EBK NUMERICAL METHODS FOR ENGINEERS
EBK NUMERICAL METHODS FOR ENGINEERS
7th Edition
ISBN: 8220100254147
Author: Chapra
Publisher: MCG
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Chapter 6, Problem 23P

Determine the roots of the simultaneous nonlinear equations

( x 4 ) 2 + ( y 4 ) 2 = 5 x 2 + y 2 = 16

Use a graphical approach to obtain your initial guesses. Determine refined estimates with the two-equation Newton-Raphson method described in Sec. 6.6.2.

Expert Solution & Answer
Check Mark
To determine

To calculate: The root of the non-linear simultaneous equations,

(x4)2+(y4)2=5x2+y2=16

By the two-equation Newton-Raphson method and find the initial guess by the graphical method.

Answer to Problem 23P

Solution:

The root of the simultaneous non-linear equations y=x2+1 and y=2cosx:

For the initial root (1.8,3.6) is x=1.80583 and y=3.56917.

For the initial root (3.6,1.8) is x=3.56917 and y=1.80583.

Explanation of Solution

Given information:

The non-linear simultaneous equation,

(x4)2+(y4)2=5x2+y2=16

Formula used:

The Newton-Raphson formula for two non-linear equation is,

xi+1=xiuiviyviuiyuixviyuiyvixyi+1=yiviuixuivixuixviyuiyvix

Calculation:

Use MATLAB to draw the graph of the equations, (x4)2+(y4)2=5 and x2+y2=16 as below,

Code:

clear;clc;

%x-coordinates spacing is defined.

x =linspace(-10,10,100);

%first function is defined.

y1_1 =(- x.^2+8.*x -11).^(1/2)+4;

y1_2 =4-(- x.^2+8.*x -11).^(1/2);

%second function is defined.

y2_1 =(4- x).^(1/2).*(x +4).^(1/2);

y2_2 =-(4- x).^(1/2).*(x +4).^(1/2);

Y1 =[y1_1,y1_2];

Y2 =[y2_1,y2_2];

X =[x,x];

%Plot command is used to draw the two function.

plot(X,Y1,X,Y2)

legend('(x-4)^2 + (y-4)^2 = 5','x^2 + y^2 = 16')

xlabel('x')

ylabel('y')

Output:

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 6, Problem 23P

From the above graph, it is observed that there are two roots at (1.8,3.6) and (3.6,1.8).

Consider theequations,

(x4)2+(y4)2=5x2+y2=16

Rewrite the equation as below,

u(x,y)=5(x4)2(y4)2v(x,y)=16x2y2

Partial differentiate the above functions with respect to x,

ux=x[5(x4)2(y4)2]=x(5)x[(x4)2]x[(y4)2]=02(x4)0=2x+8

And,

vx=x(16x2y2)=x(16)x(x2)x(y2)=02x0=2x

Now, partial differentiate the above functions with respect to y,

uy=y[5(x4)2(y4)2]=y(5)y(x4)2y(y4)2=002(y4)=2y+8

And,

vy=y(16x2y2)=y(16)y(x2)y(y2)=002y=2y

Use initial guesses x0=1.8 and y0=3.6, the first iteration is,

x0+1=x0u0v0yv0u0yu0xv0yu0yv0xx1=1.8{5((1.8)4)2((3.6)4)2}(2(3.6))(16(1.8)2(3.6)2)(2(3.6)+8)(2(1.8)+8)(2(3.6))(2(3.6)+8)(2(1.8))=1.8{5(2.2)2(0.4)2}(7.2)(163.2412.96)(7.2+8)(3.6+8)(7.2)(7.2+8)(3.6)=1.8056

And,

y0+1=y0v0u0xu0v0xu0xv0yu0yv0xy1=3.6(16(1.8)2(3.6)2)(2(1.8)+8){5((1.8)4)2((3.6)4)2}(2(1.8))(2(1.8)+8)(2(3.6))(2(3.6)+8)(2(1.8))=3.6(163.2412.96)(3.6+8){5(2.2)2(0.4)2}(3.6)(3.6+8)(7.2)(7.2+8)(3.6)=3.5694

Now, use x1=1.8056 and y1=3.5694, the second iteration is,

x1+1=x1u1v1yv1u1yu1xv1yu1yv1xx2=1.8056{5((1.8056)4)2((3.5694)4)2}(2(3.5694))(16(1.8056)2(3.5694)2)(2(3.5694)+8)(2(1.8056)+8)(2(3.5694))(2(3.5694)+8)(2(1.8056))=1.8056{0.00081}(7.14)(0.00081)(0.86)(4.389)(7.14)(0.86)(3.6112)=1.80583

