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course Mth 271

10/9 9p

09. 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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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:

-5.5/1 = (y2--6)/(10-0)

-5.5(10)=y2+6

=-61

point is (10, -61)

confidence rating #$&*:2

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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):

Your explanation makes sense, but it seems like you should be able to set it up and solve for the missing y value. I don’t understand why this doesn’t work.

Self-critique Rating:2

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You used -6, which is a slope, as a y value when you wrote y2 - -6 .

You can't subtract a slope from a y value. The slope and the y value don't measure the same thing, so they can't be added or subtracted. They are unlike terms.

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Had you used 100, which is a y value, instead of the slope -6 your equation would have read

-5.5/1 = (y2--100)/(10-0)

and you would have gotten the same result as the given solution.

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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:

y’ = .1t-6

y’ = .1(20)-6 = 2-6

2-6 = -4 slope at t=20

average slope -5-4= -9

-9/2 = -4.5

-4.5*(10)= 45

45-0 = 0

point is (20, 0)

confidence rating #$&*:2

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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):n/a

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Self-critique Rating:3

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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:

y’=.1(t)-6

.1(30)-6

slope = -3

Averge slope -3+-4 = -7

-7/2 = -3.5

-3.5*10 = -35

point is (30, -35)

confidence rating #$&*:2

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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):n/a

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Self-critique Rating:3

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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:

t slope avg. slope *10 rise point

40 -2 -2.5 -25 (40, -60)

50 -1 -3.5 -35 ( 50, -75)

60 0 -.5 -5 (60, -80)

70 1 .5 5 (70, -75)

It seems like it is easier to see the shape of the graph, or it has a more defined shape, when the averages are used.

confidence rating #$&*:1

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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):

I was sure about my points but not about the difference. I think my sketch of the original graph was off so it was hard to tell.

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Self-critique Rating:3

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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:

2at+b a= -.2 b=5

y’= (-.2)(t) + 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.

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Self-critique (if necessary):

n/a

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Self-critique Rating:3

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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:

y = -.2(30)^2+ 5(30) +100 = 70 point is (30, 70)

y’=2(-.2)(30) + 5 = -7 slope = -7

confidence rating #$&*:2

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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): n/a

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Self-critique Rating:3

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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:

y-intercept is (0,100)

y= -7x + 100

confidence rating #$&*:2

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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):I must have calculated the y-intercept wrong because it seems like it should match. I plugged in 0 for x and got 100 for the y-intercept , so the y-intercept form would be y= -7x +100. I understand why the point-slope form and subsequent y-intercept form in the solution is correct but I don’t know what I did wrong.

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Self-critique Rating:1

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If the graph was a straight line then the slope would be unchanged, and the line through every point having the slope at that point would have the same y intercept.

But the graph curves.

So you can't assume that the y intercept for this line is the same at that of the curving graph.

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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:

depth function y at y y-coord. of straight line

30 70 y= -7(30)+280=70

31 62.8 y= -7(31)+280 = 63

32 55.2 y= -7(32)+280 = 56

confidence rating #$&*:1

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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):n/a

elf-critique Rating:3

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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:

70-70 = 0

62.8 - 63 = -.2

55.2-56 =- .8

The difference is increasing at an increasing rate.

confidence rating #$&*:1

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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).

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Self-critique (if necessary):

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Self-critique rating:

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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:

depth function y at y y-coord. of straight line

30 70 y= -7(30)+280=70

31 62.8 y= -7(31)+280 = 63

32 55.2 y= -7(32)+280 = 56

confidence rating #$&*:1

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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):n/a

elf-critique Rating:3

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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:

70-70 = 0

62.8 - 63 = -.2

55.2-56 =- .8

The difference is increasing at an increasing rate.

confidence rating #$&*:1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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: `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:

depth function y at y y-coord. of straight line

30 70 y= -7(30)+280=70

31 62.8 y= -7(31)+280 = 63

32 55.2 y= -7(32)+280 = 56

confidence rating #$&*:1

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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):n/a

elf-critique Rating:3

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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:

70-70 = 0

62.8 - 63 = -.2

55.2-56 =- .8

The difference is increasing at an increasing rate.

confidence rating #$&*:1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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:

#*&!#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#