Numerical Method for engineers-chapter 17

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 Σ

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 (b) s y = = 0.703041 25 − 1 2 2 (c) s y = 0.703041 = 0.494267 (a) y =

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

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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 Σ

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 (b) s y = = 1.105751 28 − 1 2 2 (c) s y = 1.105751 = 1.222685 (a) y =

1.105751 × 100% = 3.95% 28.025 (e) t0.1/2,28–1 = 1.703288 1.105751 L = 28.025 − 1.703288 = 27.66907 28 1.105751 U = 28.025 + 1.703288 = 28.38093 28 (f) The data can be sorted and grouped. (d) c.v. =

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 Best fit equation Standard error Correlation coefficient

y versus x y = 4.851535 + 0.35247x 1.06501 0.914767

x versus y x = −9.96763 + 2.374101y 2.764026 0.914767

We can also plot both lines on the same graph

y

12 8 y y versus x x versus y

4 0 0

5

10

15

x 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 Sr =

n

∑( y

i

− a1 xi )

2

i =1

The partial derivative of this function with respect to the single parameter a1 can be determined as ∂S r = −2 ∂a1

∑ [( y

i

− a1 xi ) xi ]

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

∑y x

i i

− a1

∑x

2 i

which can be solved for a1 =

∑y x ∑x

i i 2 i

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

R2 = 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 / x = 1.306653; r = 0.956222)

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 x 2

( 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

y = 0.191x2 - 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 1 = 0.34154 + 0.36932 y x Therefore, α3 = 1/0.34154 = 2.927913 and β3 = 0.36932(2.927913) = 1.081337, and the saturation-growth-rate model is y = 2.927913

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 log10 y = 0.153296 + 0.311422 log10 x Therefore, α2 = 100.153296 = 1.423297 and β2 = 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 = 1.4233x

1

0.3114

2

R = 0.9355

0 0

3

6

9

(c) Polynomial regression can be applied to develop a best-fit parabola y = −0.03069 x 2 + 0.449901x + 0.990728 The model and the data can be plotted as 3 2 y = -0.0307x2 + 0.4499x + 0.9907

1

2

R = 0.9373 0 0

3

6

9

17.9 We regress log10(y) versus log10(x) to give log10 y = 1.325225 − 0.54029 log10 x Therefore, α2 = 101.325225 = 21.14583 and β2 = −0.54029, and the power model is y = 21.14583x −0.54029 The model and the data can be plotted as 14 12 10 8 6 4 2 0

y = 21.146x -0.5403 R2 = 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, α1 = e6.303701 = 546.5909 and β1 = 0.818651, and the exponential model is y = 546.5909e 0.818651 x The model and the data can be plotted as y = 546.59e0.8187x

4000

R2 = 0.9933

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, α5 = 102.737662 = 546.5909 and β5 = 0.355536, and the base-10 exponential model is y = 546.5909 × 10 0.355536 x 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 β1 and β5 can be developed as in e − β1t = 10 − β 5t Take the natural log of this equation to yield − β 1t = − β 5 t ln 10 or

β 1 = 2.302585β 5 This result can be verified by substituting the value of β5 into this equation to give

β 1 = 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 α 4 + β 4 x Therefore, if the model holds, a plot of ln(y/x) versus x should yield a straight line with an intercept of lnα4 and a slope of β4. 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

-2.30259 y = -2.4733x + 2.2682

3 2 1 0 -1 -2 -3

2

R = 0.9974

0

0.5

1

1.5

2

Therefore, β4 = −2.4733 and α4 = e2.2682 = 9.661786, and the fit is y = 9.661786 xe −2.4733 x 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 c 1 1 1 = s 2+ k k max c k max 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

a1 =

5( 4.399338) − 6.229444(1.758375) = 0.202489 5(18.66844) − (6.229444) 2

a0 =

1.758375 6.229444 − 0.202489 = 0.099396 5 5

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

10.06074c 2 2.037189 + c 2

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 = 6.666 2.037189 + (2) 2

17.14 (a) We regress y versus x to give y = 20.6 + 0.494545 x The model and the data can be plotted as 50 40 30 20

