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# AA Notes 2-7 and 2-8

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Fitting a Model to Data

Fitting a Model to Data

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• 1. SECTIONS 2-7 AND 2-8 FITTING A MODEL TO DATA
• 2. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k b. y = x c. y = kx 2 k d. y = 2 x
• 3. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k b. y = 20 = 10k x c. y = kx 2 k d. y = 2 x
• 4. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k b. y = 20 = 10k x k=2 c. y = kx 2 k d. y = 2 x
• 5. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k b. y = 20 = 10k x k=2 y = 2x c. y = kx 2 k d. y = 2 x
• 6. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k b. y = 20 = 10k y = 2(30) x k=2 y = 2x c. y = kx 2 k d. y = 2 x
• 7. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k b. y = 20 = 10k y = 2(30) x k=2 y = 60 y = 2x c. y = kx 2 k d. y = 2 x
• 8. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k b. y = 20 = 10k y = 2(30) x k k=2 y = 60 20 = 10 y = 2x c. y = kx 2 k d. y = 2 x
• 9. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k b. y = 20 = 10k y = 2(30) x k k=2 y = 60 20 = 10 y = 2x k = 200 c. y = kx 2 k d. y = 2 x
• 10. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k 200 b. y = y= 20 = 10k y = 2(30) x x k k=2 y = 60 20 = 10 y = 2x k = 200 c. y = kx 2 k d. y = 2 x
• 11. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k 200 b. y = y= 200 20 = 10k y = 2(30) x x y= k 30 k=2 y = 60 20 = 10 y = 2x k = 200 c. y = kx 2 k d. y = 2 x
• 12. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k 200 b. y = y= 200 20 = 10k y = 2(30) x x y= k 30 k=2 y = 60 20 = 20 10 y= y = 2x k = 200 3 c. y = kx 2 k d. y = 2 x
• 13. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k 200 b. y = y= 200 20 = 10k y = 2(30) x x y= k 30 k=2 y = 60 20 = 20 10 y= y = 2x k = 200 3 c. y = kx 2 k d. y = 2 20 = k(10) 2 x
• 14. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k 200 b. y = y= 200 20 = 10k y = 2(30) x x y= k 30 k=2 y = 60 20 = 20 10 y= y = 2x k = 200 3 c. y = kx 2 k d. y = 2 20 = k(10) 2 x k= 1 5
• 15. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k 200 b. y = y= 200 20 = 10k y = 2(30) x x y= k 30 k=2 y = 60 20 = 20 10 y= y = 2x k = 200 3 c. y = kx 2 k d. y = 2 20 = k(10) 2 x k= 1 5 y= x1 5 2
• 16. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k 200 b. y = y= 200 20 = 10k y = 2(30) x x y= k 30 k=2 y = 60 20 = 20 10 y= y = 2x k = 200 3 c. y = kx 2 k d. y = 2 20 = k(10) 2 y = (30) 1 2 x 5 k= 1 5 y= x1 5 2
• 17. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k 200 b. y = y= 200 20 = 10k y = 2(30) x x y= k 30 k=2 y = 60 20 = 20 10 y= y = 2x k = 200 3 c. y = kx 2 k d. y = 2 20 = k(10) 2 y = (30) 1 2 x 5 k= 1 5 y = 180 y= x1 5 2
• 18. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k 200 b. y = y= 200 20 = 10k y = 2(30) x x y= k 30 k=2 y = 60 20 = 20 10 y= y = 2x k = 200 3 c. y = kx 2 k d. y = 2 20 = k(10) 2 y = 5 (30) 1 2 k x 20 = 2 k= 51 y = 180 10 y= x1 5 2
• 19. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k 200 b. y = y= 200 20 = 10k y = 2(30) x x y= k 30 k=2 y = 60 20 = 20 10 y= y = 2x k = 200 3 c. y = kx 2 k d. y = 2 20 = k(10) 2 y = 5 (30) 1 2 k x 20 = 2 k= 51 y = 180 10 y= x k = 2000 1 2 5
• 20. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k 200 b. y = y= 200 20 = 10k y = 2(30) x x y= k 30 k=2 y = 60 20 = 20 10 y= y = 2x k = 200 3 c. y = kx 2 k 2000 d. y = 2 y = 2 20 = k(10) 2 y = 5 (30) 1 2 k x x 20 = 2 k= 51 y = 180 10 y= x k = 2000 1 2 5
