numerical method for engineers-chapter 20
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CHAPTER 20
20.1 A plot of log10kversus log10fcan be developed as
y = 0.4224x - 0.83
R2
= 0.9532-0.6
-0.4
-0.2
0
0.2
0.7 0.9 1.1 1.3 1.5 1.7 1.9 2.1 2.3
As shown, the best fit line is
83.0log422363.0log 1010 = kf
Therefore, 2 = 100.83 = 0.147913 and 2 = 0.422363, and the power model is
422363.00.147913xy =
The model and the data can be plotted as
y = 0.1479x0.4224
R2
= 0.9532
0
0.5
1
1.5
0 20 40 60 80 100 120 140 160
20.2 We can first try a linear fit
y = 2.5826x + 1365.9
R2
= 0.9608
1200
1400
1600
1800
-60 -30 0 30 60 90 120
As shown, the fit line is somewhat lacking. Therefore, we can use polynomial regression to
fit a parabola
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y = 0.0128x2
+ 1.8164x + 1331
R2
= 0.9934
1200
1400
1600
1800
-60 -30 0 30 60 90 120
This fit seems adequate in that it captures the general trend of the data. Note that a slightly
better fit can be attained with a cubic polynomial, but the improvement is marginal.
20.3 This problem is ideally suited for Newton interpolation. First, order the points so that they
are as close to and as centered about the unknown as possible.
x0 = 20 f(x0) = 8.17
x1 = 15 f(x1) = 9.03
x2 = 25 f(x2) = 7.46
x3 = 10 f(x3) = 10.1
x4 = 30 f(x4) = 6.85
x5 = 5 f(x5) = 11.3
x6 = 0 f(x6) = 12.9
The results of applying Newtons polynomial at T= 18 are
Order f(x) Error
0 8.17000 0.344
1 8.51400 -0.018
2 8.49600 -0.00336
3 8.49264 0.00022
4 8.49286 -0.00161
5 8.49125 -0.00298
6 8.48827
The minimum error occurs for the third-order version so we conclude that the interpolation is
8.4926.
20.4 We can use MATLAB to solve this problem,
>> format long>> T = [0 5 10 15 20 25 30];>> c = [12.9 11.3 10.1 9.03 8.17 7.46 6.85];>> p = polyfit(T,c,3)
p =-0.00006222222222 0.00652380952381 -0.34111111111111
12.88785714285715
Thus, the best-fit cubic would be
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32 20.000062220.006523810.3411111112.8878571 TTTc +=
We can generate a plot of the data along with the best-fit line as
>> Tp=[0:30];>> cp=polyval(p,Tp);
>> plot(Tp,cp,T,c,'o')
We can use the best-fit equation to generate a prediction at T= 8 as
>> y=polyval(p,8)
y =
10.54463428571429
20.5 The multiple linear regression model to evaluate is
caTaao 210 ++=
As described in Section 17.1, we can use general linear least squares to generate the best-fit
equation. The [Z] and y matrices can be set up using MATLAB commands as
>> format long>> t = [0 5 10 15 20 25 30];>> T = [t t t]';
>> c = [zeros(size(t)) 10*ones(size(t)) 20*ones(size(t))]';>> Z = [ones(size(T)) T c];>> y = [14.6 12.8 11.3 10.1 9.09 8.26 7.56 12.9 11.3 10.1 9.03 8.177.46 6.85 11.4 10.3 8.96 8.08 7.35 6.73 6.2]';
The coefficients can be evaluated as
>> a=inv(Z'*Z)*(Z'*y)
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a =13.52214285714285-0.20123809523809-0.10492857142857
Thus, the best-fit multiple regression model is
cTo 1428570.104928575238090.20123809571428513.5221428 =
We can evaluate the prediction at T= 17 and c = 5 and evaluate the percent relative error as
>> op = a(1)+a(2)*17+a(3)*5
op =9.57645238095238
We can also assess the fit by plotting the model predictions versus the data. A one-to-one
line is included to show how the predictions diverge from a perfect fit.
4
8
12
16
4 8 12 16
The cause for the discrepancy is because the dependence of oxygen concentration on the
unknowns is significantly nonlinear. It should be noted that this is particularly the case for
the dependency on temperature.
20.6 The multiple linear regression model to evaluate is
caTaTaTaao 43
32
210 ++++=
The Excel Data Analysis toolpack provides a convenient tool to fit this model. We can setup a worksheet and implement the Data Analysis Regression tool as
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When the tool is run, the following worksheet is generated
Thus, the best-fit model is
cTTTos 10492857.0000043704.000574444.0336423.0027143.1432 +=
The model can then be used to predict values of oxygen at the same values as the data.
These predictions can be plotted against the data to depict the goodness of fit.
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Thus, although there are some discrepancies, the fit is generally adequate. Finally, the
prediction can be made at T= 20 and c = 10,
19754.8)10(10492857.0)20(000043704.0)20(00574444.0)20(336423.0027143.14 32 =+=sowhich compares favorably with the true value of 8.17 mg/L.
20.7 (a) The linear fit is
y = 1.05897x + 0.81793
R2
= 0.90583
0
20
40
60
80
0 20 40 60 80
The tensile strength at t= 32 can be computed as
34.704891381793.0)32(05897.1 =+=y
(b) A straight line with zero intercept can be fit as
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y = 1.07514x
R2
= 0.90556
0
20
40
60
80
0 20 40 60 80
For this case, the tensile strength at t= 32 can be computed as
34.40452)32(07514.1 ==y
20.8 Linear regression with a zero intercept gives [note that T(K) = T(oC) + 273.15].
y = 29.728x
R2
= 0.9999
0
5000
10000
15000
0 100 200 300 400 500
Thus, the fit is
Tp 728.29=
Using the ideal gas law
n
V
T
pR
=
For our fit
728.29=T
p
For nitrogen,
g/mole28
kg1=n
Therefore,
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8.32428/10
10728.29
3=
=R
This is close to the standard value of 8.314 J/gmole.
