Chap 1 - Polar Coordinates

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CHAPTER 1

POLAR COORDINATES

1.1 Parametric equations

1.1.1 Definition

1.1.2 Sketching parametric equations

1.2 Polar Coordinates

1.2.1 Polar coordinates system

1.2.2 Relationship between Cartesian and Polar

Coordinates

1.2.3 Forming polar equations from Cartesian

equations and vice-versa

1.2.4 Sketching polar equations1.2.5 Intersection Of Curves In Polar

Coordinates



1.1 Parametric Equations

1.1.1 Definition:

Equations )(t f x = , )(t g y = that express x and y

in t is known as parametric equations, and t is called

the parameter.

How the parameter may be eliminated from the

parametric equations to obtain the Cartesianequations?

- no specific method

- use algebraic manipulation

Example: Form Cartesian equations by eliminating

parameter t in the following equations:

(a) t x 2= , 14 2 −= t y

(b) t x sin4= , t y 2cos2=

(c)t t e ye x −== ,

(d) t yt x ln3,3 ==

2



1.1.2 Curve Sketching of parametric equations

Constructing tables of x and y

3

2 ways for specific values of t.

Eliminating the parameter.

Example:

Sketch the graph of the following equations

(a) t x 2= , 14 2 −= t y

(b) 53 −= t x , 52 += t y



1.2 Polar Coordinates

1.2.1 Polar Coordinates System

Definition:

The polar coordinates of point P is written as an

ordered pair ( )θ ,r , that is ),( θ r where

r - distance from origin to P θ - angle from polar axis to the line OP

P ( r, θ )

O

r

θ

Note:

(i) θ is positive in anticlockwise direction, and it

is negative in clockwise direction.

(ii) Polar coordinate of a point is not unique.

(iii) A point ( )θ ,r − is in the opposite direction of

point ( )θ ,r .

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Example: Plot the following set of points in the

same diagram:

(a)

)

o225,3 ,

)

o225,1 ,

)

o225,3−

(b) ⎟ ⎞

⎜⎝

⎛

3,2

π , ⎟

⎞⎜⎝

⎛ −

3,2

π , ⎟

⎞⎜⎝

⎛ −

3,2

π

a)

5

b)



For every point P (r,θ ) in , there exist 3

more coordinates that represent the point P .

P ( 2, 450 )

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Example:

Find all possible polar coordinates of the points

whose polar coordinates are given as the following:

(a)o45,1 P (b)

o60,2 −Q (c)o225,1− R



1.2.2 Relationship between Cartesian and Polar

Coordinates

1) Polar Cartesian

θ cosr x = θ sinr y =

2) Cartesian Polar

22 y xr += x

y=θ tan

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Example: Find the Cartesian coordinates of the

points whose polar coordinates are given as

(a) ⎟ ⎞⎜⎝ ⎛

47,1 π (b) ⎟ ⎞⎜

⎝ ⎛ −

32,4 π (c)

o30,2 −

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Example: Find all polar coordinates of the points

whose rectangular coordinates are given as

(a) (b)( )5,11 ( )2,0 (c) ( )4,4 −−

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1.2.3 Forming polar equations from Cartesian

equations and vice-versa.

To change the equation in Cartesian coordinates to polar

coordinates, and conversely, use equation

θ cosr x = θ sinr y = 22 y xr +=

Example: Express the following rectangular equations in

polar equations.

(a) (b) (c)2 x y = 1622 =+ y x 1= xy

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Example: Express the following polar equations in

rectangular equations.

(a) θ sin2=r (b)θ θ sin5cos4

3+

=r

(c) θ θ sin4cos4 +=r (d) tan secr θ θ =

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1.2.4 Graph Sketching of Polar Equations

There are two methods to sketch a graph of ( )θ f r = :

(1) Form a table for r and .θ ( π θ 20 ≤≤ ).

From the table, plot the ( )θ ,r points.

(2) Symmetry test of the polar equation.

