SOLUCIONARIO CAPITULO 17 CHAPRA PDF

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Solution, manual solution, metodos numericos, Chapra,chapter 17...

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1

CHAPTER 17 17.1 The data can be tabulated as i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

(a) y (b) s y 2

(c) s y

y 8.8 9.4 10 9.8 10.1 9.5 10.1 10.4 9.5 9.5 9.8 9.2 7.9 8.9 9.6 9.4 11.3 10.4 8.8 10.2 10 9.4 9.8 10.6 8.9 241.3

(yi – y )2 0.725904 0.063504 0.121104 0.021904 0.200704 0.023104 0.200704 0.559504 0.023104 0.023104 0.021904 0.204304 3.069504 0.565504 0.002704 0.063504 2.715904 0.559504 0.725904 0.300304 0.121104 0.063504 0.021904 0.898704 0.565504 11.8624

241.3 9.652 25 11.8624 0.703041 25 1 0.7030412 0.494267

0.703041 (d) c.v. 100% 7.28% 9.652 (e) t0.05/2,25–1 = 2.063899 L 9.652 0.703041 2.063899 9.361799 25 U 9.652 0.703041 2.063899 9.942201 25

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2 17.2 The data can be sorted and then grouped. We assume that if a number falls on the border between bins, it is placed in the lower bin. lower 7.5 8 8.5 9 9.5 10 10.5 11

upper 8 8.5 9 9.5 10 10.5 11 11.5

Frequency 1 0 4 7 6 5 1 1

The histogram can then be constructed as

Frequency

8 6 4 2 0 7

8

9

10

11

12

Bin

17.3 The data can be tabulated as i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

y 28.65 28.65 27.65 29.25 26.55 29.65 28.45 27.65 26.65 27.85 28.65 28.65 27.65 27.05 28.45 27.65 27.35 28.25 31.65

(yi – y )2 0.390625 0.390625 0.140625 1.500625 2.175625 2.640625 0.180625 0.140625 1.890625 0.030625 0.390625 0.390625 0.140625 0.950625 0.180625 0.140625 0.455625 0.050625 13.14063

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3 20 21 22 23 24 25 26 27 28

(a) y (b) s y 2

(c) s y

28.55 28.35 28.85 26.35 27.65 26.85 26.75 27.75 27.25 784.7

0.275625 0.105625 0.680625 2.805625 0.140625 1.380625 1.625625 0.075625 0.600625 33.0125

784.7 28.025 28 33.0125 1.105751 28 1 1.1057512 1.222685

1.105751 (d) c.v. 100% 3.95% 28.025 (e) t0.1/2,28–1 = 1.703288 L 28.025 1.105751 1.703288 27.66907 28 1 . 105751 U 28.025 1.703288 28.38093 28 (f) The data can be sorted and grouped. Lower 26 26.5 27 27.5 28 28.5 29 29.5 30 30.5 31 31.5

Upper 26.5 27 27.5 28 28.5 29 29.5 30 30.5 31 31.5 32

Frequency 1 4 3 7 4 6 1 1 0 0 0 1

The histogram can then be constructed as

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4 8 7 Frequency

6 5 4 3 2 1 0 26

27

28

29

30

31

32

Bin

(g) 68% of the readings should fall between y s y and y s y . That is, between 28.025 – 1.10575096 = 26.919249 and 28.025 + 1.10575096 = 29.130751. Twenty values fall between these bounds which is equal to 20/28 = 71.4% of the values which is not that far from 68%. 17.4 The results can be summarized as y versus x y = 4.851535 + 0.35247x

Best fit equation Standard error Correlation coefficient

x versus y x = 9.96763 + 2.374101y 2.764026 0.914767

1.06501 0.914767

We can also plot both lines on the same graph

y

12 8 y y versus x x versus y

4

x

0 0

5

10

15

20

Thus, the “best” fit lines and the standard errors differ. This makes sense because different errors are being minimized depending on our choice of the dependent (ordinate) and independent (abscissa) variables. In contrast, the correlation coefficients are identical since the same amount of uncertainty is explained regardless of how the points are plotted. 17.5 The results can be summarized as y 31.0589 0.78055 x

( s /y

x

4.476306; r 0.901489)

At x = 10, the best fit equation gives 23.2543. The line and data can be plotted along with the point (10, 10).