y = 0.4945x + 20.6 R2 = 0.8385

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

log10 y = 0.99795 + 0.385077 log10 x Therefore, α2 = 100.99795 = 9.952915 and β2 = 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

y = 9.9529x 0.3851

20

R2 = 0.9553

10 0 0

10

20

30

40

50

60

(c) We regress 1/y versus 1/x to give 1 1 = 0.019963 + 0.197464 y x Therefore, α3 = 1/0.01996322 = 50.09212 and β3 = 0.19746357(50.09212) = 9.89137, and the saturation-growth-rate model is y = 50.09212

x 9.89137 + x

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

y = 50.092

20

R2

10

x x + 9.891369

= 0.98919

0 0

10

20

30

40

50

60

(d) We employ polynomial regression to fit a parabola y = −0.01606 x 2 + 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

R = 0.98

2

0 0

10

20

30

40

50

60

Comparison of fits: The linear fit is obviously inadequate. Although the power fit follows the general trend of the data, it is also inadequate because (1) the residuals do not appear to be randomly distributed around the best fit line and (2) it has a lower r2 than the saturation and parabolic models. The best fits are for the saturation-growth-rate and the parabolic models. They both have randomly distributed residuals and they have similar high coefficients of determination. The saturation model has a slightly higher r2. Although the difference is probably not statistically significant, in the absence of additional information, we can conclude that the saturation model represents the best fit. 17.15 We employ polynomial regression to fit a cubic equation to the data y = 0.046676 x 3 − 1.04121x 2 + 7.143817 x − 11.4887

(r 2 = 0.828981; s y / x = 0.570031)

The model and the data can be plotted as 6

y = 0.0467x3 - 1.0412x2 + 7.1438x - 11.489

5

R2 = 0.829

4 3 2 1 0 0

2

4

6

8

10

12

14

17.16 We employ multiple linear regression to fit the following equation to the data y = 14.46087 + 9.025217 x1 − 5.70435 x 2

(r 2 = 0.995523; s y / x = 0.888787)

The model and the data can be compared graphically by plotting the model predictions versus the data. A 1:1 line is included to indicate a perfect fit.

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15 50

model

40 30 20 1 1:

10 0 0

10

20

30

40

50

data

17.17 We employ multiple linear regression to fit the following equation to the data ( r 2 = 0.958333; s y / x = 1.414214)

y = 14.66667 − 6.66667 x1 + 2.333333 x 2

The model and the data can be compared graphically by plotting the model predictions versus the data. A 1:1 line is included to indicate a perfect fit. 25

model

20 15 10 1 1:

5 0 0

5

10

15

20

25

data

17.18 We can employ nonlinear regression to fit a parabola to the data. A simple way to do this is to use the Excel Solver to minimize the sum of the squares of the residuals as in the following worksheet,

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16

The formulas are

Thus, the best-fit equation is y = 674.001x 2 − 233.945 x + 604.087 The model and the data can be displayed graphically as 4000 3000 2000 1000 0 0

0.5

1

1.5

2

2.5

Note that if polynomial regression were used, a slightly different fit would result, y = 674.007 x 2 − 233.961x + 604.094 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.

17 17.19 We can employ nonlinear regression to fit the saturation-growth-rate equation to the data from Prob. 17.14. A simple way to do this is to use the Excel Solver to minimize the sum of the squares of the residuals as in the following worksheet,

The formulas are

Thus, the best-fit equation is y = 50.529

x 10.129 + x

The model and the data can be displayed graphically as

50 40 30 20 10 0 0

10

20

30

40

50

60

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18

Recall that for Prob. 17.14c, a slightly different fit resulted, y = 50.09212

x 9.89137 + x

17.20 MATLAB provides a very nice environment for solving this problem: (a) Prob. 17.4: First, we can enter the data >> X=[0 2 4 6 9 11 12 15 17 19]'; >> Y=[5 6 7 6 9 8 7 10 12 12]';

Then, we can create the Z matrix which consists of a column of ones and a second column of the x’s. >> Z=[ones(size(X)) X] Z =

1 1 1 1 1 1 1 1 1 1

0 2 4 6 9 11 12 15 17 19

Next we can develop the coefficients of the normal equations as >> ZTZ=Z'*Z ZTZ =

10 95

95 1277

We can compute the right-hand side of the normal equations with >> ZTY=Z'*Y ZTY = 82 911

We can then determine the coefficients for the linear regression as >> A=inv(ZTZ)*ZTY

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.

