Page 1 / 5 FACULTY OF ENGINEERING/EUROPEAN UNIVERSITY OF LEFKE MATH 224 (MATH 208/302/305) ENGINEERING MATHEMATICS (NUMERICAL METHODS) SPRING 14-15 GRADUATION MAKEUP EXAM Date/Time/Place: 24. 06. 2015/09:00-11:00/AS2XX Instructor: Prof. Dr. Hüseyin Oğuz Student Registration No: Student Name-Surname: Important Note: Your own scientific calculator is only allowable gadget to use during the exam with the prohibition of its exchange. Please choose 4 questions out of 5 by indicating in this sheet. 1. (25 p) The lateral surface area,, of a cone is given by: = +h where is the radius of the base and is the height. Determine the radius of a cone which has a surface area of 1200 m 2 and a height of 20 m by using Newton-Raphson method with =, =.. 2. (25 p) A company manufactures fitting components. The composition of these components are given according to the materials used as follows: Component Material 1 Material 2 Material 3 1 15 30 55 2 30 45 25 3 55 20 25 Supplies of these materials vary from week to week, so the company needs to determine a different manufacturing run each week. For example, one week the total amounts of materials available are 5700 units of Material 1, 5800 units of Material 2, and 4800 units of Material 3. a) Set up the system of linear algebraic equations modeling the manufacturing run, b) Use LU Factorization method with partial pivoting to solve for the number of component 1, component 2, and component 3 to be manufactured this week. 3. (25 p) The following data can be modeled by a saturation-growth-rate model of the form = x 1 3 5 7 10 y 2.2 5.0 5.5 6.1 6.6 a) Write the equation in a linear form, and use linear least-squares regression to determine the constants and for which the function best fits the data. b) Use the equation to estimate y value at a x value of 6. 4. (25 p) Determine the distance traveled for the following data by using the best combination of the trapezoidal and Simpson s rules: t, min 1 2 3.25 4.5 6 7 8 9 9.5 10 v, m/s 5 6 5.5 7 8.5 8 6 7 7 5 5. (25 p) a) Determine the value of 6+5, expressing the result in polar and rectangular forms. b) For a transmission line, the characteristic impedance is given by: = + + ; =25Ω,=510,=8010,=0.0410, Determine in polar form. Recall: =2000 / = ; = +4 + ; +3 +3 + 6 8
MATH 302 NUMERICAL METHODS SPRING 14-15 GRADUATION MAKEUP EXAM SOLUTIONS Date: 24. 06. 2015 Instructor: Prof. Dr. Hüseyin Oğuz 1. First derive the equation with the values substituted as follows: = +h = +20 1200=0 = +400 1200=0 = +400 + +400 = Applying Newton-Raphson formula given above gives with the initial guess of =17 Iteration 1: Iteration 2: = =17 1717 +400 1200 17 +400+17 17 +400 =17 201.872 117.052 =15.275 = 15.275 17 100=11.29>0.5 15.275 Page 2 / 5 = =15.275 15.27515.275 +400 1200 15.275 +400+15.275 15.275 +400 =15.275 7.6598 =15.204 108.188 = 15.204 15.275 100=0.47<0.5 15.204 =15.275 h 2 2. a) The system of linear algebraic equations modeling the production run can be set up by using table data given as follows : 15 100 + 30 100 + 55 100 =5700 0.15 +0.30 +0.55 =5700 30 100 + 45 100 + 20 100 =70 0.30 +0.45 +0.20 =5800 55 100 + 25 100 + 25 100 =4800 0.55 +0.25 +0.25 =4800 The equations with the partial pivoting are given as follows: 0.55 +0.25 +0.25 =4800 0.30 +0.45 +0.20 =5800 0.15 +0.30 +0.55 =5700 b) The system can be written in matrix form as follows: 4800 0.30 0.45 0.20 =5800 0.15 0.30 0.55 5700 The solution of the above three simultaneous linear equations by using LU Factorization method will give the value of,,