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5

35 30 25 20 15 10 5 0 0

10

20

30

40

The value of 10 is nearly 3 times the standard error away from the line, 23.2543 – 3(4.476306) = 9.824516 Thus, we can tentatively conclude that the value is probably erroneous. It should be noted that the field of statistics provides related but more rigorous methods to assess whether such points are “outliers.” 17.6 The sum of the squares of the residuals for this case can be written as n

Sr

yi

a1 xi

2

i 1

The partial derivative of this function with respect to the single parameter a1 can be determined as Sr a1

(y i

2

a 1x i )x i

Setting the derivative equal to zero and evaluating the summations gives 0

yi xi

a1

xi

which can be solved for a1

y i xi 2

xi

So the slope that minimizes the sum of the squares of the residuals for a straight line with a zero intercept is merely the ratio of the sum of the dependent variables (y) times the sum of the independent variables (x) over the sum of the independent variables squared (x2). Application to the data gives x 2 4

y 1 2

xy 2 8

x2 4 16

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6 6 7 10 11 14 17 20

5 2 8 7 6 9 12

30 14 80 77 84 153 240 688

36 49 100 121 196 289 400 1211

Therefore, the slope can be computed as 688/1211 = 0.5681. The fit along with the data can be displayed as

y = 0.5681x

12

2

R = 0.8407 8 4 0 0

5

10

15

20

17.7 (a) The results can be summarized as y

2.01389 1.458333 x

16

( s /y

1.306653; r 0.956222)

x

y = 1.4583x - 2.0139 R2 = 0.9144

12 8 4 0 0

2

4

6

8

As can be seen, although the correlation coefficient appears to be close to 1, the straight line does not describe the data trend very well. (b) The results can be summarized as y 1.488095 0.45184 x 0.191017 x2

( s y/

x

0.344771; r 0.997441)

A plot indicates that the quadratic fit does a much better job of fitting the data.

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7

16

2

y = 0.191x - 0.4518x + 1.4881 R2 = 0.9949

12 8 4 0 0

2

4

6

8

17.8 (a) We regress 1/y versus 1/x to give 1 y

0.34154 0.36932

1 x

Therefore, 3 = 1/0.34154 = 2.927913 and saturation-growth-rate model is y 2.927913

3

= 0.36932(2.927913) = 1.081337, and the

x 1.081337 x

The model and the data can be plotted as 3 2 1 0 0

3

6

9

(b) We regress log10(y) versus log10(x) to give log 10 y

0.153296 0.311422 log10 x

Therefore,

2

= 100.153296 = 1.423297 and

2

= 0.311422, and the power model is

y 1.423297x 0.311422 The model and the data can be plotted as

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8

3 2

y = 1.4233x0.3114 R2 = 0.9355

1 0 0

3

6

9

(c) Polynomial regression can be applied to develop a best-fit parabola 0.03069 x2

y

0.449901 x 0.990728

The model and the data can be plotted as 3 2 y = -0.0307x2 + 0.4499x + 0.9907 R2 = 0.9373

1 0 0

3

6

9

17.9 We regress log10(y) versus log10(x) to give log 10 y 1.325225 0.54029 log10 x Therefore,

2

= 101.325225 = 21.14583 and

y 21.14583 x

2

= 0.54029, and the power model is

0. 54029

The model and the data can be plotted as 14 12 10 8 6 4 2 0

-0.5403

y = 21.146x 2 R = 0.9951

0

5

10

15

20

The model can be used to predict a value of 21.14583(9) 0.54029 = 6.451453. 17.10 We regress ln(y) versus x to give PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

9

ln y

6.303701 0.818651x

Therefore,

1

= e6.303701 = 546.5909 and

1

= 0.818651, and the exponential model is

x y 546.5909 e 0.818651

The model and the data can be plotted as 0.8187x

y = 546.59e R2 = 0.9933

4000 3000 2000 1000 0 0

0.5

1

1.5

2

2.5

A semi-log plot can be developed by plotting the natural log versus x. As expected, both the data and the best-fit line are linear when plotted in this way. 8.5 8 7.5 7 6.5 6 0

