ENGRD 241: Engineering Computation
PS07S, Fall 2002
SOLUTION TO PROBLEM SET NUMBER 7
1. C&C Problem 17.13, page 472. Use Excel for calculations. Also use the Excel
Trendline capability (C&C Section 19.8.1) for parts (a), (b), and (d) as well as for a
logarithmic fit. Because the Trendline capability does not have an option for part (c),
you will need to calculate this separately.
The following are obtained by least-squares regression using EXCEL Trendline capability.
(a) A straight line
(b) Power equation
y = 9.6583x 0.395
R2 = 0.9309
Power-equation fit
y = 0.4958x + 20.667
R2 = 0.8201
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45
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40
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30
30
25
25
y
y
Linear fit
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20
15
15
10
10
5
5
0
0
0
10
20
30
x
40
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60
0
10
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x
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50
(c) Saturation growth rate equation. A saturation growth equation, which is typically of the form
x
1 b 1 1
1
1
y  a3
, can be linearized by expressing it as  3  . A plot of
versus
x
y a3 x a3
b3  x
y
would, therefore, be linear. So, we convert each data value into its reciprocal and use those
values in EXCEL Trendline to get the required fit, which, as shown in the below Trendline
1
1
1
 52.36 and
chart, is  0.2134  0.0191 . Therefore, a3 
0.0191
y
x
b3  a3 (0.2134)  (52.36)(0.2134)  11.17 .
1/y = 0.21341/x + 0.0191
R2 = 0.992
0.07
0.06
0.05
1/y
0.04
0.03
0.02
0.01
0
0
0.05
0.1
0.15
0.2
0.25
1/x
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ENGRD 241: Engineering Computation
PS07S, Fall 2002
(d) A parabola. A parabola is a second-order polynomial. Using EXCEL Trendline, a
second-order polynomial fit is obtained as shown below.
Parabola
y = -0.0155x 2 + 1.3458x + 12.167
R2 = 0.9476
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y
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10
5
0
0
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30
40
50
60
x
(e) Logarithmic
Logarithmic
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y 25
20
y = 11.14Ln(x) - 0.4559
R2 = 0.971
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10
5
0
0
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20
30
x
40
50
60
A look at the R2 values for each of the ‘fits’ obtained above shows that the curve obtained by
the saturation growth equation appears to be the best because it has the highest R2 value
(0.992). The standard error is another parameter that can be used to check how good a fit is.
The saturation growth equation works better for data that seem to saturate beyond a certain
point. This saturation can be observed in the given data. So, the curve obtained by the
saturation growth equation stands out as the best.
2. C&C Problem 18.11, page 506.
From the given data,
x0 = 1, x1 = 2, x2 = 3, x3 = 5, x4 = 6
f ( x0 ) = 4.75, f ( x1 ) = 4, f ( x2 ) = 5.25, f ( x3 ) = 19.75, f ( x4 ) = 36
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ENGRD 241: Engineering Computation
PS07S, Fall 2002
Substitute these values in Eq.(18.37) on C&C page 503:
For i = 1, we have,
6
6
[5.25 – 4] +
[4.75 – 4]
3 2
2 1
By one of the properties of a natural spline, f  (1) = 0. So, the above equation reduces to:
4 f  (2) + f  (3) = 12
………….(1)
(2 – 1) f  (1) + 2(3 – 1) f  (2) + (3 – 2) f  (3) =
For i = 2, we have,
(3 – 2) f  (2) + 2(5 – 2) f  (3) + (5 – 3) f  (5) =
reduces to:
f  (2) + 6 f  (3) + 2 f  (5) = 36
6
6
[19.75 – 5.25] +
[4 – 5.25], which
53
3 2
………………………..(2)
For i = 3, we have,
6
6
[36 – 19.75] +
[5.25 – 19.75]
65
53
From the properties of a natural spline, we have f  (6) = 0. So, the above equation reduces to,
2 f  (3) + 6 f  (5) = 54
………………(3)
(5 – 3) f  (3) + 2(6 – 3) f  (5) + (6 – 5) f  (6) =
The equations (1), (2) and (3) can be expressed in matrix form as,
 4 1 0   f (2)  12 

     
 1 6 2   f (3)  = 36 
 0 2 6   f (5)  54 

 
  