19 A =

4.8515 0.3525

This result is identical to that obtained in Prob. 17.4. Next, we can determine the r2 and sy/x, >> Sr=sum((Y-Z*A).^2) Sr =

9.0740

>> r2=1-Sr/sum((Y-mean(Y)).^2) r2 = 0.8368 >> syx=sqrt(Sr/(length(X)-length(A))) syx = 1.0650

In order to determine the confidence intervals we can first calculate the inverse of [Z]T[Z] as >> ZTZI=inv(ZTZ) ZTZI = 0.3410 -0.0254

-0.0254 0.0027

The standard errors of the coefficients can be computed as >> sa0=sqrt(ZTZI(1,1)*syx^2) sa0 = 0.6219 >> sa1=sqrt(ZTZI(2,2)*syx^2) sa1 = 0.0550

The t statistic can be determined as TINV(0.1, 10 – 2) = 1.8595. We can then compute the confidence intervals as >> >> >> >>

a0min=A(1)-1.8595*sa0; a0max=A(1)+1.8595*sa0; a1min=A(2)-1.8595*sa1; a1max=A(2)+1.8595*sa1;

which yields the confidence intervals for a0 and a1 as [3.6951, 6.0080] and [0.2501, 0.4548], respectively. (b) Prob. 17.15: 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.

20

First, we can determine the coefficients >> >> >> >> >> >>

X=[3 4 5 7 8 9 11 12]'; Y=[1.6 3.6 4.4 3.4 2.2 2.8 3.8 4.6]'; Z=[ones(size(X)) X X.^2 X.^3]; ZTZ=Z'*Z; ZTY=Z'*Y; A=inv(ZTZ)*ZTY

A = -11.4887 7.1438 -1.0412 0.0467

The standard error can be computed as >> Sr=sum((Y-Z*A).^2); >> syx=sqrt(Sr/(length(Y)-length(A))) syx = 0.5700

The standard errors of the coefficients can be computed as >> ZTZI=inv(ZTZ) ZTZI = 49.3468 -23.4771 3.2960 -0.1412

-23.4771 11.4162 -1.6270 0.0705

3.2960 -1.6270 0.2349 -0.0103

-0.1412 0.0705 -0.0103 0.0005

>> sa0=sqrt(ZTZI(1,1)*syx^2) sa0 = 4.0043 >> sa1=sqrt(ZTZI(2,2)*syx^2) sa1 = 1.9260 >> sa2=sqrt(ZTZI(3,3)*syx^2) sa2 = 0.2763 >> sa3=sqrt(ZTZI(4,4)*syx^2) sa3 = 0.0121 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.

21

The t statistic can be determined as TINV(0.1, 8 – 4) = 2.13185. We can then compute the confidence intervals for a0, a1, a2, and a3 as [–20.0253, –2.9521], [3.0379, 11.2498], [–1.6302, –0.45219], and [0.02078, 0.072569], respectively. 17.21 Here’s VBA code to implement linear regression: Option Explicit Sub Regres() Dim n As Integer Dim x(20) As Double, y(20) As Double, a1 As Double, a0 As Double Dim syx As Double, r2 As Double n = 7 x(1) = 1: x(2) = 2: x(3) = 3: x(4) = 4: x(5) = 5 x(6) = 6: x(7) = 7 y(1) = 0.5: y(2) = 2.5: y(3) = 2: y(4) = 4: y(5) = 3.5 y(6) = 6: y(7) = 5.5 Call Linreg(x, y, n, a1, a0, syx, r2) MsgBox "slope= " & a1 MsgBox "intercept= " & a0 MsgBox "standard error= " & syx MsgBox "coefficient of determination= " & r2 MsgBox "correlation coefficient= " & Sqr(r2) End Sub Sub Linreg(x, y, n, a1, a0, syx, r2) Dim i As Integer Dim sumx As Double, sumy As Double, sumxy As Double Dim sumx2 As Double, st As Double, sr As Double Dim xm As Double, ym As Double sumx = 0 sumy = 0 sumxy = 0 sumx2 = 0 st = 0 sr = 0 'determine summations for regression For i = 1 To n sumx = sumx + x(i) sumy = sumy + y(i) sumxy = sumxy + x(i) * y(i) sumx2 = sumx2 + x(i) ^ 2 Next i 'determine means xm = sumx / n ym = sumy / n determine coefficients a1 = (n * sumxy - sumx * sumy) / (n * sumx2 - sumx * sumx) a0 = ym - a1 * xm 'determine standard error and coefficient of determination For i = 1 To n st = st + (y(i) - ym) ^ 2 sr = sr + (y(i) - a1 * x(i) - a0) ^ 2 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.