20.9 This problem is ideally suited for Newton interpolation. First, order the points so that they
are as close to and as centered about the unknown as possible.
x0 = 740 f(x0) = 0.1406x1 = 760 f(x1) = 0.15509
x2 = 720 f(x2) = 0.12184
x3 = 780 f(x3) = 0.16643
x4 = 700 f(x4) = 0.0977
The results of applying Newtons polynomial at T= 750 are
Order f(x) Error 0 0.14060 0.007245
1 0.14785 0.000534
2 0.14838 -7E-05
3 0.14831 0.00000
4 0.14831
Note that the third-order polynomial yields an exact result, and so we conclude that the
interpolation is 0.14831.
20.10 A program can be written to fit a natural cubic spline to this data and also generate the first
and second derivatives at each knot.
Option Explicit
Sub Splines()Dim i As Integer, n As IntegerDim x(100) As Double, y(100) As Double, xu As Double, yu As DoubleDim dy As Double, d2y As DoubleDim resp As VariantRange("a5").Selectn = ActiveCell.RowSelection.End(xlDown).Selectn = ActiveCell.Row - nRange("a5").SelectFor i = 0 To nx(i) = ActiveCell.Value
ActiveCell.Offset(0, 1).Selecty(i) = ActiveCell.ValueActiveCell.Offset(1, -1).Select
Next iRange("c5").SelectRange("c5:d1005").ClearContentsFor i = 0 To nCall Spline(x(), y(), n, x(i), yu, dy, d2y)ActiveCell.Value = dy
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ActiveCell.Offset(0, 1).SelectActiveCell.Value = d2yActiveCell.Offset(1, -1).Select
Next iDoresp = MsgBox("Do you want to interpolate?", vbYesNo)If resp = vbNo Then Exit Do
xu = InputBox("z = ")Call Spline(x(), y(), n, xu, yu, dy, d2y)MsgBox "For z = " & xu & Chr(13) & "T = " & yu & Chr(13) & _
"dT/dz = " & dy & Chr(13) & "d2T/dz2 = " & d2yLoopEnd Sub
Sub Spline(x, y, n, xu, yu, dy, d2y)Dim e(100) As Double, f(100) As Double, g(100) As Double, r(100) As Double,d2x(100) As DoubleCall Tridiag(x, y, n, e, f, g, r)Call Decomp(e(), f(), g(), n - 1)Call Substit(e(), f(), g(), r(), n - 1, d2x())Call Interpol(x, y, n, d2x(), xu, yu, dy, d2y)End Sub
Sub Tridiag(x, y, n, e, f, g, r)Dim i As Integerf(1) = 2 * (x(2) - x(0))g(1) = x(2) - x(1)r(1) = 6 / (x(2) - x(1)) * (y(2) - y(1))r(1) = r(1) + 6 / (x(1) - x(0)) * (y(0) - y(1))For i = 2 To n - 2e(i) = x(i) - x(i - 1)f(i) = 2 * (x(i + 1) - x(i - 1))g(i) = x(i + 1) - x(i)r(i) = 6 / (x(i + 1) - x(i)) * (y(i + 1) - y(i))r(i) = r(i) + 6 / (x(i) - x(i - 1)) * (y(i - 1) - y(i))
Next ie(n - 1) = x(n - 1) - x(n - 2)
f(n - 1) = 2 * (x(n) - x(n - 2))r(n - 1) = 6 / (x(n) - x(n - 1)) * (y(n) - y(n - 1))r(n - 1) = r(n - 1) + 6 / (x(n - 1) - x(n - 2)) * (y(n - 2) - y(n - 1))End Sub
Sub Interpol(x, y, n, d2x, xu, yu, dy, d2y)Dim i As Integer, flag As IntegerDim c1 As Double, c2 As Double, c3 As Double, c4 As DoubleDim t1 As Double, t2 As Double, t3 As Double, t4 As Doubleflag = 0i = 1DoIf xu >= x(i - 1) And xu
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t3 = -c3t4 = c4dy = t1 + t2 + t3 + t4t1 = 6 * c1 * (x(i) - xu)t2 = 6 * c2 * (xu - x(i - 1))d2y = t1 + t2flag = 1
Elsei = i + 1End IfIf i = n + 1 Or flag = 1 Then Exit Do
LoopIf flag = 0 ThenMsgBox "outside range"End
End IfEnd Sub
Sub Decomp(e, f, g, n)Dim k As IntegerFor k = 2 To ne(k) = e(k) / f(k - 1)
f(k) = f(k) - e(k) * g(k - 1)Next kEnd Sub
Sub Substit(e, f, g, r, n, x)Dim k As IntegerFor k = 2 To nr(k) = r(k) - e(k) * r(k - 1)
Next kx(n) = r(n) / f(n)For k = n - 1 To 1 Step -1x(k) = (r(k) - g(k) * x(k + 1)) / f(k)
Next kEnd Sub
Here is the output when it is applied to the data for this problem:
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The plot suggests a zero second derivative at a little above z= 1.2 m. The program is set up toallow you to evaluate the spline prediction and the associated derivatives for as many cases as
you desire. This can be done by trial-and-error to determine that the zero second derivative
occurs at about z= 1.2315 m. At this point, the first derivative is 73.315 oC/m. This can be
substituted into Fouriers law to compute the flux as
scm
cal01466.0
cm100
m
m
C315.73
Ccms
cal02.0
2=
=J
20.11 This is an example of the saturated-growth-rate model. We can plot 1/[F] versus 1/[B] and
fit a straight line as shown below.
