An Engineers Guide to Matlab Solutions

Solution to Exercises Chapter 1 1.1 E1 = 1+5^3+3^3-153 E2 = 1+6^4+3^4+4^4-1634 E3 = 5^6+4^6+8^6+8^6+3^6+4^6-548834 E4 = 1+factorial(7)-71^2 E5 = factorial(1)+factorial(4)+factorial(5)+factorial(6)+factorial(7)+factorial(8)-215^2

1.2 pie = pi - (100-(2125^3+214^3+30^3+37^2)/82^5)^(1/4)

1.3 format long g I1 = 53453/log(53453) I2 = 613*exp(1)/37-35/991 format short

1.4 x = (49-27*sqrt(5)+3*sqrt(6)*sqrt(93-49*sqrt(5)))^(1/3) R = 0.5*sqrt((8*2^(2/3)-16*x+2^(1/3)*x^2)/(8*2^(2/3)-10*x+2^(1/3)*x^2)) Answers: x = 2.5375 + 1.9080i

R = 0.8161 - 0.0000i

1.5 E1 = cot(pi/5)-sqrt(25+10*sqrt(5))/5 E2 = sin(pi/15)-sqrt(7-sqrt(5)-sqrt(30-6*sqrt(5)))/4 E3 = pi-16*atan(1/5)+4*atan(1/239)

1.6 g = sqrt(17-sqrt(17)); b = sqrt(17+sqrt(17)); a = sqrt(34+6*sqrt(17)+sqrt(2)*(sqrt(17)-1)*g-8*b*sqrt(2)); E1 = sin(pi/17)-sqrt(2)/8*sqrt(g^2-sqrt(2)*(a+g))

1.7 a = 1; b = 2; E1 = exp(a)-sinh(a)-cosh(a) E2 = sinh(2*b)/(cosh(a+b)*cosh(a-b))-tanh(a+b)+tanh(a-b)

1.8 L = 1.5; h = 1; r = 1.6; V = L*(r^2*acos((r-h)/r)-(r-h)*sqrt(2*r*h-h^2)) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a1retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Answer: V = 3.2209

1.9 g = 60*pi/180; a = 35*pi/180; n = 4/3; D = a-g+asin(n*sin(g-asin(sin(a)/n))) Answer: D = 0.4203

1.10 k = 7; n =12; Cnk = factorial(n)/factorial(k)/factorial(n-k) Answer: Cnk = 792

1.11 r = 2.5; I = (pi/8-8/9/pi)*r^4 Answer: I = 4.2874

1.12 c = 5; K = (4*c-1)/(4*c+4)+0.615/c Answer: K = 0.9147

1.13 B = 0.6; K = 3/(1-B)^3*(0.5-2*B+B*(1.5-log(B))) Answer: K =

23.7420

1.14 R = 30; r = 12; S = 50; theta = asin((R-r)/S); L = 2*S*cos(theta)+pi*(R+r)+2*theta*(R-r) Answer: L = 238.4998

1.15 Fn = 250; f = 0.35; r = 0.4; theta = 60*pi/180; T = (4*f*r*Fn*sin(theta/2))/(theta+sin(theta)) Answer: T = 36.5875 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a2retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

1.16 A = 1.7; B = 1.2; D = 1.265*((A*B)^3/(A+B))^(1/5) Answer: D = 1.5682

1.17 n = 6; M = 1/sin(pi/n); C = (1+M^2)/4/M; alpha = acos(sqrt(C^2+2)-C); aG = M*(1-M^2)*sin(alpha)/(1+M^2-2*M*cos(alpha))^2 Answer: aG = -1.3496

1.18 L = 3000; d = 45; V = 1600; Deltap = 0.03*L/d^1.24*(V/1000)^1.84 Answer: Deltap = 1.9048

1.19 E1 = 206e9; E2 = 206e9; nu1 = 0.3; nu2 = 0.3; d1 = 0.038; d2 = 0.070; F = 450; z = 0.00025; a = (0.375*F*((1-nu1^2)/E1+(1-nu2^2)/E2)/(1/d1+1/d2))^(1/3); pmax = 1.5*F/pi/a^2; za = z/a; sxsy = -pmax*((1-za*atan(1/za))*(1-nu1)-0.5/(1+za^2)) sz = -pmax/(1+za^2) Answers: sxsy = 2.0779e+008 sz = -1.2421e+009

1.20 E1 = 206e9; E2 = 206e9; nu1 = 0.3; nu2 = 0.3; d1 = 0.038; d2 = 0.070; F = 450; z = 0.000025; L = 0.05; b = sqrt(2*F/pi/L*((1-nu1^2)/E1+(1-nu2^2)/E2 )/(1/d1+1/d2)); pmax = 2*F/pi/b/L; zb = z/b; sx = -pmax*2*nu2*(sqrt(1+zb^2)-zb) sy = -pmax*((2-1/(1+zb^2))*sqrt(1+zb^2)-2*zb) sz = -pmax/sqrt((1+zb^2)) Answers: sx = -5.0360e+007 sy = -3.5432e+007 sz = -1.3243e+008

1.21 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a3retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

e = 0.8; NL = pi*e*sqrt(pi^2*(1-e^2)+16*e^2)/(1-e^2)^2 Answer: NL = 72.0220

1.22 h = 0.03; d0 = 0.006; d1 = 0.016; E = 206e9; d2 = d1+h*tan(pi/6); k = pi*E*d0*tan(pi/6)/log((d2-d0)*(d1+d0)/(d2+d0)/(d1-d0)) Answer: k = 5.2831e+009

1.23 alpha = 2e-5; E = 206e9; nu = 0.3; Ta = 260; Tb = 150; a = 0.006; b = 0.012; r = 0.010; c1 = alpha*E*(Ta-Tb)/2/(1-nu)/log(b/a); c2 = a^2/(b^2-a^2)*log(b/a); br = b/r; sr = c1*(c2*(br^2-1)-log(br)) st = c1*(1-c2*(br^2+1)-log(br)) T = Tb+(Ta-Tb)*log(br)/log(b/a) Answers: sr = -3.7670e+007 st = 1.1859e+008 T = 178.9338

1.24 k = 1.4; pepo = 0.3; psi = sqrt(k/(k-1))*sqrt(pepo^(2/k)-pepo^((k+1)/k)) Answer: psi = 0.4271

1.25 x = 0.45; K = 1.2/x/(sqrt(16*x^2+1)+log(sqrt(16*x^2+1)+4*x)/4/x)^(2/3) Answer: K = 1.3394

1.26 c = sqrt(8)/9801; n = 0; oneopi0 = c*(1103+26390*n)*factorial(4*n)/factorial(n)^4/396^(4*n); n = 1; oneopi1 = c*(1103+26390*n)*factorial(4*n)/factorial(n)^4/396^(4*n); format long e pin0 = 1/oneopi0 diffpino = pin0-pi pin0n1 = 1/(oneopi0+oneopi1) diffpin0n1 = pin0n1-pi © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a4retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

format short Answers: pin0 = 3.141592730013306e+000 diffpino = 7.642351240733092e-008 pin0n1 = 3.141592653589794e+000 diffpin0n1 = 4.440892098500626e-016

1.27 r = 10; rc = 3; k = 1.4; eta = 1-1/r^(k-1)*((rc^k-1)/k/(rc-1)) Answer: eta = 0.4803

1.28 M = 2; k = 1.4; AA = 1/M*(2/(k+1)*(1+(k-1)/2*M^2))^((k+1)/(2*(k-1))) Answer: AA = 1.6875

1.29 n = 5; ep = 0.1; w = 0.5 T = 1/sqrt(1+(ep*cos(n*acos(w))).^2) w=1 T = 1/sqrt(1+(ep*cos(n*acos(w))).^2) w = 1.5 T = 1/sqrt(1+(ep*cosh(n*acosh(w))).^2) Answers: w = 0.5000 T = 0.9988 w = 1 T = 0.9950 w = 1.5000 T = 0.1605

