course Mth 272 012. `query 12
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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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, using the midpoint rule, we need to find the midpoints of the intervals. Because the lake is 160 ft wide and broken into intervals of 20, the midpoints will be 10, 30, 50, 70, 90, 110, 130, and 150. Now we can use the midpoint rule to find the area by multiplying these quantities with the width of 20 and adding them. So our area becomes 12000. To use the trapezoidal rule, we must first find the average heights of the trapezoids. We can do this by adding the coinciding heights and then dividing that quantity by 2 to get the average. So (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, amd (75+80)/2 = 77.5. Now we can use the trapezoidal rule to find the area. So Area = 20 (25+52+68+82+77.5+74+77.5) = 20 (456) = 9120. We can tell that the midpoint rule estimate will be greater from the shape of the lake because the rectangles it creates will be larger than the lake, giving us a greater area. Confidence rating:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I had trouble using the midpoint rule when there was no given formula and am not sure if I solved the midpoint area correctly. However, I did understand that the midpoint rule would give a greater area than the trapezoidal rule, which my answer backs up. Self-critique Rating: ********************************************* Question: `q Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):