course Mth 163
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11:54:45 query predictions Sketch your graph representing the predicted height of the low end vs. the weight on the spring.
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RESPONSE --> it should be a rising graph
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11:54:54 STUDENT RESPONSE: I predict that the spring will stretch at a greater rate as the weight is added. In my experience, springs lose their strength as they are stretched, and will not go back to their original shape. INSTRUCTOR COMMENT: ** Within their range of elasticity the graph is very nearly linear. If stretched too far the spring will lose its permanent elastic properties and will then deviate from linearity **
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RESPONSE --> ok
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11:55:13 comment on how the actual graph of the data compared with your prediction
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RESPONSE --> it was close but off by the width of the graph
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11:56:56 query linked outline discuss your experience with the Linked Outline. Did you find it helpful?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. I predicted a curvature and it was close to the actual graph...the Linked Outline was very helpful in predicting what a graph may look like without actually doing a graph
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11:57:02 ** Many students find the Linked Outline very helpful. **
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RESPONSE --> ok
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11:57:07 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> na
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course Mth 163
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11:54:45 query predictions Sketch your graph representing the predicted height of the low end vs. the weight on the spring.
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RESPONSE --> it should be a rising graph
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11:54:54 STUDENT RESPONSE: I predict that the spring will stretch at a greater rate as the weight is added. In my experience, springs lose their strength as they are stretched, and will not go back to their original shape. INSTRUCTOR COMMENT: ** Within their range of elasticity the graph is very nearly linear. If stretched too far the spring will lose its permanent elastic properties and will then deviate from linearity **
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RESPONSE --> ok
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11:55:13 comment on how the actual graph of the data compared with your prediction
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RESPONSE --> it was close but off by the width of the graph
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11:56:56 query linked outline discuss your experience with the Linked Outline. Did you find it helpful?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. I predicted a curvature and it was close to the actual graph...the Linked Outline was very helpful in predicting what a graph may look like without actually doing a graph
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11:57:02 ** Many students find the Linked Outline very helpful. **
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RESPONSE --> ok
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11:57:07 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> na
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X}zyo assignment #005 FT̩̎V Precalculus I 02-10-2006 ϟ榺j assignment #006 FT̩̎V Precalculus I 02-10-2006
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12:05:02 Query 4 basic function families What are the four basic functions? What are the generalized forms of the four basic functions?
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RESPONSE --> y= f(x) =x, y = x^2, y = f(x) = 2^x, y = f(x) = x^p y= mx+ b, y = f(x) = ax^2 + bx + c, y = f(x) = A b^x + c, f(x) = A (x-h)^p + c
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12:05:21 ** STUDENT RESPONSE: Linear is y=mx+b Quadratic is y=ax^2 + bx +c Exponential is y= A*2^ (kx)+c Power = A (x-h)^p+c INSTRUCTOR COMMENTS: These are the generalized forms. The basic functions are y = x, y = x^2, y = 2^x and y = x^p. **
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RESPONSE --> ok
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12:05:50 For a function f(x), what is the significance of the function A f(x-h) + k and how does its graph compare to that of f(x)?
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RESPONSE --> it raises the graph up or down, and it is either wider or narrower
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12:05:56 ** STUDENT RESPONSE: A designates the x strectch factor while h affects a y shift & k affects an x shift INSTRUCTOR COMMENTS: k is the y shift for a simple enough reason -- if you add k to the y value it raises the graph by k units. h is the x shift. The reason isn't quite as simple, but not that hard to understand. It's because when x is replaced by x - h the y values on the table shift 'forward' by h units. A is a multiplier. When all y values are multiplied by A that moves them all A times as far from the x axis, which is what causes the stretch. Thus A f(x-h) + k is obtained from f(x) by vertical stretch A, horizontal shift h and vertical shift k. The two aren't the same, but of course they're closely related. **
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RESPONSE --> ok
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12:09:29 query introduction to rates and slopes, problem 1 ave vel for function depth(t) = .02t^2 - 5t + 150 give the average rate of depth change from t = 20 to t = 40
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RESPONSE --> t= 20 .02(20)^2 + 5(20) + 150 = 258 t= 40 .02(40)^2 + 5(40) +150 = 382 382- 258 = 124
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12:09:51 ** depth(20) = .02(20^2) - 5(20) + 150 = 58 depth(40) = .02(40^2) - 5(40) + 150 = -18 change in depth = depth(40) - depth(20) = -18 - 58 = -76 change in clock time = 40 - 20 = 20. Ave rate of depth change = change in depth / change in clock time = -76 / 20 = -3.8 **
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RESPONSE --> ok I messed up on the signs in the equation
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12:10:13 ** depth(60) = .02(60^2) - 5(60) + 150 = -78 depth(80) = .02(80^2) - 5(80) + 150 = -122 change in depth = depth(80) - depth(60) = -122 - (-78) = -44 change in clock time = 40 - 20 = 20. Ave rate of depth change = change in depth / change in clock time = -44 / 20 = -2.2 **
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems ok.
