alexandramutch.files.wordpress.com€¦ · web viewthere is no way of calculating the exact...

Calculus II

Year 12 Mathematics C | Mr Kirkman

Alexandra Mutch

Due: 20 August 2013

Alexandra Mutch

Table of Contents

Introduction...........................................................................................................................................................................2

Section 1: Use of Simpson’s Rule..................................................................................................................................3

Task 1................................................................................................................................................................................... 3

Section 2: Precision of Simpson’s Rule......................................................................................................................4

Task 2................................................................................................................................................................................... 4

Task 3................................................................................................................................................................................... 7

Task 4................................................................................................................................................................................... 9

Task 5................................................................................................................................................................................ 11

Task 6................................................................................................................................................................................ 13

Task 7................................................................................................................................................................................ 14

Section 3: Application of Simpson’s Rule...............................................................................................................17

Task 8................................................................................................................................................................................ 17

Task 9................................................................................................................................................................................ 25

1

Alexandra Mutch

Introduction

Simpson’s Rule is an approximation for area. It can be applied to area under curves to find the definite integral. It is commonly used by naval architects to calculate volume of centre of mass of ships, however, can be used to approximate any definite integral.

Johannes Kepler had used similar formulas more than 100 earlier, but Simpson’s Rule is credited to Thomas Simpson.

Simpson’s Rule:

A≑ w3 [E+4MO+2ME ]

Where A is approximated area, w is the width of strips, E is the sum of lengths of end strips, MO

is the sum of lengths of odd middle strips and ME is the sum of lengths of even middle strips.

A restriction of Simpson’s Rule is that the number of strips must be an even number.

2

Alexandra Mutch

Section 1: Use of Simpson’s Rule

Task 1

Approximate ∫sin (x2 ) . dxbetween 0 and π4 using 6 strips with Simpson’s Rule.

n = 6 strips

Strip width, w=b−an

w=

π4−0

6

w= π24 units

Sum of end strips, E = E1 + E2

E = 0 + 0.57847

E = 0.57847

Sum of odd strips, MO = M1 + M3 + M5

MO = 0.01713 + 0.15360 + 0.41539

MO = 0.58612

Sum of even strips, ME = M2 + M4

ME = 0.06849 + 0.27073

ME = 0.33922

Simpson’s Rule:

A≑ w3 [E+4MO+2M E ]

A≑

π243

[0.57847+4×0.58612+2×0.33922 ]

3

Strip x y

E1 0 0

M1π24 0.01713

M2π12 0.06849

M3π8 0.15360

M4π6 0.27073

M55π24 0.41539

E2π4 0.57847

Alexandra Mutch

A≑ π72

×3.60139

A≑0.15714 sq .units

Section 2: Precision of Simpson’s Rule

Task 2Determine the area under the line 4 x− y−2=0 where 2≤x ≤11.

(a) Basic area formula4 x− y−2=0

y=4 x−2

Fig. 1 – Diagram of Area Under 4x-y-2=0

*Not to scale

The diagram above shows that the total area under the line y=4 x−2 can be calculated by separating the area into a triangle (A1) and a rectangle (A2).

Area of A1

Area of a triangle, A1=12×b×h

A1=12×(11−2)×(42−6)

A1=12×9×36

A1=162 sq .unit s

4

42

6A2

A1

y

x112

y=4 x−2

Alexandra Mutch

Area of A2

Area of a rectangle, A2=L×W

A2=(11−2)×(6−0)

A2=9×6

A2=54 sq .units

Total Area

A1+A2=162+54

Atot=216 sq .units

(b) Definite integral

∫2

11

(4 x−2 ) . dx

¿ [2 x2−2x+c ]211

¿ [2(11)2−2(11)+c ]−[2(2)2−2(2)+c ]

¿ [220 ]−[4 ]

¿216 sq .units

(c) Simpson’s Rule

Approximate ∫(4 x−2) . dxbetween 2 and 11 using 10 strips

n = 10 strips


w=11−210

w=0.9units


E = 6 + 42

5

Strip x y

E1 2.0 6.0

M1 2.9 9.6

M2 3.8 13.2

M3 4.7 16.8

M4 5.6 20.4

M5 6.5 24.0

M6 7.4 27.6

M7 8.3 31.2

M8 9.2 34.8

M9 10.1 38.4

E2 11.0 42

Alexandra Mutch

E = 48

Sum of odd strips, MO = M1 + M3 + … + M9

MO = 9.6 + 16.8 + … + 38.4

MO = 120

Sum of even strips, ME = M2 + M4 + M6 + M8

ME = 13.2 + 20.4 + 27.6 + 34.8

ME = 96

Simpson’s Rule:

