course mth173 ??z???????????assignment #009
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10:15:56 `qNote that there are 9 questions in this assignment. `q001. The process we used in the preceding series of exercises to approximate the graph of y corresponding to the graph of y ' can be significantly improved. Recall that we used the initial slope to calculate the change in the y graph over each interval. For example on the first interval we used the initial slope -6 for the entire interval, even though we already knew that the slope would be -5 by the time we reached the t = 10 point at the end of that interval. We would have been more accurate in our approximation if we had used the average -5.5 of these two slopes. Using the average of the two slopes, what point would we end up at when t = 10?
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RESPONSE --> starting at point (0,100) at a 'dt=10 at a slope of -5.5 you would end at point (10, 45) confidence assessment: 2
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10:16:08 If the average value slope = -5.5 is used between t = 0 and t = 10, we find that the rise is -55 and we go from (0,100) to (10,45).
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RESPONSE --> okay self critique assessment: 3
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10:18:40 `q002. We find that using the average of the two slopes we reach the point (10, 45). At this point we have y ' = -5. By the time we reach the t = 20 point, what will be the slope? What average slope should we therefore use on the interval from t = 10 to t = 20? Using this average slope what point will we end up with when t = 20?
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RESPONSE --> at t=20 the slope is -4 so the average slope between t=10 and t=20 would be -4.5 over a intrival of 10 sec. the new point would be (20, 0) confidence assessment: 3
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10:19:00 The slope at t = 20 is y ' = -4. Averaging this with the slope y ' = -5 at t = 10 we estimate an average slope of (-5 + -4) / 2 = -4.5 between t = 10 and t = 20. This would imply a rise of -45, taking the graph from (10,45) to (20, 0).
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RESPONSE --> okay self critique assessment: 3
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10:21:09 `q003. We found that we reach the point (20, 0), where the slope is -4. Following the process of the preceding two steps, what point will we reach at t = 30?
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RESPONSE --> since the slope at t=30 is -3 and the slope of t=20 is -4 the average slope between the two is -3.5 over a time intrival of ten seconds, we get the point (30, -35) confidence assessment: 3
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10:21:19 The slope at t = 20 is y ' = -3. Averaging this with the slope y ' = -4 at t = 20 we estimate an average slope of (-4 + -3) / 2 = -4.5 between t =20 and t = 30. This would imply a rise of -35, taking the graph from (20,0) to (30, -35).
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RESPONSE --> okay self critique assessment: 3
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10:28:44 `q004. Continue this process up through t = 70. How does this graph differ from the preceding graph, which was made without averaging slopes?
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RESPONSE --> This would graph a curve with a more shallow curve than the other graph, with the points {(0,100), (10, 45), (20, 0), (30, -35), (40, -60), (50, -75), (60, -80), (70, -75) confidence assessment: 3
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10:31:10 The average slopes over the next four intervals would be -2.5, -1.5, -.5 and +.5. These slopes would imply rises of -25, -15, -5 and +5, taking the graph from (30,-35) to (40, -60) then to (50, -75) then to (60, -80) and finally to (70, -75). Note that the graph decreases at a decreasing rate up to the point (60, -80) before turning around and increasing to the ponit (70, -75).
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RESPONSE --> the point at (60, -80) is the vertex of the parabola , also my average slopes match up with the ones in the answer window self critique assessment: 2
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10:33:33 `q005. Now suppose that y = -.2 t^2 + 5 t + 100 represents depth vs. clock time. What is the corresponding rate function y ' ?
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RESPONSE --> y= at^2 +bt +c y=.2t^2 +5t +c y'= 2at +b y'=2(.2)t +(5) y'= .4t +5 confidence assessment: 3
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10:33:48 The rate function corresponding to y = a t^2 + b t + c is y ' = 2 a t + b. In this example a = -.2, b = 5 and c = 100 so y ' = 2(-.2) t + 5, or y ' = -.4 t + 5.
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RESPONSE --> okay self critique assessment: 3
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10:38:24 `q006. From the depth function and the rate function we can find the coordinates of the t = 30 point of the graph, as well as the rate of change of the function at that point. The rate of change of the function is identical to the slope of the graph at that point, so we can find the slope of the graph. What are the coordinates of the t = 30 point and what is the corresponding slope of the graph? Sketch a graph of a short line segment through that point with that slope.
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RESPONSE --> using the depth function: y= .2t^2 +5t +100 and the rate function: y'= .4t +5 and plugging in for t=30 we get: the point (30, 430) with a slope of 17 confidence assessment: 2
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10:40:49 At t = 30 we have y = -.2 * 30^2 + 5 * 30 + 100 = 70, and y ' = -.4 * 30 + 5 = -7. The graph will therefore consist of the point (30, 70) and a short line segment with slope -7.
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RESPONSE --> I didn't have my - in front of my A values lets try that again: y=-.2t^2 +5t +100 and y'= -.4t +5 that gives me the point (30, 70) and a slope of -7 an exact match to the answer self critique assessment: 2
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10:43:54 `q007. What is the equation of the straight line through the t = 30 point of the depth function of the preceding example, having a slope matching the t = 30 rate of change of that function?
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RESPONSE --> y= mx +b with m being slope and b being y-intercept 70= (-7) (30) +b 70 = -210 +b b = 280 The formula for a straight line with the point (30,70) and a slope of -7 is y= -7x + 280 confidence assessment: 3
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10:44:36 A straight line through (30, 70) with slope -7 has equation y - 70 = -7 ( x - 30), found by the point-slope form of a straight line. This equation is easily rearranged to the form y = -7 x + 280.
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RESPONSE --> okay self critique assessment: 3
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10:54:37 `q008. Evaluate the depth function y at t = 30, 31, 32. Evaluate the y coordinate of the straight line of the preceding question at t = 30, 31, 32. What are your results?
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RESPONSE --> using the depth function: y= -.2t^2 +5t +100 and the straight line formula: y= -7x +280 and evaluating for t/x = 30,31,32 the depth function: {(30, 70), (31, 62.8), (32, 55.2)} Straight line: {(30, 70), (31, 63), (32, 56)} confidence assessment: 2
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10:56:11 Plugging t = 30, 31, 32 into the original function y = -.2 t^2 + 5 t + 100 we get y = 70, 62.8, 55.2 respectively. Plugging t = 30, 31, 32 into the straight-line equation y = -7 t + 280 we get y = 70, 63, 56 respectively.
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RESPONSE --> okay self critique assessment: 3
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11:00:16 `q009. By how much does the straight-line function differ from the actual depth function at each of the three points? How would you describe the pattern of these differences?
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RESPONSE --> As the parabola forms the straight line becomes a tangent at point (30,70) then the parabola starts to drop off the straight line continues at a slope -7. confidence assessment: 2
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11:02:37 At t = 30 the two functions are identical. At t = 31 the values are 62.8 and 63; the function is .2 units below the straight line. At t = 32 the values are 55.2 and 56; the function is now .8 units below the straight line. The pattern of differences is therefore 0, -.2, -.8. The function falls further and further below the straight line as we move away from the point (30, 70).
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RESPONSE --> as t/x gets farther away from 30 the differnces between the two y values continue to get greater. self critique assessment: 2
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