1. Polar Coordinates

2. Section 1
Polar Points

3. (r,θ)
= (-r,θ+π)
(-r,θ) = (r,θ+π)

4. A point which has only one rectangular representation has infinitely many polar representations.
If a point has a polar representation ( r, θ), then it has also the polar representation
( - r , θ + π )
and further more , for every k ε Z, that point has the representation
( r, θ + 2kπ ) and the representation ( - r , (θ + π ) + 2k π ) = ( - r , θ + (2k+1) π )

5. Example
The point having the polar representation ( 5 , π /3) has also all of the following polar representations:
( 5 , 7 π /3 ) , ( 5 , 13 π /3 ) , ( 5 , 19 π /3 ) , ( 5 , 25 π /3 ),….. .
( 5 , -5 π /3 ) , ( 5 , -11 π /3 ) , ( 5 , -17 π /3 ) , ( 5 , -23 π /3 ), …...
( -5 , 4 π /3 ) , ( -5 , 10 π /3 ) , ( -5 , 16 π /3 ) , ( -5 , 22 π /3 ) , …...
( - 5 , -2 π /3 ) , ( -5 , -8 π /3 ), ( -5 , -14 π /3 ) , ( -5 , -20 π /3 ) , …...

6. Homework Plot the following polar points:
1. (3,0) , ( -3, π) , ( 3, 20 π), (-3,23 π)
2. (-3,0) , ( 3, π) , ( -3, 20 π), (3,23 π)
3. (3, π/2) , ( -3, 3π/2) , ( -3, -π/2) ,( 3, 21π/2), (-3,23π/2)

7. 4. (3, -π/2) , ( 3, 3π/2) , ( -3, π/2)
,( -3, 21π/2), (3,23π/2)
5. (3, π/6) , ( -3, 7π/6) , ( 3, 61π/2)
,( -3, 19π/6), (-3,-53π/6)
6. . (-3, π/6) , ( 3, 7π/6) , ( -3, 61π/2)
,( 3, 19π/6), (3,-53π/6)

8. Relationship between Polar & Rectangular Coordinate Systems
Section 2
Relationship between Polar & Rectangular Coordinate Systems

9. Let (x,y,) є R2
A polar coordinates point corresponding to (x , y) is any pair (r , θ) satisfying:
x = r cosθ , y = r sinθ

10. Examples 1
Assuming that the each of the following represents a point in a polar coordinate system, express each of them in rectangular coordinates.
1. ( 8, π/3 )
2. ( 8, 5π/6 )
3. ( 4 √2, 5π/4)

11. Solution Let (x,y,) be the rectangular representation of the considered point. 1.
x = 8 cos (π/3) = 8 ( ½) = 4
y = 8 sin (π/3) = 8 ( √3 / 2 ) = 4 √3
Thus the rectangular representation of the given point is ( 4, 4 √3 )

12. 2. x = 8 cos (5π/6) = - 4 √3 y = 8 sin (5π/6) = 4
Thus the rectangular representation of the given point is ( - 4 √3 , 4 )

13. 3. x = 4 √2 cos (5π/4) = - 4 y = 4 √2 sin (5π/4) = - 4 .
Thus the rectangular representation of the given point is ( - 4, - 4)

14. Question
Assuming that ( - 5, 11π/6 ) represents a point in a polar coordinate system, express each it in rectangular coordinates.

15. Answer Let (x,y,) be the rectangular representation of the given point.
x = -5 cos (11π/3) = - 5 (1/2) = -5/2
y = -5 sin (11π/3) = - 5 (√3 / 2) = -5 √3 / 2
Thus the rectangular representation of the given point is ( - 4, - 4)

16. So the trigonometric values for are the same, as the those for - π/3
Now Ahmad Fayzi says “ Why?” and 2 minutes after him Khalil will say “ Why?” The following answer is for both of them!!
11π/3 = (12π - π) / 3 = 4π- π/3
So the trigonometric values for are the same, as the those for - π/3 Or
11π/3 = 4π- π/3 = 2π+ (2π- π/3)
So the trigonométrique values for are the same, as the those for (2π- π/3)

17. Examples 2 .
Assuming that the each of the following represents a point in a rectangular coordinate system, express each of them in polar coordinates.
1 ( 4, 4 √3)
2. (- 4 √3 ,4 )
3. ( - 4, - 4)

18. Solution Let ( r, θ) be a polar representation of the given point 1.
4 = 8 cosθ and √3 = 8 sinθ ,
and hence cosθ = ½ , sinθ = √3 / 2 .
Therefore, θ = π/3 +2kπ ; kεZ
Thus a representation of the point in the polar coordinates is ( 8 , π/3)

19. 2. r = √ [(4 √3)2 + (4 )2 ] = 8 , - 4 √3 = 8 cosθ and 4 = 8 sinθ , and
hence cosθ = - √3 / 2 , sinθ = ½ .
Therefore, θ = [π - (π/6)] + 2kπ
= 5π / 6 + 2kπ ; kεZ
Thus a representation of the point in the polar coordinates is ( 8 , 5π/6 )

20. 3. r = √ [(4)2 + (4)2] = 4 √2 , - 4 = 4 √2 cosθ and - 4 = 4 √2 sinθ
and hence
cosθ = - 1 / √2 , sinθ = - 1 / √2
Therefore, θ = [π + (π/4)] + 2kπ
= 5π / 4 + 2kπ ; kεZ
Thus the representation of the point in the rectangular coordinates is (4 √2 , 5π/4)

21. Question
Assuming that ( 1 , - 1 ) represents a point in a rectangular coordinate system, express it in polar coordinates

22. Answer Let (r, θ ) be the polar representation of the given point.
1 = √2 cosθ and = √2 sinθ
and hence
cosθ = 1 / √2 , sinθ = - 1 / √2
Therefore, θ = [2π - (π/4)] + 2kπ
= 7π / 4 + 2kπ ; kεZ
Thus the representation of the point in the rectangular coordinates is (√2 , 7π/4)

23. Ahmad Fauzi Wonders and 2 minutes later Khalil will wonder:
Why do not choose r = - √ [(1)2 + (-1)2] = - √2 ?
The following Remark is again for both of them!!
And also for the rest of the students

24. Remark 1
A matter that sometimes confuses the students when finding a polar representation( r, θ) of a point (x,y) from its rectangular representation. The student asks why, for instance, in example 2, when finding r, we chose the positive root of ( x2 + y2 ) and not its negative root . The answer to that is that we could have chosen the negative root as well.