22 Next i syx = (sr / (n - 2)) ^ 0.5 r2 = (st - sr) / st End Sub

17.22 A log-log plot of stress versus N suggests a linear relationship. 10000

1000

100 1

100

10000

1000000

We regress log10(stress) versus log10(N) to give log10 (stress) = 3.075442 − 0.06943 log10 N Therefore, α2 = 103.075442 = 1189.711 and β2 = –0.06943, and the power model is stress = 1189.711N −0.06943 The model and the data can be plotted on untransformed scales as 1400 1200

y = 1189.7x-0.0694

1000

R2 = 0.9658

800 600 400 200 0 0

200000

400000

600000

800000

1000000 1200000

17.23 A log-log plot of µ versus T suggests a linear relationship.

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.

23

10 1 0.1 10

100

1000

0.01 0.001 0.0001

We regress log10µ versus log10T to give log10 µ = 4.581471 − 3.01338 log10 T

( r 2 = 0.975703)

Therefore, α2 = 104.581471 = 38,147.94 and β2 = –3.01338, and the power model is

µ = 38,147.94T −3.01338 The model and the data can be plotted on untransformed scales as 2.5

-3.0134

y = 38148x

2

2

R = 0.9757

1.5 1 0.5 0 0

50

100

150

200

250

300

350

17.24 This problem was solved using an Excel spreadsheet and TrendLine. Linear regression gives 200 160 120 80 40 0

y = 5.8x + 60 R2 = 0.9755 0

5

10

15

20

Polynomial regression yields a best-fit parabola

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.

24

y = 0.15067x2 + 2.78661x + 68.03571

200 160 120 80 40 0

R2 = 0.99795

0

5

10

15

20

Exponential model: 200 160 120 80 40 0

y = 67.306e0.0503x R2 = 0.9979 0

5

10

15

20

The linear model is inadequate since it does not capture the curving trend of the data. At face value, the parabolic and exponential models appear to be equally good. However, knowledge of bacterial growth might lead you to choose the exponential model as it is commonly used to simulate the growth of microorganism populations. Interestingly, the choice matters when the models are used for prediction. If the exponential model is used, the result is B = 67.306e 0.0503( 40) = 503.3317 For the parabolic model, the prediction is B = 0.15067t 2 + 2.78661t + 68.03571 = 420.5721 Thus, even though the models would yield very similar results within the data range, they yield dramatically different results for extrapolation outside the range. 17.25 The exponential model is ideal for this problem since (1) it does not yield negative results (as could be the case with a polynomial), and (2) it always decreases with time. Further, it is known that bacterial death is well approximated by the exponential model.

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.

25

2500 y = 1978.6e-0.0532x

2000

R2 = 0.9887

1500 1000 500 0 0

10

20

30

40

50

(a) The model says that the concentration at t = 0 was 1978.6. (b) The time at which the concentration reaches 200 can be computed as 200 = 1978.6e −0.0532 t  200  ln  = −0.0532t  1978.6   200  ln  1978.6  t=  = 43.1 hr − 0.0532 17.26 (a) Linear model 2000

y = 19.47x - 234.29

1500

R2 = 0.8805

1000 500 0 -500

0

20

40

60

80

Although this model does a good job of capturing the trend of the data, it has the disadvantage that it yields a negative intercept. Since this is clearly a physically unrealistic result, another model would be preferable. (b) Power model based on log transformations. We regress log10(F) versus log10(v) to give log10 F = −0.56203 + 1.984176 log10 v −

Therefore, α2 = 10 0.56203 = 0.274137 and β2 = 1.984176, and the power model is F = 0.274137v 1.984176 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.

