Solutions to Problems in Chapter Three

Transcription

1 Solutions to Problems in Chapter Three Test Your Understanding Problems T3.1-1 The session is diary diary1.txt x = [1:5];y = [0:2:8]; x.*y save session1 clear x y load session1 x.*y diary off type diary1.txt x = [1:5];y = [0:2:8]; x.*y save session1 clear x y load session1 x.*y diary off T3.3-1 The script file is % script file rad_deg.m rad_angle = [1:5]; deg_angle = rad_angle*(180/pi); angle_table = format bank disp( radians degrees ) disp(angle_table) The session is 3-1

2 rad_deg radians degrees T3.3-2 The script file is % script file sphere_v.m format bank r = [1:.1:2]; V = 4*pi*r.^3/3; vol_table = [r,v ]; disp( radius volume ) disp(vol_table) The session is sphere_v radius volume T3.3-3 The script file is % script file sphere a.m r = input( Enter a value for the radius: ); A = 4*pi*r.^2; disp( The area of the sphere is: ) A 3-2

3 The session is sphere a Enter a value for the radius: 3 The area of the sphere is: A = T3.4-1 The session is x = [5:20:85]; y = [10:30:130]; log(x.*y)-(log(x)+log(y)) 1.0e-014 * T3.4-2 The session is x = sqrt(2+6i) x = i abs(x) angle(x) real(x) imag(x) T3.4-3 The session is x = [0:.4:2*pi]; exp(i*x)-(cos(x)+i*sin(x)) The answers are essentially zero, which demonstrates that the identity is correct. 3-3

4 T3.4-4 The session is x = [0:.4:2*pi]; asin(x)+acos(x)-pi/2 The answers are essentially zero, which demonstrates that the identity is correct. T3.4-5 The session is x = [0:.4:2*pi]; tan(2*x)-2*tan(x)./(1-tan(x).^2) The answers are essentially zero, which demonstrates that the identity is correct. T3.4-6 The session is x = [0:.1:5]; sin(i*x)-i*sinh(x) The answers are essentially zero, which demonstrates that the identity is correct. T3.5-1 The function file is function y = f5(x) y = exp(-.2*x).*sin(x+2)-.1; You can plot the function to obtain solution estimates to use with fzero, oryoucansimply try values of x between 0 and 10. The session is fzero( f5,0) fzero( f5,4) fzero( f5,6) So the solutions are x = , , and

5 T3.5-2 The function file is function y = f6(x) y = 1 + exp(-.2*x).*sin(x+2); You can plot the function to obtain solution estimates to use with fminbnd, oryoucan simply try values of x between 0 and 10. The session is fminbnd( f6,0) f6(ans) fminbnd( f6,10) f6(ans) So the solutions are (x, y) =(2.5150, ) and (x, y) =(8.7982, ). T3.5-3 Refer to Example Modify the function file given in the example to use an area of 200 ft 2 rather than 100 ft 2. The function file is function L = channel(x) L = 200./x(1)-x(1)./tan(x(2))+2*x(1)./sin(x(2)); Because this problem is similar to Example 3.5-1, we can try the same guess as in the example. The session is x = fminsearch( channel,[20,1]) x = The answer is d = ft and θ = radians, or 60. T3.6-1 Using cell indexing to create a 1 4 cell array, the script file is: A{1} = [1:4]; A{2} = [0,9,2]; A{3} = [2:5]; 3-5

6 A{4} = [6:8]; [x,y] = deal(a{1:2}); B = [x,y] C = [A{2};A{4}] [u,v] = deal(a{2:3}); D = min([u,v]) T3.7-1 The script file is student(1).name = John Smith ; student(1).ssn = ; student(1). = smithj@myschool.edu ; student(1).tests = [67,75,84]; student(2).name = Mary Jones ; student(2).ssn = ; student(2). = jonesm@myschool.edu ; student(2).tests = [84,78,93]; student(3).name = Alfred E. Newman ; student(3).ssn = ; student(3). = NewmanA@myschool.edu ; student(3).tests = [55,45,58]; T3.7-2 The session is student(3).tests(2) = 53; T3.7-3 The session is new_student = rmfield(student, SSN ) End-of-Chapter Problems 1. The session is x = [1:5]; save session2 x = [1:10]; load session2 z = 3*x z = 3-6

7 diary off type diary2.txt x = [1:5]; save session2 x = [1:10]; load session2 z = 3*x z = diary off 2. No answer is required here. 3. The script file is % File p3.m k = [1:10]; amount = 1000*(1.055).^k; table = [k,amount ]; disp( Year Amount ) format bank disp(table) The session is p3 Year Amount

8 4. The script file is % File p4.m TF = [0:10:100]; TC = 5*(TF-32)/9; table = [TF,TC ]; disp( deg F deg C ) format bank disp(table) The session is p4 deg F deg C The script file is % File p5.m disp( This computes the roots of ax^2 + bx + c = 0. ) a = input( Enter the coefficient a: ); b = input( Enter the coefficient b: ); c = input( Enter the coefficient c: ); x = roots([a,b,c]); disp( The roots are: disp(x) The session for a =5,b = 24, and c = 145 is p5 This computes the roots of ax^2 + bx + c = 0. Enter the coefficient a: 5 3-8

