Assignment 12

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course Mth 272

October 17 around 4:25

012. `query 12

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

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Your solution:

The four intervals are

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

Function values are

f(1)=0

f(2 )=0.5

f(3)=0.471

f(4)=0.433

f(5)=0.4

The average altitudes are

(0 + .5)/2=0.25

(0.5+0.471)/2=0.486

(0.471+0.433)/2=0.452

(0.433+0.4)/2=0.417

The total are is

0.25+0.486+0.452+0.417=1.605

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. **

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Self-critique (if necessary): ok

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self-critique rating #$&*: 3

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Question: `q5.6.32 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:

Altitudes

(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

(80+0)/2=40

Areas

25*20=500

52*20=1040

68*20=1360

82*20=1640

77.5*20=1550

74*20=1480

77.5*20=1550

40*20=800

Total area is

500+1040+1360+1640+1550+1480+1550+800=9920 square feet

The midpoint widths would be based on widths at 20 ft intervals starting at 10 and going to 150. Due to the convex shape of the pond these estimates will lie between the estimates made at 20 ft intervals starting at 0 and going to 160. The convex shape of the pond shows that the midpoint of each rectangle will be above the trapezoidal approximation, so the midpoint estimate will be greater than the trapezoidal estimate.

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. **

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Self-critique (if necessary): ok

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self-critique rating #$&*: 3

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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 need to review how to tell if the trapezoidal or midpoint estimate will be greater based on the shape of a point.

Self-critique Rating: 3

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&#Good responses. Let me know if you have questions. &#

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