qa _ 09

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course Mth 173

6/ 9 10 pm

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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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:

Y’ = .1 t - 5.5

= .1 (10) - 5.5

Y= -4.5

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):

I understand this logic

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Self-critique Rating:3

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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 will be -4

We should use 5 + 4 /2 = -4.5 as the average slope

Since the slope is -4.5 for a 10 second interval we find that the run is -45

If the point was (10, 45) then the resulting point is (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).

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Self-critique (if necessary):

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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:

Between t = 20 and t= 30 the slope will be -3 at t= 30

Therefore the average slope would be -4 + -3 / 2= -3.5

For a slope of -3.5 for a time interval of 6 seconds the rise would be -35

The point when then go from (20,0) to (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):

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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:

From t= 30 to t= 40 would be slope of -2

then the average slope would be -2.5

then from (30, -35) to (40, -60)

from t= 40 to t= 50 the slope would be -1

the average slope would be -2 + -2 / 2= -1.5

then from (40, -55) to (50, -75)

from t= 50 to t= 60 the slope is 0

the average slope would then be -1 + 0 / 2 = -.5

then from (50, -70) to (60, -80)

from t= 60 to t= 70 the slope would be +1

the average slope would be 0 + 1 / 2 = .5

the from (60, -75) to (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).

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Self-critique (if necessary):

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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:

Y’ =2 at + b

Y’ = (2) -.2 (t) + 5

Y’ = -.4 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):

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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:

y = -.2 t^2 + 5 t + 100

y= -.2 (30)^2 + 5(30) + 100

y= 70

(30, 70)

Y’= -.4t + 5

Y’ = -.4 (30) + 5

= - 7 depth/ time

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):

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Self-critique Rating:

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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 = mx + b

Y = (30) x + 30

confidence rating #$&*:

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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):

This makes sense

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Self-critique Rating:3

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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:

y = -.2 t^2 + 5 t + 100

for t = 30, 31, 32

y = 70, 62.8, 55.2

for

y = -7 x + 280

t= 30, 31, 32

y= 70, 63, 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):

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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:

0, .2, .8

The pattern continues to fall further from the line

confidence rating #$&*:

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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:

0, .2, .8

The pattern continues to fall further from the line

confidence rating #$&*:

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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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qa _ 09

&#Your work looks good. Let me know if you have any questions. &#