1.30 m = 1; n = 2; nu = 0.3; hR = 0.05; Rl = 0.1; Lm = pi/4*Rl*(4*m+1); Om = (1-nu^2)*Lm^4/(Lm^2+n^2+1.78*n^2*Lm^2); Om = Om+hR^2/12*(Lm^4+n^4+1.78*m^2*Lm^2) Answer: Om = 7.5159e-003

1.31 V = 2.9; D = 0.3; ep = 0.00025; nu = 1.012e-6; Re = V*D/nu; f = ((64/Re)^8+9.5*(log(ep/3.7/D+5.74/Re^0.9)-(2500/Re)^6)^-16)^(1/8) Answer: f = 0.0193

1.32 n = 3; th = pi/7; a = cos(th)+(n^2*cos(2*th)+sin(th)^4)/(n^2-sin(th)^2)^(3/2) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a5retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Answer: a = 1.1168

1.33 [Note: These solutions are obtained with a large amount of user interaction, almost on a line by line basis.] (a) x = a*sin(t); y = a*cos(t)^2*(2+cos(t))/(3+sin(t)^2); xp = diff(x,t,1); xpp = diff(x,t,2); yp = diff(y,t,1); ypp = diff(y,t,2); [Nnum Nden] = numden(xp*ypp-yp*xpp); Nden = factor(Nden) Nnum = simple(Nnum) [Dnum Dden] = numden(simple(xp^2+yp^2)) Partial answers: Nden = (sin(t)^2 + 3)^3 Nnum = -a^2*cos(t)^2*(cos(t) + 2)^3*(9*cos(t) - 6) Dnum = a^2*(cos(2*t) + 1)*(9*cos(2*t) - 80*cos(t) + 73) Dden = 4*(cos(t) - 2)^4 At this point, we place these results in standard notation and make a few trigonometric substitutions as follows: 3'2

3

! =

a 2 cos 2 t (cos t + 2 ) (9cos(t ) 6 )(4(cos(t ) 2) 4 )

(sin

2

3

3/ 2

t + 3) (a 2 (cos(2t ) + 1)(9cos(2t ) 80cos(t ) + 73)) 3

=

6

24a 2 cos 2 t (cos t + 2 ) (3cos(t ) 2 )(cos(t ) 2 )

(4

3

3/ 2

cos 2 t ) (2a 2 cos 2 t )

3/ 2

(9cos(2t )

80cos(t ) + 73) 3

=

3

24a 2 cos 2 t #%(cos t + 2 )(cos(t ) 2 )& (3cos(t ) 2 )(cos(t ) 2 ) 3 3/ 2 23 / 2 a 3 cos3 t (4 cos 2 t ) (9cos(2t ) 80cos(t ) + 73) 3

3

=

(4 cos(t ) ) (3cos(t ) 2 )(cos(t ) 2 ) 6 2 a cos t (4 cos 2 t )3 (9cos(2t ) 80cos(t ) + 73)3 / 2

=

6 2 sec t (3cos(t ) 2 )(cos(t ) 2 ) a (9cos(2t ) 80cos(t ) + 73)3 / 2

3

(b) syms a t x = 3*t*a/(1+t^3); y = x*t; xp = diff(x,t,1); xpp = diff(x,t,2); yp = diff(y,t,1); ypp = diff(y,t,2); kn = simple(xp*ypp-yp*xpp); kd = simple(xp^2+yp^2); k = simple(kn/kd^(3/2)) Answer: k = (2*a^2)/(3*(t^3 + 1)^2*((a^2*(t^8 + 4*t^6 - 4*t^5 - 4*t^3 + 4*t^2 + 1))/(t^3 + 1)^4)^(3/2))

© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a6retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(c) syms a t x = a*(t-tanh(t)); y = a*sech(t); xp = diff(x,t,1); xpp = diff(x,t,2); yp = diff(y,t,1); ypp = diff(y,t,2); kn = simple(xp*ypp-yp*xpp); kn = subs(kn, cosh(t)^2 - 1, sinh(t)^2); kd = collect(simple(xp^2+yp^2), a); kd = subs(kd,1 - 1/cosh(t)^2, sinh(t)^2/cosh(t)^2); kd = subs(kd, (1/cosh(t)^2)-1, -sinh(t)^2/cosh(t)^2); k = simple(kn/kd^(3/2)) Answer: k = (a^2*sinh(t)^2)/(cosh(t)^3*((a^2*sinh(t)^2)/cosh(t)^2)^(3/2))

1.34 (a) z = vpa('cos(pi*cos(pi*cos(log(pi+20))))+1', 40) Answer: 0.000000000000000000000000000000000039321609261272240194541

(b) z = vpa('exp(pi*sqrt(163))', 35) Answer: z = 262537412640768743.99999999999925007

1.35 syms x tay1 = taylor(x*(7*x+1)/(1-x)^3, 5, x, 0) tay2 = taylor((1+x-sqrt(1-6*x+x^2))/4,5,x,0) tay3 = taylor(2*x/(1-x)^3,5,x,0) tay4 = taylor(x*(x^2+4*x+1)/(1-x)^3,5,x,0) Answers: tay1 = 52*x^4 + 27*x^3 + 10*x^2 + x tay2 = 11*x^4 + 3*x^3 + x^2 + x tay3 = 20*x^4 + 12*x^3 + 6*x^2 + 2*x tay4 = 37*x^4 + 19*x^3 + 7*x^2 + x

1.36 syms x n e a t L1 = limit((xê-1)/e, e, 0) L2 = limit((1-sin(2*x))^(1/x), x, 0) L3 = limit(((exp(t)-1)/t)^(-a), t, 0) L4 = limit(log(x^n)/(1-x^2), x, 1) Answers: L1 = log(x)

L2 = 1/exp(2)

L3 = 1 L4 = -n/2


1.37 syms s t rD = roots(sym2poly(s^4+0.282*s^3+4.573*s^2+0.4792*s+2.889)); rN = roots(sym2poly(0.1*s^3+0.0282*s^2-0.0427*s+0.0076)); Xt = vpa(ilaplace(((s-rN(1))*(s-rN(2))*(s-rN(3)))/((s-rD(1))*(s-rD(2))*(s-rD(3))*(s-rD(4))), s, t),5) Answer: Xt = (1.3974*(cos(1.9456*t)+0.050033*sin(1.9456*t)))/exp(0.097401*t)(0.39736*(cos(0.87145*t)+0.049799*sin(0.87145*t)))/exp(0.043599*t)


Solution to Exercises Chapter 2 2.1 n = 0:7; a = 2*n-1; b = 2*n+1; aplusb = a+b aminusb = a-b atimesb = a'*b detatimesb = det(atimesb) atimesbtranspose = a*b' Answers: aplusb = 0 4 8 12 16 20 24 28 aminusb = -2 -2 -2 -2 -2 -2 -2 -2 atimesb = -1 -3 -5 -7 -9 -11 -13 -15 1 3 5 7 9 11 13 15 3 9 15 21 27 33 39 45 5 15 25 35 45 55 65 75 7 21 35 49 63 77 91 105 9 27 45 63 81 99 117 135 11 33 55 77 99 121 143 165 13 39 65 91 117 143 169 195 detatimesb = 0 atimesbtranspose = 552

2.2 x = sort( [17 -3 -47 5 29 -37 51 -7 19], 'descend'); xs = [x(find(sort(x, 'descend')<=0)), x(find(sort(x, 'descend')>0))] Answer: w = -3

