course Mth 272 ?????????\Q????assignment #012
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16:18:10 5.6.26 (was 5.6.24 trap rule n=4, `sqrt(x-1) / x on [1,5]
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RESPONSE --> first find the new intervals = [1,2], [2,3],[3,4],[4,5] since (5-1)/4= 1 plugging these into the function we get 0, .5, .471, .433, .4 then find the average altitudes (0+.5)/2= .25 (.5+.471)/2= .486 (.471+.433)/2= .452 (.433+.4)/2= .417 1(.25+.486+.452+.417) = 1.604 approximately confidence assessment: 3
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16:18:16 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 --> got it self critique assessment: 3
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17:04:52 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 --> you would use trapezoids for these altitudes. The average altitudes are (0+50)/2 = 25 (50+54) /2= 52 (54+82)/2= 68 (82+82)/2= 82 (82+73)/2=77.5 (73+75)/2=74 (75+80)/2= 77.5 total area now 20(25+52+68+82+77.7+74+77.5) 9920 sqft midpoint rule [0,160] the new intervals [0,20],[20,40],[40,60] and so on. the midpoints would be 10,30,50,70,90,110,130,150 the area when using the midpoint rule is 12800. This is becuase the rectanglestake up area outside of the ponds accurate limits due to its irregular shape. This makes the area greater for the midpoin rule that for the trap rule. confidence assessment: 3
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17:04:59 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 --> understood self critique assessment: 3
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17:05:40 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The last one was pretty difficult, but i was able to figure it out, so i think I understand this material well. confidence assessment: 3
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