y = 0.008442x + 0.010246
R2
= 0.999932
0
0.02
0.04
0.06
0.08
0.1
0 2 4 6 8 10
The model coefficients can then be computed as
60128.97010246.0
13 ==
823968.0)008442.0(60128.973 ==
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Therefore, the best-fit model is
][823968.0
][60128.97][
F
FB
+=
A plot of the original data along with the best-fit curve can be developed as
0
20
40
60
80
100
120
0 2 4 6 8 10
20.12 Nonlinear regression can be used to estimate the model parameters. This can be done using
the Excel Solver
Here are the formulas:
Therefore, we estimate k01 = 7,526.842 and E1 = 4.478896.
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20.13 Nonlinear regression can be used to estimate the model parameters. This can be done using
the Excel Solver. To do this, we set up columns holding each side of the equation: PV/(RT)
and 1 + A1/V+ A2/V. We then form the square of the difference between the two sides and
minimize this difference by adjusting the parameters.
Here are the formulas that underly the worksheet cells:
Therefore, we estimate A1 = -237.531 and A2 = 1.192283.
20.14 The standard errors can be computed via Eq. 17.9
pn
Ss rxy
=/
n = 15
Model A Model B Model C
Sr 135 105 100
Number of model parameters fit (p) 2 3 5
sy/x 3.222517 2.95804 3.162278
Thus, Model B seems best because its standard error is lower.
20.15 A plot of the natural log of cells versus time indicates two straight lines with a sharp break
at 2. The Excel Trendline tool can be used to fit each range separately with the exponential
model as shown in the second plot.
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-3
-2
-1
0
1
2
0 2 4 6
y = 0.1000e1.1999x
R2
= 1.0000 y = 0.4951e0.4001x
R2
= 1.0000
0
2
4
6
0 2 4 6 8
20.16 This problem can be solved with Excel,
20.17 This problem can be solved with a power model,
21
0aaWHaA =
This equation can be linearized by taking the logarithm
WaHaaA 10210101010 loglogloglog ++=
This model can be fit with the Excel Data Analysis Regression tool,
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The results are
Therefore the best-fit coefficients are a0 = 101.86458 = 0.013659, a1 = 0.5218, and a2 = 0.5190.
Therefore, the model is
5190.05218.00.013659 WHA =
We can assess the fit by plotting the model predictions versus the data. We have included the
perfect 1:1 line for comparison.
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1.9
2.1
2.3
1.9 2.1 2.3
The prediction at H= 187 cm and W= 78 kg can be computed as
008646.2)78(87)0.013659(1 5190.05218.0 ==A
20.18 The Excel Trend Line tool can be used to fit a power law to the data:
Note that although the model seems to represent a good fit, the performance of the lower-
mass animals is not evident because of the scale. A log-log plot provides a better perspective
to assess the fit:
y = 3.3893x0.7266
R2
= 0.9982
0.1
1
10
100
1000
0.1 1 10 100 1000
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20.19 For the Casson Region, linear regression yields
99985.0180922.0065818.0 2 =+= r
For the Newton Region, linear regression with zero intercept gives
99992.0032447.0 2 == r
On a Casson plot, this function becomes
180131.0=
We can plot both functions along with the data on a Casson plot
0
1
2
3
4
0 5 10 15 20
180922.0065818.0 += 180922.0065818.0 +=
180131.0 180131.0
20.20 Recall from Sec. 4.1.3, that finite divided differences can be used to estimate the
derivatives. For the first point, we can use a forward difference (Eq. 4.17)
1725.0153204
8.876.96=
=
d
d
For the intermediate points, we can use centered differences. For example, for the second
point
8647.0153255
8.87176=
=
d
d
For the last point, we can use a backward difference
2157.17714765
33804258=
=
d
d
All the values can be tabulated as
d/d87.8 153 0.1725
96.6 204 0.8647
176 255 1.6314
263 306 1.7157
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351 357 3.0196
571 408 4.7353
834 459 6.4510
1229 510 7.7451
1624 561 8.6078
2107 612 10.3333
2678 663 12.48043380 714 15.4902
4258 765 17.2157
We can plot these results and after discarding the first few points, we can fit a straight line as
shown (note that the discarded points are displayed as open circles).
y = 0.003557x + 2.833492
R2
= 0.984354
0
5
10
15
20
0 1000 2000 3000 4000
Therefore, the parameter estimates are Eo = 2.833492 and a = 0.003557. The data along with
the first equation can be plotted as
0
4000
8000
12000
0 200 400 600 800
As described in the problem statement, the fit is not very good. We therefore pick a point near
the midpoint of the data interval
)2107,612( ==
We can then plot the second model
)1(5.269)1(1
2107 003557.0003557.0)612(003557.0
=
= ee
e
The result is a much better fit
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0
1000
2000
3000
4000
5000
0 100 200 300 400 500 600 700 800
20.21 The problem is set up as the following Excel Solver application. Notice that we have
assumed that the model consists of a constant plus two bell-shaped curves:
22
22
21
21 )(
2)(
1)(axkaxk ekek
cxf
++=
The resulting solution is
Thus, the retina thickness is estimated as 0.312 0.24 = 0.072.
