Assignment 12

#$&*

course MTH272

If your solution to a stated problem does not match the given solution, you should self-critique per instructions at 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.

012. `query 12

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Question: `q5.6.26 (was 5.6.24 trap rule n=4, `sqrt(x-1) / x on [1,5]

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

sqrt(x-1) / x on [1,5] n=4

(5-1)/4=1

With the trap rule uses the end point values from each interval unlike the midpoint that use the center of the intervals for the values.

There are 4 separate intervals spaced 1 unit apart and measuring at each end of every unit for a total of 8 points.

(1, 2), (2, 3), (3, 4), (4, 5) the sets of points along the x axis. Plug each of these into the given function.

(0+.5+.5+.4714+.4714+.43301+.43301+.4) = 3.20882 multiplying the outcome of the 8 separate point.

Then multiply 3.20882 by (1/2) with the 1 representing the width of the units divided by the number of units (5-1)/4=1. The 2 is to split what has been doubled to increase the accuracy of our measurements.

3.20882(1/2) =1.60441

confidence rating #$&*:3

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Given Solution:

`a Dividing [1, 5] into four intervals each will have length ( 5 - 1 ) / 4 = 1. The four intervals are therefore

[1, 2], [2, 3], [3, 4], [4,5].

The function values at the endpoints f(1) = 0, f(2) = .5, f(3) = .471, f(4) = .433 and f(5) = .4. The average altitudes of the trapezoids are therefore

(0 + .5) / 2 = 0.25, (.5 + .471)/2 = 0.486, (.471 + .433) / 2 = 0.452 and (.433 + .4) / 2 = 0.417.

Trapeoid areas are equal to ave height * width; since the width of each is 1 the areas will match the average heights 0.25, 0.486, 0.452, 0.417.

The sum of these areas is the trapezoidal approximation 1.604. **

DER

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Self-critique (if necessary):NA

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Self-critique Rating:

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Question: `q5.6.32 (was 5.6.24 est pond area by trap and midpt (20 ft intervals, widths 50, 54, 82, 82, 73, 75, 80 ft).

How can you tell from the shape of the point whether the trapezoidal or midpoint estimate will be greater?

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Your solution:

(20 ft intervals, widths 50, 54, 82, 82, 73, 75, 80 ft)

Trapezoidal

(50+54+54, 82+82+82+82+73+73+75+75+80)=862 taking each of the points and laying them out as they would on each side of the interval.

862*20= 17240 multiplying the interval by the width

17240/2=8620 dividing by 2 gives the area in square feet

Midpoint

If these are all midpoint widths at interval of 20

{(20*50)(20*54)+(20*82)+(20*82)+(20*73)+(20*75)+(20*80)}=9920

Analysis

It would appear that the midpoint has greater area than the Trapezoidal, but if you draw out the graph of the points the Trapezoidal is more accurate of estimation of area. I would say the Trapezoidal has more area from looking at the graph.

confidence rating #$&*:3

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Given Solution:

`a Since widths at 20-ft intervals are 50, 54, 82, 82, 73, 75, 80 ft the pond area can be approximated by a series of trapezoids with these altitudes. The average altitudes are therefore respectively

(0 + 50 / 2) = 25

(50 + 54) / 2 = 52

(54 + 82) / 2 = 68

(82 + 82) / 2 = 82

etc., with corresponding areas

25 * 20 = 500

52 * 20 = 1040

etc., all areas in ft^2.

The total area, according to the trapezoidal approximation, will therefore be

20 ft (25+52+68+82+77.5+74+77.5+40) ft = 9920 square feet.

The midpoint widths would be calculated based on widths at positions 10, 30, 50, 70, ..., 150 ft. Due to the convex shape of the pond these estimates and will lie between the estimates made at 0, 20, 40, ..., 160 feet.

The convex shape of the pond also ensures that the midpoint of each rectangle will 'hump above' the trapezoidal approximation, so the midpoint estimate will exceed the trapezoidal estimate. **

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Self-critique (if necessary):

So we need to assume a starting point of 0 on each side of the given points. I am noy sure why, but I can program that into my brain for recovery for the same kind of problem. My ending assumption is correct with the midpoint point verses the trapezoidal type measurements is that it gives a much tighter measurement there for more accurate. So getting the same numbers using the 2 rules the trapezoidal is bring us closer to the actual area.

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Self-critique Rating:3

@& At each end of the pond, its width shrinks to zero.*@

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Question: `q Add comments on any surprises or insights you experienced as a result of this assignment.

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Self-critique (if necessary):

Just where we are using 0 as an assumed position at the beginning and end of the set of numbers on problem 2 when using the trap rule.

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Self-critique Rating:

@& Let me know if my brief note doesn't clarify that. I think you'll be OK with this.*@

Assignment 12

#$&*

course MTH272

If your solution to a stated problem does not match the given solution, you should self-critique per instructions at 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.

012. `query 12

*********************************************

Question: `q5.6.26 (was 5.6.24 trap rule n=4, `sqrt(x-1) / x on [1,5]

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

sqrt(x-1) / x on [1,5] n=4

(5-1)/4=1

With the trap rule uses the end point values from each interval unlike the midpoint that use the center of the intervals for the values.

There are 4 separate intervals spaced 1 unit apart and measuring at each end of every unit for a total of 8 points.

(1, 2), (2, 3), (3, 4), (4, 5) the sets of points along the x axis. Plug each of these into the given function.

(0+.5+.5+.4714+.4714+.43301+.43301+.4) = 3.20882 multiplying the outcome of the 8 separate point.

Then multiply 3.20882 by (1/2) with the 1 representing the width of the units divided by the number of units (5-1)/4=1. The 2 is to split what has been doubled to increase the accuracy of our measurements.

3.20882(1/2) =1.60441

confidence rating #$&*:3

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

.............................................

Given Solution:

`a Dividing [1, 5] into four intervals each will have length ( 5 - 1 ) / 4 = 1. The four intervals are therefore

[1, 2], [2, 3], [3, 4], [4,5].

The function values at the endpoints f(1) = 0, f(2) = .5, f(3) = .471, f(4) = .433 and f(5) = .4. The average altitudes of the trapezoids are therefore

(0 + .5) / 2 = 0.25, (.5 + .471)/2 = 0.486, (.471 + .433) / 2 = 0.452 and (.433 + .4) / 2 = 0.417.

Trapeoid areas are equal to ave height * width; since the width of each is 1 the areas will match the average heights 0.25, 0.486, 0.452, 0.417.

The sum of these areas is the trapezoidal approximation 1.604. **

DER

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Self-critique (if necessary):NA

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Self-critique Rating: