Engineering Mathematics Test 10

Concept:

Trapezoidal rule states that for a function y = f(x)

xn = x0 + nh, where n = Number of sub-intervals

h = step-size

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{h}{2}\left[ {\left( {{y_0} + {y_n}} \right) + 2\left( {{y_1} + {y_2} + {y_3} + \ldots + {y_{n - 1}}} \right)} \right]\)

For a trapezoidal rule, a number of sub-intervals must be a multiple of 1.

Calculation:

N = 3

\(h = \frac{{{x_n} - {x_p}}}{n} = \frac{{1 - \left( 1 \right)}}{3} = \frac{2}{3}\)

By applying Trapezoidal rule, \(\mathop \smallint \limits_{ - 1}^1 \left| x \right|dx = \frac{h}{2}\left[ {{y_n} - {y_o}} \right] + 2\left( {{y_1} + {y_2} \ldots .. + {y_{n - 1)}}} \right]\)

\(= \frac{2}{{\frac{3}{2}}}\left[ {\left( {1 + 1} \right) + 2\left( {\frac{1}{3} + \frac{1}{3}} \right)} \right] = \frac{{10}}{9} = 1.1111\)

P1: For Numerical Integration, the methods used are:

1) Simpson’s 1/3 rule

2) Simpson 3/8 rule

3) Trapezoidal rule

∴ P₁ → M₃

P₂: Solution to a transcendental equation:

The method used are:

1) Bisection method

2) Newton Raplason method

3) Secant method

4) Regular – Falsi method

∴ P₂ → M₁

P₃: Solution to a system of the linear equation: Gauss elimination method

∴ P₃ → M₄

P₄: Solution to a differential equation.

The methods used are:

1) Euler’s method

2) Runge – Kutta method

∴ P₄ – M₂

Concept:

According to Newton-Raphson Method

\({X_{n + 1}} = {X_n} - \frac{{f\left( {{X_n}} \right)}}{{f'\left( {{X_n}} \right)}}\)

Calculation:

For function f(x) = x³+ x – 1 at x₀ = 1

\({X_1} = {X_0} - \frac{{f\left( {{X_0}} \right)}}{{f'\left( {{X_0}} \right)}} = 1 - \frac{{\left( {1 + 1 - 1} \right)}}{{\left[ {3 \times {{\left( 1 \right)}^2} + 1} \right]}} = 1 - \frac{1}{4} = \frac{3}{4}\)

Concept:

In Bisection method, for f(x) = 0 in the interval (a, b), the interval width is reduced by a factor of one one-half at each step, at the end of the n^th step, the new interval will be [a_n, b_n] of length \(\frac{{\left| {b - a} \right|}}{{{2^n}}}\).

The error, \(\varepsilon \ge \frac{{\left| {b - a} \right|}}{{{2^n}}}\) 

Calculation:

In the given question,

a = 2, b = 4

ε = 0.2% = 0.002

\( \Rightarrow 0.002 \ge \frac{{\left| {4 - 2} \right|}}{{{2^n}}}\) 

\(\Rightarrow 0.002 \ge \frac{2}{{{2^n}}}\) 

⇒ 2ⁿ ≥ 1000

⇒ n ≥ 10

The minimum number of iterations required are, n = 10

Concept:

\(\frac{{dy}}{{dx}} = f\left( {x,y} \right),\;y\left( {{x_0}} \right) = {y_0}\)

Runge-Kutta Method of first order:

k₁ = h f(x₀, y₀)

y₁ = y₀ + k where k = k₁ = h f (x₀, y₀)

y₂ = y₁ + k where k = h f (x₁, y₁)

Runge-Kutta Method of Second order:

k₁ = h f(x₀, y₀)

k₂ = h f (x₀ + h, y₀ + k₁)

\(k = \frac{{{k_1} + {k_2}}}{2}\)

y₁ = y₀ + k

Runge-Kutta Method of Third order:

k₁ = h f (x₀, y₀)

\(\begin{array}{l} {k_2} = hf\left( {{x_0} + \frac{h}{2},{y_0} + \frac{{{k_1}}}{2}} \right)\\ k' = hf\left( {{x_0} + h,{y_0} + {k_1}} \right) \end{array}\)  

\(\begin{array}{l} {k_3} = hf\left( {{x_0} + h,{y_0} + k'} \right)\\ k = \frac{1}{6}\left( {{k_1} + 4{k_2} + {k_3}} \right) \end{array}\)  

y₁ = y₀ + k

Runge-Kutta Method of fourth order:

k₁ = h f (x₀, y₀)

\(\begin{array}{l} {k_2} = hf\left( {{x_0} + \frac{h}{2},{y_0} + \frac{{{k_1}}}{2}} \right)\\ {k_3} = hf\left( {{x_0} + \frac{h}{2},{y_0} + \frac{{{k_2}}}{2}} \right) \end{array}\)  

\(\begin{array}{l} {k_4} = hf\left( {{x_0} + h,{y_0} + {k_3}} \right)\\ k = \frac{1}{6}\left( {{k_1} + 2{k_2} + 2{k_3} + {k_4}} \right) \end{array}\)  

y₁ = y₀ + k

Calculation:

Question is about third order Runge Kutta method with step-size h = 0.1.

