Lecture 9 (Polar Coordinates and Polar Curves)

Polar Coordinates

Polar Curves

Institute of Mathematics, University of the Philippines Diliman

Mathematics 54 (Elementary Analysis 2)

Polar Curves 1/ 39

Polar Coordinates Graphs in Polar Coordinates Special Curves in Polar Coordinates

The Polar Coordinate System

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Polar Coordinates

Example.

Plot the following points:

1 A = (1, π/4)

2 B = (2,−π/4)

3 C = (−2, π/6)

4 D = (−3,−π/3)

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Polar Coordinates

Example.

1 A = (1, π/4)

2 B = (2,−π/4)

3 C = (−2, π/6)

4 D = (−3,−π/3)

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Polar Coordinates

Example.

1 A = (1, π/4)

2 B = (2,−π/4)

3 C = (−2, π/6)

4 D = (−3,−π/3)

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Polar Coordinates

Example.

1 A = (1, π/4)

2 B = (2,−π/4)

3 C = (−2, π/6)

4 D = (−3,−π/3)

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Polar Coordinates

Example.

1 A = (1, π/4)

2 B = (2,−π/4)

3 C = (−2, π/6)

4 D = (−3,−π/3)

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Polar Coordinates

Example.

1 A = (1, π/4)

2 B = (2,−π/4)

3 C = (−2, π/6)

4 D = (−3,−π/3)

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Polar Coordinates

Example.

1 A = (1, π/4)

2 B = (2,−π/4)

3 C = (−2, π/6)

4 D = (−3,−π/3)

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Polar Coordinates

Example.

1 A = (1, π/4)

2 B = (2,−π/4)

3 C = (−2, π/6)

4 D = (−3,−π/3)

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Polar Coordinates

Example.

1 A = (1, π/4)

2 B = (2,−π/4)

3 C = (−2, π/6)

4 D = (−3,−π/3)

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Polar Coordinates

Example.

1 A = (1, π4

)2 B = (

2,−π4

) 3 C = (−2, π6)

4 D = (−3,−π3

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Polar Coordinates

Example.

1 A = (1, π4

)= (1, 9π/4)

2 B = (2,−π

) 3 C = (−2, π6)

4 D = (−3,−π3

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Polar Coordinates

Example.

1 A = (1, π4

)= (1, 9π/4) = (−1, 5π/4)

2 B = (2,−π

) 3 C = (−2, π6)

4 D = (−3,−π3

Polar Curves 4/ 39

Polar Coordinates

Example.

1 A = (1, π4

)= (1, 9π/4) = (−1, 5π/4)

2 B = (2,−π

) = (2, 7π/4)

3 C = (−2, π6)

4 D = (−3,−π3

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Polar Coordinates

Example.

1 A = (1, π4

)= (1, 9π/4) = (−1, 5π/4)

2 B = (2,−π

) = (2, 7π/4)

3 C = (−2, π6) = (2, 7π/6)

4 D = (−3,−π3

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Polar Coordinates

Example.

1 A = (1, π4

)= (1, 9π/4) = (−1, 5π/4)

2 B = (2,−π

) = (2, 7π/4)

3 C = (−2, π6) = (2, 7π/6)

4 D = (−3,−π3

) = (3, 2π/3)

Polar Curves 4/ 39

Polar Coordinates

Conversion Equations

Polar to Cartesian

1 x = r cosθ

2 y = r sinθ

Cartesian to Polar

1 r2 = x2 +y2

2 tanθ = y

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Polar Coordinates

Polar to Cartesian

1 x = r cosθ

2 y = r sinθ

Cartesian to Polar

1 r2 = x2 +y2

2 tanθ = y

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Polar Coordinates

Polar to Cartesian

1 x = r cosθ

2 y = r sinθ

Cartesian to Polar

1 r2 = x2 +y2

2 tanθ = y

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Polar Coordinates

Polar to Cartesian

1 x = r cosθ

2 y = r sinθ

Cartesian to Polar

1 r2 = x2 +y2

2 tanθ = y

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Polar Coordinates

Polar to Cartesian

1 x = r cosθ

2 y = r sinθ

Cartesian to Polar

1 r2 = x2 +y2

2 tanθ = y

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Polar Coordinates

Example 1.

