Numerical Methods for Engineers
Numerical Methods for Engineers
7th Edition
ISBN: 9780073397924
Author: Steven C. Chapra Dr., Raymond P. Canale
Publisher: McGraw-Hill Education
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Chapter 20, Problem 60P

Temperatures are measured at various points on a heatedplate (Table P20.60). Estimate the temperature at (a) x = 4 , y = 3.2 , and (b) x = 4.3 , y = 2.7 .

TABLE P20.60 Temperatures ( ° C ) at various points on a square heated plate.

x = 0 x = 2 x = 4 x = 6 x = 8
y = 0 100.00 90.00 80.00 70.00 60.00
y = 2 85.00 64.49 53.50 48.15 50.00
y = 4 70.00 48.90 38.43 35.03 40.00
y = 6 55.00 38.78 30.39 27.07 30.00
y = 8 40.00 35.00 30.00 25.00 20.00

(a)

Expert Solution
Check Mark
To determine

To calculate: The value of temperature at x=4,y=3.2 from the given table if the temperatures in (°C) are recorded at various points on a heated plate.

x=0 x=2 x=4 x=6 x=8
y=0 100.00 90.00 80.00 70.00 60.00
y=2 85.00 64.49 53.50 48.15 50.00
y=4 70.00 48.90 38.43 35.03 40.00
y=6 55.00 38.78 30.39 27.07 30.00
y=8 40.00 35.00 30.00 25.00 20.00

Answer to Problem 60P

Solution:

The value of temperature at x=4,y=3.2 is 43.368°C.

Explanation of Solution

Given Information:

The data is provided as,

x=0 x=2 x=4 x=6 x=8
y=0 100.00 90.00 80.00 70.00 60.00
y=2 85.00 64.49 53.50 48.15 50.00
y=4 70.00 48.90 38.43 35.03 40.00
y=6 55.00 38.78 30.39 27.07 30.00
y=8 40.00 35.00 30.00 25.00 20.00

Formula used:

The zero-order Newton’s interpolation formula:

f0(x)=b0

The first-order/linear Newton’s interpolation formula:

f1(x)=b0+b1(xx0)

The second- order/quadratic Newton’s interpolating polynomial is given by,

f2(x)=b0+b1(xx0)+b2(xx0)(xx1)

Where,

b0=f(x0)b1=f[x1,x0]b2=f[x2,x1,x0]

The first finite divided difference is,

f[xi,xj]=f(xi)f(xj)xixj

And, the n th finite divided difference is,

f[xn,xn1,...,x1,x0]=f[xn,xn1,...,x1]f[xn1,...,x1,x0]xnx0

Calculation:

To calculate the temperature x=4,y=3.2, Newton interpolation is the perfect choice.

First use the linear interpolation formula and arrange the points as close to about x=4,

f1(y)=b0+b1(yy0)

The values are,

y0=4,f(y0)=38.43y1=2,f(y1)=53.5y2=6,f(y2)=30.39y3=0,f(y3)=80

And,

y4=8,f(y4)=30y=3.2

First calculate b1,

b1=53.538.4324=7.535

Put in above equation,

f1(y)=38.43+(7.535)(3.24)=44.458

Similarly for quadratic interpolation,

f2(y)=b0+b1(yy0)+b2(yy0)(yy1)

Now calculate b2,

b2=30.3953.56253.538.432464=5.7775+7.5352=0.87875

Put in quadratic interpolation equation,

f2(y)=44.458+(0.87875)(3.24)(3.22)=44.4580.8436=43.6144

Now, do it for cubic interpolation by the use of the formula,

f3(y)=b0+b1(yy0)+b2(yy0)(yy1)+b3(yy0)(yy1)(yy2)

Now calculate b3,

b3=(8030.390630.3953.562)02(30.3953.56253.538.4324)6404=(8.26833+5.7775)2(5.7775+7.535)24=(1.2454)(0.87875)4=0.0916625

Put in cubic interpolation equation,

f3(y)=43.61440.0916625(3.24)(3.22)(3.26)=43.61440.2463888=43.3680112

And, the error is calculated as,

Error=f2(y)f1(y)=43.614444.458=0.8436

Similarly the other dividend can be calculated as shown above,

Therefore, the difference table can be summarized for y=3.2 as,

Order f(3.2) Error
0 38.43 6.028
1 44.458 0.8436
2 43.6144 0.2464
3 43.368 0.112442
4 43.48045

Since the minimum error for order third, therefore, it can be concluded that the value of temperature at x=4,y=3.2 is 43.368.