26

The model and the data can be plotted as y = 0.2741x1.9842

2000

R2 = 0.9481

1500 1000 500 0 0

20

40

60

80

This model represents a superior fit of the data as it fits the data nicely (the r2 is superior to that obtained with the linear model in (a)) while maintaining a physically realistic zero intercept. (c) Power model based on nonlinear regression. We can use the Excel Solver to determine the fit.

The cell formulas are

Therefore, the best-fit model is F = 2.53842v 1.43585 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.

27

The model and the data can be plotted as 2000 1500 1000 500 0 0

20

40

60

80

This model also represents a superior fit of the data as it fits the data nicely while maintaining a physically realistic zero intercept. However, it is very interesting to note that the fit is quite different than that obtained with log transforms in (b). 17.27 We can develop a power equation based on natural logarithms. To do this, we regress ln(F) versus ln(v) to give ln F = −1.29413 + 1.984176 ln v −

Therefore, α2 = e 1.29413 = 0.274137 and β2 = 1.984176, and the power model is F = 0.274137v 1.984176 The model and the data can be plotted as y = 0.2741x1.9842

2000

R2 = 0.9481

1500 1000 500 0 0

20

40

60

80

Note that this result is identical to that obtained with common logarithms in Prob. 17.26(b). Thus, we can conclude that any base logarithm would yield the same power model. 17.28 The sum of the squares of the residuals for this case can be written as Sr =

∑(y n

i =1

i

− a1 xi − a 2 xi2

)

2

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.

28 The partial derivatives of this function with respect to the unknown parameters can be determined as ∂S r = −2 ∂a1

∑ [( y

∂S r = −2 ∂a 2

∑ [( y

]

i

− a1 xi − a 2 xi2 ) xi

i

− a1 xi − a 2 xi2 ) xi2

]

Setting the derivative equal to zero and evaluating the summations gives

( ∑ x )a + (∑ x )a = ∑ x y 2 i

3 i

1

2

i

( ∑ x )a + ( ∑ x )a = ∑ x 3 i

4 i

1

2

i

2 i yi

which can be solved for a1 =

∑x y ∑x −∑x y ∑x ∑ x ∑ x − (∑ x )

a2 =

∑x ∑x y −∑x y ∑x ∑ x ∑ x − (∑ x )

i

4 i

i

2 i

2 i

2 i

2 i

i

i

4 i

3 i

3 2 i

4 i

2 i

i

i

3 i

3 2 i

The model can be tested for the data from Table 12.1. x 10 20 30 40 50 60 70 80 Σ

y 25 70 380 550 610 1220 830 1450

x2 100 400 900 1600 2500 3600 4900 6400 20400

x3 1000 8000 27000 64000 125000 216000 343000 512000 1296000

x4 10000 160000 810000 2560000 6250000 12960000 24010000 40960000 87720000

a1 =

312850(87720000) − 20516500(1296000) = 7.771024 20400(87720000) − (1296000) 2

a2 =

20400( 20516500) − 312850(1296000) = 0.119075 20400(87720000) − (1296000)

xy 250 1400 11400 22000 30500 73200 58100 116000 312850

x2y 2500 28000 342000 880000 1525000 4392000 4067000 9280000 20516500

Therefore, the best-fit model is

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29 y = 7.771024 x + 0.119075 x 2 The fit, along with the original data can be plotted as 2500 2000 1500 1000 500 0 0

20

40

60

80

100

17.29 We can use the Excel Solver to determine the fit.

The cell formulas are

Therefore, the best-fit model is y = 9.8974 xe −2.53187 x 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.

30

2 1.5 1 0.5 0 0

0.5

1

1.5

2

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