-7 -37 -47

51

29

19

17

5

2.3 y = [0, -0.2, 0.4, -0.6, 0.8, -1.0, -1.2, -1.4, 1.6]; sy = sin(y); % (a) mx = max(sy(find(sy<0))) mn = min(sy(find(sy<0))) % (b) mp = sqrt(sy(find(sy>0))) Answers: mx = -0.1987 mn = -0.9854 mp = 0.6240

0.8470 0.9998

2.4 v = logspace(log10(6), log10(106), 8) v5 = v(5) z = v(1:2:7) Answers: v5 = 30.9607 z = 6.0000 13.6295 30.9607 70.3300


2.5 z = magic(5) z(:,2) = z(:,2)/sqrt(3); z(5,:) = z(5,:)+z(3,:); z(:,1) = z(:,1).*z(:,4); q = z-diag(diag(z))+diag([2 2 2 2 2]); format long g dq = diag(q*q') format short qmax = max(max(q)) qmin = min(min(q)) Answers: z = 17 24 1 8 15 23 5 7 14 16 4 6 13 20 22 10 12 19 21 3 11 18 25 2 9 dq = [486 104189 7300 qmax = 330 qmin = 1

44522 111024]

2.6 w = magic(2); wa = repmat(w, 2, 2) wb = repmat(w, 3, 1) wc = [repmat(w, 3, 1), repmat(w', 3, 1)] wad = [w, w; w, w] wbd = [w; w; w] wcd = [w, w'; w, w'; w, w'] Answers: wa = % or wad 1 3 1 4 2 4 1 3 1 4 2 4 wb = % or wbd 1 3 4 2 1 3 4 2 1 3 4 2 wc = % or wcd 1 3 1 4 2 3 1 3 1 4 2 3 1 3 1 4 2 3

3 2 3 2

4 2 4 2 4 2


(a) x = magic(3); t = x(1,:); x(1:2,:) = x(2:3,:); x(3,:) = t; x Answers: x = 3 4 8

5 9 1

7 2 6

8 3 4

1 5 9

(b) x = magic(3) t = x(:,3); x(:,2:3) = x(:,1:2); x(:,1) = t; x Answers: x = 6 7 2

2.8 x = magic(5); xf = fliplr(x); xf = fliplr(xf-diag(diag(xf))) Answer: xf = 17 24 1 8 0 23 5 7 0 16 4 6 0 20 22 10 0 19 21 3 0 18 25 2 9

2.9 a = 1; b = 1.5; e = 0.3; phi = linspace(0, 2*pi); s = a*cos(phi)+sqrt(b^2-(a*sin(phi)-e).^2); plot(phi*180/pi, s) Answer:


2.5

2

1.5

1

0.5

0 0

50

100

150

200

250

300

350

400

2.10 n = 1:25; a = n*pi/sqrt(19); P = 100*(1+2*cumsum((sin(a)./a).^2))/4.3589; plot(n(2:25), P(2:25)) Answer: 100

98

96

94

92

90

88

86

84

82 0

5

10

15

20

25

2.11 a = sqrt(2.8); n = 1:100; % (1x100) x = 1:0.5:5; % (1x9) exact = a./sqrt(a^2+x.^2).*sin(pi*sqrt(a^2+x.^2))/sin(pi*a); e = 100*(prod(1-(1./(n.^2-a^2))'*x.^2)-exact)./exact % prod works on columns Answer: e = 1.0001

2.2643

4.0610

6.4176

9.3707 12.9670 17.2642 22.3330 28.2588

2.12 x = [72, 82, 97, 103, 113, 117, 126, 127, 127, 139, 154, 159, 199, 207]; beta = 3.644; delta = (sum(x.^beta)/length(x))^(1/beta) Answer: delta = 144.2741


2.13 b = 2; phi = linspace(0, pi/2, 10); the = linspace(0, 2*pi, 24); [p, t] = meshgrid(phi, the); mesh(b*sin(p).*cos(t), b*sin(p).*sin(t), b*cos(p)) Answer:

2

1.5

1

0.5

0 2 1

2 1

0

1.5

0.5 0 -1

-1 -2

-0.5

-1.5 -2

2.14 x = linspace(0.1, 1, 5); n = -25:25; y = pi*x*sqrt(2); exact = 2*pi^4./y.^3.*(sinh(y)+sin(y))./(cosh(y)-cos(y)); [xx, nn] = meshgrid(x, n); apx = sum(1./(nn.^4+xx.^4)); format long g comp = [apx' exact'] format short Answer: comp = 10002.1644054971 91.7751642291914 12.9244097924593 4.40350416999114 2.15691498449604

10002.1644456719 91.7752044040698 12.9244499673344 4.40354434485533 2.15695515933427

2.15 k = 0:25; n = 2; x = linspace(1, 6, 6); [xx, kk] = meshgrid(x, k); Jn = sum((-1).^kk.*(xx/2).^(2*kk+n)./factorial(kk)./gamma(kk+1+n)); format long g Comp = [Jn' besselj(n, x)'] format short Answer: Comp = 0.1149034849319 0.352834028615638

0.114903484931901 0.352834028615638


0.486091260585891 0.364128145852073 0.0465651162777518 -0.242873209960186

0.486091260585891 0.364128145852073 0.0465651162777523 -0.242873209960185

2.16 n = 7; k = 1:(2*n-1); ser = sum(cos(k*pi/n)) Answer: ser = -1.0000

2.17 n = 6; k = 1:n; N1 = n*(2*n^2+3*n+1)/6 S1 = sum(k.^2) N2 = n*(6*n^4+15*n^3+10*n^2-1)/30 S2 = sum(k.^4) N3 = n^2*(2*n^4+6*n^3+5*n^2-1)/12 S3 = sum(k.^5) Answers: N1 = 91 S1 = 91 N2 = 2275 S2 = 2275 N3 = 12201 S3 = 12201

2.18 N = 10; n = 0:N; PieDiff = sum((4./(8*n+1)-2./(8*n+4)-1./(8*n+5)-1./(8*n+6)).*(1/16).^n) -pi Answer: PieDiff = 0

2.19 k=1:300; dn=0.5*cumsum(1./k); On = find(dn>1); d1 = On(1) tw = find(dn>2); d2 = tw(1) th = find(dn>3); d3 = th(1) Answer: d1 = 4 d2 = 31 d3 = 227

2.20 N = 10; k = 0:N; e = sum(1./factorial(k))-exp(1) Jo = sum((-1).^k./factorial(k).^2)-besselj(0,2) co = sum((-1).^k./factorial(2*k))-cos(1) Answers: e = -2.7313e-008 Jo = 6.3838e-016 co = -1.1102e-016


2.21 K = 41; k = 1:2:K; y = pi/3; x = 0.75; SKdiff = sum(exp(-k*x).*sin(k*y)./k)-0.5*atan(sin(y)/sinh(x)) Answer: SKdiff = -2.2204e-016

2.22 K = 14; k = 0:K; a = 3; x = -2; axdiff = sum((x*log(a)).^k./factorial(k))-a^x Answer: axdiff = 9.0230e-008

2.23 N = 500; k = 1:N; z = 10; c = pi*coth(pi*z)/(2*z)-1/(2*z^2) S1 = cumsum(1./(k.^2+z^2)); K = find(abs(S1-c)<3e-3); N =K(1) Answer: N = 333

2.24 w = [-1 -1;1 -1;-1 1;1 1]/2; orthw = w'*w q = [1 -1 -1 1;1 1 -1 -1;1 -1 1 -1;1 1 1 1]/2; orthq = q'*q Answer: orthw = 1 0

