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Oil Pump Problem

Oil Pump Problem. The height the end of the pump arm is above the ground is given by the equation H = 6 + 4Sin( π t). 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?.

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Oil Pump Problem

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  1. Oil Pump Problem 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 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 Method D: accurate graph Height ‘H’ Notes

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

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

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

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

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

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