A≑ w3 [E+4MO+2M E ]

A≑ 0.93

[48+4×120+2×96 ]

A≑0.3×720

A≑216 sq .units

6

Alexandra Mutch

Task 3 y=x2+5 −1≤x ≤1

(a) Definite integral

∫−1

1

(x2+5 ) . dx

¿ [ x33 +5 x+c ]−1

1

¿ [(1)33 +5 (1)+c ]−[ (−1)33 +5 (−1)+c]¿ [5 13 ]−[−5 13 ]¿10 2

3sq .units


Approximate ∫(x2+5) . dxbetween -1 and 1 using 10 strips

n = 10 strips


w=1−(−1)10

w=0.2units


E = 6 + 6

E = 12


7

Strip x y

E1 -1.0 6.00

M1 -0.8 5.64

M2 -0.6 5.36

M3 -0.4 5.16

M4 -0.2 5.04

M5 0.0 5.00

M6 0.2 5.04

M7 0.4 5.16

M8 0.6 5.36

M9 0.8 5.64

E2 1.0 6.00

Alexandra Mutch

MO = 5.64 + 5.16 + … + 5.64

MO = 26.6


ME = 5.36 + 5.04 + 5.04 + 5.36

ME = 20.8

Simpson’s Rule:

A≑ w3 [E+4MO+2M E ]

A≑ 0.23

[12+4×26.6+2×20.8 ]

A≑ 115

×160

A≑10 23sq .units

8

Alexandra Mutch

Task 4y=(x+1)(x−1)(x−2) −1≤x ≤1

y=( x2−1 ) ( x−2 )

y=x3−2 x2−x+2


∫−1

1

(x3−2 x2−x+2 ) .dx

¿ [ x44 −2x3

3 −x2

2 +2x+c ]−1

1

¿ [(1)44 −2 (1 )3

3− (1 )2

2+2(1)+c ]−[ (−1)44 −2 (−1 )3

3− (−1 )2

2+2(−1)+c]

¿ [1 112 ]−[−1 712 ]¿2 23sq .units


Approximate ∫(x3−2 x2−x+2) . dxbetween -1 and 1 using 10 strips

n = 10 strips


w=1−(−1)10

w=0.2units


E = 0 + 0

E = 0

9

Strip x y

E1 -1.0 0.000

M1 -0.8 1.008

M2 -0.6 1.664

M3 -0.4 2.016

M4 -0.2 2.112

M5 0.0 2.000

M6 0.2 1.728

M7 0.4 1.344

M8 0.6 0.896

M9 0.8 0.432

E2 1.0 0.000

Alexandra Mutch


MO = 1.008 + 2.016 + … + 0.432

MO = 6.8


ME = 1.664 + 2.112 + 1.728 + 0.896

ME = 6.4

Simpson’s Rule:

A≑ w3 [E+4MO+2M E ]

A≑ 0.23

[0+4×6.8+2×6.4 ]

A≑ 115

×40

A≑2 23sq .units

10

Alexandra Mutch

Task 5y= (x−2 ) ( x+1 )2(x−4) −1≤x ≤2

y= (x−2 )(x2+2 x+1)(x−4)

y=(x3−3 x−2)(x−4 )

y=x4−4 x3−3 x2+10 x+8


∫−1

2

(x4−4 x3−3 x2+10 x+8 ) .dx

¿ [ x55 −x4−x3+5x2+8 x+c ]−1

2

¿ [(2)55 −(2 )4−(2 )3+5 (2 )2+8 (2)+c]−[(−1)55 −(−1 )4−(−1 )3+5 (−1 )2+8(−1)+c ]¿ [18.4 ]− [−3.2 ]

¿21.6 sq .units


Approximate ∫ (x4−4 x3−3 x2+10 x+8 ) . dxbetween -1 and 2

using 10 strips

n = 10 strips


w=2−(−1)10

w=0.3units


E = 0 + 0

E = 0

11

Strip x y

E1 -1.0 0.0000

M1 -0.7 1.1421

M2 -0.4 3.8016

M3 -0.1 6.9741

M4 0.2 9.8496

M5 0.5 11.8125

M6 0.8 12.4416

M7 1.1 11.5100

M8 1.4 8.9856

M9 1.7 5.0301

E2 2.0 0.0000

Alexandra Mutch


MO = 1.1421 + 6.9741 + … + 5.0301

MO = 36.4688


ME = 3.8016 + 9.8496 + 12.4416 + 8.9856

ME = 35.0784

Simpson’s Rule:

A≑ w3 [E+4MO+2M E ]

A≑ 0.33

[0+4×36.4688+2×35.0784 ]

A≑ 110

×216.032

A≑21.6032 sq .units

12

Alexandra Mutch

Task 6

The area under the functions in tasks 2 – 5 were approximated using Simpson’s Rule with significant increases in the number of strips from 10 to 100 and 1,000. The table below shows the calculated area and other values.