4800 =0.30 0.45 0.20;=5800 0.15 0.30 0.55 5700 Page 3 / 5 Forward Elimination of Unknowns: Since there are three equations, there will be two steps of forward elimination of unknowns. First step: Divide Row 1 by 0.55 and then multiply it by 0.30 ( =. =0.5455 and subtract the result from Row 2:. 2 1 0.55 0.30= 0 0.3136 0.0636 Divide Row 1 by 0.55 and then multiply it by 0.15 =. =0.2727 and subtract the results from Row 3:. 3 1 0.55 0.15= 0 0.23182 0.48182 = 0 0.3136 0.0636 0 0.2318 0.4818 Second step: We now divide Row 2 by (0.3136) and then multiply by (0.2318) =. =0.7392 and subtract the results from Row 3:. 3 2 0.3136 0.2318= 0 0 0.4348 Second step (Forward substitution): Third step (Back substitution): = 0 0.3136 0.0636 0 0 0.4348 1 0 0 1 0 0 = 1 0=0.5455 1 0 1 0.2727 0.7392 1 = 1 0 0 4800 0.5455 1 0 =5800 0.2727 0.7392 1 5700 =4800 =5800 0.54554800=3181.6 =5700 0.27274800 0.73923181.6=2039.2 = 0 0.3136 0.0636 0 0 0.4348 4800 =3181.6 2039.2 = 2039.2 =4689.97 0.4348
= 3181.6 0.06364689.97 0.3136 = 4800 0.259194.3 0.254689.97 0.55 =9194.3 =2416.2 Page 4 / 5 The solution vector is: 2416 =9194 4690 3. a) Employing saturation growth-rate model requires transformation (linearization) of the model to use least-square method as follows: = = = + = + = ; = ; = ;= = 1 = 1 ; = = Apply Least-Square method for the transformed form of saturation-growth-rate model to find regression coefficients as follows: = = Table. Summations of data to calculate coefficients of the linearized model: = = 1 1 2.2 0.4546 1 0.4546 1 2 3 5.0 0.2 0.3333 0.0667 0.1111 3 5 5.5 0.1818 0.2 0.03636 0.04 4 7 6.1 0.1639 0.1429 0.02342 0.02042 5 10 6.6 0.1515 0.1 0.01515 0.01 1.1518 1.7762 0.5962 1.1815 By substituting the corresponding values at the bottom of the table to the least square equations given above, we have: = 1.1518 5 = 50.5962 1.77621.1518 51.1815 1.7762 =0.3397 0.3397 1.7762 =0.1097= 1 5 = 1 0.1097 =9.1158;= = 0.3397 0.1097 =3.0966 b) Saturation-growth-rate model to estimate y value for a given x value of 6: = + = 9.1158 3.0966+ = 9.11586 3.0966+6 =6.013 4. Best combination of trapezoidal and Simpson s rules to the data is given as follows:
Page 5 / 5 2 1 5+6 2 +4.5 26+45.5+7 +6 4.5 7+8.5 +9 6 8.5+38+6+7 6 2 8 +10 9 7+47+5 =59.9375 6 =59.9375 60 =3596.25 1 5. a) 6+5=6 +5 tan 5 = 61 39.81 1 6 Applying De Moivre s theorem: 6+5 = 61 39.81 = 61 339.81 =476.4 119.43 For rectangular form: =cos+sin 476.4 119.43 =476.4cos119.43 +sin119.43 = 234.1+415 b) Determine the moduli and arguments of the complex roots: = + / + = / = = The roots are symmetrically displaced from one another = =180 apart round an Argand diagram as follows: + + = 25+2000510 8010 +20000.0410 = 25+31.4159 8010 +2.51310 By multiplying numerator and denominator with the complex conjugate of the denominator as follows : 25+31.4159 8010 2.51310 =9.894810 3.769510 8010 +2.51310 8010 2.51310 6.95510 =142268.87 54198.42 142268.87 54198.42=142268.87 + 54198.42 tan 54198.42 142268.87 Applying De Moivre s theorem: =152242.9 20.86 142268.87 54198.42 =152242.9 20.86 =152242.9 20.86 = 390.2 10.43 1 4h, 390.2 10.43 +180 = 390.2 169.57 2 2