20.22 Simple linear regression can be applied to yield the following fit
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y = 7.6491x - 1.5459
R2
= 0.8817
0
20
40
60
80
100
0 2 4 6 8 10 12
The shear stress at the depth of 4.5 m can be computed as
875.325459.1)5.4(6491.7 ==
20.23 (a) and (b) Simple linear regression can be applied to yield the following fit
y = 0.7335x + 0.7167
R2
= 0.8374
0
1
2
3
4
0 0.5 1 1.5 2 2.5 3 3.5
(c) The minimum lane width corresponding to a bike-car distance of 2 m can be computed as
m1837.27167.0)2(7335.0 =+=y
20.24 (a) and (b) Simple linear regression can be applied to yield the following fit
y = 0.1519x + 0.8428
R2
= 0.894
12
16
20
24
80 90 100 110 120 130 140 150
(c) The flow corresponding to the precipitation of 120 cm can be computed as
067.198428.0)120(1519.0 =+=Q
(d) We can redo the regression, but with a zero intercept
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y = 0.1594x
R2
= 0.8917
12
16
20
24
80 90 100 110 120 130 140 150
Thus, the model is
PQ 1594.0=
where Q = flow and P= precipitation. Now, if there are no water losses, the maximum flow,
Qm, that could occur for a level of precipitation should be equal to the product of the annual
precipitation and the drainage area. This is expressed by the following equation.
=
yr
cm)km( 2 PAQm
For an area of 1100 km2 and applying conversions so that the flow has units of m3/s
d365
yr
s400,86
d
cm100
m1
km
m10
yr
cmkm100,1
2
262
= PQm
Collecting terms gives
PQm 348808.0=
Using the slope from the linear regression with zero intercept, we can compute the fraction of
the total flow that is lost to evaporation and other consumptive uses can be computed as
543.0348808.0
1594.0348808.0=
=F
20.25 The data can be computed in a number of ways. First, we can use linear regression.
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y = 0.2879x - 0.629
R2
= 0.9855
0
2
4
6
8
10
12
0 5 10 15 20 25 30 35 40
For this model, the chlorophyll level in western Lake Erie corresponding to a phosphorus
concentration of 10 mg/m3 is
2495.2629.0)10(2879.0 ==c
One problem with this result is that the model yields a physically unrealistic negative
intercept. In fact, because chlorophyll cannot exist without phosphorus, a fit with a zero
intercept would be preferable. Such a fit can be developed as
y = 0.2617x
R2
= 0.9736
0
2
4
6
8
10
12
0 5 10 15 20 25 30 35 40
For this model, the chlorophyll level in western Lake Erie corresponding to a phosphorus
concentration of 10 mg/m3 is
617.2)10(2617.0 ==c
Thus, as expected, the result differs from that obtained with a nonzero intercept.
Finally, it should be noted that in practice, such data is usually fit with a power model. Thismodel is often adopted because it has a zero intercept while not constraining the model to be
linear. When this is done, the result is
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y = 0.1606x1.1426
R2
= 0.9878
0
2
4
6
8
10
12
0 5 10 15 20 25 30 35 40
For this model, the chlorophyll level in western Lake Erie corresponding to a phosphorus
concentration of 10 mg/m3 is
2302.2)10(1606.0 1426.1 ==c
20.26
46.010
6.4
==m 4.11014
==n
Use the following values
x0 = 0.3 f(x0) = 0.08561
x1 = 0.4 f(x1) = 0.10941
x2 = 0.5 f(x2) = 0.13003
x3 = 0.6 f(x3) = 0.14749
MATLAB can be used to perform the interpolation
>> format long
>> x=[0.3 0.4 0.5 0.6];>> y=[0.08561 0.10941 0.13003 0.14749];
>> a=polyfit(x,y,3)
a =
Columns 1 through 3
0.00333333333331 -0.16299999999997 0.35086666666665
Column 4
-0.00507000000000
>> polyval(a,0.46)
ans =
0.12216232000000
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This result can be used to compute the vertical stress
552795.1)14(6.4
100==q
189693.0)1221623.0(552795.1 ==z
20.27 This is an ideal problem for general linear least squares. The problem can be solved with
MATLAB:
>> t=[0.5 1 2 3 4 5 6 7 9]';>> p=[6 4.4 3.2 2.7 2.2 1.9 1.7 1.4 1.1]';>> Z=[exp(-1.5*t) exp(-0.3*t) exp(-0.05*t)];>> a=inv(Z'*Z)*(Z'*p)
a =4.00462.9213
1.5647
Therefore,A = 4.0046, B = 2.9213, and C= 1.5647.
20.28 First, we can determine the stress
418.347,265.10
25000==
We can then try to fit the data to obtain a mathematical relationship between strain and stress.
First, we can try linear regression:
y = 1.37124E-06x - 2.28849E-03
R2
= 0.856845
-0.002
0
0.002
0.004
0.006
0.008
0.01
0 2000 4000 6000 8000
This is not a particularly good fit as the r2 is relatively low. We therefore try a best-fit
parabola,
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y = 2.8177E-10x2
- 8.4598E-07x + 1.0766E-03
R2
= 0.98053
0
0.002
0.004
0.006
0.008
0.01
0 2000 4000 6000 8000
We can use this model to compute the strain as
437210 104341.6100766.1)4178.2347(104598.8)4178.2347(108177.2 =+=
The deflection can be computed as
m0057907.0)9(104341.6 4 == L
20.29 Clearly the linear model is not adequate. The second model can be fit with the ExcelSolver:
Notice that we have reexpressed the initial rates by multiplying them by 1 106. We did thisso that the sum of the squares of the residuals would not be miniscule. Sometimes this will
lead the Solver to conclude that it is at the minimum, even though the fit is poor. The solution
is:
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Although the fit might appear to be OK, it is biased in that it underestimates the low values
and overestimates the high ones. The poorness of the fit is really obvious if we display the
results as a log-log plot:
1.E-06
1.E-04
1.E-02
1.E+00
1.E+02
0.01 1 100
v0
v0mod
Notice that this view illustrates that the model actually overpredicts the very lowest values.