Task Degree a b n w E Mo Me A

2 x1 2 1110 0.9 48 120 96 216

100 0.09 48 1,200 1,176 2161,000 0.009 48 12,000 11,976 216

3 x2 -1 110 0.2 12 26.6 20.8 10.666667

100 0.02 12 266.66 260.68 10.6666671,000 0.002 12 2,666.67 2,660.67 10.666667

4 x3 -1 110 0.2 0 6.8 6.4 2.6666667

100 0.02 0 66.68 66.64 2.66666671,000 0.002 0 666.668 666.664 2.6666667

5 x4 -1 210 0.3 0 36.4693 35.0788 21.60348

100 0.03 0 360.045 359.91 21.61,000 0.003 0 3,600.00 3,599.99 21.6

The sum(seq(…)) function on graphics calculator was used to calculate the MO odd and ME

values.

For polynomials of degree 3 of less, the approximation was equal to the definite integral, no matter what the number of strips was. This shows that, for the given numbers of strips, the approximation was always 100% accurate for polynomials of degree 3 or less.

For the polynomial of degree 4, the approximated area using 10 strips had an error of approximately 0.016%. This is a relatively small error percentage, but was the only approximation that was not equal to the definite integral. When the number of strips was increased to 100, the approximation was equal to the definite integral. This indicates that the accuracy of the approximation increases as the number of strips increases with polynomials of degree 4 or higher.

This means that Simpson’s Rule is accurate for approximating area under curves of polynomial degree 3 or less. For approximating area under curves of polynomial degree 4 or more, a greater number of strips should be used for improved accuracy.

13

Alexandra Mutch

Task 7The error involved in using Simpson’s Rule varies as the fourth power of the width of strips. For

example, when the strip width is halved, the error is reduced by a factor of 116 . The validity of

this statement will be tested in this task.

Definite Integral

I=∫2

6

(7 x4−3 x3 ) . dx

I=[7 x55 −3 x4

4 +c ]2

6

I=[7(6)55 −3(6)4

4+c ]−[ 7(2)55 −3 (2)

4

4+c ]

I¿ [10,886.4−972 ]−[44.8−12 ]

I=9,881.6 sq .units

Simpson’s Rule – 2 Stripsn = 2 strips


w=6−22

w=2 units


E = 88 + 8,424

E = 8,512

Sum of odd strips, MO = M1

MO = 1,600

14

Strip x yE1 2 88M1 4 1,600E2 6 8,424

Alexandra Mutch

Sum of even strips, No even strips, therefore ME = 0

Simpson’s Rule:

I 2≑w3 [E+4MO+2M E ]

I 2≑23

[8,512+4×1,600+2×0 ]

I 2≑23×14,912

I 2≑9,94113sq .units

Simpson’s Rule – 4 Stripsn = 4 strips


w=6−24

w=1 unit


E = 88 + 8,424

E = 8,512

Sum of odd strips, MO = M1 + M3

MO = 486 + 4,000

MO = 4,486

Sum of even strips, ME = M2

ME = 1,600

15

Strip x yE1 2 88M1 3 486M2 4 1,600M3 5 4,000E2 6 8,424

Alexandra Mutch

Simpson’s Rule:

I 4≑w3 [E+4MO+2ME ]

I 4≑13

[8,512+4×4,486+2×1,600 ]

I 4≑13×29,656

I 4≑9,88513sq .units

Error

%Error=( Approximated areaActual area−1)×100

Area:

I = 9,881.6, I2 = 9,941 13 , I4 = 9,885 13

2 Strips

%Error I 2=( 9,941 139,881.6−1)×100

%Error I 2≈350579

%

4 Strips

%Error I 4=( 9,885 139,881.6−1)×100

%Error I 4=1754,632

%

When the strip width was halved (I2 to I4), the I4 proportion error of I2:

¿%Error I 4%Error I 2

¿( 1754,632 )( 350579 )

16

Alexandra Mutch

¿ 116

DiscussionTherefore the original statement that the error of the approximation varies as the fourth power of the width of strips is correct for this case.

