course Mth 173 “´ÉƒžÃôfÀÁ{«ÁÖ©ü¸¼‡ŒÙ«z»¹assignment #009
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13:14:05 `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 --> I do not recall which points or equations I might have to refer to for this question. confidence assessment: 0
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13:15:37 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 --> Oh well, i guess there is no previous equation for me to refer to. I did not think to go from -5.5 and multiply by -10 to get the original rise. self critique assessment: 2
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13:21:22 `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 --> If the slope at t = 0 is -6 and the slope at t = 10 is -5, then we could think that the slope at t = 20 is -4. The average slope would then be -4.5. Using -4.5 as the average slope, the point when t = 20 is (20, 0) confidence assessment: 1
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13:21:59 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 --> I was not very sure if the methods I took were the correct. self critique assessment: 2
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13:23:41 `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 --> The slope when t = 30 is -3. The average slope between t = 20 and t = 30 is -3.5. The point when t = 30 is (30, -35) confidence assessment: 3
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13:23:47 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 --> self critique assessment:
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13:31:41 `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 --> t = 0 m = -6 t = 10 m = -5 t = 20 m = -4 If this trend (when t + 10, then m + 1) the slope when t = 70 is 1, making the point (70, -75) If the slope never becomes positve then the point at t = 70 is (70, -80) confidence assessment: 1
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13:31:55 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 --> self critique assessment:
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13:32:52 `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 --> If y = -.2t^2 + 5t + 100 then y' = -.4t + 5 confidence assessment: 2
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13:32:59 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 --> self critique assessment:
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13:43:51 `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 --> y = -.2(30)^2 + 5(30) + 100 When t = 30 the coordinates are (30, 70) y' = -.4(30) + 5 = -7 When t = 30 the slope is -7 In my graph the x - intercept is (40, 0) confidence assessment: 2
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13:43:57 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 --> self critique assessment:
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13:48:25 `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 - 70 = -7 (x - 30) y - 70 = -7x + 210 y = -7x + 280 confidence assessment: 3
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13:48:28 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 --> self critique assessment:
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13:54:26 `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 --> t = 30 y = 70 t = 31 y = 62.8 t = 32 y = 55.2 t = 30 y = -7x + 280 t = 31 y = -7.4x + 299.6 t = 32 y = -7.8x + 304.8 confidence assessment: 2
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13:55:40 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 --> I did not plug the times into the straight line equation. I used the points and slopes at those three times to find three seperate straight line equations. self critique assessment: 2
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13:59:02 `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 --> The straight line function only differs slitghtly at the three points, but I still trust the depth function more. The straight line produced y values of 70, 63, and 56 which follow in the pattern of y - 7 = the next y. The depth function y values of 70, 62.8, and 55.2 follow not visible pattern (yet). confidence assessment: 2
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13:59:36 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 --> Another question I was not sure exactly how to answer, I know for next time. self critique assessment:
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