• 21. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k 200 b. y = y= 200 20 = 10k y = 2(30) x x y= k 30 k=2 y = 60 20 = 20 10 y= y = 2x k = 200 3 c. y = kx 2 k 2000 d. y = 2 y = 2 20 = k(10) 2 y = 5 (30) 1 2 k x x 20 = 2 y = 2000 k= 51 y = 180 10 2 30 y= x k = 2000 1 2 5
• 22. WARM-UP The point with coordinates (10, 20) appears on the graph for each variation function below. Determine the value of k in each case, and then determine the value of y when x is 30. a. y = kx k 200 b. y = y= 200 20 = 10k y = 2(30) x x y= k 30 k=2 y = 60 20 = 20 10 y= y = 2x k = 200 3 c. y = kx 2 k 2000 d. y = 2 y = 2 20 = k(10) 2 y = 5 (30) 1 2 k x x 20 = 2 y = 2000 k= 51 y = 180 10 2 30 20 y= y= 5x1 2 k = 2000 9
• 23. MATHEMATICAL MODEL
• 24. MATHEMATICAL MODEL A mathematical representation of a real-world situation
• 25. MATHEMATICAL MODEL A mathematical representation of a real-world situation A good model ﬁts all possibilities as best it can; often is just an approximation
• 26. EXAMPLE 1 Matt Mitarnowski and Fuzzy Jeff measured the intensity of light at various distances from a lamp and got the data where d = distance in meters and I = intensity in watts/m2. d I 2 560 2.5 360 3 250 3.5 185 4 140
• 27. EXAMPLE 1 Matt Mitarnowski and Fuzzy Jeff measured the intensity of light at various distances from a lamp and got the data where d = distance in meters and I = intensity in watts/m2. d I 600 2 560 450 2.5 360 3 250 300 3.5 185 150 4 140 0 0 1.25 2.50 3.75 5.00
• 28. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations.
• 29. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k I= d
• 30. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k k I= 560 = d 2
• 31. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k k I= 560 = k = 1120 d 2
• 32. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k k 1120 I= 560 = k = 1120 I= d 2 d
• 33. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k k 1120 1120 I= 560 = k = 1120 I= I= d 2 d 4
• 34. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k k 1120 1120 I= 560 = k = 1120 I= I= d 2 d 4 I = 280
• 35. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k k 1120 1120 I= 560 = k = 1120 I= I= d 2 d 4 I = 280 Not close
• 36. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k k 1120 1120 I= 560 = k = 1120 I= I= d 2 d 4 I = 280 Not close k I= 2 d
• 37. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k k 1120 1120 I= 560 = k = 1120 I= I= d 2 d 4 I = 280 Not close k k I= 2 560 = 2 d 2
• 38. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k k 1120 1120 I= 560 = k = 1120 I= I= d 2 d 4 I = 280 Not close k k I= 2 560 = 2 k = 2240 d 2
• 39. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k k 1120 1120 I= 560 = k = 1120 I= I= d 2 d 4 I = 280 Not close k k 2240 I= 2 560 = 2 k = 2240 I= 2 d 2 d
• 40. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k k 1120 1120 I= 560 = k = 1120 I= I= d 2 d 4 I = 280 Not close k k 2240 2240 I= 2 560 = 2 k = 2240 I= 2 I= d 2 d 16
• 41. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k k 1120 1120 I= 560 = k = 1120 I= I= d 2 d 4 I = 280 Not close k k 2240 2240 I= 2 560 = 2 k = 2240 I= 2 I= d 2 d 16 I = 140
• 42. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k k 1120 1120 I= 560 = k = 1120 I= I= d 2 d 4 I = 280 Not close k k 2240 2240 I= 2 560 = 2 k = 2240 I= 2 I= d 2 d 16 I = 140 Right on!
