1. RATE of CHANGE

2. SLOPE or m1m1 m2m2 m 1 = m 2 Slope of a line is constant Slope is known as the change in ‘y’ with respect to ‘x’    m is - m is + m is 0 Slope of a curve varies Slope of the tangent line = slope of the curve at the point of contact

3. One application of slope is velocity velocity is the change in position with respect to time. Two aspects of velocity I. Average velocity Time (h) Distance (km) 00 1105 2197 3315 4402 5500 What is the average velocity between the 2 nd and 4 th hours? (2, 197) ; (4, 402) velocity is 102.5 km/h

4. One application of slope is velocity velocity is the change in position with respect to time. Two aspects of velocity I. Average velocity Time (h) Distance (km) 00 1105 2197 3315 4402 5500 What is the average velocity for the first 3 hours of the trip? (0, 0) ; (3, 315) velocity is 105 km/h

5. Distance (km) Time (minutes) Tangent of the curve Instantaneous rate of change at the given moment in time.

6. Kim and Duncan took a trip and recorded information about time and distance travelled. 00 1015 2030 45 4060 5080 60100 70105 80110 90135 100150 110155 Time (minutes) Number of kilometres traveled Distance (km) Time (minutes) 0 10 20 30 40 50 60 70 80 90 100 110 120 130 10 20 30 40 50 60 70 80 90 100 110 120 130 150 140 160

7. Kim and Duncan took a trip and recorded information about time and distance travelled. 00 1015 2030 45 4060 5080 60100 70105 80110 90135 100150 110155 Time (minutes) Number of kilometres traveled Distance (km) Time (minutes) 0 10 20 30 40 50 60 70 80 90 100 110 120 130 10 20 30 40 50 60 70 80 90 100 110 120 130 150 140 160

8. 0 10 20 30 40 50 60 70 80 90 100 110 120 minutes 10 150 140 130 120 110 100 30 20 50 40 70 60 90 80 160km Find their average speed (velocity) over the first 50 min of the trip. Express your answer in km/hour Starting values? Ending values? (0 min, 0 km ) (50 min, 80 km ) 1.6km/min x 60 min = 96km/h

9. y2y2 y1y1 x2x2 x2x2 distance time m = slope Rate of change is really a calculation of the slope of a line between any two given points.

10. 0 10 20 30 40 50 60 70 80 90 100 110 120 minutes 10 150 140 130 120 110 100 30 20 50 40 70 60 90 80 160km Find their average velocity between 40 and 80 min into the trip. Express your answer in km/hour Starting values? Ending values? ( 40 min, 60 km ) (80 min, 110 km) 1km/min x 60 min = 75km/h

11. 0 10 20 30 40 50 60 70 80 90 100 110 120 minutes 10 150 140 130 120 110 100 30 20 50 40 70 60 90 80 160km Find their average velocity between 30 and 90 min into the trip. Express your answer in km/hour Starting values? Ending values? ( 30 min, 45 km ) (90 min, 135 km) 1.5km/min x 60 min = 90km/h

12. 0 10 20 30 40 50 60 70 80 90 100 110 120 minutes 10 150 140 130 120 110 100 30 20 50 40 70 60 90 80 160km If the average velocity the entire trip was 80km/h what would the graph look like 60 min – 80 km 120 min – 160 km

13. minkm 55 10 20 4030 6040 8050 9060 10080 Distance (km) 20 40 60 80 100 120 0 10 20 30 40 50 60 70 80 90 100 Time (min) CAR TRIP

14. Distance (km) minkm 55 10 20 4030 6040 8050 9060 10080 A B C D 0 10 20 30 40 50 60 70 80 90 100 110 Time (min) 20 40 60 80 100 120 Calculate the average velocity from A to BB to C C to DA to D A B C D 60km/h30km/h 90km/h47.5km/h

15. A toy rocket is launched from the ground and its flight path is described as y = -x 2 + 13 x – 30. h = - (t) 2 + 13t - 30 h represents the height (in meters) of the rocket ‘t’ represents the time (in seconds)of the rocket flight h =-t 2 t

16. h = - (t) 2 + 13t - 30 10 5 15 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 seconds metres Fuse is lit but nothing happens for 3 seconds It strikes the ground 10 seconds after the fuse is lit Vertex tells how high the rocket reaches

17. h = - (t) 2 + 13t - 30 10 5 15 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 seconds metres Velocity or speed of the rocket can now be determine