0 1

1 0 0 0

0 1 0 0

orthq = 0 0 1 0

0 0 0 1

2.25 A = [2 -2 -4; -1 3 4; 1 -2 -3]; T = A*A-A Answer: T = 0 0 0

0 0 0

0 0 0


A = [1 2 2; 2 1 2; 2 2 1]; T = A*A-4*A-5*eye(3) Answer: T = 0 0 0

0 0 0

0 0 0

2.27 A = [1 -1; 2 -1]; B = [1 1; 4 -1]; T = (A+B)^2-A^2-B^2 Answer: T = 0 0

0 0

2.28 A = [7 -2 1; -2 10 -2; 1 -2 7]; P = [1/sqrt(2) 1/sqrt(3) 1/sqrt(6); 0 1/sqrt(3) -2/sqrt(6); -1/sqrt(2) 1/sqrt(3) 1/sqrt(6)]; L = eig(A) LL = diag(inv(P)*A*P) Answers: L = 6.0000 6.0000 12.0000 LL = 6.0000 6.0000 12.0000

2.29 c = pi/180; th1 = 30*c; th2 = 30*c; th3 = 30*c; a1 = 1; a2 = 2; a3= 3; a01 = [cos(th1) -sin(th1) 0 a1*cos(th1); sin(th1) cos(th1) 0 a1*sin(th1);0 0 1 0; 0 0 0 1]; a12 = [cos(th2) -sin(th2) 0 a2*cos(th2); sin(th2) cos(th2) 0 a2*sin(th2);0 0 1 0; 0 0 0 1]; a23 = [cos(th3) -sin(th3) 0 a3*cos(th3); sin(th3) cos(th3) 0 a3*sin(th3);0 0 1 0; 0 0 0 1]; t = a01*a12*a23; thx3 = atan2(t(2,1), t(1,1))/c thy3 = atan2(t(2,2), t(1,2))/c qx = t(1,4) qy = t(2,4) Answers: thx3 = 90.0000 thy3 = 180 qx = 1.8660 qy = 5.2321

2.30 c = pi/180; th1 = 15*c; th2 = 105*c; th3 = 145*c;


th4 = 0; th5 = 215*c; th6 = 55*c; a2 = 0.431; a3 = 0.019; r4 = a2; r3 = 0.125; T12 = [cos(th1) 0 sin(th1) 0 ; sin(th1) 0 -cos(th1) 0 ; 0 1 0 0; 0 0 0 1]; T23 = [cos(th2) -sin(th2) 0 a2*cos(th2); sin(th2) cos(th2) 0 a2*sin(th2); 0 0 1 0; 0 0 0 1]; T34 = [cos(th3) 0 sin(th3) a3*cos(th3); sin(th3) 0 -cos(th3) a3*sin(th3); 0 1 0 r3; 0 0 0 1]; T45 = [cos(th4) 0 sin(th4) 0; sin(th4) 0 -cos(th4) 0; 0 1 0 r4; 0 0 0 1]; T56 = [cos(th5) 0 sin(th5) 0; sin(th5) 0 -cos(th5) 0; 0 1 0 0; 0 0 0 1]; T67 = [cos(th6) 0 sin(th6) 0; sin(th6) 0 -cos(th6) 0; 0 1 0 0; 0 0 0 1]; T17 =T12*T23*T34*T45*T56*T67 Answer: T17 = 0.2418 -0.5540 0.7966 -0.4729 0.9128 -0.1485 -0.3804 -0.2561 0.3290 0.8192 0.4698 0.5459 0 0 0 1.0000

2.31 x = [17 31 5;6 5 4;19 28 9;12 11 10]; H = diag(x*inv(x'*x)*x')' Answer: H = 0.7294

0.9041

0.4477

0.9188

2.32 % square n = 1:2:401; tau = linspace(-0.5, 0.5, 200); plot(tau, 4/pi*(1./n)*sin(2*pi*n'*tau)) % sawtooth n = 1:200; tau = linspace(-1, 1, 200); plot(tau, 0.5+1/pi*(1./n)*sin(2*pi*n'*tau)) % sawtooth n = 1:200; tau = linspace(-1, 1, 200); plot(tau, 0.5-1/pi*(1./n)*sin(2*pi*n'*tau)) % triangular n = 1:200; tau = linspace(-1, 1, 200); plot(tau, 0.5*pi-4/pi*((1./(2*n-1).^2)*cos(pi*(2*n'-1)*tau))) % recitified sine wave n = 1:200; tau = linspace(-1, 1, 200); plot(tau, 2/pi+4/pi*((1./(1-4*n.^2))*cos(2*pi*n'*tau))) % half sine wave n = 2:2:106; tau = linspace(-2, 2, 200); plot(tau, 1/pi+.5*sin(pi*tau)-2/pi*(1./(n.^2-1)*cos(pi*n'*tau))) % exponential n = 1:250; tau = linspace(0, 4*pi, 350); plot(tau, (exp(2*pi)-1)/pi*(0.5+((1./(1+n.^2))*cos(n'*tau)-(n./(1+n.^2))*sin(n'*tau)))) % trapezoidal n = 1:2:105; © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a9retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

tau = linspace(-2, 2, 200); plot(tau, 64*((sin(n*pi/4)./(pi*n).^2)*sin(pi*n'*tau)))

2.33 a = sqrt(3); n = 1:25; t = (10:10:80)*pi/180; c = pi/sinh(pi*a)/2; b = 1./(n.^2+a^2); e1 = 100*(b*cos(n'*t)./(c*cosh(a*(pi-t))/a-0.5/a^2)-1) e2 = 100*(n.*b*sin(n'*t)./(c*sinh(a*(pi-t)))-1) Answers: e1 = -1.2435 0.8565 0.8728 -1.9417 -0.9579 -8.1206 0.7239 1.1661 e2 = 8.0538 10.4192 -8.9135 -5.4994 12.9734 -0.5090 -17.2259 11.2961

2.34 n = (1:2:51)*pi; a = 2; eta = 0:1/14:1; xi = 0:1/14:1; temp = meshgrid(1./(n.*sinh(n*a)), eta); z = 4*(temp.*sinh(a*n'*eta)')*sin(n'*xi); mesh(xi, eta, z) Answer:

1.4 1.2 1 0.8 0.6 0.4 0.2 0 1 0.8

1 0.6

0.8 0.6

0.4 0.4 0.2

0.2 0

0

2.35 n = (1:50)*pi; eta = 0:0.05:1; tau = 0:0.05:2; a = 0.25; temp = meshgrid(sin(a*n)./n.^2, eta); z = 2*pi/a/(1-a)*temp.*sin(n'*eta)'*cos(n'*tau); mesh(tau, eta, z) Answers:

© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a10 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

4 2 0 -2

-4 1 0.9 0.8 0.7

2 0.6 0.5

1.5 0.4 1

0.3 0.2

0.5

0.1 0

0

2.36 A = [1 3 4; 31 67 9; 7 5 9]; B = [11 34 6; 7 13 8; 43 10 53]; DADB = det(A)*det(B) DAB = det(A*B) Answer: DADB = -3832062 DAB = -3832062

2.37 syms x y x1 = -1; y1 = 1; x2 = 0; y2 = 0; x3 = 1; y3 = 1; p = det([x^2+y^2 x y 1; x1^2+y1^2 x1 y1 1; x2^2+y2^2 x2 y2 1; x3^2+y3^2 x3 y3 1]); p = factor(p) Answer: p = 2*(x^2 + y^2 - 2*y)

2.38 syms x y x1 = -1; y1 = 1; x2 = 1; y2 = 1; x3 = 2; y3 = 2; p = det([x^2 x y 1; x1^2 x1 y1 1; x2^2 x2 y2 1; x3^2 x3 y3 1]); p = factor(p) Answer: p = 2*(x^2 - 3*y + 2)

