course Mth 163
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18:24:32 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 --> The four basic functions are : y = x y = x^2 y = 2^x y = x^
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18:24:37 ** 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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18:26:41 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 --> the x shift is h the multiplier is A K+y value raise the graph by k units.
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18:27:11 ** 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 --> I should of went into a little more detail. However I do understand the question
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18:27:55 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 --> change in clock time = 40 - 20 = 20. change in depth = depth(40) - depth(20) = -18 - 58 = -76
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18:28:27 ** 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 --> I didn't understand the first part of the problem therefore I didn't put the actual velocity.
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18:30:06 What is the average rate of depth change from t = 60 to t = 80?
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RESPONSE --> change in clock time = 40-20=20. change in depth = depth(80)-depth(60)=-122-(-78) =-44 depth(60) = .02(60^2) - 5(60)+150 = -78 depth(80) = .02(80^2) - 5(80)+150 = -122
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18:30:34 ** 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 --> I forgot to put the actual change.
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18:32:54 describe your graph of y = .02t^2 - 5t + 150
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RESPONSE --> first of all, it is a parabola that opens up. Its vertex is at x = -b / (2a) = -(-5) / (2 * .02) = 125 it intersects the x axis at t = 35 and t = 215 the graph is decreasing at a decreasing rate
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18:32:57 02-13-2006 18:32:57 ** 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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NOTES ------->
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18:33:02 ** 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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18:34:10 describe the pattern to the depth change rates
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RESPONSE --> the rate changes by positive .8 rates are 3.8, -3 and -2.2
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18:34:21 ** 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 --> I should have used more specifics.
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18:35:01 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 --> [depth(50.5) - depth(49.5)]/(50.5 - 49.5) = -3 / 1 = -3
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18:35:05 ** 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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18:35:32 what is the average rate of change for the six-second time interval centered at the midpoint.
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RESPONSE --> [depth(53) - depth(47)]/(53 - 47) = -18 / 6 = -3.
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18:35:36 ** 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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18:36:17 What did you observe about your two results?
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RESPONSE --> they match eachother and match the average rate for the interval 40 < t < 60
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18:36:55 ** 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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18:37:23 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 --> .4595 degrees/minute
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18:37:30 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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18:38:12 what is the average rate of change for the six-second time interval centered at the midpoint.
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RESPONSE --> .4603 degrees/minute
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18:38:17 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"