Section 3: Application of Simpson’s Rule

Task 8The selected number of strips for this task is n = 4. This will be the simplest for calculations while including odd and even middle strips, along with end strips. 4 strips also satisfies the restriction that the number of strips must be even.

y = x

Definite Integral

∫c

d

x .dx

¿ [ x22 +k ]c

d

¿ [d22 +k ]−[ c22 +k ]¿ d

2−c2

2

Simpson’s Rulen = 4 strips

As the function is y = x, the y value of each strip will be equal to the x value.

The middle strips were calculated by adding one strip width each time. I.e. M1 = c + w, M2 = c + 2w, etc.

17

Strip x yE1 c c

M1 c+ d−c4

c+ d−c4

M2 c+ d−c2

c+ d−c2

M3 c+ 3 (d−c )4

c+ 3 (d−c )4

E2 d d

Alexandra Mutch


w=d−c4


E = c+d


MO = (c+ d−c4 )+(c+ 3 (d−c )

4 )MO = 2c+(d−c)

MO = c+d


ME = c+d−c2

Simpson’s Rule:

A≑ w3 [E+4MO+2M E ]

A≑( d−c4 )3 [(c+d)+4×(c+d)+2×(c+ d−c

2 )]A≑ d−c

12[c+d+4c+4 d+2c+d−c ]

A≑ d−c12

[6 c+6d ]

A≑ c (d−c )2

+d (d−c )2

A≑ cd−c2

2+ d

2−cd2

18

Alexandra Mutch

A≑ d2−c2

2

19

Alexandra Mutch

y = x2

Definite Integral

∫c

d

x2 . dx

¿ [ x33 +k ]c

d

¿ [d33 +k ]−[ c33 +k ]¿ d

3−c3

3

Simpson’s Rule

n = 4 strips


w=d−c4

y-values of strips

M1, M 1=(c+ d−c4 )

2

M 1=c2+2( cd−c2

4 )+ (d−c )2

42

M 1=c2+ cd−c2

2+ d

2−2cd+c2

16

M 1=16c2

16+ 8cd−8c

2

16+ d

2−2cd+c2

16

20

Strip x yE1 c c2

M1 c+ d−c4

9c2+d2+6cd16

M2 c+ d−c2

c2+d2+2cd4

M3 c+ 3 (d−c )4

c2+9d2+6cd16

E2 d d2

Alexandra Mutch

M 1=9c2+d2+6cd

16

M2, M 2=(c+ d−c2 )

2

M 2=c2+2( cd−c2

2 )+ (d−c )2

22

M 2=c2+cd−c2+ d2−2cd+c2

4

M 2=4 cd4

+ d2−2cd+c2

4

M 2=c2+d2+2cd

4

M3, M 3=(c+3 (d−c )4 )

2

M 3=c2+2(3 cd−3c24 )+ (3d−3c )2

42

M 3=c2+ 3cd−3 c2

2+

(3d−3c )2

42

M 3=16c2

16+ 24cd−24c

2

16+ 9d

2−18cd+9c2

16

M 3=c2+9d2+6cd

16


E = c2+d2


21

Alexandra Mutch

MO = ( 9c2+d2+6cd16 )+( c2+9d2+6cd16 )MO = 10c

2+10d2+12cd16

MO = 5c2+5d2+6cd

8


ME = c2+d2+2cd

4

Simpson’s Rule:

A≑ w3 [E+4MO+2M E ]

A≑( d−c4 )3 [(c2+d2)+4×( 5c

2+5d2+6 cd8 )+2×( c

2+d2+2 cd4 )]

A≑ d−c12 [ 2c2+2d22

+5c2+5d2+6cd

2+c2+d2+2cd

2 ]A≑ d−c

12 [ 8c2+8d2+8cd2 ]A≑ d−c

12[ 4c2+4 d2+4cd ]

A≑ c2d−c3

3+ d

3−c d2

3+ c d

2+c2d3

A≑ d3−c3

3

22

Alexandra Mutch

y = x3

Definite Integral

∫c

d

x3 . dx

¿ [ x44 +k ]c

d

¿ [ d44 +k ]−[ c44 +k ]¿ d

4−c4

4

Simpson’s Rulen = 4 strips


w=d−c4

y-values of strips

M1, M 1=(c+ d−c4 )

