Graphing in polar coordinate. Reminder (1) The following polar representations represent the same...

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Graphing in polar coordinate

Reminder (1)

The following polar representations represent the same point:

( r , θ ) ( r , θ + 2n π ) ; nεZ ( - r , θ + (2n+1)π ) ; nεZ

Examples (1)

The following polar representations represent the same point

( 5 , π/2 ) ( 5 , π/2 + 2 π ) = ( 5 , 5π/2) ( 5 , π/2 - 2 π ) = ( 5 , - 3π/2) ( - 5 , π/2 + π ) = ( - 5 , 3π/2 ) ( - 5 , π/2 – π ) = ( - 5 , -π/2 )

Reminder (2)

All of the polar representations of the form (0, θ) represent the pole ( the origin):

Examples ( 0 , 0 ) ( 0 , π ) ( 0 , π/6 )

The Graph of θ = θ0 For any value of r, θ is always θ0

Any point of the form ( r , θ0 ), where r is any real number, belongs to the graph of the curve

All the following points belongs to that graph: ( 1 , θ0 ), ( 2 , θ0 ), ( 3 , θ0 ) ( -1 , θ0 ) ≡(1, θ0 + π), ( -2 , θ0 ) ≡ (2 , θ0 + π) ( √3/5 , θ0 )

The graph is a straight line making an angle equal to θ0 with the polar axis

Example θ = π/4

The graph is a straight line making an angle equal to π/4 with the polar axis

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Reminder

The equation of this straight line is y = x

We can arrive at this equation, as follows:We have,θ = π/4→ tan θ = 1→ sin θ / cos θ = 1→ r sin θ / r cos θ = 1→ y / x = 1 → y = x

The Graph of r = cθ , where c is nonzero constant Example r = θ

θr = θ( r , θ )

00 ( 0 , 0 )

π/2π/2(π/2 , π/2 )

ππ(π , π)

3π/23π/2(3π/2 , 3π/2 )

2π2π(2π , 2π )

2π+π/22π+π/2(2π+ π/2 , 2π+ π/2 )

3π3π(3π , 3π )

The graph is a spiral

The Graph of r = r0 , where are is a positive number For any value of θ, r is always r0

Any point of the form (r0 , θ ), where θ is any angle ( or any real number ), belongs to the graph of the curve

All the following points belongs to that graph: (r0 , 0 ), (r0 , π/2 ), (r0 , π) , (r0 , π/6)

( - r0 , π ) ≡ (r0 , 0 ), (- r0 , 3π/2 ), ≡ ( r0 , π/2 )

The graph is a circle centered at the origin and of radius equal to r0

Example: r = 5

The graph is a circle centered at the origin and of radius equal to 5

r = 5

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5

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The graph of:r = c sinθ

Or

r = c cosθ

Where c is a nonzero constant

Example

r = 6 sin θ

r = 6 sin θ

θr = 6 sin θ

00(0,0)

π/26(6 , π/2)

π0(0 , π)

3π/2- 6(- 6 , 3π/2)≡ (6 , π/2)

2π0(0 , 2π) ≡ (0,0)

r = 6sin θ

The graph is a circle centered at (0,3) of radius equal to 6/2 = 3

The graph of:r = a sinθ + br = a cosθ + b

Where,a/b = 1 or - 1

Cardioids ( Heart)( special case of the limaςon(

Example

r = 5 - 5sinθ

The Graph of r = 5 - 5sinθ = 5(1-sin θ)

θr = 5(1-sin θ)( r , θ )

05(5,0)

π/20(0 , π/2)

π5(5 , π)

3π/210(10 , 3π/2)

2π5(5 , 2π)

The Graph of r = 5 - 5sinθ

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The graph of:r = a sinθ + br = a cosθ + bWhere a and b are nonzero.