(b)

Expert Solution
Check Mark
To determine

To calculate: The value of temperature at x=4.3,y=2.7 from the given table if the temperatures in (°C) are recorded at various points on a heated plate.

x=0 x=2 x=4 x=6 x=8
y=0 100.00 90.00 80.00 70.00 60.00
y=2 85.00 64.49 53.50 48.15 50.00
y=4 70.00 48.90 38.43 35.03 40.00
y=6 55.00 38.78 30.39 27.07 30.00
y=8 40.00 35.00 30.00 25.00 20.00

Answer to Problem 60P

Solution:

The value of temperature at x=4.3,y=2.7 is 46.14°C.

Explanation of Solution

Given Information:

The data is provided as,

x=0 x=2 x=4 x=6 x=8
y=0 100.00 90.00 80.00 70.00 60.00
y=2 85.00 64.49 53.50 48.15 50.00
y=4 70.00 48.90 38.43 35.03 40.00
y=6 55.00 38.78 30.39 27.07 30.00
y=8 40.00 35.00 30.00 25.00 20.00

Formula used:

The zero-order Newton’s interpolation formula:

f0(x)=b0

The first-order/linear Newton’s interpolation formula:

f1(x)=b0+b1(xx0)

The second- order/quadratic Newton’s interpolating polynomial is given by,

f2(x)=b0+b1(xx0)+b2(xx0)(xx1)

Where,

b0=f(x0)b1=f[x1,x0]b2=f[x2,x1,x0]

The first finite divided difference is,

f[xi,xj]=f(xi)f(xj)xixj

And, the n th finite divided difference is,

f[xn,xn1,...,x1,x0]=f[xn,xn1,...,x1]f[xn1,...,x1,x0]xnx0

Calculation:

To calculate the temperature x=4.3,y=2.7, Newton interpolation is the perfect choice.

Since, this is a two-dimensional interpolation, therefore one way is to use cubic interpolation along the y direction for specific values of x and then go along the x direction for values of y obtained from the previous analysis.

First use the linear interpolation formula and arrange the points as close to about x=2,

f1(y)=b0+b1(yy0)

The values are,

y0=2,f(y0)=64.49y1=4,f(y1)=48.90y2=6,f(y2)=38.78y3=0,f(y3)=90.00

And,

y4=8,f(y4)=35.00y=2.7

First calculate b1,

b1=48.964.4942=7.795

Put in above equation,

f1(y)=64.49+(7.795)(2.72)=59.0335

Similarly for quadratic interpolation,

f2(y)=b0+b1(yy0)+b2(yy0)(yy1)

Now calculate b2,

b2=38.7848.906448.9064.494262=5.06+7.7954=0.68375

Put in quadratic interpolation equation,

f2(y)=59.0335+(0.68375)(2.72)(2.74)=59.03350.6222125=58.4112875

Now, do it for cubic interpolation by the use of the formula,

f3(y)=b0+b1(yy0)+b2(yy0)(yy1)+b3(yy0)(yy1)(yy2)

Now calculate b3,

b3=(90.0038.780638.7848.9064)04(38.7848.906448.9064.4942)6202=(8.5366+5.06)6(5.06+7.535)42=(0.57943)(0.61875)2=0.01965833

Put in cubic interpolation equation,

f3(y)=58.4112875+0.01965833(2.72)(2.74)(2.76)=58.4112875+0.059033=58.47032

And, the error is calculated as,

Error=f2(y)f1(y)=58.411287559.0335=0.6222125

Similarly the other dividend can be calculated as shown above,

Therefore, the difference table can be summarized for y=2.7 as,

Order f(2.7) Error
0 64.49 5.4565
1 59.0335 0.6225
2 58.411 0.05932
3 58.47032

Now, do this for x=4,

f1(y)=b0+b1(yy0)

