Your SlideShare is downloading. ×
  • Like
AA Notes 2-7 and 2-8
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×

Now you can save presentations on your phone or tablet

Available for both IPhone and Android

Text the download link to your phone

Standard text messaging rates apply

AA Notes 2-7 and 2-8

  • 424 views
Published

Fitting a Model to Data

Fitting a Model to Data

Published in Education , Technology
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Be the first to comment
    Be the first to like this
No Downloads

Views

Total Views
424
On SlideShare
0
From Embeds
0
Number of Embeds
1

Actions

Shares
Downloads
6
Comments
0
Likes
0

Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
    No notes for slide

Transcript

  • 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