\(\frac{{dy}}{{dx}} = - y = f\left( {x,y} \right);{x_0} = 0,\;{y_0} = 1\)

f(x₀, y₀ ) = f(0, 1) = -1

k₁ = h f (x₀, y₀) = 0.1 × (-1) = -0.1

\({k_2} = hf\left( {{x_0} + \frac{h}{2},{y_0} + \frac{{{k_1}}}{2}} \right) = hf\left( {0.05,\;0.95} \right)\)

= 0.1 [-0.95] = -0.095

k’ = h f (x₀ + h, y₀ + k₁) = h f (0.1, 0.9) = =0.9 h

k’ = -0.09

k₃ = h f (x₀ + h, y₀ + k’) = h f (0.1, 0.91) = -0.91 h

k₃ = -0.091

\(\begin{array}{l} k = \frac{1}{6}\left( {{k_1} + 4{k_2} + {k_3}} \right)\\ = \frac{{ - 1}}{6}\left( {0.1 + 4\left( {0.095} \right) + 0.091} \right) = - 0.095167 \end{array}\)  

f(x) = 2x³ + 5x - 12

f^I(x) = 6x² + 5

\(\begin{array}{l} {{\rm{x}}_{{\rm{k}} + 1}} = {{\rm{x}}_{\rm{k}}} - \frac{{{\rm{f}}\left( {{{\rm{x}}_{\rm{k}}}} \right)}}{{{\rm{f'}}\left( {{{\rm{x}}_{\rm{k}}}} \right)}}\\ 1.02{{\rm{x}}_{\rm{k}}} = {{\rm{x}}_{\rm{k}}} - \frac{{(2{{\rm{x}}^3}_{\rm{k}} + 5{{\rm{x}}_{\rm{k}}} - 12)}}{{6{{\rm{x}}^2}_{\rm{k}} + 5}} \end{array}\)

\(\begin{array}{l} 1.02{{\rm{x}}_{\rm{k}}} = \frac{{{{\rm{x}}_{\rm{k}}}{\rm{\;}}(6{{\rm{x}}^2}_{\rm{k}} + 5) - 2{{\rm{x}}^3}_{\rm{k}} - 5{{\rm{x}}_{\rm{k}}} + 12}}{{6{{\rm{x}}^2}_{\rm{k}} + 5}}\\ 1.02{{\rm{x}}_{\rm{k}}}(6{{\rm{x}}^2}_{\rm{k}} + 5) = {\rm{\;}}6{{\rm{x}}^3}_{\rm{k}} + 5{{\rm{x}}_{\rm{k}}} - 2{{\rm{x}}^3}_{\rm{k}} - 5{{\rm{x}}_{\rm{k}}} + 12 \end{array}\)

\(\begin{array}{l} 6.12{{\rm{x}}^3}_{\rm{k}} + 5.1{{\rm{x}}_{\rm{k}}} = 6{{\rm{x}}^3}_{\rm{k}} - 2{{\rm{x}}^3}_{\rm{k}} + 12\\ 0.12{{\rm{x}}^3}_{\rm{k}} + 5.1{{\rm{x}}_{\rm{k}}} + 2{{\rm{x}}^3}_{\rm{k}} - 12 = 0\\ 2.12{{\rm{x}}^3}_{\rm{k}} + 5.1{{\rm{x}}_{\rm{k}}} - 12 = 0 \end{array}\)

Concept:

\({\rm{Number\;of\;interval}} = \frac{{{\rm{b}} - {\rm{a}}}}{{\rm{h}}}{\rm{\;}}\)

Where,

b is the upper limit, a is the lower limit, h is the step size

According to Simpson's \(\frac{1}{3}\) rule

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{3}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{n}}} + 2\left( {{{\rm{y}}_2} + {{\rm{y}}_4} + {\rm{\;}} \ldots } \right) + 4\left( {{{\rm{y}}_1} + {{\rm{y}}_3} + {\rm{\;}} \ldots } \right)} \right]\)

Calculation:

Given:

f(x) = x|x|

a = -1, b = 1.4, h = 0.6

\({\rm{Number\;of\;interval}} = \frac{{1.4 - \left( { - 1} \right)}}{{0.6}} = 4\)

According to Simpson's 1/3^rd rule

\(\mathop \smallint \limits_{ - 1}^{1.4} x\left| x \right|dx = \frac{{\rm{h}}}{3}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_4} + 2\left( {{{\rm{y}}_2}} \right) + 4\left( {{{\rm{y}}_1} + {{\rm{y}}_3}} \right)} \right]\)

\(\mathop \smallint \limits_{ - 1}^{1.4} x\left| x \right|dx = \frac{{0.6}}{3}\left[ { - 1 + 1.96 + 2\left( {0.04} \right) + 4\left( { - 0.16 + 0.64} \right)} \right]{\rm{\;}}\)

