Polar Coordinates

Section 1 Polar Points

(r,θ) = (-r,θ+π) (-r,θ) = (r,θ+π)

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 )

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 ) , …...

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)

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)

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

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

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)

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 )

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 )

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)

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

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)

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)

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)

Solution Let ( r, θ) be a polar representation of the given point 1. 4 = 8 cosθ and 4 √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)

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 )

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)

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

Answer Let (r, θ ) be the polar representation of the given point. 1 = √2 cosθ and - 1 = √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)

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

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.

Starting from the rectangular representation ( 4, 4 √3) of the point, let’s choose r to be - √(16 + 48) = - 8 . Then, we have 4 = - 8 cosθ and 4 √3 = - 8 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)

Examples 3 Express each of the following equations (Given in the rectangular form) in the polar coordinates: 1. x2 + y2 - 4 = 0 2. x2 - y2 - 4 = 0 3. 4 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

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

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

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

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.

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

5. The given equation 4(x + y) 2 = 3x – 4y Which is equivalent to: 4(x2 + y2) +8xy = 3x – 4y, becomes : 4r2 + 8 r2cos θ sin θ = 3 rcosθ - 4 rsinθ or 4r2 ( 1 + 2 cos θ sin θ ) = r(3 cosθ - 4 sinθ) Which leads to r = 0 Or r = (3 cosθ - 4 sinθ) / 4( 1 + 2 cosθ sinθ )

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

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

Examples 4 Express each of the following equations (Given in the rectangular form) in the polar coordinates: 1. r = 3 2. r = 6sinθ 3. r = 6cosθ 4. r = 1 – cosθ 5. r = cosθ – 5sinθ 6. θ = π/4

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

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

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

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

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

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

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

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