = sin2θ

Replacing the point (r , θ) by the point (r , - θ) does not result into an equation equivalent to the original one. This, however this does not indicate a lack of symmetry about the polar axis. The point ( -r , π – θ) is equivalent polar representation for the same point with the polar representation (r , - θ)Replacing the point (r , θ) by ( -r , π – θ) in the original equation, we get: - r = f(π – θ) = sin[2(π – θ)] = sin[2π –2θ] = sin (–2θ) = - sin2θ

Which is equivalent to the original equation r = sin2θ. This show that the curve is symmetric about the polar axis.

Replacing the point (r , θ) by the point (r , π – θ) does not result into an equation equivalent to the original one. This however this does not indicate a lack of symmetry about the y-axis. The point ( -r , – θ) is equivalent polar representation for the same point with the polar representation (r , π – θ).Replacing the point (r , θ) by ( -r , – θ) in the original equation, we get: - r = sin(– 2θ) = - sin2θ

Which is equivalent to the original equation r = sin2θ. This show that the curve is symmetric about the y-axis.

Replacing the point (r , θ) by the point ( -r , θ) does not result into an equation equivalent to the original one. However, replacing the point (r , θ) by the equivalent representation ( r , π + θ) in the original equation, we get: r = sin[2(π + θ)] = sin2θ

Which is equivalent to the original equation r = sin2θ. This show that the curve is symmetric about the pole.

The rose curve r = sin2θ exhibits all of the three types of polar symmetry

Sec 6

Points of Intersection

Consider the curves r = f(θ) & r = g(θ)

To find the points of intersection of these two curves, we find the simultaneous solution for these equations, and at the same time remind ourselves of the following:

The simultaneous solution may fail to yields all points of intersection. For instance the pole (origin) has different representations. On one curve it might have one or more representation, while on the other curve a completely distinct representation or set of representations from those on the first curve. Thus although it might be a common point, there is no common representation for it that satisfies both equation simultaneously.

A solution to both equations may cease to be as such, when replaced by another representation for the same point.

Examples

Find all points of intersection of the given two curves:

r = 4 , r = 8sinθ r = 8cosθ , r = 8sinθ r = cos2θ , r = sinθ

Solutions

1. r = 4 & r = 8sinθ

Solving the equations simultaneously, we

get:

sinθ = 1/2 → θ = π/6 , θ = 5π/6

Thus the points are (4 , π/6 ) and (4 , 5π/6 )

Graph the two curves and indicate the points of intersection!

The intersection of the curvesr = 4 & r = 8sinθ

2. r = 8cosθ & r = 8sinθ

Solving the equations simultaneously, we get:

tan θ = 1 → θ = π/4 , θ = 5π/4, which correspond to the representations( 4√2 , π/4 ) and ( - 4√2 , 5π/4 ). These representations represent the same point. Why? What's the Cartesian coordinates of this point.Graph the curves and notice that they intersect also at the pole.Thus the curves intersect at two point:

The pole and the point ( 4√2 , π/4 ).

The intersection of the curvesr = 8cosθ & r = 8sinθ

3. r = cos2θ , r = sinθ

Solving the equations simultaneously, we get:cos2θ = sinθ→ 1 – 2sin2θ = sinθ→ 2sin2θ + sinθ – 1 = 0→ ( 2sinθ – 1 ) ( sin θ + 1) = 0→ sin θ = 1/2 Or sin θ = -1→ θ = π/6, θ = 5π/6 Or θ = 3π/2→ The point of intersection are:

( 1/2 , π/6 ) and ( 1/2 , 5π/6 ) and ( -1 , 3π/2)Notice that while ( -1 , 3π/2) is a representation of the same point who has also the representation ( 1 , π/2), the latter satisfies r = sinθ but not r = cos2θ.

Graph the two curves and indicate the points of intersection.

The intersection of the curvesr = cos2θ & r = sinθ

Sec 7

Arch Length

Let r = f(θ), and let dr/dθ be continuous on

[θ1 , θ2 ]. Then, the arc length L of the

curve from θ = θ1 to θ = θ2 is:

Provided no part of the graph is traced more than once

on the interval [θ1 , θ2 ].

dffL 2

1

22 )]([)(

Examples

Find the length of the curve : r = 2 – 2cosθ r = 2 + 2cosθ

Solutions

1. r = 2 – 2cosθ

16]11[8])cos2[(4

?....sin4

sin4sin222

]cos1[22

]cos22[2

]sincoscos21[2

]sin2[)cos22(

202

2

02

2

02

2

02

2

2

0

2

0

222

0

222

21

21

21

0

Whyd

dd

d

d

d

dL

curvegiventheoflengtharchthebeLLet

r = 2 – 2cosθ

2. r = 2 + 2cosθ

16)2(8

)]10()01[(8

])sin2()sin2[(4

?]....)cos(cos[4

cos4cos222

]cos1[22

]cos22[2

]sincoscos21[2

]sin2[)cos22(

2202

2

20

2

2

02

2

02

2

2

0

2

0

222

0

222

21

21

21

0

Whydd

dd

d

d

d

dL

curvegiventheoflengtharchthebeLLet

Another Method

16

]01[16

)sin2(8

?...cos8

cos8cos224

]cos1[24

]cos22[4

]sincoscos21[4

]sin2[)cos22(2

,,,

02

02

02

02

2

0

0

22

0

22

21

21

21

0

Whyd

dd

d

d

d

dL

havewesymmetrybygraphttheFrom

2. r = 2 + 2cosθ

Sec 8

Area

Let r = f(θ) and 0 < θ2 – θ1 ≤ 2π Let r = f(θ) be continuous and either

f(θ) ≥ 0 or f(θ) ≤ 0 on [θ1 , θ2 ] Then the area A of the region enclosed by

the curve r = f(θ) and the lines θ = θ1

and θ = θ2 is:

df )(221

2

1

Examples

Find the area enclosed by : The curve r = 1 – cosθ , the positive x-

axis, the y-axis. The curve r = cos2θ The curve r = 4 + 4cosθ, but outside the

circle r=6

Solutions

1. The area enclosed by r = 1 – cosθ,

the positive x-axis and the y-axis

!

)coscos21(

)cos1(

,2

0

2

0

21

2

0

21

2

2

solutiontheFinish

d

dA

soand

andlinestheand

curvegiventheenclosedbyareatheisAarearequiredThe

2. The area enclosed by the rose curver = cos2θ

4

!

2cos8

,4

0

,

24

021

solutiontheFinish

dA

soand

andlinestheand

curvegiventhebyenclosedareathetimeseightisAarearequiredthe

symmetryBy

1. The area enclosed by r = 4 +4 cosθ,

but outside the circle r = 6

3

3

The area A is the difference of the area A1 enclosed by r = 4 + 4cosθ and the lines θ=-π/3 and θ =π/3 and the area A2 enclosed by r = 6 and the same lines.

How do we know that? Thus,

ddA 2212

21 )6()cos44(

3

3

3

3