course MTH 271 10/04,1515 assignment #009009. Finding the average value of the rate using a predicted point
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21:17:38 `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 --> y-100=-5.5(x-0) y-100=-5.5x y=-5.5x + 100 y=-5.5(10) + 100 y=-55 + 100 y=45 (10,45) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:21:42 `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 --> y'(20)=.1(20)-6 y'(20)=-4 Average of slopes: (-5 + -4)/2=-4.5 Rise=-45 (20,0) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:24:24 `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: y'=.1(30) - 6 y'=-3 Average slope: (-4 + -3)/2 = -3.5 Rise=-35 (30,-35) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:44:15 `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 --> y'(40)=.1(40)-6 y'(40)=-2 Average slopes:(-3+-2)/2=-2.5 Rise=-25 (40, -60) y'(50)=.1(50)-6 y'(50)=-1 Average of slopes: (-2 + -1)/2=-1.5 Rise=-15 (50, -75) y'(60)=.1(60) -6 y'(60)=0 Average of slopes: (-1.5 + 0)/2 =-.75 Rise=-7.5 (60,-82.5) y'(70)=.1(70) - 6 y'(70)=1 Average of slopes=(0 + 1)/2 =.5 Rise=5 (70, -77.5) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:46:09 `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=-.2t^2 + 5t + 100 a=-.2 b=5 y'=2at + b y'=2(-.2)t + 5 y'=-.4t + 5 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:49:00 `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(30)=-.2(30^2) +5(30)+100 y(30)=70 (30,70) y'(30)=-.4(30) + 5 y'(30)=-7 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:53:09 `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 --> 'dy/'dx=-7 point=(30,70) 'dy=(y-70) 'dx=(x-30) [(y-70)/(x-30)]=-7 y-70=-7(x-30) y-70=-7x + 210 y=-7x + 280 confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:59:16 `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 --> y=-.2t^2 + 5t + 100 y(30)=-.2(30^2) + 5(30) + 100 y(30)=70 y(31)=-.2(31^2) + 5(31) + 100 y(31)=62.8 y(32)=-.2(32^2) + 5(32) + 100 y(32)=55.2 y=-7x + 280 y(30)=-7(30) + 280 y(30)=70 y(31)=-7(31) + 280 y(31)=63 y(32)=-7(32) + 280 y(32)=56 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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22:01:32 `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 --> t=30, deviation=0 t=31, deviation=.2 t=32, deviation=.8 The deviation between the values becomes greater as you get further from the point to which the line is tangent, 30. So, the tangent line approximates the values most accurately to those points closest to the original point. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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