Determine the polar coordinates of the point having Cartesian coordinates(−p3,1).

Solution. Recall that r2 = x2 +y2 and tanθ = yx . Thus,

r2 = (−p3)2 +12 =⇒ r = 2

tanθ = 1

−p3=⇒ θ = 5π

Hence, the polar coordinates are(2, 5π

(−2, 11π

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Polar Coordinates

Example 1.

Solution. Recall that r2 = x2 +y2 and tanθ = yx .

r2 = (−p3)2 +12 =⇒ r = 2

tanθ = 1

−p3=⇒ θ = 5π

(−2, 11π

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Polar Coordinates

Example 1.

r2 = (−p3)2 +12

=⇒ r = 2

tanθ = 1

−p3=⇒ θ = 5π

(−2, 11π

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Polar Coordinates

Example 1.

r2 = (−p3)2 +12 =⇒ r = 2

tanθ = 1

−p3=⇒ θ = 5π

(−2, 11π

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Polar Coordinates

Example 1.

r2 = (−p3)2 +12 =⇒ r = 2

tanθ = 1

=⇒ θ = 5π

(−2, 11π

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Polar Coordinates

Example 1.

r2 = (−p3)2 +12 =⇒ r = 2

tanθ = 1

−p3=⇒ θ = 5π

(−2, 11π

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Polar Coordinates

Example 1.

r2 = (−p3)2 +12 =⇒ r = 2

tanθ = 1

−p3=⇒ θ = 5π

or(−2, 11π

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Polar Coordinates

Example 1.

r2 = (−p3)2 +12 =⇒ r = 2

tanθ = 1

−p3=⇒ θ = 5π

(−2, 11π

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Polar Coordinates

Example 1.

r2 = (−p3)2 +12 =⇒ r = 2

tanθ = 1

−p3=⇒ θ = 5π

(−2, 11π

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Polar Coordinates

Example 2.

Determine the Cartesian coordinates of the point having polar coordinates(−5,−π3

Solution. Recall that x = r cosθ and y = r sinθ. Thus,

x =−5cos(−π

) =− 52

y =−5sin(−π

) = 5p

Hence, the Cartesian coordinates are(− 5

2 , 5p

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Polar Coordinates

Example 2.

Solution. Recall that x = r cosθ and y = r sinθ.

x =−5cos(−π

) =− 52

y =−5sin(−π

) = 5p

2 , 5p

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Polar Coordinates

Example 2.

x =−5cos(−π

=− 52

y =−5sin(−π

) = 5p

2 , 5p

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Polar Coordinates

Example 2.

x =−5cos(−π

) =− 52

y =−5sin(−π

) = 5p

2 , 5p

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Polar Coordinates

Example 2.

x =−5cos(−π

) =− 52

y =−5sin(−π

2 , 5p

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Polar Coordinates

Example 2.

x =−5cos(−π

) =− 52

y =−5sin(−π

) = 5p

2 , 5p

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Polar Coordinates

Example 2.

x =−5cos(−π

) =− 52

y =−5sin(−π

) = 5p

2 , 5p

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Polar Equations and Polar Curves

Example 1.

Sketch r = 2.

Remark. In general, the graph of the equation r = k is a circle centered at the poleof radius |k|. Note that r = k and r =−k represent the same curve.

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Example 1.

Sketch r = 2.

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Example 1.

Sketch r = 2.

Remark. In general, the graph of the equation r = k is a circle centered at the poleof radius |k|.

Note that r = k and r =−k represent the same curve.

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Example 1.

Sketch r = 2.

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Example 2.

Sketch θ = π

Remark. In general, the graph of the equation θ = k is a line passing through thepole making an angle k with the polar axis. Also, its Cartesian form is y = (tank)x,when non-vertical, or x = 0, when vertical.

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Example 2.

