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# Chapter 2 Solution

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### Chapter 2 Solution

1. 1. Step 3: SOLVING THE PROBLEMS
2. 2. The first thing we must do to solve this problem is find how long Wilfred and John will be on the ride before Francis and Kieran get off.
3. 3. The first thing we must do to solve this problem is find how long Wilfred and John will be on the ride before Francis and Kieran get off. We can do this by subtracting the time of Wilfred and John’s boarding of the ride from the time when Francis and Kieran get off.
4. 4. The first thing we must do to solve this problem is find how long Wilfred and John will be on the ride before Francis and Kieran get off. We can do this by subtracting the time of Wilfred and John’s boarding of the ride from the time when Francis and Kieran get off. 35 Time that Francis and Kieran will get off
5. 5. The first thing we must do to solve this problem is find how long Wilfred and John will be on the ride before Francis and Kieran get off. We can do this by subtracting the time of Wilfred and John’s boarding of the ride from the time when Francis and Kieran get off. 35 - 21 Time that Francis and How long ‘F’ and ‘K’ Kieran will get off have been on the ride when ‘W’ and ‘J’ get on
6. 6. The first thing we must do to solve this problem is find how long Wilfred and John will be on the ride before Francis and Kieran get off. We can do this by subtracting the time of Wilfred and John’s boarding of the ride from the time when Francis and Kieran get off. 35 - 21 = 14 Time that Francis and How long ‘F’ and ‘K’ Kieran will get off have been on the ride when ‘W’ and ‘J’ get on
7. 7. This tells us that Francis and Kieran will be getting off the ferris wheel in 14 minutes.
8. 8. This tells us that Francis and Kieran will be getting off the ferris wheel in 14 minutes. Therefore we can conclude that Wilfred and John will have been on the ferris wheel for 14 minutes.
9. 9. Now that we know how long they have been on the ride, we must find what height Wilfred and John are at when Francis and Kieran get off.
10. 10. Now that we know how long they have been on the ride, we must find what height Wilfred and John are at when Francis and Kieran get off. We can do this by solving one of the equations that we derived a few minutes ago with t=14 substituted into it.
11. 11. Now that we know how long they have been on the ride, we must find what height Wilfred and John are at when Francis and Kieran get off. We can do this by solving one of the equations that we derived a few minutes ago with t=14 substituted into it. ***I will be using both the sine and cosine equation for this part.***
12. 12. Sine Equation
13. 13. Sine Equation The Original Equation: H(t)=75sin((2π/35)(t-10.75))+76
14. 14. Sine Equation Substitute t=14 into the equation. H(14)=75sin((2π/35)(14-10.75))+76
15. 15. Sine Equation Simplify the brackets. H(14)=75sin((2π/35)3.25)+76
16. 16. Sine Equation Simplify the brackets. H(14)=75sin(6.5π/35)+76
17. 17. Sine Equation Calculate the sine of the value inside the brackets. [Accurate to 4 decimal places.] H(14)=75(0.5509)+76
18. 18. Sine Equation Simplify the equation. H(14)=41.3173+76
19. 19. Sine Equation Simplify the equation. H(14)=117.3173
20. 20. Sine Equation H=117.3173m
21. 21. Cosine Equation
22. 22. Cosine Equation The Original Equation: H(t)=-75cos((2π/35)(t-2.0))+76
23. 23. Cosine Equation Substitute t=14 into the equation. H(14)=-75cos((2π/35)(14-2.0))+76
24. 24. Cosine Equation Simplify the brackets. H(14)=-75cos((2π/35)12.0)+76
25. 25. Cosine Equation Simplify the brackets. H(14)=-75cos(24.0π/35)+76
26. 26. Cosine Equation Calculate the cosine of the value inside the brackets. [Accurate to 4 decimal places.] H(14)=-75(-0.5509)+76
27. 27. Cosine Equation Simplify the equation. H(14)=41.3173+76
28. 28. Cosine Equation Simplify the equation. H(14)=117.3173
29. 29. Cosine Equation H=117.3173m
30. 30. This means that Wilfred and John were approximately 117 m off the ground when Francis and Kieran got off.
31. 31. We have just answered the first part of the question.
32. 32. We have just answered the first part of the question. Now we must use this to answer the second part of the question:
33. 33. We have just answered the first part of the question. Now we must use this to answer the second part of the question: How much longer will it be until they reach the same height again?
34. 34. Some people might be confused by this question, so lets take a look at it.
35. 35. Some people might be confused by this question, so lets take a look at it. This is the graph that we constructed earlier.
36. 36. Some people might be confused by this question, so lets take a look at it. The point on the graph is the point that we just found. (t=14)
37. 37. Some people might be confused by this question, so lets take a look at it. The height of this point is 117m.
38. 38. Some people might be confused by this question, so lets take a look at it. However, there is another point with a height of 117m.
39. 39. Some people might be confused by this question, so lets take a look at it. Therefore there is another time that they will reach a height of 117m.