3

M 1=( 9 c2+d2+6cd16 )×(c+ d−c4 )

M 1=9c3+cd2+6c2d

16+ 9c

2d−9 c3+d3−cd2+6c d2−6c2d64

M 1=36c3+4cd2+24 c2d

64+−9c3+d3−cd2+3c2d+5c d2

64

M 1=27c3+d3+27c2d+9c d2

64

23

Strip x yE1 c c3

M1 c+ d−c4

27c3+d3+27c2d+9c d2

64

M2 c+ d−c2

c3+d3+3 c2d+3c d2

8

M3 c+ 3 (d−c )4

c3+27d3+9c2d+27c d2

64E2 d d3

Alexandra Mutch

M2, M 2=(c+ d−c2 )

3

M 2=( c2+d2+2cd4 )×(c+ d−c2 )

M 2=c3+dc2+2c2d

4+ c

2d−c3+d3−c d2+2cd2−2c2d8

M 2=2c3+2dc2+4 c2d

8+−c3+d3−c2d+c d2

8

M 2=c3+d3+3c2d+3 c d2

8

M3, M 3=(c+3 (d−c )4 )

3

M 3=( c2+9 d2+6cd16 )×(c+ 3 (d−c )4 )

M 3=c3+9c d2+6 c2d

16+ 3c

2d−3 c3+27d3−27cd2+18c d2−18c2d64

M 3=4 c3+24 cd2+36c2d

64+−3c3+27d3−15c2d2−9c d2

64

M 3=c3+27d3+9c2d+27 cd2

64


E = c3+d3


24

Alexandra Mutch

MO = ( 27c3+d3+27c2d+9cd2

64 )+( c3+27 d3+9c2d+27c d2

64 )MO = 28 c

3+28d3+36c2d+36cd2

64

MO = 7c3+7 d3+9c2d+9c d2

16


ME = c3+d3+3 c2d+3c d2

8

Simpson’s Rule:

A≑ w3 [E+4MO+2M E ]

A≑( d−c4 )3 [(c3+d3)+4×( 7c

3+7d3+9c2d+9 c d2

16 )+2×( c3+d3+3c2d+3c d2

8 )]A≑ d−c

12 [c3+d3+ 7c3+7d3+9c2d+9 cd24+c3+d3+3c2d+3cd2

4 ]A≑ d−c

12 [ 4 c3+4 d34+7 c3+7 d3+9c2d+9c d2

4+c3+d3+3c2d+3c d2

4 ]A≑ d−c

12 [ 12c3+12d3+12d+12c d24 ]A≑ d−c

12[3c3+3d3+3d+3cd2 ]

A≑ c3d−c4+d4−c d3+c2d2−c3d+c d3−c2d2

4

A≑ d4−c4

4

25

Alexandra Mutch

DiscussionThis shows that Simpson’s Rule is 100% accurate for polynomials of degree 3 or less because the general form for the approximated area was equal to the general form for the definite integral in each of the three cases.

26

Alexandra Mutch

Task 9

It is proposed that a level road be constructed roughly North-South for the entire length of the map shown below. Justification of the selection of the road path will be outlined below along with mathematical reasoning for the optimum height of the road to minimise filling and removal costs.

Fig. 2 - Map of Area for Proposed Road

There is no way of calculating the exact height of the mountain between the topographical lines. Therefore it is assumed that the gradient of each of the areas between topographical lines is constant. The actual slope of the mountain would not be constant between the topographical lines, but rough and uneven, so any further calculations should be used as estimation only.

27

Alexandra Mutch

Fig. 3 - Selected Road Path

The road path (indicated in Fig. 3 in blue) was selected because it does not intersect the Dark Sea and does not go over the peaks of the mountain. It also follows relatively gradual mountain slopes which means dirt filling/removal would be easier (and probably cost less) and there would be less chance of landslide etc. during the dirt moving process.

The blue path would be a better choice than the red path as the red one goes right over the peak of the mountain. This means more dirt would need to be moved in order to create a level road which would be less cost effective. In addition, the red path crosses the river at less of an angle which means it will cover a greater area of the river than the blue path. This means the error (due to the requirement of a bridge) will be greater and any calculations would be less accurate.

The optimum height for the selected road will be determined by:

Calculating the cross sectional area underneath the selected road path. Multiplying the area by one ‘road width’ to find the total volume of dirt on the road path Calculating the point of equilibrium between the height of peaks and valleys of the

mountain

It is assumed that any rivers crossed by the road do not affect the height of the mountain at their locations. In reality a bridge would be built over the water or the river would be excavated further downward and out of the way of the road. This would mean less dirt would be required to fill the valley at the location of the river. Therefore calculations of the volume of dirt to be moved will be slightly more than the actual volume required.