2.39 A = [16, 32, 33, 13; 5, 11, 10, 8; 9, 7, 6, 12; 34, 14, 15, 1]; c = [91, 16, 5, 43]; supw = A\c' % or supw = linsolve(A, c') determin = det(A) invA = inv(A) Answers: supw = [-0.1258 -8.7133 11.2875 -0.0500] determin = 7680 invA = -0.0177 -0.0023 0.0180 0.0333 -0.3344 1.2352 -0.4695 0.1000 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a11 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

0.3500 -1.1375 0.0333 -0.1500

0.3875 -0.1000 0.1500 -0.0333

2.40 a = 0.005; b = 0.0064; c = 0.008; E1 = 2.1e9; E2 = 0.21e9; nu1 = 0.4; nu2 = 0.4; Uo = 0.00025; Matrix = [1, a^2, 0, 0; 1, b^2, -1, -b^2; ... -(1+nu1), (1-nu1)*b^2, (1+nu2)*E1/E2, -(1-nu2)*b^2*E1/E2;... 0, 0, -(1+nu2), (1-nu2)*c^2]; y = [0, 0, 0, -Uo*E2*c]'; AB = Matrix\y; hoop1 = -AB(1)/b^2+AB(2) hoop2 = -AB(3)/b^2+AB(4) % check urc = -(1+nu2)*AB(3)/E2/c+(1-nu2)*c*AB(4)/E2 Answers: hoop1 = -6.3013e+007 hoop2 = -1.1790e+007 urc = -2.5000e-004

2.41 syms i1 i2 i3 R V1 V2 V3 Q1 = 'V1-6*R*i1+4*R*(i2-i1)=0,'; Q2 = 'V2+2*R*(i3-i2)-3*R*i2-4*R*(i2-i1)=0,'; Q3 = '-V3-R*i3-2*R*(i3-i2y)=0'; [i1 i2 i3] = solve([Q1 Q2 Q3]) Answer: i1 = 1/182/R*(23*V1 + 12*V2 - 8*V3) i2 = 1/91/R*(6*V1 + 15*V2 - 10*V3) i3 = 1/91/R*(4*V1 + 10*V2 - 37*V3)

2.42 Either syms s t V1 V2 V3 U real A = [2*s+t -s 0; -s 2*s+t -s; 0 -s s+t]; B = [s*U 0 0]'; V=inv(A)*B or syms s t V1 V2 V3 U real [V1 V2 V3] = solve((2*s+t)*V1-s*V2-s*U, -s*V1+(2*s+t)*V2-s*V3, -s*V2+(s+t)*V3, V1, V2, V3); V1 = factor(V1) V2 = factor(V2) V3 = factor(V3) gives Answer: V1 = (U*s*(s^2 + 3*s*t + t^2))/(s^3 + 6*s^2*t + 5*s*t^2 + t^3) V2 = (U*s^2*(s + t))/(s^3 + 6*s^2*t + 5*s*t^2 + t^3) V3 = (U*s^3)/(s^3 + 6*s^2*t + 5*s*t^2 + t^3)


Solution to Exercises Chapter 3 3.1 ft = input('Enter the value of length in feet: '); disp([num2str(ft) ' ft = ' num2str(ft*0.3048) ' m']) Answer: Enter the value of length in feet: 11.4 11.4 ft = 3.4747 m

3.2 acre = input('Enter the number of acres: '); disp([num2str(acre) ' acres = ' num2str(acre*43560*0.092903) ' sq. m']) Answer: Enter the number of acres: 2.4 2.4 acres = 9712.4512 sq. m

3.3 Decimal = input('Enter a positive integer < 4.5x10^15: '); disp(['The binary representation of ' num2str(Decimal) ' is ' num2str(dec2bin(Decimal))]) Answer: Enter a positive integer < 4.5x10^15: 37 The binary representation of 37 is 100101

3.4 R = input('Enter the real part of a complex number: '); I = input('Enter the imaginary part of a complex number: '); z = complex(R,I); disp(['The magnitude and phase of ' num2str(z) ' is']) disp(['Magnitude = ' num2str(abs(z)) ' Phase angle = ' num2str(angle(z)*180/pi) ' degrees']) Answer: Enter the real part of a complex number: -7 Enter the imaginary part of a complex number: 13 The magnitude and phase of -7+13i is Magnitude = 14.7648 Phase angle = 118.3008 degrees

3.5 n = 0:15; f = (((1+sqrt(5))/2).^n-((1-sqrt(5))/2).^n)/sqrt(5); disp([repmat('F', length(n),1) num2str(n',2) repmat(' = ', length(n),1) num2str(f')]) % or fprintf(1, 'F%2.0f = %3.0f \n', [n;f]) Answers: F 0 = 0 F1= 1 F2= 1 F3= 2 … F15 = 610 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a1retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

3.6 n = 9:-1:2; L = length(n); A = [repmat('cos(pi/', L, 1) int2str(n') repmat(') = ', L, 1) num2str(cos(pi./n'),'%1.5f')] B = [repmat(' pi/', L, 1) int2str(n') repmat(' = ', L, 1) num2str(180./n','%2.1f') repmat(' degrees',L,1)] disp([A B]) fprintf(1,'cos(pi/%1.0f) = %1.5f pi/%1.0f = %2.1f degrees\n', [n; cos(pi./n); n; 180./n]) Answer: cos(pi/9) = 0.93969 cos(pi/8) = 0.92388 cos(pi/7) = 0.90097 cos(pi/6) = 0.86603 cos(pi/5) = 0.80902 cos(pi/4) = 0.70711 cos(pi/3) = 0.50000 cos(pi/2) = 0.00000

pi/9 = 20.0 degrees pi/8 = 22.5 degrees pi/7 = 25.7 degrees pi/6 = 30.0 degrees pi/5 = 36.0 degrees pi/4 = 45.0 degrees pi/3 = 60.0 degrees pi/2 = 90.0 degrees

3.7 Mon ={'January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', ... 'September', 'October', 'November', 'December'}; Srt = char(sort(Mon)); disp([Srt(1:6,:) repmat(' ',6,1) Srt(7:12,:)]) Answer: April June August March December May February November January October July September

3.8 N = input('Enter an integer < 12: '); n = 1:N; disp(' ') disp([repmat('For n = ', N, 1) int2str(n') repmat(', ', N, 1) int2str(n') repmat('! = ', N, 1) int2str(factorial(n)')]) disp(' ') disp(['The sum of these ' int2str(N) ' factorials = ' int2str(sum(factorial(n)))]) Answer: Enter an integer < 12: 4 For n = 1, For n = 2, For n = 3, For n = 4,

1! = 1 2! = 2 3! = 6 4! = 24

The sum of these 4 factorials = 33


Solution to Exercises Chapter 4 4.1 day = char('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday'); nonleapyear = [31 28 31 30 31 30 31 31 30 31 30 31]; leapyear = [31 29 31 30 31 30 31 31 30 31 30 31]; datee = input('Enter month, day and year in the form xx/xx/xxxx for 2010, 2011, or 2012: ','s'); f = findstr(datee,'/'); monthno = str2num(datee(1:f(1)-1)); dayno = str2num(datee(f(1)+1:f(2)-1)); yearno = str2num(datee(f(2)+1:f(2)+4)); if yearno == 2010 offse t= 5; nodaypermonth = nonleapyear; elseif yearno == 2011 offset = 6; nodaypermonth = nonleapyear; elseif yearno == 2012 offset = 7; nodaypermonth = leapyear; else error('Not programmed for this year') end if monthno>1 dayofyear = sum(nodaypermonth(1:monthno-1))+dayno; else dayofyear = dayno; end dayofweek = offset+dayofyear-fix(dayofyear/7)*7; if dayofweek>7 dayofweek = dayofweek-7; end disp(['The date ' datee(1:f(2)+4) ' is the ' num2str(dayofyear) ' day of the year and falls on a ' deblank(day(dayofweek,: )) '.']) Answer: Enter month, day and year in the form xx/xx/xxxx for 2010, 2011, or 2012: 08/31/2011 The date 08/31/2011 is the 243 day of the year and falls on a Wednesday.