\(\therefore \mathop \smallint \limits_{ - 1}^{1.4} x\left| x \right|dx = 0.592{\rm{\;}}\)       

Given that, \(y' = \frac{{dy}}{{dx}} = \sqrt {{x^2} + {y^2}} \)

y(1) = 2 ⇒ y₀ = 2, x₀ = 1

h = 0.1

According to Euler’s method, \({y_{n + 1}} = {y_n} + hy'\left( {{x_n},{y_n}} \right)\)

First iteration:

y₀ = 2, x₀ = 1, h = 0.1

\(y'\left( {{x_0},{y_0}} \right) = \sqrt {{1^2} + {2^2}} = \sqrt 5 \)

\({y_1} = {y_0} + hy'\left( {{x_0},{y_0}} \right) = 2 + 0.1\left( {\sqrt 5 } \right) = 2.22\)

Second iteration:

y₁ = 2.22, x₁ = x₀ + h = 1 + 0.1 = 1.1, h = 0.1

\(y'\left( {{x_1},{y_1}} \right) = \sqrt {{{1.1}^2} + {{2.2}^2}} = \sqrt 6 \)

let f(x) = 2x – log x – 7

Taking x₀ = 3.5, x₁ = 4, in the method of false position, we get

\({x_2}\; = \;{x_0} - \frac{{\left( {{x_1} - {x_0}} \right)}}{{f\left( {{x_1}} \right) - f\left( {{x_0}} \right)}}f\left( {{x_0}} \right)\; = \;3.5 - \frac{{0.5}}{{0.3979 + 0.5441}}\left( { - 0.5441} \right)\)

Since f(3.7888) = - 0.0009 and f(u) = 0.3979 therefore the root lies between 3.7888 and 4

Taking x₀ = 3.7888, x₁ = 4 we obtain

\({x_3}\; = \;3.7888 - \frac{{0.2112}}{{0.3988}}\left( { - 0.009} \right)\; = \;3.7893\)

Hence the required root correct to three places of decimal is 3.789

Concept:

Partial pivoting and Complete pivoting: the numerically largest coefficient of x is chosen from all the equation and brought as the first pivot by interchanging the first equation with the equation having largest coefficient of x.

In the second step numerically, largest coefficient of y is chosen from the remaining equation (leaving the first equation) and brought as the second pivot by interchanging the second equation with the equation having the largest coefficient of y’

If we chose numerically largest coefficient of the entire matrix of coefficient, this requires not only an interchange of equations but also an interchange of the position of the variables. This method of elimination is called complete pivoting

Calculation:

In matrix form, the given system of equation can be written as

Ax = B

Augmented matrix for Gauss- eliminated is

\(\left[ {\begin{array}{*{20}{c}} 1&4&{ - 1}\\ 1&1&{ - 6}\\ 3&{ - 1}&{ - 1} \end{array}\left| {\begin{array}{*{20}{c}} { - 5}\\ { - 12}\\ 4 \end{array}} \right.} \right]\)

\(\left[ {\begin{array}{*{20}{c}} 1&4&{ - 1}\\ 1&1&{ - 6}\\ 3&{ - 1}&{ - 1} \end{array}\left| {\begin{array}{*{20}{c}} { - 5}\\ { - 12}\\ 4 \end{array}} \right.} \right]\)

In first column maximum element is 3.

\(\left[ {\begin{array}{*{20}{c}} 3&{ - 1}&{ - 1}\\ 1&1&{ - 6}\\ 1&4&{ - 1} \end{array}\left| {\begin{array}{*{20}{c}} 4\\ { - 12}\\ { - 5} \end{array}} \right.} \right]\)

\(\left[ {\begin{array}{*{20}{c}} 3&{ - 1}&{ - 1\;}\\ 0&{4/3}&{\frac{{ - 17}}{3}}\\ 0&{13/3}&{ - 2/3} \end{array}\left| {\begin{array}{*{20}{c}} 4\\ { - 40/3}\\ { - 19/3} \end{array}} \right.} \right]\)

Now to get the second pivot, find the largest coefficient in the second column which is 13/3

\(\left[ {\begin{array}{*{20}{c}} 3&{ - 1}&{ - 1\;}\\ 0&{13/3}&{ - 2/3}\\ 0&0&{\frac{{ - 71}}{{13}}} \end{array}\left| {\begin{array}{*{20}{c}} 4\\ { - 19/3}\\ {\frac{{ - 178}}{{13}}} \end{array}} \right.} \right]\)

\(\begin{array}{l} \frac{{ - 71}}{{13}}z = \frac{{ - 148}}{{13}} \Rightarrow z = 2.0845\\ \frac{{13}}{3}y - \frac{2}{3}z = \frac{{ - 19}}{3} \Rightarrow y = - 1.1408 \end{array}\)  

3x – y – z = 4 ⇒ x = 1.6479