Sketch θ = π

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Example 2.

Sketch θ = π

Remark. In general, the graph of the equation θ = k is a line passing through thepole making an angle k with the polar axis.

Also, its Cartesian form is y = (tank)x,when non-vertical, or x = 0, when vertical.

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Example 2.

Sketch θ = π

Remark. In general, the graph of the equation θ = k is a line passing through thepole making an angle k with the polar axis. Also, its Cartesian form is y = (tank)x,when non-vertical,

or x = 0, when vertical.

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Example 2.

Sketch θ = π

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Example 3.

Sketch r = 4cosθ.

Remark. In general, the graph of the equation r = acosθ is a circle of radius |a/2|tangent to the line θ =π/2 at the pole. We only need to vary θ on [0,π] to trace outthe curve. Exercise: Find its Cartesian form.

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Example 3.

Sketch r = 4cosθ.

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Example 3.

Sketch r = 4cosθ.

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Example 3.

Sketch r = 4cosθ.

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Example 3.

Sketch r = 4cosθ.

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Example 3.

Sketch r = 4cosθ.

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Example 3.

Sketch r = 4cosθ.

Polar Curves 10/ 39

Example 3.

Sketch r = 4cosθ.

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Example 3.

Sketch r = 4cosθ.

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Example 3.

Sketch r = 4cosθ.

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Example 3.

Sketch r = 4cosθ.

Remark. In general, the graph of the equation r = acosθ is a circle of radius |a/2|tangent to the line θ =π/2 at the pole.

We only need to vary θ on [0,π] to trace outthe curve. Exercise: Find its Cartesian form.

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Example 3.

Sketch r = 4cosθ.

Remark. In general, the graph of the equation r = acosθ is a circle of radius |a/2|tangent to the line θ =π/2 at the pole. We only need to vary θ on [0,π] to trace outthe curve.

Exercise: Find its Cartesian form.

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Example 3.

Sketch r = 4cosθ.

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Example 3.

Sketch r =−5sinθ.

Remark. In general, the graph of the equation r = asinθ is a circle of radius |a/2|tangent to the polar axis at the pole. We only need to vary θ on [0,π] to trace outthe curve. Exercise: Find its Cartesian form.

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Example 3.

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Example 3.

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Example 3.

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Example 3.

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Example 3.

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Example 3.

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Example 3.

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Example 3.

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Example 3.

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Example 3.

Remark. In general, the graph of the equation r = asinθ is a circle of radius |a/2|tangent to the polar axis at the pole.

We only need to vary θ on [0,π] to trace outthe curve. Exercise: Find its Cartesian form.

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Example 3.

Remark. In general, the graph of the equation r = asinθ is a circle of radius |a/2|tangent to the polar axis at the pole. We only need to vary θ on [0,π] to trace outthe curve.

Exercise: Find its Cartesian form.

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Example 3.

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Polar Region

Illustration.

Let R be the set of points satisfying the conditions

1 É r É 2π

6É θ É π

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Polar Region

Illustration.

1 É r É 2π

6É θ É π

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Polar Region

Illustration.

1 É r É 2π

6É θ É π

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Polar Region

Illustration.

1 É r É 2π

6É θ É π

Polar Curves 12/ 39

Polar Region

Illustration.

1 É r É 2π

6É θ É π

Polar Curves 12/ 39

Polar Region

Illustration.

1 É r É 2π

6É θ É π

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Polar Regions

Exercises.

Graph the following set of points:

1 1 É |r| É 2, 2π/3 É θ É 5π/6

2 4 É r É 5

3 π/3 É θ É 2π/3

Exercises.

Find the polar equivalent of the following:

1 x = 2

2 xy = 1

3 x2 + (y−3)2 = 9

4 x = e2t cos t,y = e2t sin t, t ∈RFind the Cartesian form of the following:

1 r2 = 4r cosθ

2 r = 4

2cosθ− sinθ

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Symmetry in the Polar Plane

Symmetry About θ = 0

A polar curve is symmetric about the line θ = 0 (or x−axis)

if whenever (r,θ), in itsequation, is replaced by (r,−θ) or by (−r,π−θ), equivalent equation is obtained.

