1. The height the end of the pump arm is above the ground is given by the equation H = 6 + 4Sin(πt) Oil Pump Problem where t = time in minutes And H is the vertical height in metres How long is the height of the pump arm above 7 metres from the ground in one complete cycle? Method A: unit circle Method B: unit circle Method C: graph Height ‘H’ Method D: accurate graph Notes

2. Oil Pump method A Solve 7 = 6 + 4Sin(A) 0.25 = Sin(A) A = 0.253 A = π – 0.253 Calculate Sin -1 0.25 Home The height the end of the pump arm is above the ground is given by the equation H = 6 + 4Sin(πt) where t = time in minutes And H is the vertical height in metres 1 = 4Sin(A) π 0, 2π A TC S Positive Sin so Quadrant 1 & 2 = 0.92 min Quadrant 2 A = 2.889 Let A = πt And t = A ÷ π t = 2.889 ÷ π Solve 7 = 6 + 4Sin(πt) RADIANS! t = 0.253 ÷ π = 0.08 min How long is the height of the pump arm above 7 metres from the ground in one complete cycle? Time = 0.92 – 0.08 = 0.84 min

3. Oil Pump method B Solve 7 = 6 + 4Sin(πt) 0.25 = Sin(πt) πt = 0.253 πt = π – 0.253 Calculate Sin -1 0.25 The height the end of the pump arm is above the ground is given by the equation H = 6 + 4Sin(πt) where t = time in minutes And H is the vertical height in metres Height ‘H’ 1 = 4Sin(πt) π 0, 2π A TC S Positive Sin so Quadrant 1 & 2 2 nd time = 0.92 min (2dp) Quadrant 2 πt = 2.889 RADIANS! 1 st time = 0.253 ÷ π = 0.08 min How long is the height of the pump arm above 7 metres from the ground in one complete cycle? Time = 0.92 – 0.08 = 0.84 min Home

4. Oil Pump method C Solve 7 = 6 + 4Sin(πt) 0.25 = Sin(A) (A) = Sin -1 0.25 The height the end of the pump arm is above the ground is given by the equation H = 6 + 4Sin(πt) where t = time in minutes And H is the vertical height in metres 1 = 4Sin(A) Sketch the graph A = 0.253 Let A = πt Solve 7 = 6 + 4Sin(A) RADIANS! Or A = π – 0.253 Or A = 2.889 = 0.92 min And t = A ÷ π t = 2.889 ÷ π t = 0.253 ÷ π = 0.08 min How long is the height of the pump arm above 7 metres from the ground in one complete cycle? Time = 0.92 – 0.08 = 0.84 min Home

5. Oil Pump method D Solve 7 = 6 + 4Sin(πt) 0.25 = Sin(πt) t = 1 – 0.08 t = 0.92 min (πt) = Sin -1 0.25 The height the end of the pump arm is above the ground is given by the equation H = 6 + 4Sin(πt) where t = time in minutes And H is the vertical height in metres Height ‘H’ 1 = 4Sin(πt) Sketch the graph πt = 0.253 Length of 1 cycle is 2 π ÷ π = 2 Midpoint = 1 t = 0.08 min (2dp) RADIANS! How long is the height of the pump arm above 7 metres from the ground in one complete cycle? Time = 0.92 – 0.08 = 0.84 min Home

6. Oil Pump Notes Solve 7 = 6 + 4Sin(A) 0.25 = Sin(A) A = 0.253 A = π – 0.253 Calculate Sin -1 0.25 The height the end of the pump arm is above the ground is given by the equation H = 6 + 4Sin(πt) where t = time in minutes And H is the vertical height in metres 1 = 4Sin(A) π 0, 2π A TC S Positive Sin so Quadrant 1 & 2 = 0.92 min Quadrant 2 A = 2.889 Let A = πt And t = A ÷ π t = 2.889 ÷ π Solve 7 = 6 + 4Sin(πt) RADIANS! t = 0.253 ÷ π = 0.08 min How long is the height of the pump arm above 7 metres from the ground in one complete cycle? Time = 0.92 – 0.08 = 0.84 min Home