The third and fourth models provide a means to rectify this problem. Because they raise [S] topowers, they have more degrees of freedom to follow the underlying pattern of the data. For
example, the third model gives:
Finally, the cubic model results in a perfect fit:
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Thus, the best fit is
][385.0
][1045.23
35
0S
Sv
+
=
20.30 As described in Section 17.1, we can use general linear least squares to generate the best-fit
equation. We can use a number of different software tools to do this. For example, the [Z] and
y matrices can be set up using MATLAB commands as
>> format long>> x = [273.15 283.15 293.15 303.15 313.15]';>> Kw = [1.164e-15 2.950e-15 6.846e-15 1.467e-14 2.929e-14]';>> y=-log10(Kw);
>> Z = [(1./x) log10(x) x ones(size(x))];
The coefficients can be evaluated as
>> a=inv(Z'*Z)*(Z'*y)Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 6.457873e-020.
a =1.0e+003 *5.180671875000000.013424560546880.00000562858582
-0.03827636718750
Note the warning that the results are ill-conditioned. According to this calculation, the best-fit
model is
276367.3800562859.0log42456.1367.5180
log 1010 ++= aaa
w TTT
K
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28
We can check the results by using the model to make predictions at the values of the original
data
>> yp=10.^-(a(1)./x+a(2)*log10(x)+a(3)*x+a(4))
yp =
1.0e-013 *0.011611933082420.029432357145510.068284617294940.146363305750490.29218444886852
These results agree to about 2 or 3 significant digits with the original data.
20.31 Using MATLAB:
>> t=0:2*pi/128:2*pi;>> f=4*cos(5*t)-7*sin(3*t)+6;>> y=fft(f);>> y(1)=[ ];>> n=length(y);>> power=abs(y(1:n/2)).^2;>> nyquist=1/2;>> freq=n*(1:n/2)/(n/2)*nyquist;>> plot(freq,power);
20.32 Since we do not know the proper order of the interpolating polynomial, this problem issuited for Newton interpolation. First, order the points so that they are as close to and as
centered about the unknown as possible.
x0 = 1.25 f(x0) = 0.7
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x1 = 0.75 f(x1) = 0.6x2 = 1.5 f(x2) = 1.88
x3 = 0.25 f(x3) = 0.45x4 = 2 f(x4) = 6
The results of applying Newtons polynomial at i = 1.15 are
Order f(x) Error
0 0.7 -0.26
1 0.44 -0.11307
2 0.326933 -0.00082
3 0.326112 0.011174
4 0.337286
The minimum error occurs for the second-order version so we conclude that the interpolation
is 0.3269.
20.33 Here are the results of first through fourth-order regression:
y = 3.54685x - 2.57288
R2
= 0.78500
-2
0
2
4
6
8
0 0.5 1 1.5 2 2.5
y = 3.3945x2
- 4.0093x + 0.3886
R2
= 0.9988
-2
0
2
4
6
8
0 0.5 1 1.5 2 2.5
y = 0.5663x3
+ 1.4536x2
- 2.1693x - 0.0113
R2
= 0.9999
-2
0
2
4
6
8
0 0.5 1 1.5 2 2.5
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y = 0.88686x4
- 3.38438x3
+ 7.30000x2
- 5.40448x +
0.49429
R2
= 1.00000
-2
0
2
4
6
8
0 0.5 1 1.5 2 2.5
The 2nd through 4th-order polynomials all seem to capture the general trend of the data. Each
of the polynomials can be used to make the prediction at i = 1.15 with the results tabulatedbelow:
Order Prediction
1 1.50602 0.26703 0.2776
4 0.3373
Thus, although the 2nd through 4th-order polynomials all seem to follow a similar trend, theyyield quite different predictions. The results are so sensitive because there are few data
points.
20.34 Since we do not know the proper order of the interpolating polynomial, this problem issuited for Newton interpolation. First, order the points so that they are as close to and as
centered about the unknown as possible.
x0 = 0.25 f(x0) = 7.75
x1 = 0.125 f(x1) = 6.24
x2 = 0.375 f(x2) = 4.85x3 = 0 f(x3) = 0
x4 = 0.5 f(x4) = 0
The results of applying Newtons polynomial at t= 0.23 are
Order f(x) Error
0 7.75 -0.2416
1 7.5084 0.296352
2 7.804752 0.008315
3 7.813067 0.025579
4 7.838646
The minimum error occurs for the second-order version so we conclude that the interpolation
is 7.805.
20.35(a) The linear fit is
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y = 2.8082x - 0.5922
R2
= 0.9991
0
10
20
30
0 2 4 6 8 10 12
The current for a voltage of 3.5 V can be computed as
2364.95922.0)5.3(8082.2 ==y
Both the graph and the r2 indicate that the fit is good.
