course Mth 272 ]јSbmTassignment #012
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15:04:20 5.6.26 (was 5.6.24 trap rule n=4, `sqrt(x-1) / x on [1,5]
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RESPONSE --> 5-1= 4 intervals (1,2) (2,3) (3,4) (4,5) sqrt(x-1)/x x=1 =0 x=2 =0.5 x=3 =0.471 x=4 =0.433 x=5 =0.4 x+ f(x)/2= trap height (and also equals area) = 0.25, 0.486. 0.452. 0.417 added up= 1.604 total area 5-1= 4 intervals (1,2) (2,3) (3,4) (4,5) sqrt(x-1)/x x=1 =0 x=2 =0.5 x=3 =0.471 x=4 =0.433 x=5 =0.4 x+ f(x)/2= trap height (and also equals area) = 0.25, 0.486. 0.452. 0.417 added up= 1.604 total area confidence assessment: 3
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15:04:26 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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RESPONSE --> ok self critique assessment: 3
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15:04:46 5.6.32 (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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RESPONSE --> x+y/2= avg altitudes =25, 52, 68, 82, 77.5, 77.5, 74, 40 (total altidudes)*20 (width)= total area =9920 ft^2 with offsetting 20-value interval widths makes clear that the midpoint of rectangles will exceed the trap estimate due to the obvious convex shape of the pond. confidence assessment: 2
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15:05:59 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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RESPONSE --> OK, i felt it really hard to describe what I was trying to get out as my answer to the problem. I am still unsure how to put it in words. I can find the areas, total areas, etc. That's not the problem. The problem is attempting to describe the overall picture of the situation. self critique assessment: 2
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15:06:12 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> see previous response confidence assessment: 3
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