The polar equations is symmetrical

(a) about the x-axis if ( ) ( )θ θ f f =−

- consider in range [0, 1800] only

(b) about the y-axis if ( ) ( )θ θ π f f =−

- consider in range [0, 900] and [2700, 3600]

(c) at the origin if ( ) ( )θ θ π f f =+

- consider in range [0, 1800] or [180

0, 360

0]

* if symmetry at all, consider in range [0, 900] only.

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Example 1: Sketch the graph of θ sin2=r

Solution: (Method 1)

Here is the complete table

0 30 60 90 120 150 180 210

0 1.0 1.732 2 1.732 1 0 -1.0

240 270 300 330 360

-1.732 -2 -1.732 -1 0

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Method 2

Symmetrical test for θ θ sin2)( =

Symmetry Symmetrical test

About x-axis

( ) ( ) ?θ θ f f =−

About y-axis

( ) ( ) ?θ θ π f f =−

About origin

( ) ( ) ?θ θ π f f =+

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Since r symmetry at y-axis, consider in range

[0, 900] and [270

0, 360

0]

0 30 60 90 270 300 330 360 0 1.0 1.732 2 -2 -1.732 -1 0

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Example 2: Sketch the graph of θ cos2

3−=r


About x-axis

( ) ( ) ?θ θ f f =−

About y-axis

( ) ( ) ?θ θ π f f =−

About origin

( ) ( ) ?θ θ π f f =+

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Since r symmetry at x-axis, consider in range

[0, 1800] only.

0 30 60 90 120 150 180

θ cos2

3−=r

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Example 3: Sketch the graph of θ 2sin2=r


About x-axis

( ) ( ) ?θ θ f f =−

About y-axis

( ) ( ) ?θ θ π f f =−

About origin

( ) ( ) ?θ θ π f f =+

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Since r symmetry at ____________, consider in

range ____________ only.

θ 2sin2=r

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1.2.5 Intersection Of Curves In Polar Coordinates.

In Cartesian, we normally either solve two

simultaneous equations or plot the graph in order to findthe point of intersection between two curves.

But in polar, method for finding points of intersection

must include both solving simultaneous equations and

sketching the curves.

Example 1:

Find the points of intersection of the circle θ cos2=r and

θ sin2=r for π θ ≤≤0

Solution:

♦ Solve simultaneous equations

4

5,

4sin2cos2

π π θ θ θ =⇒=

when 24

cos2,4

===π π

θ r

when 24

5cos2,4

5 −=== π π θ r

So, point of intersection: ⎟ ⎞

⎜⎝

⎛

4,2

π

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Note: ⎟ ⎠

⎞⎜⎝

⎛

4,2

π and ⎟

⎞⎜⎝

⎛ −

4

5,2

π is the same point.

♦ Draw the curves:

From graph, (0,0) is also a point of intersection.

Conclusion: Points of intersection are ⎟ ⎠

⎞⎜⎝

⎛

4,2

π and (0,0).

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Example 2:

Find the points of intersection of the curves

θ cos2

3

−=r and 3

2π

θ = .

Solution:

♦ Solves simultaneous equations

2

2

1

2

3

3

2cos

2

3=⎟

⎞⎜

⎝

⎛ −−=−=π

r

From the solution of equations, we get only one point.

i.e ⎟ ⎞

⎜⎝

⎛

3

2,2

π .

♦ Draw the curves:

22



From the graph,3

5π θ = is another solution.

When3

5π θ = , 1

2

1

2

3

3

5cos

2

3=⎟

⎠

⎞⎜⎝

⎛ −=−=

π r

Thus ⎟ ⎞

⎜⎝

⎛

3

5,1

π is a point of intersection.

Conclusion: The points of intersection are ⎟

⎠

⎞⎜

⎝

⎛

3

2,2

π and

⎟ ⎞

⎜⎝

⎛

3

5,1

π .

Chap 1 - Polar Coordinates

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