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12:10:41 describe your graph of y = .02t^2 - 5t + 150
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RESPONSE --> it will continue to raise as the depth increases
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12:10:46 ** The graph is a parabola. y = .02t^2 - 5t + 150 has vertex at x = -b / (2a) = -(-5) / (2 * .02) = 125, at which point y = .02 (125^2) - 5(125) + 150 = -162.5. The graph opens upward, intercepting the x axis at about t = 35 and t = 215. Up to t = 125 the graph is decreasing at a decreasing rate. That is, it's decreasing but the slopes, which are negative, are increasing toward 0.**
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RESPONSE --> ok
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12:11:00 describe the pattern to the depth change rates
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RESPONSE --> they continue to go up
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12:11:06 ** Rates are -3.8, -3 and -2.2 for the three intervals (20,40), (40,60) and (60,80). For each interval of `dt = 20 the rate changes by +.8. **
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RESPONSE --> ok
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12:11:18 query problem 2. ave rates at midpoint times what is the average rate of depth change for the 1-second time interval centered at the 50 sec midpoint?
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RESPONSE --> I dont know
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12:11:25 ** The 1-sec interval centered at t = 50 is 49.5 < t < 50.5. For depth(t) = .02t^2 - 5t + 150 we have ave rate = [depth(50.5) - depth(49.5)]/(50.5 - 49.5) = -3 / 1 = -3. **
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RESPONSE --> ok
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12:11:53 what is the average rate of change for the six-second time interval centered at the midpoint.
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RESPONSE --> about the 10 seconds
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12:11:58 ** The 6-sec interval centered at t = 50 is 47 < t < 53. For depth(t) = .02t^2 - 5t + 150 we have ave rate = [depth(53) - depth(47)]/(53 - 47) = -18 / 6 = -3. **
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RESPONSE --> ok
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12:12:14 What did you observe about your two results?
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RESPONSE --> that the depth change was very close
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12:12:19 ** The two rates match each other, and they also match the average rate for the interval 40 < t < 60, which is also centered at t = 50. For a quadratic function, and only for a quadratic function, the rate is the same for all intervals having the same midpoint. This is a property unique to quadratic functions. **
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RESPONSE --> ok
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12:13:18 query problem 3. ave rates at midpt times for temperature function Temperature(t) = 75(2^(-.05t)) + 25. What is the average rate of depth change for the 1-second time interval centered at the 50 sec midpoint?
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RESPONSE --> I dont know
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12:13:27 STUDENT RESPONSE: .46 degrees/minute INSTRUCTOR COMMENT: More precisely .4595 deg/min, and this does not agree exactly with the result for the 6-second interval. Remember that you need to look at enough significant figures to see if there is a difference between apparently identical results.**
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RESPONSE --> ok
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12:13:37 what is the average rate of change for the six-second time interval centered at the midpoint.
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RESPONSE --> they are the same
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12:13:44 STUDENT RESPONSE: .46 degrees/minute INSTRUCTOR COMMENT: The 1- and 6-second results might possibly be the same to two significant figures, but they aren't the same. Be sure to recalculate these according to my notes above and verify this for yourself. The average rate for the 6-second interval is .4603 deg/min. It differs from the average rate .4595 deg/min, calculated over the 1-second interval, by almost .001 deg/min. This differs from the behavior of the quadratic. For a quadratic that the results for all intervals centered at the same point will all agree. This is not the case for the present function, which is exponential. Exact agreement is a characteristic of quadratic functions, and of no other type. **
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RESPONSE --> ok
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