qa_09

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course Mth 173

10/7 2:00

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

009. Finding the average value of the rate using a predicted point

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Question: `qNote that there are 9 questions in this assignment.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q001. The process we used in the preceding qa to approximate the graph of y corresponding to the graph of y ' or the y ' function (the function is y ' = .1 t - 6 for t = 0 to t = 100) 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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Your solution:

The value of the slope at t = 0 is -6 and the value of the slope at t = 10 is -5. Thus the average of the slope will be -5.5.

Now from the point (0, 100) we construct a line of slope = -5.5 to reach a point for t = 10.

Now the run for the line = 20 - 10 = 10 and the slope = -5.5

The slope = rise / run, thus rise = slope * run

Thus rise = -5.5 * 10 = -55

Given that the y co-ordinate corresponding to t = 0 is 100 and the rise = -55, the y co-ordinate corresponding to t = 10 thus will be = 100 - 55 = 45.

Thus from the point (0,100) and with a slope of -5.5 we reach the point (10, 45)

confidence rating #$&*: 3

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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).

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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?

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Your solution:

The slope of the function at t = 20 will be calculated as y’(20) = 0.1*20 - 6 = -4.

The slope of the function at t = 10 is -5, thus the average of the slope for the t = 10 to t = 20 interval will thus be -4.5. Using this average slope we can construct a line from the point (10, 45) to reach a point on t = 20. The run for the line = 20 - 10 = 10 and the slope = -4.5. slope = rise / run, thus rise = slope * run = -4.5 * 10 = -45.

The y co-ordinate corresponding to t = 10 is 45 and the rise = -45, thus the y co-ordinate corresponding to t = 20 will thus be = 45 - 45 = 0. Thus using the average slope of -4.5 we end up with the point (20, 0) when t = 20.

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).

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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?

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Your solution:

Following the process done in the previous questions, the slope at t = 20 is -4 and at t = 30 is -3, thus the average slope = -3.5 and the run for the line = 10. Thus the rise for the line = run * average slope = 10 * -3.5 = -35. Thus the point (20, 0) with a rise of -35 will reach the point (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 = -3.5 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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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `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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Your solution:

The slope at t = 30 is -3 and slope at t = 40 is -2, thus the average slope is -2.5 for the t = 30 to t = 40 interval. The run is 10 and the average slope = -2.5, thus the rise = -25. Thus the point (30, -35) reaches the point (40, -60)

The slope at t = 40 is -2 and slope at t = 50 is -1, thus the average slope is -1.5 for the t = 40 to t = 50 interval. The run is 10 and the average slope = -1.5, thus the rise = -15. Thus the point (40, -60) reaches the point (50, -75)

The slope at t = 50 is -1 and slope at t = 60 is 0, thus the average slope is -.5 for the t = 50 to t = 60 interval. The run is 10 and the average slope =-.5, thus the rise = -5. Thus the point (50, -75) reaches the point (60, -80)

And finally the slope at t = 60 is 0 and slope at t = 70 is 1, thus the average slope is .5 for the t = 60 to t = 70 interval. The run is 10 and the average slope = .5, thus the rise = 5. Thus the point (60, -80) reaches the point (70, -75)

The points thus obtained for the graph are (30 , -35) , (40 , -60) , (50, -75) , (60, -80) , (70, -75) whereas for the preceding question the points obtained were ( 30 , -50) , (40, -80) , (50 , -100) , (60 , -110) and ( 70, -110). Thus we see that y coordinates for the average slope graph differ from the normal slope graph by 15, 20, 25, 30, 35 units respectively for t = 30, 40, 50, 60, 70. Thus as the value of t increases the normal slope graph grows much and much lower from the average slope graph. The y value for the average slope graph decreases by 25, 15, 5 and -5 units respectively which further decreases whereas for normal slope graph the y value decreases by 30, 20, 10 and 0 units respectively which further keeps decreasing. We can thus say that the average slope graph is more accurate approximation as compared to the normal slope graph.

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).

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `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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Your solution:

Given the function y = -.2 * t^2 + 5t + 100 representing the depth vs. clock time.

The corresponding y’ function for the same is given by differentiating the given y function. Thus y’ = -.4t + 5, is the corresponding rate function.

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.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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.

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Your solution:

The given depth function is y = -.2 t^2 + 5 t + 100, thus the coordinates of the t = 30 point on the graph can be calculated as y(30) = -.2 * 30 * 30 + 5 * 30 + 100

= -180 + 150 + 100 = 70

The corresponding rate function is y’ = -.4t + 5

Thus the rate at t = 30 will be y’(30) = -.4 * 30 + 5 = -7

The graph thus is a small short straight line segment from the point (30, 70) with the slope = -7

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.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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?

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Your solution:

The point corresponding to t = 30 on the graph is (30, 70) and the rate of change of depth at t = 30 is -7, thus the slope of the line = -7. The line passing through (30, 70) and with a slope of -7 can be calculated as

(y - 70) / (x - 30) = -7 , (y - 70) = -7x + 210

Rearranged to

y = -7x + 280

Thus 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 is 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.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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?

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Your solution:

The given depth function is y = -.2 t^2 + 5 t + 100

Thus the value of the function at t = 30 is -.2 * 30 * 30 + 5 * 30 + 100 = 70

The value of the function at t = 31 is -.2 * 31 * 31 + 5 * 31 + 100 = 62.8

The value of the function at t = 32 is -.2 * 32 * 32 + 5 * 32 + 100 = 55.2

The given straight line function is y = -7x + 280

Thus the value of the function at t = 30 is = -7 * 30 + 280 = 70

The value of the function at t = 31 is = -7 * 31 + 280 = 63

The value of the function at t = 32 is = -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.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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?

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Your solution:

The value of the depth function and the straight line function is the same at t = 30 that is 70.

The value of the straight line function at t = 31 is greater than the value of the actual depth function. Thus the actual depth function is 63 - 62.8 = 0.2 units below the straight line function.

The value of the straight line function at t = 32 is greater than the value of the actual depth function. Thus the actual depth function is 56 - 55.2 = 0.8 units below the straight line function.

The difference of the straight line function from the actual depth function is thus 0, -.2 and -.8 for t = 30, 31, 32 respectively.

Thus the pattern in the differences can be seen, that is the magnitude of the difference of the straight line function and the actual depth function increases as the value of x increases. We see that the magnitude of the difference increases from 0 to .2 and further increases to .8 as t increases. In terms of the graph we see that the graph moves farther and farther away from the straight line (that is the actual depth function falls more and more away from the straight line function) as the value of t increases.

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).

Self-critique (if necessary):

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Self-critique rating:

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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?

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Your solution:

The value of the depth function and the straight line function is the same at t = 30 that is 70.

The value of the straight line function at t = 31 is greater than the value of the actual depth function. Thus the actual depth function is 63 - 62.8 = 0.2 units below the straight line function.

The value of the straight line function at t = 32 is greater than the value of the actual depth function. Thus the actual depth function is 56 - 55.2 = 0.8 units below the straight line function.

The difference of the straight line function from the actual depth function is thus 0, -.2 and -.8 for t = 30, 31, 32 respectively.

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).

Self-critique (if necessary):

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Self-critique rating:

#*&!

qa_09

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