qa 9

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08:40:49 Note 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 --> If we use -5.5 it would be 10 * -5.5 = -55. Then you take 100 - 55 = 45. So it goes from (0, 100) to (10, 45)

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08:40:59 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 --> OK

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08:42:22 `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 --> The slope would be .1(20) - 6 = -4. The average would be (-4 + -5) / 2 = -4.5. So to get the point at t = 20 you would first take -4.5 * 10 = -45. So it goes from (10, 45) to (20, 0)

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08:42:31 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 --> OK

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08:43:28 `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 at t = 30 is .1(30) - 6 = -3. The average would be (-4 +-3) / 2 = -3.5. So to get the point you would take -3.5 * 10 = -35. So it goes from (20, 0) to (30, -35)

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08:43:42 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 --> OK

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09:09:18 `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 --> you would get the following order pairs: (40, -60) (50, -75) (60, -80) (70, -75)

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09:09:28 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 --> OK

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09:10:30 `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' = -.4t + 5

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09:10:37 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 --> OK

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09:26:43 `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 --> The point would be (30, 70). The slope of the graph at t = 30 is -.4(30) + 5 = -7 cm / s.

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09:26:50 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 --> OK

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09:27:50 `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 --> So the equation to solve is y - 70 = -7(x - 30). If you simplify this you get y - 70 = -7x + 210. Add 7 to both sides to get the equation which is y = -7x + 280.

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09:32:18 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 --> OK

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09:43:30 `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 --> From the depth function. If you put in t = 30 you would have y = -.2(30)^2 + 5(30) + 100 = 70 cm. For t = 31 you would have y = -.2(31)^2 + 5(30) + 100 = 62.8 cm. If you put in t = 32 you would have y = -.2(32)^2 + 5(32) + 100 = 55.2 cm. For the stright line function. if you have t = 30 you would have y = -7(30) + 280 = 70. If you have t = 31 you would have y = -7(31) + 280 = 63. If you have t = 32 you would have y = -7(32) + 280 = 56. As you can see you get just about the same answer using either equation.

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09:43:42 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 --> OK

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09:44:28 `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 --> they are almost the same

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09:44:41 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 --> OK

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