course mth 163 f¿îBknD~exassignment #006
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19:44:41 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 families are: Linear y = mx + b Quadratic y = ax^2 + bx + c Exponential y = A * 2^(kx) + c Power y = A (x-h)^p +c confidence assessment: 3
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19:44:49 ** 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 --> self critique assessment: 3
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19:47:14 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 --> k will tell the vertical shift h tells us how far to the left or right the graph will go and A tells us how much the graph is stretched. confidence assessment: 2
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19:47:28 ** 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 --> self critique assessment: 3
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20:01:55 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 --> After substituting into the equation to find the y coordinates: y= .02(20)^2 - 5(20) + 150 y= .02(40)^2 - 5(40) + 150 y= 8 - 100 +150 y= 32 - 200 +150 y= 58 y= -18 (20, 58) (40, -18) You can use the slope formula: (y2-y1)/(x2-x1) slope = (-18 - 58) / (40 - 20) slope = (-76) / (20) slope = -19/5 or -3.8 confidence assessment: 2
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20:02:03 ** 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 --> self critique assessment: 3
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20:12:20 02-05-2007 20:12:20 What is the average rate of depth change from t = 60 to t = 80?
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NOTES -------> y= .02(60)^2 - 5(60) + 150 y= .02(80)^2 - 5(80) + 150 y= 72 - 300 + 150 y= 128 - 400 +150 y= -78 y= -122 (60, -78) (80, -122) Using the slope formula: (-122 - (-78)) / (80-60) (-122 + 78) / 20 (-44) / 20 -2.2
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20:12:22 What is the average rate of depth change from t = 60 to t = 80?
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RESPONSE --> confidence assessment: 2
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20:12:28 ** 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 --> self critique assessment: 3
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20:24:02 describe your graph of y = .02t^2 - 5t + 150
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RESPONSE --> The graph begins at (150, 0) and decreases at a decreasing rate ending at (-150, 100) confidence assessment: 2
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20:27:37 ** 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 --> The graph is a parabola, but since this is about depth and time, I didn't not see the need to make it a parabola. self critique assessment: 2
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20:32:22 describe the pattern to the depth change rates
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RESPONSE --> Aside from the initial change everything decreases at a rate of .8 confidence assessment: 2
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20:32:36 ** 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 --> self critique assessment: 3
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20:36:07 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 --> If 50 sec is exactly half way between 40 & 60 so we can take the average rate and divide it by two.: the rate change is -3 so divided by 2 is -1.5 confidence assessment: 2
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20:41:02 ** 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, so the rate would be the same? Even if it is a midpoint. I did not realize this. self critique assessment: 2
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20:45:03 what is the average rate of change for the six-second time interval centered at the midpoint.
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RESPONSE --> To be honest, I am ot sure what data I am supposed to use...I am sure once I read the explaination, I will understand. confidence assessment: 1
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20:46:09 ** 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 --> I do not understand where this data come from. self critique assessment: 2
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20:47:08 What did you observe about your two results?
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RESPONSE --> Based on the data, the average rate is the same. confidence assessment: 1
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20:47:19 ** 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 --> self critique assessment: 3
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21:08:01 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 am still not fully understanding the midpoint functions. I will look back over the notes to see if I can ""get it"" confidence assessment: 0
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21:08:10 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 --> self critique assessment: 3
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21:09:07 what is the average rate of change for the six-second time interval centered at the midpoint.
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RESPONSE --> This fits in the same catergory. I will have to look over the notes again. I did not understand where the data came from in the last group like this. confidence assessment: 0
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21:10:16 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 --> I should have remembered that it would be the same as the previous question. self critique assessment: 1
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