25. Starting from the rectangular representation ( 4, 4 √3) of the point, let’s
choose r to be - √( ) =
Then, we have
4 = cosθ and
4 √3 = Sinθ
→ cosθ = -1/2 and sinθ = - √3 / 2 .
Thus, θ = π + π / 3 = 4π / 3.
This give us the polar representation ( - 8 , 4π/3 ) which is another polar representation for the same point represented by the polar representation ( 8 , π/3).
Notice that (- 8 , 4π/3 ) = (- 8 , π+π/3)

26. Examples 3
Express each of the following equations (Given in the rectangular form) in the polar coordinates:
1. x2 + y = 0
2. x2 - y = 0
y - x2 = 0
4. x = 3
5. 4(x + y) 2 = 3x - 4y
6. 2x + 3y – 5 = 0
7. x2 - 4x + y2 – 6y + 13 = 0

27. Solution Let ( r, θ) be a polar representation of the Cartesian point(x , y ) We substitute rcosθ and rsinθ for x and y respectively and keep in mind that x2 + y2 = r2

28. 1. The given equation . x2 + y2 - 4 = 0 becomes r2 = 4
or equivalently r = 2 or r = -2

29. 2. The given equation . x2 - y2 - 4 = 0 becomes
r2 cos2 θ r2 sin2 θ = which leads to;
r2 (cos2 θ sin2 θ) = which leads to;
r2 cos 2θ = 4
r2 = 4 / cos2θ
r = 2 √ sec2θ or r = √ sec2θ

30. 3. The given equation . 4y = x2 becomes
4r sinθ = r2 cos2 θ ,which leads to;
Either r = 0 or r = 4sinθ / cos2 θ = 4 secθ tanθ
The equation r = 0 representing the pole (origin) and included in the equation r = 4 secθ tanθ may be discarded.

31. 4.
The given equation x = 3
becomes
r cosθ = 3
Or r = 3secθ

32. 5. The given equation 4(x + y) 2 = 3x – 4y Which is equivalent to:
4(x2 + y2) +8xy = 3x – 4y, becomes : 4r r2cos θ sin θ = 3 rcosθ - 4 rsinθ
or 4r2 ( cos θ sin θ ) = r(3 cosθ - 4 sinθ)
Which leads to
r = Or
r = (3 cosθ - 4 sinθ) / 4( cosθ sinθ )

33. 6. The given equation 2x + 3y – 5 = 0 , becomes :
2r cos θ + 3 rsinθ = 5
Or r = 5 / (2cos θ + 3sin θ)

34. 7. The given equation x2 - 4x + y2 – 6y + 13 = 0,
Which is equivalent to:
x2 + y2 -4x – 6y + 13 = 0
, becomes : r2 – 4 r cosθ – 6r sinθ + 13 = 0 Or
r2 – r(4 cosθ – 6 sinθ) + 13 = 0

35. Examples 4 Express each of the following equations (Given in the
rectangular form) in the polar coordinates:
r = 3
r = 6sinθ
r = 6cosθ
r = 1 – cosθ
r = cosθ – 5sinθ
θ = π/4

36. Solution Let ( x, y) be a polar representation of the Cartesian point (r , θ ) We rplace rcosθ and rsinθ by x and y respectively and keep in mind that x2 + y2 = r2

37. 1.
We have; r = 3
or r2 = 9
Or equivalently x2 + y2 = 9

38. 2. We have; r = 6sinθ or r2 = 6 r sinθ Or equivalently x2 + y2 = 6y
Or x2 + ( y– 3 ) = 9

39. 3. We have; r = 6cosθ or r2 = 6 r cosθ Or equivalently x2 + y2 = 6x
Or ( x – 3 )2 + y2 = 9

40. 4.
We have; r = 1 – cosθ
multiplying both sides of the equation by r , we get;
r2 = r - r cosθ
Or r = r2 + r cosθ
Squaring , we get:
r2 = ( r2 + r cosθ)2 Or
x2 + y2 =( x2 + y2 + x )2

41. Checking, for unintentionally introduced points:
1. Multiplying by r introduce only the pole r = 0, which is already included in the original equation ( for θ = 0, for example)
2. Squaring introduced the equation r = - ( r2 + r cosθ), which is equivalent to 1 = - ( r + cosθ) or 1 = - r - cosθ), which has the same graph as the original curve 1 = -(-r) – cos(π+θ) = r + cos θ . Why? Replac (r, θ) by ( -r, θ+π ) in the original equation

42. 5.
We have; r = cosθ – 5sinθ
multiplying both sides of the equation by r , we
get;
r2 = rcosθ – 5rsinθ
Or x2 + y2 = x - 5y

43. 6.
We have; θ = π/4
Which can be replaced by tanθ = 1; 0 < θ < π/2
Or sinθ / cosθ = 1; 0 < θ < π/2
Or sinθ = cosθ ; 0 < θ < π/2
( cos θ is not zero, since 0 < θ < π/2)
Or rsinθ = rcosθ
Which becomes: y = x