0.5

1

1.5

2

2.5

17.11 For the data from Prob. 17.10, we regress log10(y) versus x to give log10 y

2.737662 0.355536 x

Therefore,

5

= 102.737662 = 546.5909 and

5

= 0.355536, and the base-10 exponential model is

y 546.5909 10 0.355536x The model and the data can be plotted as

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10

5000 4000 3000 2000 1000 0 0

0.5

1

1.5

2

2.5

This plot is identical to the graph that was generated with the base-e model derived in Prob. 17.10. Thus, although the models have a different base, they yield identical results. The relationship between 1t

e

10

1

and

5

can be developed as in

5t

Take the natural log of this equation to yield 1t

5t

ln 10

or 1

2.302585

5

This result can be verified by substituting the value of 1

5

into this equation to give

2.302585(0.355536) 0.818651

This is identical to the result derived in Prob. 17.10. 17.12 The function can be linearized by dividing it by x and taking the natural logarithm to yield ln( y / x) ln

4

4x

Therefore, if the model holds, a plot of ln(y/x) versus x should yield a straight line with an intercept of ln 4 and a slope of 4. x 0.1 0.2 0.4 0.6 0.9 1.3 1.5 1.7

y 0.75 1.25 1.45 1.25 0.85 0.55 0.35 0.28

ln(y/x) 2.014903 1.832581 1.287854 0.733969 -0.05716 -0.8602 -1.45529 -1.80359

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11 1.8

0.18

y = -2.4733x + 2.2682

3 2 1 0 -1 -2 -3

R 2 = 0.9974

0

0.5

Therefore,

4

1

= 2.4733 and

1.5

4

2

= e2.2682 = 9.661786, and the fit is

2.4733 x

9.661786xe

y

-2.30259

This equation can be plotted together with the data: 2

1

0 0

0.5

1

1.5

2

17.13 The equation can be linearized by inverting it to yield 1 k

cs k max

1 c2

1 kmax

Consequently, a plot of 1/k versus 1/c should yield a straight line with an intercept of 1/kmax and a slope of cs/kmax c, mg/L 0.5 0.8 1.5 2.5 4

k, /d 1.1 2.4 5.3 7.6 8.9 Sum

1/c2 4.000000 1.562500 0.444444 0.160000 0.062500 6.229444

1/k 0.909091 0.416667 0.188679 0.131579 0.112360 1.758375

1/c2 1/k 3.636364 0.651042 0.083857 0.021053 0.007022 4.399338

(1/c2)2 16.000000 2.441406 0.197531 0.025600 0.003906 18.66844

The slope and the intercept can be computed as

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12 5(4.399338) 6.229444(1.758375)

a1

0.202489

5(18.66844) (6.229444) 2 1.758375 5

a0

0.202489

6.229444 5

0.099396

Therefore, kmax = 1/0.099396 = 10.06074 and cs = 10.06074(0.202489) = 2.037189, and the fit is 10.06074 c 2 2.037189 c 2

k

This equation can be plotted together with the data: 10 8 6 4 2 0 0

1

2

3

4

5

The equation can be used to compute k

10.06074(2) 2 2.037189 (2) 2

6.666

17.14 (a) We regress y versus x to give y

20.6

0.494545x

The model and the data can be plotted as 50 40 30 y = 0.4945x + 20.6 R2 = 0.8385

20 10 0 0

10

20

30

40

50

60

(b) We regress log10y versus log10x to give PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

13

log 10 y

0.99795 0.385077 log10 x

Therefore,

2

= 100.99795 = 9.952915 and

2

= 0.385077, and the power model is

y 9.952915 x 0.385077 The model and the data can be plotted as 50 40 30

0.3851

y = 9.9529x

20

2

R = 0.9553

10 0 0

10

20

30

40

50

60

(c) We regress 1/y versus 1/x to give 1 y

0.019963 0.197464

1 x

Therefore, 3 = 1/0.01996322 = 50.09212 and saturation-growth-rate model is y 50.09212

x 9.89137

3

= 0.19746357(50.09212) = 9.89137, and the

x

The model and the data can be plotted as 50 40 30

y = 50.092

20

x x + 9.891369

R2 = 0.98919

10 0 0

10

20

30

40

50

60

(d) We employ polynomial regression to fit a parabola y

0.01606 x2

1.377879 x 11.76667

The model and the data can be plotted as PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

14

50 40 30 2

20

y = -0.0161x + 1.3779x + 11.767

10

R2 = 0.98

0 0

10

20