9 Enter the coefficient b: 24 Enter the coefficient c: 145 The roots are: -2e e+000-2e e The script file is labor1 = input( Enter the unit labor cost for product 1: ); labor2 = input( Enter the unit labor cost for product 2: ); labor3 = input( Enter the unit labor cost for product 3: ); labor4 = input( Enter the unit labor cost for product 4: ); u1 = [6, 2, 4, 9];u3 = [1, 4, 2, 3]; u2 = [labor1,labor2,labor3,labor4]; U = [u1, u2, u3 ]; P = [10, 12, 13, 15;8, 7, 6, 4;12, 10, 13, 9;6, 4, 11,5]; C = U *P; Quarterly_Costs = sum(u *P); Category_Costs = sum((u *P) ); disp( The cost for quarter 1 is: ) disp(quarterly_costs(1)) disp( The cost for quarter 2 is: ) disp(quarterly_costs(2)) disp( The cost for quarter 3 is: ) disp(quarterly_costs(3)) disp( The cost for quarter 4 is: ) disp(quarterly_costs(4)) disp( The materials cost is: ) disp(category_costs(1)) disp( The labor cost is: ) disp(category_costs(2)) disp( The transportation cost is: ) disp(category_costs(3)) The results are (in thousands of dollars): The cost for quarter 1 is: 444 The cost for quarter 2 is: 391 The cost for quarter 3 is: 558 The cost for quarter 4 is: 392 The materials cost is: 760 The labor cost is: 709 The transportation cost is:

10 7. The script file is alloy_2024 = input( Enter the desired amount of alloy 2024: ) alloy_6061 = input( Enter the desired amount of alloy 6061: ) alloy_7005 = input( Enter the desired amount of alloy 7005: ) alloy_7075 = input( Enter the desired amount of alloy 7075: ) alloy_356 = input( Enter the desired amount of alloy 356.0: ) alloy = [alloy_2024,alloy_6061,alloy_7005,alloy_7075,alloy_356]; composition = [4.4,1.5,.6,0,0;0,1,0,.6,0;0,1.4,0,0,4.5; 1.6,2.5,0,0,5.6;0,.3,0,7,0]; raw_material = alloy*composition; disp( The amount of copper required is ) disp(raw_material(1)) disp( The amount of magnesium required is ) disp(raw_material(2)) disp( The amount of manganese required is ) disp(raw_material(3)) disp( The amount of silicon required is ) disp(raw_material(4)) disp( The amount of zinc required is ) disp(raw_material(5)) 8. The session is x = [0:2]; y = -3+i*x; abs(y) sqrt(y) i i i (-5-7i)*y i i i y/(6-3i) i i i 3-10

11 9. The session is x = -5-8i;y = 10-5i; abs(x*y) angle(x*y) abs(x/y) angle(x/y) The session is (180/pi)*atan2(8,5) (180/pi)*atan2(8,-5) (180/pi)*atan2(-8,5) (180/pi)*atan2(-8,-5) The answers are in degrees. 11. The session is x = [0:.5:5]; sinh(x)-((exp(x)-exp(-x))/2) Because the results are all zero, the identity is correct for the given values of x. 3-11

12 12. The session is x=[-100:10:100]; asinh(x)-log(x+sqrt(x.^2+1)) 1.0e-011 * Columns 1 through Columns 8 through Because the results are all essentially zero, the identity is correct for the given values of x. 13. The session is x = [0:.1:1]; y = cos(x); z = y.^3-3*y.^2+4*y+10; table=[x,z ]; disp( x z ),disp(table) x z The session for part (a) is epsilon = 8.854e-12; d =.001*[3, 4, 5, 10]; L = 1;r =.001; C = pi*epsilon*l./log((d-r)./r)*1.0e+11; table = [d,c ]; disp( d (m) C (F) X 10^11 ),disp(table) d (m) C (F) X 10^

13 Thescriptfileforpart(b)is epsilon = 8.854e-12; L = 1;r =.001; d = input( Enter a value for d in meters: ) C = pi*epsilon*l./log((d-r)./r); disp( The capacitance in farads is: ) disp(c) 15. The script file is beta_deg = input( Enter a value for beta in degrees: ); % Convert to radians. beta = (pi/180)*beta_deg; mu = input( Enter a value for mu: ); F2 = input( Enter a value for force F2: ); F1 = F2*exp(mu*beta); disp( The force F1 is: ) disp(f1) The answer for β = 130, F 2 = 100, and µ =0.3 isf 1 = newtons. 16. The function file is function y = sind(x) % Computes sin(x) for x in degrees. y = sin((pi/180)*x); A test session follows. sind([45,90]) The function file is function y = temp(x) % Converts from Fahrenheit to Celsius. y = (5/9)*(x-32); 3-13