18. Distance (km) Time (minutes) 0 10 20 30 40 50 60 70 80 90 100 110 120 130 10 20 30 40 50 60 70 80 90 100 110 120 130 150 140 160 Along a curve the rate of change is continually shifting To find the average rate of change between two points we determine a straight line distance between them

19. A constant positive rate of change (acceleration) A constant negative rate of change (negative acceleration) deceleration

20. (6s, 9m) (10s, 17m)

21. (18s, 9m) (16s, 14m)

22. Formula for rate of change a represents the initial time b represents the ending time f(a) represents the height calculated at time ‘a’ f(b) represents the height calculated at time ‘b’

23. Formula for rate of change h = - (t) 2 + 13t - 30 (a)f(a) h = - (t) 2 + 13t - 30 (b)f(b)

24. Formula for rate of change a = 1s b = 3s f(a) = - 4.9(1) 2 + 15(1) + 10 A projectile is traveling along a path determined by h = - 4.9x 2 + 15x + 10. What is its average rate of speed between 1 and 3 seconds” f(b) = - 4.9(3) 2 + 15(3) + 10 f(a) = - 4.9 + 25 f(a) = 20.5m f(b) = - 44.1 + 55 f(b) = 10.9m

25. Formula for rate of change a = 1s b = 3s f(a) = - 4.9(1) 2 + 15(1) + 10 f(b) = - 4.9(3) 2 + 15(3) + 10 f(a) = - 4.9 + 25 f(a) = 20.1m f(b) = - 44.1 + 55 f(b) = 10.9m Falling at an average sped of 4.8m/s

26. Given the function h = – 4.9t 2 + 20t + 15 What was the initial height of the projectile? When did the projectile reach maximum height? 15m Find ‘t’ h = – 4.9t 2 + 20t + 15 h = - 4.9(2.04) 2 + 20(2.04) + 15 h = 35.41m Maximum height ( 2.04s, 35.41m) 2.04s 35.41m

27. Distance (km) Time (minutes) Tangent of the curve Instantaneous rate of change at the given moment in time.

28. I. Instantaneous velocity – how far am I traveling at one specific moment? Need the slope of the tangent at the specific point.

29. I. Instantaneous velocity – how far am I traveling at one specific moment? Need the slope of the tangent at the specific point. (15.01s, ?m) (14.99s, ?m) ● (15s, 16m)

30. Formula for rate of change a represents the initial time b represents the ending time f(a) represents the height calculated at time ‘a’ f(b) represents the height calculated at time ‘b’ (15.01s, ?m) (14.99s, ?m)

31. Two people have been stranded at sea in a life raft. Their only chance of being rescued is if somebody sees the flare shot from their flare gun. Assume the flare follows the path given by h = -t 2 + 10t + 25 (h = height in metres and t = time in seconds) What is the maximum height the flare will reach? How long will it be before the flare falls back into the sea?? How high is the flare after 3 seconds? What is the average velocity of the flare between 2 and 3 seconds? At 6 seconds what is the velocity of the flare? 12.07s 50m 46m 5m/s -2m/s

32. A giant yo-yo is flung electronically and it follows a path h = t 2 –10t +24 (h = height in metres and t = time in seconds) What is the minimum depth the yo-yo will reach? What was the height of the yo-yo when it was launched?? How high is the yo-yo after 9 seconds? What is the average velocity of the yo-yo between 8 and 10 seconds? At 7 seconds what is the velocity of the yo-yo? 24m -1m 15m 8m/s 3m/s

33. Little Jimmy is strolling around the pool when he drops his favorite toy into the deep end!! The deep end is 8 meters deep. His babysitter jumps into the pool and swims along the path d = x 2 – 10x + 15. (h = height in metres and t = time in seconds) How deep can the babysitter swim? Will she succeed in reaching the toy? How long before she returns to the surface? What is the average velocity of the swimmer between 2 and 5 seconds? At 7 seconds what is the velocity of the swimmer? yes 10m 8.16s 3.7m/s 4m/s

34. The height of a football after is is thrown can be determined by the function h = - 4.9t 2 + 28t + 2. (h = height in metres and t = time in seconds) What is the maximum height the ball will reach After how many seconds does it reach the max height? How long before the ball returns to the ground? What is the average velocity of the football between 1 and 3 seconds? At precisely 4 seconds what is the velocity of the football? 2.9s 42m 5.78s 8.4m/s -11.2m/s

35. Distance (km) Time (minutes) 0 10 20 30 40 50 60 70 80 90 100 110 120 130 10 20 30 40 50 60 70 80 90 100 110 120 130 150 140 160