• 43. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k k 1120 1120 I= 560 = k = 1120 I= I= d 2 d 4 I = 280 Not close k k 2240 2240 I= 2 560 = 2 k = 2240 I= 2 I= d 2 d 16 I = 140 Right on! 2240 I= 2 d
• 44. EXAMPLE 1 This looks like either a hyperbola or an inverse-square curve. Let’s test them out by checking some points in our equations. k k 1120 1120 I= 560 = k = 1120 I= I= d 2 d 4 I = 280 Not close k k 2240 2240 I= 2 560 = 2 k = 2240 I= 2 I= d 2 d 16 I = 140 Right on! 2240 I= 2 Make sure to check that all points are close d
• 45. STEPS TO FINDING A MODEL FROM DATA
• 46. STEPS TO FINDING A MODEL FROM DATA 1. Graph the data
• 47. STEPS TO FINDING A MODEL FROM DATA 1. Graph the data 2. Describe the graph
• 48. STEPS TO FINDING A MODEL FROM DATA 1. Graph the data 2. Describe the graph 3. State the equation
• 49. STEPS TO FINDING A MODEL FROM DATA 1. Graph the data 2. Describe the graph 3. State the equation 4. Find k, plug it in
• 50. STEPS TO FINDING A MODEL FROM DATA 1. Graph the data 2. Describe the graph 3. State the equation 4. Find k, plug it in 5. Check other points
• 51. STEPS TO FINDING A MODEL FROM DATA 1. Graph the data 2. Describe the graph 3. State the equation 4. Find k, plug it in 5. Check other points 6. State the model
• 52. EXAMPLE 2 Matt Mitarnowski measured how far a marble rolled down a ramp over a period of time. He obtained the following data where t is time in seconds and D is distance in inches. t D 1 0.6 2 2.4 3 5.2 4 9.8 5 15.2
• 53. EXAMPLE 2 Matt Mitarnowski measured how far a marble rolled down a ramp over a period of time. He obtained the following data where t is time in seconds and D is distance in inches. 20 t D 15 1 0.6 2 2.4 10 3 5.2 5 4 9.8 5 15.2 0 0 1.25 2.50 3.75 5.00
• 54. EXAMPLE 2 Matt Mitarnowski measured how far a marble rolled down a ramp over a period of time. He obtained the following data where t is time in seconds and D is distance in inches. 20 t D 15 1 0.6 2 2.4 10 3 5.2 5 4 9.8 5 15.2 0 0 1.25 2.50 3.75 5.00 Should be a parabola
• 55. D = kt 2
• 56. D = kt 2 .6 = k(1) 2
• 57. D = kt 2 .6 = k(1) 2 k = .6
• 58. D = kt 2 .6 = k(1) 2 k = .6 D = .6t 2
• 59. D = kt 2 .6 = k(1) 2 k = .6 D = .6t 2 D = .6(2) 2
• 60. D = kt 2 .6 = k(1) 2 k = .6 D = .6t 2 D = .6(2) 2 D = 2.4
• 61. D = kt 2 .6 = k(1) 2 k = .6 D = .6t 2 D = .6(2) 2 D = 2.4 D = .6t 2
• 62. D = kt 2 .6 = k(1) 2 k = .6 D = .6t 2 D = .6(2) 2 D = 2.4 D = .6t 2 Check the rest of the points to make sure satisfy the model
• 63. CONVERSE OF THE FUNDAMENTAL THEOREM OF VARIATION
• 64. CONVERSE OF THE FUNDAMENTAL THEOREM OF VARIATION a. If multiplying every x-value of a function by c results in multiplying the corresponding y-value by cn, then y varies directly as the nth power of x, or y = kxn.
• 65. CONVERSE OF THE FUNDAMENTAL THEOREM OF VARIATION a. If multiplying every x-value of a function by c results in multiplying the corresponding y-value by cn, then y varies directly as the nth power of x, or y = kxn. b. If multiplying every x-value of a function by c results in dividing the corresponding y-value by cn, then y varies inversely as the n th power of x, or k . y= n x
• 66. HOMEWORK
• 67. HOMEWORK Make sure to read Sections 2-7 and 2-8 p. 113 #1-11, p. 119 #1-11