(b) A straight line with zero intercept can be fit as
y = 2.7177x
R2
= 0.9978
0
10
20
30
0 2 4 6 8 10 12
For this case, the current at V= 3.5 can be computed as
512.9)5.3(7177.2 ==y
20.36 The linear fit is
y = 4.7384x + 1.2685
R2
= 0.9873
0
10
20
30
40
50
60
0 2 4 6 8 10 12
Therefore, an estimate forL is 4.7384. However, because there is a non-zero intercept, a
better approach would be to fit the data with a linear model with a zero intercept
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32
y = 4.9163x
R2
= 0.9854
010
20
30
40
50
60
0 2 4 6 8 10 12
This fit is almost as good as the first case, but has the advantage that it has the physically
more realistic zero intercept. Thus, a superior estimate ofL is 4.9163.
20.37 Since we do not know the proper order of the interpolating polynomial, this problem is
suited for Newton interpolation. First, order the points so that they are as close to and as
centered about the unknown as possible.
x0 = 0.5 f(x0) = 20.5
x1 = 0.5 f(x1) = 20.5x2 = 1 f(x2) = 96.5
x3 = 1 f(x3) = 96.5x4 = 2 f(x4) = 637
x5 = 2 f(x5) = 637
The results of applying Newtons polynomial at i = 0.1 are
Order f(x) Error
0 20.5 -16.4
1 4.1 -17.76
2 -13.66 15.984
3 2.324 0
4 2.324 0
5 2.324
Thus, we can see that the data was generated with a cubic polynomial.
20.38 Because there are 6 points, we can fit a 5 th-order polynomial. This can be done with Eq.
18.26 or with a software package like Excel or MATLAB that is capable of evaluating the
coefficients. For example, using MATLAB,
>> x=[-2 -1 -0.5 0.5 1 2];>> y=[-637 -96.5 -20.5 20.5 96.5 637];
>> a=polyfit(x,y,5)
a =0.0000 0.0000 74.0000 -0.0000 22.5000 0.0000
Thus, we see that the polynomial is
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iiV 5.2274 3 +=
20.39 Linear regression yields
y = 0.0195x + 3.2895
R2 = 0.9768
0
5
10
15
20
0 100 200 300 400 500 600 700
The percent elongation for a temperature of 400 can be computed as
072.112895.3)400(0195.0elongation% =+=
20.40(a) A 4th-order interpolating polynomial can be generated as
y = 0.0182x4 - 0.1365x3 + 0.1818x2 + 0.263x + 0.1237
R2 = 1
0.4
0.45
0.5
0.55
0.6
1.5 1.7 1.9 2.1 2.3 2.5 2.7
The polynomial can be used to compute
568304.00.1237+.1)0.262958(2+.1)0.181771(2+(2.1)13646.0.1)0.018229(2)12( 2341 ==.JThe relative error is
%0021.0%100568292.0
568304.0568292.0=
=t
Thus, the interpolating polynomial yields an excellent result.
(b) A program can be developed to fit natural cubic splines through data based on Fig. 18.18.If this program is run with the data for this problem, the interpolation at 2.1 is 0.56846 which
has a relative error oft= 0.0295%.
A spline can also be fit with MATLAB. It should be noted that MATLAB does not use a
natural spline. Rather, it uses a so-called not-a-knot spline. Thus, as shown below, although
it also yields a very good prediction, the result differs from the one generated with the natural
spline,
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34
>> format long>> x=[1.8 2 2.2 2.4 2.6];>> y=[0.5815 0.5767 0.556 0.5202 0.4708];>> spline(x,y,2.1)
ans =0.56829843750000
This result has a relative error oft= 0.0011%.
20.41 The fit of the exponential model is
y = 97.915e0.151x
R2
= 0.9996
0500
1000
1500
2000
2500
0 5 10 15 20
The model can be used to predict the population 5 years in the future as
426891484.97 )25(150992.0 == ep
20.42 The prediction using the model from Sec. 20.4 is
30655.2)001.0()23.1(9.55 54.062.2 ==Q
The linear prediction is
x0 = 1 f(x0) = 1.4
x1 = 2 f(x1) = 8.3
987.2)123.1(12
4.13.84.1 =
+=Q
The quadratic prediction is
x0 = 1 f(x0) = 1.4x1 = 2 f(x1) = 8.3
x2 = 3 f(x2) = 24.2
19005.2)223.1)(123.1(13
12
4.13.8
23
3.82.24
987.2 =
+=Q
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20.43 The model to be fit is
21
0bb
QDbS=
Taking the common logarithm gives
QbDbbS 10210101010 loglogloglog ++=
We can use a number of different approaches to fit this model. The following shows how it
can be done with MATLAB,
>> D=[1 2 3 1 2 3 1 2 3]';>> S=[0.001 0.001 0.001 0.01 0.01 0.01 0.05 0.05 0.05]';>> Q=[1.4 8.3 24.2 4.7 28.9 84 11.1 69 200]';>> Z=[ones(size(S)) log10(D) log10(Q)];>> y=log10(S);>> a=inv(Z'*Z)*(Z'*y)
a =-3.2563-4.87261.8627
Therefore, the result is
QDS 101010 log862733.1log8726.42563.3log +=
or in untransformed format
862733.18726.4000554.0 QDS = (1)
We can compare this equation with the original model developed in Sec. 20.4 by solving Eq.1 forQ,
5368.06158.29897.55 SDQ =
This is very close to the original model.