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A polar curve is symmetric about the line θ = 0 (or x−axis) if whenever (r,θ), in itsequation, is replaced by (r,−θ) or by (−r,π−θ), equivalent equation is obtained.

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Symmetry About θ = π2

A polar curve is symmetric about the line θ = π2 (or y−axis)

if whenever (r,θ), in itsequation, is replaced by (r,π−θ) or by (−r,−θ), equivalent equation is obtained.

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A polar curve is symmetric about the line θ = π2 (or y−axis) if whenever (r,θ), in its

equation, is replaced by (r,π−θ) or by (−r,−θ), equivalent equation is obtained.

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Symmetry About the Pole

A polar curve is symmetric about the pole

if whenever (r,θ), in its equation, isreplaced by (−r,θ) or by (r,θ+π), an equivalent equation is obtained.

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A polar curve is symmetric about the pole if whenever (r,θ), in its equation, isreplaced by (−r,θ) or by (r,θ+π), an equivalent equation is obtained.

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Special Curves

Limaçons

Limaçons are curves whose equations are of the form

r = a±bcosθ or

r = a±bsinθ where a,b > 0

Testing for symmetry,

r = a±bcosθr = a±bcos(−θ) =⇒ r = a±bcosθ

thus, symmetric with respect to the x−axis

r = a±bsinθr = a±bsin(π−θ) =⇒ r = a±bsinθ

thus, symmetric with respect to the y−axis

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Special Curves

Limaçons

r = a±bcosθ or

r = a±bcosθ

r = a±bcos(−θ) =⇒ r = a±bcosθ

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Special Curves

Limaçons

r = a±bcosθ or

r = a±bcosθr = a±bcos(−θ)

=⇒ r = a±bcosθ

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Special Curves

Limaçons

r = a±bcosθ or

Polar Curves 17/ 39

Special Curves

Limaçons

r = a±bcosθ or

Polar Curves 17/ 39

Special Curves

Limaçons

r = a±bcosθ or

r = a±bsinθ

r = a±bsin(π−θ) =⇒ r = a±bsinθ

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Special Curves

Limaçons

r = a±bcosθ or

r = a±bsinθr = a±bsin(π−θ)

=⇒ r = a±bsinθ

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Special Curves

Limaçons

r = a±bcosθ or

Polar Curves 17/ 39

Special Curves

Limaçons

r = a±bcosθ or

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Limaçons

Example.

Sketch r = 1+2cosθ.

The graph is called a limaçon with a loop.The type of limaçon depends on the ratio a

b . Here, it’s ab = 1

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

The graph is called a limaçon with a loop.

The type of limaçon depends on the ratio ab . Here, it’s a

b = 12 .

Polar Curves 18/ 39

Limaçons

Example.

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Limaçons

Example.

Sketch r = 1+cosθ.

The graph is called a cardioid. Note that ab = 1.

Polar Curves 19/ 39

Limaçons

Example.

Sketch r = 1+cosθ.

Polar Curves 19/ 39

Limaçons

Example.

Sketch r = 1+cosθ.

Polar Curves 19/ 39

Limaçons

Example.

Sketch r = 1+cosθ.

Polar Curves 19/ 39

Limaçons

Example.

Sketch r = 1+cosθ.

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Limaçons

Example.

Sketch r = 1+cosθ.

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Limaçons

Example.

Sketch r = 1+cosθ.

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Limaçons

Example.

Sketch r = 1+cosθ.

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Limaçons

Example.

Sketch r = 1+cosθ.

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Limaçons

Example.

Sketch r = 1+cosθ.

The graph is called a cardioid.

Note that ab = 1.

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Limaçons

Example.

Sketch r = 1+cosθ.

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Limaçons

Example.

The graph is called a limaçon with a dent. Note that ab = 3

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

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Limaçons

Example.

The graph is called a limaçon with a dent.