4.2 dat = [45 38 47 41 35 43]; L = length(dat); vvar = zeros(L,1); for n = 2:L vvar(n) = var(dat(1:n)); end disp([repmat('var(n=', L-1, 1) int2str((2:L)') repmat('): ', L-1,1) num2str(vvar(2:L),4)]) Answer: var(n=2): 24.5 var(n=3): 22.33 var(n=4): 16.25 var(n=5): 24.2 var(n=6): 19.9


4.3 m = 6; N = 100; Lend = (N-m+1); a = 5*(1+rand(N,1)); A = zeros(Lend,1); Avg(1) = mean(a(1:m)); for k = 2:Lend Avg(k) = Avg(k-1)+(a(m+k-1)-a(k-1))/m; end plot(1:Lend, Avg, 'ks-') Answer: 8.5

8

7.5

7

6.5

6 0

10

20

30

40

50

60

70

80

90

100

4.4 xold = 1.5; fx = 1; cnt = 0; while fx>1e-8 xnew = xold-(cos(xold)+sin(xold))/(-sin(xold)+cos(xold)); fx = abs(cos(xnew)+sin(xnew)); cnt = cnt+1; xold = xnew; end disp(['At x = ' num2str(xold,8) ', f(x) = ' num2str(fx) ' after ' int2str(cnt) ' iterations']) Answer: At x = 2.3561945, f(x) = 1.1102e-016 after 4 iterations

4.5 ir = 0.045; P = 250000; A = 25000; disp('Principal Annuity Interest No. Remain') disp(' ($) ($/yr) (%) years ($)') R = P-A; n = 1; while R>A n = n+1; R = R*(1+ir)-A; end disp([num2str(P,6) blanks(6) num2str(A,5) blanks(8) num2str(ir*100,3) blanks(9) num2str(n,3) blanks(6) num2str(R,5)]) Answer: Principal Annuity Interest

No.

Remain


($) 250000

($/yr) 25000

(%) 4.5

years 12

($) 19112

4.6 alp = pi/4; tol = 1e-5; ao = 1; bo = cos(alp); co = sin(alp); while abs(co)>tol an = (ao+bo)/2; bn = sqrt(ao*bo); co = (ao-bo)/2; ao = an; bo = bn; end format long e Kalp = pi/ao/2 K = ellipke(sin(alp)^2) format short Answers: Kalp = 1.854074677301372e+000 K = 1.854074677301372e+000

4.7 h = [1 3 6 -7 -45 12 17 9]; a = 3; b=13; n = zeros(length(h),1); for k = 1:length(h) if h(k)<=a||h(k)>=b n(k) = 0; else n(k) = 1; end end disp(['n = ' int2str(n')]) Answer: n = 0 0 1 0 0 1 0 1

4.8 N = input('Enter a positive integer < 10: '); h = zeros(N,N); for m = 1:N for n = 1:N if (n+m-1) <= N h(n, m) = n+m-1; end end end disp(h) Answer: Enter a positive integer < 10: 4 1 2 3 4 2 3 4 0 3 4 0 0 4 0 0 0 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a3retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

4.9 num = 1.5; while (num < 1)||(num> 19)||(rem(num,1)~=0) num = input(' Enter any integer from 1 to 19: '); end

4.10 x = zeros(1000,1); x(1) = 3; cnt = 1; while x(cnt)~=1 cnt = cnt+1; if rem(x(cnt-1), 2) > 0 x(cnt) = 3*x(cnt-1)+1; else x(cnt) = x(cnt-1)/2; end end disp(['x = ' num2str(x(1:cnt)')]) Answer: x = 3 10 5 16 8 4 2 1

4.11 x(1) = 0; n = 201; x = zeros(n,1); for k = 2:n x(k) = x(k-1)^2+0.25; end plot(0:5:n-1, x(1:5:n), 'ks') m = 1; xw(1) = 0; xw = zeros(201, 1); while m<201 xw(m+1) = xw(m)^2+0.25; m = m+1; end figure(2) plot(0:5:m-1, xw(1:5:m), 'ks') Answers:


0.5

0.45

0.4

0.35

0.3

0.25

0.2

0.15

0.1

0.05

0 0

20

40

60

80

100

120

140

160

180

200

4.12 a = 7; x = [3 100]; for k=1:2 xold = x(k); test = 1; cnt = 0; while test>1e-6 xnew = 0.5*(xold+a/xold); test = abs(xnew-xold); xold = xnew; cnt = cnt+1; end disp(['For xo = ' int2str(x(k)) ' the number of iterations = ' num2str(cnt)]) end Answers: For xo = 3 the number of iterations = 4 For xo = 100 the number of iterations = 10

4.13 for k = 1:3 switch k case 1 p=[1 2 3 4]; s = [10 20 30 40]; case 2 p = [11 12 13 14]; s = [101 102]; case 3 p = [43 54 55]; s = [77 66 88 44 33]; end n = length(p); m = length(s); d = n-m; if d==0 h = p+s elseif d<0 h = [s(1:-d) p+s(-d+1:m)] else h = [p(1:d) s+p(d+1:n)] end end Answers: h = 11

22

33

44


h = 11 h = 77

12 114 116 66 131 98 88

4.14 a = [1.45, 2.75, 3.2, 4]; N = 200; for k = 1:length(a) xold = 0.1; cnt = 1; test = 1; while (test>1e-4) cnt = cnt+1; xnew = xold*a(k)*(1-xold); test = abs((xnew-xold)/xnew); xold = xnew; if cnt>N break end end disp(['For a = ' num2str(a(k)) ', x(n=' int2str(cnt) ') = ' num2str(xold)]) end Answers: For a = 1.45, x(n=19) = 0.31031 For a = 2.75, x(n=31) = 0.63634 For a = 3.2, x(n=201) = 0.51304 For a = 4, x(n=201) = 0.08737

4.15 Nu = 10; for k = 1:2 test = -1; while test < 0 NM(k) = input(['Enter a positive integer<' int2str(Nu) ': ']); if (NM(k) < 1)||(NM(k)> Nu)||(rem(NM(k),1)~=0) else test = 1; Nu = NM(k)^2+1; end end end N = NM(1) M = NM(2) Answer: Enter a positive integer<10: -1 Enter a positive integer<10: 14 Enter a positive integer<10: 3.6 Enter a positive integer<10: 8 Enter a positive integer<65: 0.5 Enter a positive integer<65: 66 Enter a positive integer<65: 45.03 Enter a positive integer<65: 38 N = 8 M = 38


for k = 1:3 xon = []; eon = []; switch k case 1 xo=[1 7 8 6 5 7 3 5 4]; eo=[2 6 10 4 3 6 1 2 3]; case 2 xo=[7 11 13 6]; eo=[6 10 15 7]; case 3 xo=[3 14 20 25 14 6 2 0 1 0]; eo=[4 12 19 19 14 8 4 2 1 1]; end cutoff = 5; n = 1; nend = length(eo); cnt = 1; while n<= nend z = cumsum(eo(n:nend)); zx = cumsum(xo(n:nend)); ind = find(z 0 if n+mxind<= nend eon(cnt) = z(mxind+1); xon(cnt) = zx(mxind+1); else eon(cnt-1) = eon(cnt-1)+z(mxind); xon(cnt-1) = xon(cnt-1)+zx(mxind); end n = n+mxind+1; else eon(cnt)=eo(n); xon(cnt)=xo(n); n=n+1; end cnt=cnt+1; end disp(['Case ' int2str(k)]) ve = eon(1:length(eon)); vx = xon(1:length(eon)); x2 = sum((vx-ve).^2./ve); disp(['e = ' num2str(ve) ' x = ' num2str(vx) ' chi^2 = ' num2str(x2)]) end Answers: Case 1 e = 8 10 7 6 6 x = 8 8 11 7 12 chi^2 = 8.8524 Case 2 e = 6 10 15 7 x = 7 11 13 6 chi^2 = 0.67619 Case 3 e = 16 19 19 14 8 8 x = 17 20 25 14 6 3 chi^2 = 5.6349