20.44(a) The data can be plotted as
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0.0E+00
5.0E-04
1.0E-03
1.5E-03
2.0E-03
0 10 20 30 40
(b) This part of the problem is well-suited for Newton interpolation. First, order the points so
that they are as close to and as centered about the unknown as possible.
x0 = 5 f(x0) = 1.519 103
x1 = 10 f(x1) = 1.307 103
x2 = 0 f(x2) = 1.787
103
x3 = 20 f(x3) = 1.002 103
x4 = 30 f(x4) = 0.7975 103
x5 = 40 f(x5) = 0.6529 103
The results of applying Newtons polynomial at T= 7.5 are
Order f(x) Error
0 1.51900E-03 -0.00011
1 1.41300E-03 -7E-06
2 1.40600E-03 7.66E-07
3 1.40677E-03 9.18E-08
4 1.40686E-03 5.81E-095 1.40686E-03
The minimum error occurs for the fourth-order version so we conclude that the interpolation
is 1.40686 103.
(b) Polynomial regression yields
y = 5.48342E-07x2
- 4.94934E-05x + 1.76724E-03
R2
= 9.98282E-01
0.0E+00
5.0E-04
1.0E-03
1.5E-03
2.0E-03
0 10 20 30 40
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37
This leads to a prediction of
33527 104269.11076724.1)5.7(1094934.4)5.7(1048342.5 =+=
20.45 For the first four points, linear regression yields
y = 117.16x - 1.0682
R2
= 0.9959
0
10
20
30
40
50
0 0.1 0.2 0.3 0.4
We can fit a power equation to the last four points.
y = 1665x3.7517
R2
= 0.9395
0
20
40
60
80
100
0.38 0.39 0.4 0.41 0.42 0.43 0.44 0.45
Both curves can be plotted along with the data.
0
20
40
60
80
100
0 0.1 0.2 0.3 0.4 0.5
20.46 A power fit to all the data can be implemented as in the following plot.
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y = 185.45x1.2876
R2
= 0.9637
0
20
40
60
80
100
0 0.1 0.2 0.3 0.4 0.5
Although the coefficient of determination is relatively high (r2 = 0.9637), this fit is not very
acceptable. In contrast, the piecewise fit from Prob. 20.45 does a much better job of tracking
on the trend of the data.
20.47 The Thinking curve can be fit with a linear model whereas the Braking curve can be fit
with a quadratic model as in the following plot.
y = 0.1865x + 0.0800
R2
= 0.9986
y = 5.8783E-03x2
+ 9.2063E-04x - 9.5000E-02
R2
= 9.9993E-01
0
20
40
60
80
100
0 20 40 60 80 100 120 140
A prediction of the total stopping distance for a car traveling at 110 km/hr can be computed
as
m726.91109.5000(110)109.2063(110)105.87830.0800)0.1865(110 2423 =+++= d20.48 Using linear regression gives
y = -0.6x + 39.75
R2
= 0.5856
0
10
20
30
40
50
0 10 20 30 40
A prediction of the fracture time for an applied stress of 20 can be computed as
75.2775.39)20(6.0 =+=t
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Some students might consider the point at (20, 40) as an outlier and discard it. If this is done,
4th-order polynomial regression gives the following fit
y = -1.3951E-04x4
+ 1.0603E-02x3
- 2.1959E-01x2
+ 3.6477E-02x +
4.3707E+01
R
2
= 9.8817E-01
0
10
20
30
40
50
0 10 20 30 40
A prediction of the fracture time for an applied stress of 20 can be computed as
103.1943.707+0)0.036477(2+)0.21959(200)0.010603(2+(20)0.00013951 234 ==t
20.49 Using linear regression gives
y = -2.015000E-06x + 9.809420E+00
R2
= 9.999671E-01
9.5
9.6
9.7
9.8
9.9
0 20000 40000 60000 80000 100000 120000
A prediction ofgat 55,000 m can be made as
6986.980942.9)000,55(10015.2)000,55( 6 =+= g
Note that we can also use linear interpolation to give
698033.9)000,55(1 =g
Quadratic interpolation yields
697985.9)000,55(2 =g
Cubic interpolation gives
69799.9)000,55(3 =g
Based on all these estimates, we can be confident that the result to 3 significant digits is
9.698.
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20.50 A plot of 10log versus log10can be developed as
y = 0.3792x + 2.0518
R2
= 0.9997
0
0.2
0.4
0.6
0.8
1
1.2
-3.5 -3.3 -3.1 -2.9 -2.7 -2.5
As shown, the best fit line is
0518.2log3792.0log 1010 +=
Therefore,B = 102.0518 = 112.67 and m = 0.3792, and the power model is
3792.067.121 xy =
The model and the data can be plotted as
y = 112.67x0.3792
R2
= 0.9997
0
2
4
68
10
12
14
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035
20.51 Using linear regression gives
y = 0.688x + 2.79
R2
= 0.9773
0
2
4
6
0 1 2 3 4 5 6
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Therefore, y = 2.79 and= 0.688.
20.52 A power fit can be developed as
y = 0.7374x0.5408
R2
= 0.9908
0
2
4
6
8
10
12
0 20 40 60 80 100 120 140
Therefore,= 0.7374 and n = 0.5408.
20.53 We fit a number of curves to this data and obtained the best fit with a second-order
polynomial with zero intercept
y = 1628.1x2
+ 140.05x
R2
= 1
0
1
2
3
4
5
0 0.004 0.008 0.012 0.016 0.02 0.024
Therefore, the best-fit curve is
yyu 05.1401.1628 2 +=
We can differentiate this function
05.1402.3256 += ydy
du
Therefore, the derivative at the surface is 140.05 and the shear stress can be computed as
1.8 105(140.05) = 0.002521 N/m2.
20.54 We can use transformations to linearize the model as
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aTBD
1lnln +=
Thus, we can plot the natural log ofversus 1/Ta and use linear regression to determine theparameters. Here is the data showing the transformations.