Note that ab = 3

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Limaçons

Example.

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Limaçons

Example.

Sketch r = 2+cosθ.

The graph is called a convex limaçon. Note that ab = 2.

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Limaçons

Example.

Sketch r = 2+cosθ.

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Limaçons

Example.

Sketch r = 2+cosθ.

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Limaçons

Example.

Sketch r = 2+cosθ.

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Limaçons

Example.

Sketch r = 2+cosθ.

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Limaçons

Example.

Sketch r = 2+cosθ.

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Limaçons

Example.

Sketch r = 2+cosθ.

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Limaçons

Example.

Sketch r = 2+cosθ.

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Limaçons

Example.

Sketch r = 2+cosθ.

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Limaçons

Example.

Sketch r = 2+cosθ.

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Limaçons

Example.

Sketch r = 2+cosθ.

The graph is called a convex limaçon.

Note that ab = 2.

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Limaçons

Example.

Sketch r = 2+cosθ.

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Limaçons

Types of Limaçons

In summary, for r = a±bcosθ, where a,b > 0, we have

i.) 0 < ab < 1 limaçon with a loop

ii.) ab = 1 cardioid

iii.) 1 < ab < 2 limaçon with a dent

iv.) 2 É ab convex limaçon

Remark.

The graph of r =−a±bcosθ is the same as the graph of r = a±bcosθ

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Limaçons

Types of Limaçons

Remark.

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Limaçons

Types of Limaçons

Remark.

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Limaçons

Types of Limaçons

Remark.

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Limaçons

Types of Limaçons

Remark.

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Limaçons

Types of Limaçons

Remark.

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Limaçons

Types of Limaçons

Remark.

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Limaçons

The graph of r = a±bcosθ is a limaçon oriented horizontally, i.e. symmetric alongx−axis.

r = a+bcosθ r = a−bcosθ

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Limaçons

The graph of r = a±bsinθ is a limaçon oriented vertically, i.e. symmetric alongy−axis.

r = a+bsinθ r = a−bsinθ

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Special Curves

Roses are curves whose equations are of the form

r = acosnθ;

r = asinnθ where a > 0, n ∈N

r = acosnθr = acos(n(−θ)) =⇒ r = acosnθthus, symmetric along the x−axis.additionally, symmetric along y−axis for an even n

r = asinnθ−r = asin(n(−θ)) =⇒ −r =−asinnθ =⇒ r = asinnθthus, symmetric along the y−axis.additionally, symmetric along x−axis for an even n

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Special Curves

r = acosnθ; or

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Special Curves

r = acosnθ; or

r = acosnθ

r = acos(n(−θ)) =⇒ r = acosnθthus, symmetric along the x−axis.additionally, symmetric along y−axis for an even n

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Special Curves

r = acosnθ; or

r = acosnθr = acos(n(−θ))

=⇒ r = acosnθthus, symmetric along the x−axis.additionally, symmetric along y−axis for an even n

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Special Curves

r = acosnθ; or

r = acosnθr = acos(n(−θ)) =⇒ r = acosnθ

thus, symmetric along the x−axis.additionally, symmetric along y−axis for an even n

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Special Curves

r = acosnθ; or

r = acosnθr = acos(n(−θ)) =⇒ r = acosnθthus, symmetric along the x−axis.

additionally, symmetric along y−axis for an even n

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Special Curves

r = acosnθ; or

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Special Curves

r = acosnθ; or

r = asinnθ−r = asin(n(−θ))

=⇒ −r =−asinnθ =⇒ r = asinnθthus, symmetric along the y−axis.additionally, symmetric along x−axis for an even n

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Special Curves

r = acosnθ; or

r = asinnθ−r = asin(n(−θ)) =⇒ −r =−asinnθ

=⇒ r = asinnθthus, symmetric along the y−axis.additionally, symmetric along x−axis for an even n

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Special Curves

r = acosnθ; or

r = asinnθ−r = asin(n(−θ)) =⇒ −r =−asinnθ =⇒ r = asinnθ

thus, symmetric along the y−axis.additionally, symmetric along x−axis for an even n

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Special Curves

r = acosnθ; or

r = asinnθ−r = asin(n(−θ)) =⇒ −r =−asinnθ =⇒ r = asinnθthus, symmetric along the y−axis.

additionally, symmetric along x−axis for an even n

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Special Curves

r = acosnθ; or

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Special Curves

r = acosnθ; or

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Example.

