1. Becca Ceremuga AP Calculus Period 2

3.  Acapulco, Mexico  35 meters (115 feet)

5. 115 ft Scale 1 cm= 5.75 ft Distance from Starting Point= 0 ft Time= 0 sec

6. Distance from Starting Point= 12.5 ft Scale 1 cm= 5.75 ft Time= 1 sec 102.5 ft

7. Scale 1 cm= 5.75 ft Distance from Starting Point= 35.8 ft Time= 2 sec 79.2 ft

8. Scale 1 cm= 5.75 ft Distance from Starting Point= 51.6 ft Time= 2.5 sec 63.4 ft

9. Scale 1 cm= 5.75 ft Distance from Starting Point=70.1 ft Time= 3 sec 44.9 ft

10. Scale 1 cm= 5.75 ft Time= 3.5 sec Distance from Starting Point=91.4 ft 23.6 ft

11. Scale 1 cm= 5.75 ft Time= 4 sec Distance from Starting Point= 115 ft 0 ft

12. Data Points on the VideoData from Quadratic Regression Time (s)Height (ft) 0115 1100.6 284.8 2.558.9 Time (s)Height (ft) 0115 1102.5 279.2 2.563.4 344.9 3.523.6 40 General Equation d= ax 2 +bx + c d= -5.51x 2 -6.74x+115 ft

13. 4 115 Distance (ft) Time (seconds) d= -5.51x 2 -6.74x+115

14. The average velocity from 0 seconds to 4 seconds v= d(t f )-d(t i ) t f - t i v= d(4)- d(0) 4s- 0s v= 0m–115ft 4s v= -28.75ft/s

15. d= -5.51x 2 -6.74x+115 The derivative of the distance formula= velocity v= -11.02x-6.74 ft/s Time (s)Velocity (ft/s) 0-6.74 1-17.77 2-28.8 2.5-34.31 3-39.83 3.5-45.34 4-50.85 50.85 ft/s= 35mph

16. d v Distance (ft) Time (seconds) 115 4 d= -5.51x 2 -6.74x+115 v= -11.02x-6.74

17. Due to the acceleration of gravity, the diver’s velocity increases as he makes his descent. Therefore, the highest speed he travels is at the moment before he breaks the surface of the water. -50.85 ft1 m15.5m s3.28 fts The diver never reaches a speed of 50 m/s.

18. v= -11.02x-6.74 The derivative of the velocity formula= acceleration a= -11.02 ft/s 2 In reality, the acceleration of all objects in free fall have an acceleration of -9.8m/s 2 (-32.2 ft/s 2 ). However, because the video is in slow motion, the relationship between distance and time cannot be accurately found in a realistic manner. Also, the approximation error when finding the distance vs. time of the diver, contributes to the skewed acceleration.

19. Distance (ft) Time (seconds) 115 4 d v a d= -5.51x 2 -6.74x+115 v= -11.02x-6.74 a= -11.02

21. 1. Write the general equation for the instantaneous rate of change of f(x) with respect to x at x=c 2. Determine the point at which the rate is to be found (c) and a point very close to c (x). 3. Find the values for f(x) and f(c) and substitute into the equation. 4. Calculate the rate of change at the specific point. f’c= lim f(x)- f(c) x-c x c f’c= lim f(2.01)- f(2) (2.01-2) x 2 f’c= lim 78.892 ft-79.18 ft.01 s x 2 f’c= lim -28.8 ft/s x 2

22. Line of the tangent line at x=2 (y-y 1 )= m(x-x 1 ) Point-slope Form (y-79.18)= -28.8 (x-2) Use the point: (2,78.18) and the Instantaneous velocity: (-28.8) y= -28.8x+136.78

23. d y Time (seconds) Distance (ft) 2 78.18 y= -28.8x+136.78 d= -5.51x 2 -6.74x+115

25.  A function defined by different rules for different intervals of its domain.  Example: f(x) -1.5x 2 -4x+10 if x ≤ 2 2.5x-9 if 2≤x ≥6 2.5x -6 if x>6

26. -1.5x 2 -4x+10 if x ≤ 2 2.5x-9 if 2≤x ≤6 2.5x -6 if x>6 f(x) 6 2

27. 1. Value f(c) exists *Filled in circle *0 cannot be in the denominator 2. lim(fx) exists (general limit) * lim(fx)= lim(fx) 3. f(c)= lim (fx) x c x c - x c + x c

28. at x=6 1. f(x)=2.5x-9if 2≤x≤6 f(6)= 2.5(6)-9*use the equation that has the f(6)=6 x value in its domain 2. lim(fx)= lim(fx) 2.5x-9=2.5x -6*a general limit doesn’t exist 2.5(6)-9=2.5(6) -6 therefore #2 and #3 on the list 6= 2of requirements cannot be met x c - x c +

29. 6

30.  Both equations need to have the same y value when x=6 #2 y= 2.5x -6 if x>6 #1 y= 2.5x-9 if 2≤x ≥6 y= 2.5x -b 6= 2.5(6) - b 6=8-b b= 2 y=2.5x-9 y= 2.5(6)-9 y= 6 y= 2.5x -2 3. New Equation 1. Solve for the y value of #1 2. Use this y value and the x value to solve for b

31. 6

32. *Differentiable means continuous but continuous does not mean differentiable at x=2 Left Side Slope f(x)= -1.5x 2 -4x+10 lim f’(x)=-3x-4 lim f’(x)= -3(2)-4 lim f’(x)=-10 x 2 - Right Side Slope f(x)= 2.5x-9 lim f’(x)=2.5 x 2 + x 2 - The slopes must be the same on the left and the right in order for a point to be differentiable.

33. 2 f’(x)= -10 f’(x)= 2.5

35. End Behavior Model f(x)=x 3 +5x-3 8 f(x)= x 3 lim +8 +8 x lim f(x) = 3 lim f(x)= 8 +8 +8 x +8 +8 x lim f(x) = 3 lim f(x) = x 3 lim f(x) = - x - x - x - 8 8 8 8 8

36. f(x)=x 3 +5x-3

37. f(x)= x+2 x 2 -4 End Behavior Model lim f(x)= x = 1 x 2 x lim f( )= 1 lim f( )= 0 8 8 f(x)= x+2 (x+2)(x-2) f(x)= 1 (x-2) Vertical Asymptote f(x)= x+2 x+2 Removable Discontinuity Horizontal Asymptote +8 +8 x +8 +8 x +8 +8 x 8 y= 0 x=2 x= -2 (same as) 8x -

38. Horizontal Asymptote at y=0 Vertical Asymptote at x=2 Removable Discontinuity at x=-2 f(x)= x+2 x 2 -4

39. f(x)= x 3 +2x+3 4x 3 -4 ( ( lim f(x)= x 3 4x 3 lim f(x)= 1 4 +8 +8 x x x x x +8 +8 x Horizontal Asymptote y= 1 4 lim f(x)= 6 tiny + # lim f(x)= 1 - 1+1+ 8 lim f(x)= 6 tiny - # lim f(x)= - 8 Vertical Asymptote 1+1+ x=1 (same as) 8x - End Behavior Model

40. f(x)= x 3 +2x+3 4x 3 -4 ( ( 1 1414