T Ta 1/Ta ln
0 1.787 273.15 0.003661 0.580538
5 1.519 278.15 0.003595 0.418052
10 1.307 283.15 0.003532 0.267734
20 1.002 293.15 0.003411 0.001998
30 0.7975 303.15 0.003299 -0.22627
40 0.6529 313.15 0.003193 -0.42633
Here is the fit:
y = 2150.8x - 7.3143
R2
= 0.998
-0.6
-0.4
-0.2
0
0.2
0.4
0.60.8
0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037
Thus, the parameters are estimated as D = e7.3143 = 6.65941 104 andB = 2150.8, and theAndrade equation is
Tae /8.215041065941.6 =
This equation can be plotted along with the data
0
0.5
1
1.5
2
0 10 20 30 40
Note that this model can also be fit with nonlinear regression. If this is done, the result is
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Tae /66.221041039872.5 =
Although it is difficult to discern graphically, this fit is slightly superior (r2 = 0.99816) to that
obtained with the transformed model (r2 = 0.99757).
20.55 Hydrogen: Fifth-order polynomial regression provides a good fit to the data,
y = 4.3705E-14x5
- 1.5442E-10x4
+ 2.1410E-07x3
- 1.4394E-04x2
+ 4.7032E-02x + 8.5079E+00
R2
= 9.9956E-01
14
14.2
14.4
14.6
14.8
15
15.2
200 300 400 500 600 700 800 900 1000
Carbon dioxide: Third-order polynomial regression provides a good fit to the data,
y = 4.1778E-10x3
- 1.3278E-06x2
+ 1.7013E-03x + 4.4317E-01
R2
= 9.9998E-01
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
200 300 400 500 600 700 800 900 1000
Oxygen: Fifth-order polynomial regression provides a good fit to the data,
y = -1.3483E-15x5
+ 5.1360E-12x4
- 7.7364E-09x3
+ 5.5693E-
06x2
- 1.5969E-03x + 1.0663E+00
R2
= 1.0000E+00
0.9
0.95
1
1.05
1.1
200 300 400 500 600 700 800 900 1000
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Nitrogen: Third-order polynomial regression provides a good fit to the data,
y = -3.0895E-10x3
+ 7.4268E-07x2
- 3.5269E-04x + 1.0860E+00
R2
= 9.9994E-01
1
1.04
1.08
1.12
1.16
1.2
200 300 400 500 600 700 800 900 1000
20.56 This part of the problem is well-suited for Newton interpolation. First, order the points so
that they are as close to and as centered about the unknown as possible, forx = 4.
y0 = 4 f(y0) = 38.43y1 = 2 f(y1) = 53.5
y2 = 6 f(y2) = 30.39
y3 = 0 f(y3) = 80
y4 = 8 f(y4) = 30
The results of applying Newtons polynomial at y = 3.2 are
Order f(y) Error
0 38.43 6.028
1 44.458 -0.8436
2 43.6144 -0.2464
3 43.368 0.1124484 43.48045
The minimum error occurs for the third-order version so we conclude that the interpolation is
43.368.
(b) This is an example of two-dimensional interpolation. One way to approach it is to use
cubic interpolation along the y dimension for values at specific values ofx that bracket the
unknown. For example, we can utilize the following points at x = 2.
y0 = 0 f(y0) = 90
y1 = 2 f(y1) = 64.49
y2 = 4 f(y2) = 48.9y3 = 6 f(y3) = 38.78
T(x = 2, y = 2.7) = 58.13288438
All the values can be tabulated as
T(x = 2, y = 2.7) = 58.13288438
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T(x = 4, y = 2.7) = 47.1505625T(x = 6, y = 2.7) = 42.74770188
T(x = 8, y = 2.7) = 46.5
These values can then be used to interpolate at x = 4.3 to yield
T(x = 4.3, y = 2.7) = 46.03218664
Note that some software packages allow you to perform such multi-dimensional
interpolations very efficiently. For example, MATLAB has a function interp2 that provides
numerous options for how the interpolation is implemented. Here is an example of how it can
be implemented using linear interpolation,
>> Z=[100 90 80 70 60;85 64.49 53.5 48.15 50;70 48.9 38.43 35.03 40;55 38.78 30.39 27.07 30;40 35 30 25 20];>> X=[0 2 4 6 8];>> Y=[0 2 4 6 8];>> T=interp2(X,Y,Z,4.3,2.7)
T =47.5254
It can also perform the same interpolation but using bicubic interpolation,
>> T=interp2(X,Y,Z,4.3,2.7,'cubic')
T =46.0062
Finally, the interpolation can be implemented using splines,
>> T=interp2(X,Y,Z,4.3,2.7,'spline')
T =46.1507
20.57 This problem was solved using an Excel spreadsheet and TrendLine. Linear regression
gives
y = 0.0455x + 0.107
R2
= 0.9986
0
0.2
0.4
0.6
0 2 4 6 8 10
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Forcing a zero intercept yields
y = 0.061x
R2
= 0.841
0
0.2
0.4
0.6
0.8
0 2 4 6 8 10
One alternative that would force a zero intercept is a power fit
y = 0.1821x0.4087
R2
= 0.8992
0
0.2
0.4
0.6
0.8
0 2 4 6 8 10
However, this seems to represent a poor compromise since it misses the linear trend in the data.
An alternative approach would to assume that the physically-unrealistic non-zero intercept is an
artifact of the measurement method. Therefore, if the linear slope is valid, we might try y =
0.0455x.
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