Sketch the graph of r = 2cos2θ.

The graph is a rose with 4 petals.In fact, the number of petals is 2n if n is even. And it’s n if n is odd.

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Example.

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Example.

Polar Curves 26/ 39

Example.

Polar Curves 26/ 39

Example.

Polar Curves 26/ 39

Example.

Polar Curves 26/ 39

Example.

Polar Curves 26/ 39

Example.

Polar Curves 26/ 39

Example.

Polar Curves 26/ 39

Example.

The graph is a rose with 4 petals.

In fact, the number of petals is 2n if n is even. And it’s n if n is odd.

Polar Curves 26/ 39

Example.

Polar Curves 26/ 39

Example.

Sketch the graph of r = 2sin3θ.

The graph is a rose with 3 petals.Here, n = 3 is odd. Hence, n = 3 is the number of petals.

Polar Curves 27/ 39

Example.

Polar Curves 27/ 39

Example.

Polar Curves 27/ 39

Example.

Polar Curves 27/ 39

Example.

Polar Curves 27/ 39

Example.

Polar Curves 27/ 39

Example.

Polar Curves 27/ 39

Example.

The graph is a rose with 3 petals.

Here, n = 3 is odd. Hence, n = 3 is the number of petals.

Polar Curves 27/ 39

Example.

Polar Curves 27/ 39

Special Curves

r = 2cos4θ

r = 2sin4θ

Polar Curves 28/ 39

Special Curves

r = 2cos4θ

r = 2sin4θ

Polar Curves 28/ 39

Special Curves

r = 2cos4θ r = 2sin4θ

Polar Curves 28/ 39

Special Curves

Polar Curves 28/ 39

Special Curves

r = 2cos9θ

r = 2sin9θ

Polar Curves 29/ 39

Special Curves

r = 2cos9θ

r = 2sin9θ

Polar Curves 29/ 39

Special Curves

Polar Curves 29/ 39

Special Curves

Polar Curves 29/ 39

Special Curves

Exercises

Graph the following:

1 r = 2cosθ

2 r =−3cos2θ

3 r = sin4θ

4 r = 5cos5θ

Polar Curves 30/ 39

Interesting Curves

Example.

The graph of r2 = 6cos2θ is a lemniscate.

Polar Curves 31/ 39

Interesting Curves

Example.

Polar Curves 31/ 39

Interesting Curves

Example.

Polar Curves 31/ 39

Interesting Curves

Example.

Polar Curves 31/ 39

Interesting Curves

Example.

Polar Curves 31/ 39

Interesting Curves

Example.

The graph of r = θ,θ Ê 0 is the Archimedian spiral.

Polar Curves 32/ 39

Interesting Curves

Example.

r = 1+4cos5θ

Polar Curves 33/ 39

Interesting Curves

Example.

r = sin

Polar Curves 34/ 39

Interesting Curves

Example.

r = esinθ −2cos4θ

Polar Curves 35/ 39

Interesting Curves

Example.

r = sin2(2.4θ)+cos4(2θ)

Polar Curves 36/ 39

Interesting Curves

Example.

r = sin2(1.2θ)+cos3(6θ)

Polar Curves 37/ 39

Interesting Curves

Example.

r = ecosθ −2cos4θ+ sin3(θ

Polar Curves 38/ 39

Interesting Curves

Example.

The "cannabis" curver =

10 cos8θ)(

1+ 110 cos24θ

10 + 110 cos200θ

)(1+ sinθ)

Polar Curves 39/ 39

Lecture 9 (Polar Coordinates and Polar Curves)

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Transcript of Lecture 9 (Polar Coordinates and Polar Curves)