14 A test session follows. temp([32,212]) The function file is function t = time(h,v0,g) % Computes time t to reach a specified height h, with initial speed v0. roots([.5*g,-v0,h]) A test session follows. time(100,50,9.81) The smaller value is the time to reach the height while ascending; the larger value is the time to reach the height while descending. 19. The function files are function h = height(v,r) h = (V-2*pi*r.^3/3)./(pi*r.^2); function cost = tower(r) h = height(500,r); cost = 600*pi*r.*h+800*pi*r.^2; 3-14

15 The session is: optimum r = fminbnd( tower,0,100) optimum r = min cost = tower(optimum r) min cost = e+4 optimum h = height(500,optimum r) optimum h = So the optimum radius is meters; the optimum height is meters, and the minimum cost is $91, The function file is function [L, total] = field(w,a) L = (A-W.^2/8)./W; total = 2*L+W+2*W/sqrt(2); The session is: [L, total] = field(6,80) The results are L = meters and total = meters. 21. The function file is function cost = fence(r) A = 1600; L = (A-0.5*pi*R.^2)./(2*R); cost = 30*(2*R+2*L)+40*pi*R; The session is optimum R = fminbnd(,0,100) optimum R = min cost = fence(optimum R) min cost = e+3 optimum L = (A-0.5*pi*optimum R.^2)./(2*optimum R) 3-15

16 optimum L = The optimum radius is feet, and the corresponding length is feet. minimum cost is $5, The 22. The function files are function deltav = volume(t) global r x V = 1e+9+1e+8*(1-exp(-t/100))-r*t; deltav = V-0.01*x*1e+9; function time = decrease(x,r) global r x time = fzero( volume,1); The session is time = decrease(50,1e+7) time = Thus it will take about 54 days. 23. The script file is % Using cell indexing: A(1,1) = { Motor 28C }; A(1,2) = { Test ID 6 }; A(2,1) = {[3,9;7,2]}; A(2,2) = {[6,5,1];} % Using content indexing: B{1,1} = Motor 28C ; B{1,2} = Test ID 6 ; B{2,1} = [3,9;7,2]; B{2,2} = [6,5,1]; disp(b{2,1}(1,1)) When this file is run, it displays the answer

17 24. See Section 2.3 in the text for a discussion of how to evaluate multivariable functions. Here we create a three dimensional cell array. The three layers correspond to the values L = 1, 2, and 3 respectively. Some combinations of r and d values result in negative or infinite values for the capacitance C, and thus are invalid cases. The script file is epsilon = 8.854e-12; r1 =.001*[1,2,3]; r = [r1,r1,r1,r1 ]; d1 =.001*[3,4,5,10]; d=[d1;d1;d1];l = 1 C{1,1,1} = L = 1 ; C{1,2,1} = d ; C{2,1,1} = r ; C{2,2,1} = pi*epsilon*l./log((d-r)./r); L = 2; C{1,1,2} = L = 2 ; C{1,2,2} = d ; C{2,1,2} = r ; C{2,2,2} = pi*epsilon*l./log((d-r)./r); L = 3; C{1,1,3} = L = 3 ; C{1,2,3} = d ; C{2,1,3} = r ; C{2,2,3} = pi*epsilon*l./log((d-r)./r); C{2,2,2}(1,3) The structure of the array C can be seen by typing C. For the first layer you will see displayed: C(:,:,1) = L = 1 d r [3x4 double] The other two layers have a similar structure. To see the entire array, type celldisp(c). The capacitance values are in the (2,2) cell in each layer. The values for L =2are in the second layer (because L = 2 is the second value of L). The capacitance value C ij corresponds to the values r i and d j. Thus the values for r =0.001 and d =0.005 are in the (1,3) element of the array in cell (2,2) (because r =0.001 is the first value of r and d =0.005 is the third value of d). The capacitance value is (a) One slug is the same as kg. One pound is the same as newtons. One foot is the same as meter. The script file is named convert.m and is %Convert slugs to kg convert(1).mass = ; 3-17

18 %Convert kg to slugs convert(2).mass = 1/convert(1).mass; %Convert lbs to newtons convert(1).force = ; %Convert newtons to lbs convert(2).force = 1/convert(1).force; %Convert feet to meters convert(1).length =.3048; %Convert meters to feet convert(2).length = 1/convert(1).length; (b) The session is convert(1).length* convert(2).length* convert(2).force* convert(1).force* convert(1).mass* convert(2).mass* The script file is bridge(1).location = Smith St ; bridge(1).maxload = 80; bridge(1).yearbuilt = 1928; bridge(1).duemaint = 1997; bridge(2).location = Hope Ave ; bridge(2).maxload = 90; bridge(2).yearbuilt = 1950; bridge(2).duemaint = 1999; 3-18

19 bridge(3).location = Clark St ; bridge(3).maxload = 85; bridge(3).yearbuilt = 1933; bridge(3).duemaint = 1998; bridge(4).location = North Rd ; bridge(4).maxload = 85; bridge(4).yearbuilt = 1960; bridge(4).duemaint = 1998; 27. The session is bridge(3).duemaint = 2000; 28. The session is bridge(5).location = Shore Rd ; bridge(5).maxload = 85; bridge(5).yearbuilt = 1997; bridge(5).duemaint = 2002; 3-19