And,

y1+1=y1v1u1xu1v1xu1xv1yu1yv1xy2=3.5694(16(1.8056)2(3.5694)2)(2(1.8056)+8){5((1.8056)4)2((3.5694)4)2}(2(1.8056))(2(1.8056)+8)(2(3.5694))(2(3.5694)+8)(2(1.8056))=3.5694{0.00081}(4.389)(0.00081)(3.6112)(4.389)(7.14)(0.86)(3.6112)=3.56917

Now, use x2=1.80583 and y2=3.56917, the third iteration is,

x2+1=x2u2v2yv2u2yu2xv2yu2yv2xx3=1.80583{5((1.80583)4)2((3.56917)4)2}(2(3.56917))(16(1.80583)2(3.56917)2)(2(3.56917)+8)(2(1.80583)+8)(2(3.56917))(2(3.56917)+8)(2(1.80583))=1.80583{0.0000035}(7.13834)(0.0000035)(0.86166)(4.38834)(7.13834)(0.86166)(3.61166)=1.80583

And,

y2+1=y2v2u2xu2v2xu2xv2yu2yv2xy2=3.56917(16(1.80583)2(3.56917)2)(2(1.80583)+8){5((1.80583)4)2((3.56917)4)2}(2(1.80583))(2(1.80583)+8)(2(3.56917))(2(3.56917)+8)(2(1.80583))=3.56917{0.0000035}(4.38834)(0.0000035)(3.61166)(4.38834)(7.13834)(0.86166)(3.61166)=3.56917

Thus, all the iteration can be summarized as below,

i xi yi
0 1.8 3.6
1 1.8056 3.5694
2 1.80583 3.56917
3 1.80583 3.56917

Hence, the root of the simultaneous non-linear equations y=x2+1 and y=2cosx is x=1.80583 and y=3.56917.

Use initial guesses x0=3.6 and y0=1.8, the first iteration is,

x0+1=x0u0v0yv0u0yu0xv0yu0yv0xx1=3.6{5((3.6)4)2((1.8)4)2}(2(1.8))(16(3.6)2(1.8)2)(2(1.8)+8)(2(3.6)+8)(2(1.8))(2(1.8)+8)(2(3.6))=3.6{0}(3.6)(0.2)(4.4)(0.8)(3.6)(4.4)(7.2)=3.56944

And,

y0+1=y0v0u0xu0v0xu0xv0yu0yv0xy1=1.8(16(3.6)2(1.8)2)(2(3.6)+8){5((3.6)4)2((1.8)4)2}(2(3.6))(2(3.6)+8)(2(1.8))(2(1.8)+8)(2(3.6))=1.8{0.2}(0.8)(0)(7.2)(0.8)(3.6)(4.4)(7.2)=1.80556

Now, use x1=3.56944 and y1=1.80556, the second iteration is,

x1+1=x1u1v1yv1u1yu1xv1yu1yv1xx2=3.56944{5((3.56944)4)2((1.80556)4)2}(2(1.80556))(16(3.56944)2(1.80556)2)(2(1.80556)+8)(2(3.56944)+8)(2(1.80556))(2(1.80556)+8)(2(3.56944))=3.56944{0.00095}(3.61112)(0.00095)(4.38888)(0.86112)(3.61112)(4.38888)(7.13888)=3.56917

And,

y1+1=y1v1u1xu1v1xu1xv1yu1yv1xy2=1.80556(16(3.56944)2(1.80556)2)(2(3.56944)+8){5((3.56944)4)2((1.80556)4)2}(2(3.56944))(2(3.56944)+8)(2(1.80556))(2(1.80556)+8)(2(3.56944))=1.80556{0.00095}(0.86112)(0.00095)(7.13888)(0.86112)(3.61112)(4.38888)(7.13888)=1.80583

Now, use x2=3.56917 and y2=1.80583, the third iteration is,

x2+1=x2u2v2yv2u2yu2xv2yu2yv2xx3=3.56917{5((3.56917)4)2((1.80583)4)2}(2(1.80583))(16(3.56917)2(1.80583)2)(2(1.80583)+8)(2(3.56917)+8)(2(1.80583))(2(1.80583)+8)(2(3.56917))=3.56917{0.0000035}(3.61166)(0.0000035)(4.38834)(0.86166)(3.61166)(4.38834)(7.13834)=3.56917

And,

y2+1=y2v2u2xu2v2xu2xv2yu2yv2xy2=1.80583(16(3.56917)2(1.80583)2)(2(3.56917)+8){5((3.56917)4)2((1.80583)4)2}(2(3.56917))(2(3.56917)+8)(2(1.80583))(2(1.80583)+8)(2(3.56917))=1.80583{0.0000035}(0.86166)(0.0000035)(7.13834)(0.86166)(3.61166)(4.38834)(7.13834)=1.80583

Thus, all the iteration can be summarized as below,

i xi yi
0 3.6 1.8
1 3.56944 1.80556
2 3.56917 1.80583
3 3.56917 1.80583

Hence, the root of the simultaneous non-linear equations y=x2+1 and y=2cosx is x=3.56917 and y=1.80583.

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