40. 40. So, we are looking to find that second time which has a height of 117.3173m. We can do this by making the equation equal to 117.3173m and solving it.
41. 41. So, we are looking to find that second time which has a height of 117.3173m. We can do this by making the equation equal to 117.3173m and solving it. ***This time I will only use the Sine equation and explain the difference between the two methods afterwards.***
42. 42. Sine Equation
43. 43. Sine Equation The Original Equation: H(t)=75sin((2π/35)(t-10.75))+76
44. 44. Sine Equation Substitute H(t)=117.3173 into the equation. 117.3173=75sin((2π/35)(t-10.75))+76
45. 45. Sine Equation Let ((2π/35)(t-10.7))=Ø to make it a little easier to work with. 117.3173=75sinØ+76
46. 46. Sine Equation Simplify the equation. 41.3173=75sinØ
47. 47. Sine Equation Simplify the equation. 0.5509=sinØ
48. 48. Sine Equation Calculate the arc sine of 0.5509 [to four decimal places]. 0.5834=Ø
49. 49. Sine Equation Now we have an angle. The measure of this angle is 0.5834 radians Another thing we know about this angle is that it’s sine is positive. WHY IS THIS HELPFUL???
50. 50. Sine Equation Well, when the sine of an angle is positive, it must either be in Quadrant 1 or 2. This particular angle is in Quadrant 1. HOW DO WE KNOW THIS???
51. 51. Sine Equation Here we have the Unit Circle. QII QI !/2 0 ! 2! 3!/2 QIII QIV
52. 52. Sine Equation The angle that separates Quadrants I and II is π/2. [or 1.5707...] QII QI !/2 0 ! 2! 3!/2 QIII QIV
53. 53. Sine Equation 0.5834 is much less than 1.5708 so it lies in Quadrant I. QII QI !/2 0 ! 2! 3!/2 QIII QIV
54. 54. Sine Equation This means that to find the other point with the exact same sine value, it’s angle would have to be in Quadrant II.
55. 55. Sine Equation This means that to find the other point with the exact same sine value, it’s angle would have to be in Quadrant II. To find this angle we simply subtract our angle from π (End of Quadrant II).
56. 56. Sine Equation This means that to find the other point with the exact same sine value, it’s angle would have to be in Quadrant II. To find this angle we simply subtract our angle from π (End of Quadrant II). π - Ø = Ø’
57. 57. Sine Equation Here we have our Quadrant II angle. 2.5582=Ø’
58. 58. Sine Equation Now the two angles we have are related angles. This means their sine values are equivalent. sinØ=sinØ’
59. 59. Sine Equation And if we take the arc sine of both sides and we see that: Ø=Ø’
60. 60. Sine Equation Now we can continue where we left off. Since... Ø=2.5582
61. 61. Sine Equation ...AND... Ø= ((2π/35)(t-10.7))
62. 62. Sine Equation ...Then. 2.5582= ((2π/35)(t-10.7))
63. 63. Sine Equation Now just simplify. 14.25= (t-10.7)
64. 64. Sine Equation Now just simplify. 25= t
65. 65. Sine Equation Cosine Equation t=25
66. 66. Cosine Equation The Cosine equation works quite similarly to that of Sine.
67. 67. Cosine Equation The Cosine equation works quite similarly to that of Sine. The equation is a little bit different as we saw earlier but that’s about all that applies to the first part of the procedure.
68. 68. Cosine Equation One of the major differences is when you simplify down to cosØ.
69. 69. Cosine Equation One of the major differences is when you simplify down to cosØ. First of all, this value is negative which means the angle is either in Quadrant II or III.
70. 70. Cosine Equation One of the major differences is when you simplify down to cosØ. First of all, this value is negative which means the angle is either in Quadrant II or III. This one is in Quadrant II because it is 2.1542 which is less than π (3.1415), the end of Quadrant II.
71. 71. Cosine Equation This means Ø’ is in Quadrant III and we must subtract Ø (angle that passes Q I and into II) from the full circle (2π) so it passes through Q IV and into III.
72. 72. Cosine Equation This means Ø’ is in Quadrant III and we must subtract Ø (angle that passes Q I and into II) from the full circle (2π) so it passes through Q IV and into III. After that, continue as shown in the Sine procedure until you get t=25 as your answer.
73. 73. From here, we now can finish the problem off. t=25 minutes
74. 74. Since it takes 14 minutes to get from the start to 117.3173 m the first time. 14
75. 75. Since it takes 14 minutes to get from the start to 117.3173 m the first time... ...And it takes 25 minutes to get from the start to 117.3173 m the second time... 14 - 25
76. 76. ...Then the time it takes to get from 117.3173 the first time to 117.3173 m the second time must equal... 14 - 25 =?
77. 77. 11 minutes
78. 78. When Francis and Kieran get off “The Wheel”, Wilfred and John will be 117.3173 m from the ground. It will take them 11 more minutes to get to that exact same height again.