The above set of equations is solved by a suitable method such as the Gauss elimination method
and we obtain: f  (2) = 2.2623; f  (3) = 2.9508; f  (5) = 8.0164
Using these values in Eq.(18.36) on page 503 C&C, we obtain:
2.2623
4.75
2.2623(2  1) 
 4
( x  1)3 
(2  x)  

For the 1st interval, f1(x) = 0 +
 ( x  1) ,
6(2  1)
2 1
6
 2 1
which reduces to: f1(x) = 0.37705(x – 1)3 + 4.75(2 – x) + 3.6229(x – 1)
2.2623(3  2) 
2.2623
2.9508
 4

(3  x)3 
( x  2)3  
 (3  x)
6
6(3  2)
6(3  2)
3  2
 5.25 2.9508(3  2) 

+
 ( x  2) , which reduces to:
6
3  2
f2(x) = 0.37705(3 – x)3 + 0.4918(x – 2)3 + 3.6229(3 – x) + 4.7582(x – 2)
For the 2nd interval, f2(x) =
2.9508
8.0164
 5.25 2.9508(5  3) 
(5  x)3 
( x  3)3  

 (5  x)
6(5  3)
6(5  3)
6
5  3
19.75 8.0164(5  3) 


 ( x  3) , which reduces to:
6
 53
f3(x) = 0.2459(5 – x)3 + 0.668(x – 3)3 + 1.6414(5 – x) + 7.2029(x – 3)
For the 3rd interval,
f3(x) =
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ENGRD 241: Engineering Computation
PS07S, Fall 2002
For the 4th interval,
8.0164
19.75 8.0164(6  5) 
 36

(6  x)3  0  

(6  x)  
 0  ( x  5) ,
f4(x) =

6(6  5)
6
 65

6  5 
3
which reduces to:
f4(x) = 1.3361(6 – x) + 18.4139(6 – x) + 36(x – 5)
(a) To predict f(4), we use the equation for the 3rd interval:
f3(4) = 0.2459(5 – 4)3 + 0.668(4 – 3)3 + 1.6414(5 – 4) + 7.2029(4 – 3) = 9.7582
To predict f(2.5), we use the equation for the 2nd interval.
f2(2.5) = 0.37705(3 – 2.5)3 + 0.4918(2.5 – 2)3 + 3.6229(3 – 2.5) + 4.7582(2.5 – 2) = 4.2992
(b) Using the equation for the 2nd interval,
f2(3) = 0.37705(3 – 3)3 + 0.4918(3 – 2)3 + 3.6229(3 – 3) + 4.7582(3 – 2) = 5.25
Using the equation for the 3rd interval,
f3(3) = 0.2459(5 – 3)3 + 0.668(3 – 3)3 + 1.6414(5 – 3) + 7.2029(3 – 3) = 5.25
The above results verify that f(3) calculated by using the equation for the 2nd interval is the
same as that calculated using the equation for the 3rd interval.
3. C&C Problems 20.44 and 20.45, pages 560-561. Use Excel Trendline capabilities.
20.44) We can use EXCEL Trendline capability to get a linear fit for the first four data
points, and thereby find the value of k from the regression equation:
y = 117.16x - 1.0682
R2 = 0.9959
Linear fit
45
Force (*10000 N)
40
From the regression equation,
we find that the value of the
spring constant k is 117.16 x 104
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0.4
Displacem ent (m )
Any nonlinear fit can be made for the nonlinear portion (the last four data points) of the
data. One of the fits could be a logarithmic fit, which is shown below.
Logarithm ic fit
y = 234.85Ln(x) + 268.97
R2 = 0.9051
90
Force (*10000 N)
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70
60
50
40
30
20
10
0
0.38
0.4
0.42
Displacem ent (m )
0.44
0.46
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ENGRD 241: Engineering Computation
PS07S, Fall 2002
20.45) Using EXCEL Trendline, a power curve can be fitted to the data as shown below,
y = 185.45x 1.2876
R2 = 0.9637
Pow er fit
90
Force (*10000 N)
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10
0
0
0.1
0.2
0.3
0.4
0.5
Displacem ent (m )
Generally power curves fit well for data in which one parameter saturates as the other one
increases. In the given data, beyond 60  104 N, the displacement appears to get saturated.
So, the power curve fits well for the given data.
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