28

Alexandra Mutch

Cross Sectional AreaFirst the height of the mountain at particular distances along the road (from the Northern border) will be measured and graphed. The distance from the Northern border of the topographical lined was measured to find the height of the mountain at each point.

Height of Mountain

Measured

Distance (mm)

Converted

Distance (km)

Height (km)

0 0 0.27.5 0.75 0.2

14.0 1.40 0.330.0 3.00 0.334.0 3.40 0.262.5 6.25 0.176.0 7.60 0.182.0 8.20 0.2

107.5 10.75 0.3131.0 13.10 0.3

It is assumed that the height of the mountain at the Northern and Southern edges of the map is equal to the nearest measurement. For example, the nearest measurement to the Northern border is height = 0.3km at 0.75km from the border. Therefore it is assumed that the height of the mountain at the Northern border is 0.3km and the height at the Southern border is 0.3km.

0 2 4 6 8 10 12 140

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

Graph 1 - Height vs Distance - Cross Section of Path

Distance from Northern Border (km)

Hei

ght

abov

e Se

a Le

vel (

km

)

29

Alexandra Mutch

In order to calculate the total cross sectional area underneath the road path, it has been sectioned into separate rectangles and triangles (shown in Fig. 4) which will be calculated separately and then summated.

0 2 4 6 8 10 12 140

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

Fig. 4 - Areas to be Calculated

A1

For areas 1 – 5, rectangle area calculations were used and for areas 6 – 10 triangle area calculations were used.

Calculation of A1

A=L×W

A=13.1×0.1

A=1.31 km2

The same process was used to find areas 2 – 5.

Calculation of A6

A=0.5b×h

A=0.5×(6.25−3.4)×0.1

A=0.1425k m2

The same process was used to find areas 7 – 10.

30

A2 A3

A4 A5

A6

A9

A8

A7

A

10

Alexandra Mutch

The areas were calculated to be:

SectionArea

(km2)A1 1.3100A2 0.3400A3 0.4900A4 0.1600A5 0.2350A6 0.1425A7 0.0300A8 0.0325A9 0.0200A10 0.1275Total 2.8875

The total cross sectional area under the road path was calculated to be 2.8875km2.

Volume of DirtIn order to calculate the volume of dirt under the selected road path, the cross sectional area underneath the path will be multiplied by one ‘road width’, or 11m. It is assumed that the road will consist of two lanes (one for each direction) each of width 3.5m and a 2m shoulder on either side of the lanes, totalling 11m.

It is assumed that the volume of dirt under the selected road path is a prism. Due to the small scale of the map, the height of the mountain one road width (0.11mm on the scaled diagram) from the original measurements would be the same height. The actual mountain would be uneven so calculations will be incorrect. However, due to the large size of the mountain relative to the road, this error is negligible.

Volume of dirt underneath road path:

V=Area×Width

V=2.8875×0.011

V=0.0317625 k m3

Height Optimisation The optimum height for the road is a height where all dirt taken off peaks can fill the valleys to create a flat, level road. This minimises dirt filling and removal costs.

31

Alexandra Mutch

It is assumed that there is no height restriction for the edges of the map. I.e. the height of the road does not have to match up with another road beyond the limits of the map.

Optimum Height Calculation

The volume of dirt must be moved to create a flat surface on the top. This will result in one rectangular prism of dirt or a rectangle of length 13.1km when the cross section is viewed. Therefore the optimum height will be the area divided by the length of the road.

Height= AreaLength

Height=2.887513.1

Height ≈0.2204 km

Therefore the road should be constructed at a height of approximately 220.4m above sea level to minimise filling and removal costs.

DiscussionDue to the assumptions made, the calculated cross sectional area, volume of dirt and optimum height are not completely accurate.

The assumption that the gradient of each of the areas between topographical lines is constant creates a large margin for error. The area between topographical lines has an error of ±50m

which is 16 of the largest height measurement used in calculations. If all of the assumed heights

of areas between topographical lines were out by +50m then this would add approximately 0.655km2 to the calculated cross sectional area or about 22.69%. This is a relatively large percentage of error and the calculations should be taken as a rough guide only.

32

alexandramutch.files.wordpress.com€¦ · web viewthere is no way of calculating the exact...

Documents