The values are,

y0=4,f(y0)=38.43y1=2,f(y1)=53.5y2=6,f(y2)=30.39y3=0,f(y3)=80

And,

y4=8,f(y4)=30y=2.7

First calculate b1,

b1=53.538.4324=7.535

Put in above equation,

f1(y)=38.43+(7.535)(2.74)=48.2255

Similarly for quadratic interpolation,

f2(y)=b0+b1(yy0)+b2(yy0)(yy1)

Now calculate b2,

b2=30.3953.56253.538.432464=5.7775+7.5352=0.87875

Put in quadratic interpolation equation,

f2(y)=48.2255+(0.87875)(2.74)(2.72)=48.22550.7996625=47.4253375

Now, do it for cubic interpolation by the use of the formula,

f3(y)=b0+b1(yy0)+b2(yy0)(yy1)+b3(yy0)(yy1)(yy2)

Now calculate b3,

b3=(8030.390630.3953.562)02(30.3953.56253.538.4324)6404=(8.26833+5.7775)2(5.7775+7.535)24=(1.2454)(0.87875)4=0.0916625

Put in cubic interpolation equation,

f3(y)=47.42533750.0916625(2.74)(2.72)(2.76)=47.42533750.2752624875=47.1505625

Similarly, for x=6andx=8, the above process can be repeated and the following values is obtained.

T(x=2,y=2.7)=58.47032T(x=4,y=2.7)=47.1505625T(x=6,y=2.7)=42.74770188T(x=8,y=2.7)=46.5

Now for the calculation for x=4.3,

f1(x)=b0+b1(xx0)

The values are,

x0=4,f(y0)=47.15x1=6,f(y1)=42.74x2=2,f(y2)=58.47x3=8,f(y3)=46.5

And,

x=4.3

First calculate b1,

b1=42.7447.1564=2.205

Put in above equation,

f1(x)=47.15+(2.205)(4.34)=46.4885

Similarly for quadratic interpolation,

f2(x)=b0+b1(xx0)+b2(xx0)(xx1)

Now calculate b2,

b2=58.4742.742642.7447.156424=3.9325+2.2052=0.86375

Put in quadratic interpolation equation,

f2(x)=46.4885+(0.86375)(4.34)(4.36)=46.48850.4405125=46.0479875

Now, do it for cubic interpolation by the use of the formula,

f3(x)=b0+b1(xx0)+b2(xx0)(xx1)+b3(xx0)(xx1)(xx2)

Now calculate b3,

b3=(46.558.478258.4742.7426)86(58.4742.742642.7447.1564)2484=(1.995+3.9325)2(3.93252.205)24=(0.96875)(3.06875)4=0.525

Put in cubic interpolation equation,

f3(x)=46.48850.525(4.34)(4.36)(4.32)=46.48850.348075=46.140425

And, the error is calculated as,

Error=f2(y)f1(y)=46.047987546.4885=0.4405125

Similarly the other dividend can be calculated as shown above,

Therefore, the difference table can be summarized for x=4.3 as,

Order f(4.3) Error
0 47.15 0.6615
1 46.4885 0.4405125
2 46.0479875 0.0924375
3 46.140425

Hence, the value of temperature at x=4.3,y=2.7 is 46.14°C.

This problem can also be solved with MATLAB as it contains the predefined function interp2.

The MATLAB code is as shown below,

% the given table values are stored in variable Z.

Z=[10090807060;

8564.4953.548.1550;

7048.938.4335.0340;

5538.7830.3927.0730;

4035302520];

% The X values are given.

X=[02468];

% The Y values are given.

Y=[02468];

The output in the command window is,

Numerical Methods for Engineers, Chapter 20, Problem 60P , additional homework tip  1

For more accuracy, the result can also be obtained from the bicubic interpolation as shown below,

Numerical Methods for Engineers, Chapter 20, Problem 60P , additional homework tip  2

Finally, the interpolation can also be implemented with the use of splines as shown below,

Numerical Methods for Engineers, Chapter 20, Problem 60P , additional homework tip  3

Hence, it can be concluded that the result is similar to that obtained from the calculation.

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Chapter 20 Solutions

Numerical Methods for Engineers

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