Solution to Exercises Chapter 5 5.1 Dod = [6 3 2 1.5 1.2 1.1 1.07 1.05 1.03 1.01]; c = [.88 .89 .91 .94 .97 .95 .98 .98 .98 .92]; a = [.33 .31 .29 .26 .22 .24 .21 .20 .18 .17]; Kspline = spline(Dod, c, Dod).*((Dod-1)/2).^-spline(Dod, a, Dod); Kpoly = polyval(polyfit(Dod, c, 5), Dod).*((Dod-1)/2).^-polyval(polyfit(Dod, a, 5), Dod); Kt = c.*((Dod-1)/2).^-a; disp(' Kt Kspline Kpoly') disp([Kt' Kspline' Kpoly']) Answers:

Kt Kspline 0.6504 0.6504 0.8900 0.8900 1.1126 1.1126 1.3479 1.3479 1.6098 1.6098 1.9497 1.9497 1.9814 1.9814 2.0495 2.0495 2.0870 2.0870 2.2645 2.2645

Kpoly 0.6504 0.8900 1.1135 1.3369 1.6970 1.8546 1.9221 1.9886 2.1037 2.4324

5.2 function Exercise5_2 N = 2^10; wn = 2*pi*[5, 9, 9.4, 20]; dT = 2*pi/wn(end)/4; T = N*dT; df = 1/T; t = (0:N-1)*T/N; NHSignal = foft(t, N, wn); whamm = 0.54-0.46*cos(2*pi*t/T); k1 = sum(whamm.*NHSignal)/sum(whamm); k2 = sqrt(N/sum(whamm.^2)); HSignal = whamm.*(NHSignal-k1)*k2; subplot(2,1,1) ft = fft(NHSignal, N); Sig = 2*abs(ft(1:N/2))/N; plot((0:N/2-1)*df, Sig) temp = [1 find(diff(Sig)<=0)]; % add 1 because diff shifts array down 1 temp1 = find(diff(temp)~=1); freqpeak = temp(temp1+1)*df; An = abs(ft)/N; disp([' Average power = ' num2str(sum(An.^2))]) disp([' Peaks occur at ' num2str(freqpeak) ' Hz']) subplot(2,1,2) ft = fft(HSignal, N); Sig = 2*abs(ft(1:N/2))/N; plot((0:N/2-1)*df,Sig) An = abs(ft)/N; disp([' Average power = ' num2str(sum(An.^2))]) temp = [1 find(diff(Sig)<=0)]; temp1 = find(diff(temp)~=1); % add 1 because diff shifts array down 1


freqpeak = temp(temp1+1)*df; disp([' Peaks occur at ' num2str(freqpeak) ' Hz']) function NHSignal = foft(t, N, wn) x = [0.1, 0.04, 0.04, 0.03]; Ho = [1 1.3 1.3 1.8]; NHSignal = 0; for k = 1:length(x) NHSignal = NHSignal+Ho(k)*Hw(t,wn(k),x(k)); end function hw = Hw(t, wn, x) hw = exp(-x*wn*t).*sin(sqrt(1-x^2)*wn*t); Answers: Average power = 0.07529 Peaks occur at 4.84375 9.14063 Average power = 0.0013038 Peaks occur at 4.92188 9.0625

20.0781 Hz 9.45313

20.0781 Hz

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5.3 sx = 100; sy = -60; sz = 80; txy = -40; tyz = 50; tzx = 70; c2 = sx+sy+sz; c1 = txy^2+tyz^2+tzx^2-sx*sy-sy*sz-sz*sx; c0 = sx*sy*sz+2*txy*tyz*tzx-sx*tyz^2-sy*tzx^2-sz*txy^2; ps = sort(roots([1 -c2 -c1 -c0]), 'descend') t12 = 0.5*(ps(1)-ps(2)) t23 = 0.5*(ps(2)-ps(3)) t13 = 0.5*(ps(1)-ps(3)) Answers: ps = 160.7444 54.8980 -95.6424 t12 = 52.9232 t23 = 75.2702 t13 = 128.1934

In Exercises 5.4 to 5.12 we have used the following function: function Rt = FindZeros(FunName, Nroot, x, w) f = feval(FunName, x, w); indx = find(f(1:end-1).*f(2:end)<0); L = length(indx); if L
Rt = zeros(Nroot, 1); for k = 1:Nroot Rt(k) = fzero(FunName, [x(indx(k)), x(indx(k)+1)], [], w); end

5.4 Ex4 = @(x) (sin(x)-x.*cos(x)) Rt = FindZeros(Ex4, 5, linspace(0,10*pi,100), []) Answers: Rt = 4.4934

7.7253 10.9041 14.0662 17.2208

5.5 Ex5 = @(x,p) (2*cos(x)-(x/p-p./x).*sin(x)); p = [0.1 1]; for k = 1:2 Rt = FindZeros(Ex5, 5, linspace(0,10*pi,100), p(k))' end Answers: Rt = Rt =

0.4435 3.2040 1.3065 3.6732

6.3149 6.5846

9.4460 12.5823 9.6317 12.7232

5.6 Ex6 = @(x,b) (besselj(0,x).*bessely(0,b*x)-besselj(0,b*x).*bessely(0,x)); Rt = FindZeros(Ex6, 5, linspace(0,10*pi,100), 2)' Answers: Rt = 3.1230

6.2734

9.4182 12.5614 15.7040

5.7 Ex7 = @(x, m) (m*x.*(cos(x).*sinh(x)-sin(x).*cosh(x))+cos(x).*cosh(x)+1); m = [0, 0.2, 1]; for k = 1:3 Rt = FindZeros(Ex7, 5, linspace(0,10*pi,100), m(k)); disp(['m = ' num2str(Rt')]) end Answers: m = 0: 1.8751 4.69409 7.85476 10.9955 14.1372 m = 0.2: 1.6164 4.26706 7.31837 10.4016 13.5067 m = 1: 1.24792 4.03114 7.13413 10.2566 13.3878

5.8 Ex8 = @(x,w) (cos(x).*tanh(x)-sin(x)); Rt = FindZeros(Ex8, 5, linspace(0,10*pi,100), [])' Answers: Rt = 3.9266 7.0686 10.2102 13.3518 16.4934

5.9 Ex9 = @(x,m) (besselj(m,x).*besseli(m+1,x)+besseli(m,x).*besselj(m+1,x)); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a3retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

for m = 0:2 Rt = FindZeros(Ex9, 3, linspace(0,10*pi,100), m); disp(['m = ' int2str(m) ': ' num2str(Rt')]) end Answers: m = 0: 3.1962 m = 1: 4.6109 m = 2: 5.90568

6.3064 9.4395 7.79927 10.9581 9.19688 12.4022

5.10 Ex10 = @(x,L2) (sin(x)-(x-4*x.^3/L2).*cos(x)); L2 = [2, 4, 8]*pi^2; for k = 1:3 Rt = FindZeros(Ex10, 1, linspace(0,2*pi,100), L2(k)); disp(['(Lambda/pi)^2 = ' int2str(L2(k)/pi^2) ': ' num2str(Rt')]) end Answers: (Lambda/pi)^2 = 2: 2.5185 (Lambda/pi)^2 = 4: 3.1416 (Lambda/pi)^2 = 8: 3.8834

