course Mth 174 I accidently labeled Assignment 8 as assignment9. The real assignment 9 is this one, labeled Assignment9_Real. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aIf 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q002. We find that using the average of the two slopes we reach the point (10, 45). At this point we have y ' = -5. By the time we reach the t = 20 point, what will be the slope? What average slope should we therefore use on the interval from t = 10 to t = 20? Using this average slope what point will we end up with when t = 20? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If you plug in 20 into y’=.1t-6, then you will end up with -4, which is the slope at t=20. The average slope of t=10 and t=20 is the average of -5 and -4, which is -4.5. If you go from (10, 45), then the point will end up being (20, 0). confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q003. We found that we reach the point (20, 0), where the slope is -4. Following the process of the preceding two steps, what point will we reach at t = 30? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At t=30, the slope -3, so if you average this with t=20 (-4), which is -3.5, then the new point will be at (30, -35) confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q004. Continue this process up through t = 70. How does this graph differ from the preceding graph, which was made without averaging slopes? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At t=40, the slope is -2. The average of that and t=30 is -2.5. So the new coordinate will be (40, -60) At t=50, the slope is -1. The average of that and t=40 is -1.5. So the new coordinate will be (50, -75) At t=60, the slope is 0. The average of that and t=50 is -.5. So the new coordinate will be (60, -80) At t=70, the slope is 1 The average of that and t=60 is .5. So the new coordinate will be (70, -75) confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q005. Now suppose that y = -.2 t^2 + 5 t + 100 represents depth vs. clock time. What is the corresponding rate function y ' ? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The derivative of = -.2 t^2 + 5 t + 100 is -.4t+5. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got my answer a different way, but it still ended up being the same. ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q006. From the depth function and the rate function we can find the coordinates of the t = 30 point of the graph, as well as the rate of change of the function at that point. The rate of change of the function is identical to the slope of the graph at that point, so we can find the slope of the graph. What are the coordinates of the t = 30 point and what is the corresponding slope of the graph? Sketch a graph of a short line segment through that point with that slope. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If y= -.2 t^2 + 5 t + 100, then = -.2 (30)^2 + 5 (30) + 100=70. The slope at this point is y ' = -.4 (30) + 5= -7, so the coordinates would be (30, 70) confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAt 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&*ok ********************************************* Question: `q007. What is the equation of the straight line through the t = 30 point of the depth function of the preceding example, having a slope matching the t = 30 rate of change of that function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You can use the formula y=yo+m(x-xo), so y=70-7(x-30), y=70-7x+210, which equals y=-7x+280 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aA 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q008. Evaluate the depth function y at t = 30, 31, 32. Evaluate the y coordinate of the straight line of the preceding question at t = 30, 31, 32. What are your results? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If y= -.2 t^2 + 5 t + 100, then = -.2 (30)^2 + 5 (30) + 100=70, for t=31: -.2 (31)^2 + 5 (31) + 100=62.8. For t=32, -.2 (32)^2 + 5 (32) + 100=55.2 For the y coordinate of the straight line, at t=30: y = -7 (30) + 280=70. For t=31, y = -7 (31) + 280= 63. And for t= 32, y = -7 (32)+ 280= 56. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aPlugging 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q009. By how much does the straight-line function differ from the actual depth function at each of the three points? How would you describe the pattern of these differences? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The differences of the two are not that outstanding. It seems like the bigger the x value got, the more of the difference came between the two values from the two formulas. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAt 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). "