Limaςon ( Snail)

Example

r = 2 - 4sinθ

= 2 ( 1 – 2sinθ)

The graph of r = 2 – 4sinθ = 2 ( 1 – 2 sinθ)

r = 0 if sinθ = ½

Or θ = π/6 , θ= π - π/6 = 5π/6

Thus, we must pay attention to these

values, when graphing

The Graph of r = 2( 1 - 2sinθ(

θr = θ( r , θ )

02 ( 2 , 0 )

π/60(0 , π/6 )

π/2-2(-2 , π/2) = (2, 3π/2)

5π/60(0 , 5π/6 )

π2(2 , π )

3π/26(6 , 3π/2 )

2π2(2 , 2π )

The Graph of r = 2( 1 - 2sinθ)

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The graph of:r = a sinnθr = a cosnθWhere a is positive and n a natural

number

Rose Curves

Example

r = 4sin3θ

The graph of r = 4sin3θ

r = 0 if sin3θ = 0Or 3θ = 0, π , 2π, 3π , 4π, 5π,…Or θ = 0, π/3 , 2π/3, π , 4π/3, 5π/3,…

r = 4, which is maximum if sin3θ = 1 Or 3θ = π/2 , 5π/2 , 9π/2 Or θ = π/6 , 5π/6 , 3π/2

r = - 4, which is minimum if sin3θ = -1 Or 3θ = 3π/2 , 7π/2 , 11π/2Or θ = π/2 , 7π/6 , 11π/6

The graph of r = 4sin3θ

θr = 4sin3θ( r , θ )

00 ( 0 , 0 )

π/64(4 , π/6 )

π/30(0, π/3) ≡ (0,0)

π/2- 4( - 4 , π/2 ) ≡ ( 4 , 3π/2 )

2π/30(0 , 2π/3 ) = (0,0)

5π/64(4 , 5π/6 )

π0(0 , π ) = (0,0)

The graph of r = 4sin3θValues of v grater than v will not result in new pointsIt will just trace the curve again and again

θr = 4sin3θ( r , θ )

π0(0 , π ) = (0,0)

7π/6- 4(- 4 , 7π/6 ) ≡ ( 4 , π/6 )

4π/30(0, 4π/3) = (0,0)

3π/24( 4 , 3π/2 )

5π/30(0 , 5π/3 ) ≡ (0,0)

11π/6- 4 (- 4 , 11π/6) ≡ (4, 5π/6)

r = 4 sin3θ

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sin4r

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The graph of:r = a sinnθOrr = a cosnθA rose curve consists of:n leaves, if n is odd2n leaves, if n is even

Remark

Example

r = 4sin2θ

The graph of r = 4sin2θ

r = 0 if sin2θ = 0Or 2θ = 0, π , 2π, 3π , 4π, 5π, 6π ,…Or θ = 0, π/2 , π , 3π/2, 2π, 5π/2, 3π …

r = 4, which is maximum if sin2θ = 1 Or 2θ = π/2 , 5π/2 , 9π/2, 13π/2,…Or θ = π/4 , 5π/4 , 9π/4, 13π/4,…

r = - 4, which is minimum if sin2θ = -1 Or 2θ = 3π/2 , 7π/2 , 11π/2, ..Or θ = 3π/4 , 7π/4 , 11π/4,..

The graph of r = 4sin2θ

θr = 4sin2θ( r , θ )

00 ( 0 , 0 )

π/44(4 , π/4 )

π/20(0, π/2)

3π/4- 4( - 4 , 3π/4 ) ≡ ( 4 , 7π/4 )

π0(0 , π )

5π/44(4 , 5π/4 )

3π/20(0 , π ) = (0,0)

The graph of r = 4sin2θValues of θ grater than 2π will not result in new pointsIt will just trace the curve again and again

θr = 4sin2θ( r , θ )

7π/4- 4(- 4 , 7π/4 ) = (4, 3π/4 )

2π0(0 , 2π )

9π/44(4, 9π/4)

5π/20( 0 , 5π/2 )

11π/4- 4(- 4 , 11π/4 ) ≡ (4, 7π/4 )

3π0 (0 , 3π)

The graph of r = 4sin2θ

Homework:Graph each of the following Curves

r = π/2 r = 2θ θ = π/4 r = 3sinθ r = 3+3sinθ r = 3sin4θ

r = 2cosθ r = 4 + 4cosθ r = 4 - 4cosθ r = 3 - 6cosθ r = 3 +6cosθ r = 5cos2θ r = 5cos3θ