assignment 12

course MTH 272

7/5 11 pm

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]

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Your solution:

N=4 so there are 4 intervals.

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

endpoints f(1) = 0, f(2) = .5, f(3) = .471, f(4) = .433 and f(5) = .4.

solve and get: 0.25, 0.486, 0.452 and 0.417.

Confidence rating:

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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.

height times width (*1=same as other values) 0.25, 0.486, 0.452, 0.417.

add them all up = about 1.6

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Self-critique (if necessary): I know it for the most part.

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: I did not know how to solve this problem, unfortunately.

Confidence rating: 0

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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):

&#You did not answer the given question. You need to always at least explain what you do and do not understand about the question. A phrase-by-phrase analysis is generally required when you cannot otherwise answer a question.

&#

Self-critique Rating:

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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): I don’t know how to solve this type of problem. The given response helped me to learn it a little better, though.

Self-critique Rating:

&#See my notes and let me know if you have questions. &#