1. Unit 6: Applications of Trigonometry
Part 3: Polar Coordinate System
Lesson 4: Polar Coordinates and Equations

2. Key Point 1: Plotting Polar Coordinates
Polar coordinate system – a plane with the following qualities
Pole – point O (polar coordinates 0, 𝜃 )
Polar axis – a ray from O
Polar coordinates 𝑟, 𝜃
r is the directed distance from O to P
Θ is the directed angle whose initial side is the polar axis and whose terminal side is the ray 𝑂𝑃

3. Example 1: Plotting Points in the Polar Coordinate System
Plot the point 𝑃 2, 𝜋 3 . Solution: r = 2, θ = π/3

4. Key Point 2: Finding all Polar Coordinates
Let P have polar coordinates 𝑟, 𝜃 . Any other polar coordinate of P must be of the form 𝑟, 𝜃+2𝑛𝜋 or −𝑟, 𝜃+(2𝑛+1)𝜋 −𝑟, 𝜃+(2𝑛+1)𝜋 , where n is any integer.

5. Example 2: Finding all Polar Coordinates for a Point
If the point P has polar coordinates 3, 𝜋 3 , find all polar coordinates for P. Solution: From the first formula, 3, 𝜋 3 = 3, 𝜋 3 +2𝑛𝜋 = 3, 𝜋 3 + 6𝑛𝜋 3 3, 𝜋 3 + 6𝑛𝜋 3 = 3, 6𝑛𝜋+𝜋 3 = 3, (6𝑛+1)𝜋 3 From the second formula, 3, 𝜋 3 = −3, 𝜋 3 +(2𝑛+1)𝜋 −3, 𝜋 3 +(2𝑛+1)𝜋 = −3, 𝜋 3 + (6𝑛+3)𝜋 3 = −3, (6𝑛+3)𝜋+𝜋 3 = −3, (6𝑛+4)𝜋 3

6. Key Point 3: Coordinate and Equation Conversion
Polar to Rectangular: 𝑥=𝑟 cos 𝜃 𝑦=𝑟 sin 𝜃
Rectangular to Polar: 𝑟 2 = 𝑥 2 + 𝑦 2 tan 𝜃 = 𝑦 𝑥

7. Example 3: Converting from Polar to Rectangular Coordinates
Use an algebraic method to find the rectangular coordinates of the point with polar coordinates 3, 5𝜋 6 . Approximate exact solutions with a calculator when appropriate. Solution: 𝑥=𝑟 cos 𝜃 =3 cos 5𝜋 6 =3 3 2 = 𝑂𝑅 𝑦=𝑟 sin 𝜃 =3 sin 5𝜋 6 =3 1 2 = 3 2 𝑂𝑅 1.5 Acceptable Answers: , 3 2 , , 1.5 , 2.598, 3 2 , 2.598, 1.5

8. Example 4: Converting from Rectangular to Polar Coordinates
Find two polar coordinate pairs for the point with rectangular coordinates −1, 1 . Solution: 𝑟 2 = 𝑥 2 + 𝑦 2 = (−1) 2 + (1) 2 =1+1=2 𝑟=± 2 tan 𝜃 = 𝑦 𝑥 = 1 −1 =−1 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 tan −1 tan −1 =135° 𝑂𝑅 3𝜋 4

9. Example 5: Converting from a Polar to Rectangular Equation
Convert 𝑟=4 sec 𝜃 to rectangular form. Solution: 𝑟=4 sec 𝜃 𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑏𝑦 sec 𝜃 𝑟 sec 𝜃 =4 𝑟𝑒𝑐𝑖𝑝𝑟𝑜𝑐𝑎𝑙 𝑖𝑑𝑒𝑛𝑡𝑖𝑡𝑦 𝑟 cos 𝜃 =4 𝑥=𝑟 cos 𝜃 𝑥=4

10. Example 6: Converting from a Rectangular to Polar Equation
Convert (𝑥−3) 2 + (𝑦−2) 2 =13 to polar form. Solution: (𝑥−3) 2 + (𝑦−2) 2 =13 𝑠𝑞𝑢𝑎𝑟𝑒 𝑡ℎ𝑒 𝑝𝑎𝑟𝑒𝑛𝑡ℎ𝑒𝑠𝑒𝑠 𝑥 2 −6𝑥+9+ 𝑦 2 −4𝑦+4 =13 𝑐𝑜𝑚𝑏𝑖𝑛𝑒 𝑙𝑖𝑘𝑒 𝑡𝑒𝑟𝑚𝑠 𝑥 2 + 𝑦 2 −6𝑥−4𝑦+13 =13 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 13 𝑓𝑟𝑜𝑚 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑥 2 + 𝑦 2 −6𝑥−4𝑦 =0 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑖𝑛 𝑓𝑜𝑟𝑚𝑢𝑙𝑎𝑠 𝑟 2 −6𝑟 cos 𝜃 −4𝑟 sin 𝜃 =0 𝐺𝐶𝐹 𝑟 𝑟−6 cos 𝜃 −4 sin 𝜃 =0 r = 0 is one point, so the equation is 𝑟−6 cos 𝜃 −4 sin 𝜃

11. Unit 6 HL 4
TB p. 492 #3, 8, 16, 25, 37 OR MathXL Unit 6 HL 4