5.11 channel = inline('(1+c(1)*x).^2.*(x.^2-x.^3)-c(2)','x','c'); c0 = [0.4, 7]; c1 = [0.2, 4]; for k = 1:length(c0) r = roots([-c0(k)^2 c0(k)^2-2*c0(k) 2*c0(k)-1 1 0 -c1(k)]); val = r(find((real(r)>0)&(real(r)<1))); disp(['c0 = ' num2str(c0(k)) ' c1 = ' num2str(c1(k))]) disp(['From roots: ' num2str(sort(val'))]) Rt = FindZeros(channel, 2, linspace(0,1,30), [c0(k), c1(k)])'; disp(['From fzero: ' num2str(Rt)]) end Answers: c0 = 0.4 c1 = 0.2 From roots: 0.54385 From fzero: 0.54385 c0 = 7 c1 = 4 From roots: 0.60777 From fzero: 0.60777

0.84226 0.84226 0.91168 0.91168

5.12 wavechannel = inline('2*(t(2)*sin(beta).*tan(beta-t(1))).^2-tan(beta).*tan(beta-t(1))-tan(beta).^2','beta','t'); w = [35*pi/180, 5]; Rt = FindZeros(wavechannel, 2, linspace(w(1), pi/2-0.05,40), w)'*180/pi Answers: Rt = 48.1088 81.7999

5.13 weibullbeta = inline('b-1/(sum(x.^b.*log(x))/sum(x.^b)-sum(log(x))/length(x))', 'b', 'x');


x = [72 82 97 103 113 117 126 127 127 139 154 159 199 207]; beta = fzero(weibullbeta, [2 4], [], x) Answer: beta = 3.6437

5.14 bearingstress = inline('x.*log(sqrt(x.^2-1)+x)-sqrt(x.^2-1)-c*x', 'x', 'c'); r = fzero(bearingstress, [1 4], [], 0.5) Answer: r = 2.1155

5.15 jeffcooper = inline('x.^4-2.^x', 'x','w'); Rt = FindZeros(jeffcooper, 3, linspace(-2, 20, 100), []) Answers: Rt = -0.8613

1.2396 16.0000

5.16 function Exercise5_16 format long e Zi = zstanford(1,1) Zii = zstanford(1/0.3,1) Ziii = zstanford(2.5,0.5) format short % (b) p = 0.6; t = 1/1.05; trecip = 1.05; r = fzero(@pvrtr, 5, [], p, t); z = zstanford(r,t); disp(['p = ' num2str(p) ' 1/tau = ' num2str(trecip) ' r = ' num2str(r) ' Z = ' num2str(z)]) p = 2.18; t = 1/1.2; trecip = 1.2; r = fzero(@pvrtr, 5, [], p, t); z = zstanford(r, t); disp(['p = ' num2str(p) ' 1/tau = ' num2str(trecip) ' r = ' num2str(r) ' Z = ' num2str(z)]) % (c) p = 0.6; r = 1/1.4; rrecip = 1.4; t = fzero(@pvrtt, 0.8, [], p, r); z = zstanford(r,t); disp(['p = ' num2str(p) ' 1/r = ' num2str(rrecip) ' t = ' num2str(t) ' Z = ' num2str(z)]) p = 2.18; r = 1/0.6; rrecip = 0.6; t = fzero(@pvrtt, 0.8, [], p, r); z = zstanford(r,t); disp(['p = ' num2str(p) ' 1/r = ' num2str(rrecip) ' t = ' num2str(t) ' Z = ' num2str(z)]) function z = zstanford(rr, tt) a = [0.062432384 0.12721477 -0.93633233 0.70184411 -0.35160896 0.056450032 ... 0.0299561469907 -0.0318174367647 -0.0168211055516 1.6020406008 ... -0.001099967407467 -0.000727155024312992 -0.0045245465261 0.001304687241 ... -0.00022216512840926 -0.0019814053565598 5.9757397292e-5 -3.6413534970217e-6 ... 8.413648453856e-6 -9.828688588219e-9 -1.57683056810249 0.040072898890756 ... -0.08451944938128 -0.003409313119283 -0.0019512704990109 4.938999109783e-5 ... -4.9326461293046e-5 8.856665723816e-7 5.347880295527e-8 -5.934205591923e-8 ... © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a5retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

-9.068133269285e-9 1.6182240726495e-9 -3.320447939146e-10]; g = 0.0588; k = 1:12; r = rr.^k; t = tt.^k(1:6); z1 = r(1)*sum(a(1:6).*t(1:6)/t(1))+r(2)*sum(a(7:10).*t(1:4)/t(1))+r(3)*sum(a(11:13).*t(1:3)/t(1)); z2 = r(4)*a(14)*t(1)+r(5)*(a(15)*t(2)+a(16)*t(3))+r(6)*a(17)*t(2)+r(7)*(a(18)*t(1)+a(19)*t(3)) ... +r(8)*a(20)*t(3); z31 = r(2)*(a(21)*t(3)+a(22)*t(4))+r(4)*(a(23)*t(3)+a(24)*t(5))+r(6)*(a(25)*t(3)+a(26)*t(4)); z32 = r(8)*(a(27)*t(3)+a(28)*t(5))+r(10)*(a(29)*t(3)+a(30)*t(4)); z33 = r(12)*(a(31)*t(3)+a(32)*t(4)+a(33)*t(5)); z = 1+z1+z2+(z31+z32+z33)*exp(-g*r(2)); function s = pvrtr(r, p, t) s = zstanford(r,t)-p*t/r; function s = pvrtt(t, p, r) s = zstanford(r,t)-p*t/r; Answers: Zi = 7.024239692767296e-001 Zii = 2.999999998525347e-001 Ziii = 9.922185392802401e-001 p = 0.6 1/tau = 1.05 r = 0.71309 Z = 0.80134 p = 2.18 1/tau = 1.2 r = 3.3567 Z = 0.54121 p = 0.6 1/r = 1.4 t = 0.95319 Z = 0.80068 p = 2.18 1/r = 0.6 t = 0.65049 Z = 0.85084

5.17 function Exercise5_17 re = 100000; dk = [200, 200000]; for n =1:2 lambda = fzero(@pipefrictioncoeff, [.008 .08], [], re, dk(n)); disp(['Re = ' num2str(re) ' d/k = ' num2str(dk(n)) ': lambda = ' num2str(lambda)]) end function x = pipefrictioncoeff(el, re, dk) if (dk>100000)||(dk==0) x = el-(2*log10(re*sqrt(el)/2.51))^-2; else x = el-(2*log10(2.51/re/sqrt(el)+0.27/dk))^-2; end Answers: Re = 100000 d/k = 200: lambda = 0.031298 Re = 100000 d/k = 200000: lambda = 0.01799

5.18 rts = roots([10 0 0 -75 0 -190 21]); rr = rts(find(imag(rts)==0))' Answer: rr = 2.2323

0.1100


ro = roots([3*pi*sqrt(5) -60 0 0 0 0 0 0 0 20]); posroot = ro(find(real(ro)>1)) Answer: posroot = 2.8468

5.20 function Exercise5_20 tto = fzero(@Tto, [0.2 0.9]) function t = Tto(x) t = Thq(exp(-x*pi^2))-pi/2/sqrt(5); function qq = Thq(q) k = 0:10; qq = 2*q^(1/4).*sum((-1).^k./(2*k+1).*q.^(k.*(k+1))); Answer: tto = 0.4240

5.21 fx =@(x) (x-(1+1./x).^x); r = fzero(fx, [0 10]) Anwser: r = 2.2932

5.22 are = inline('sin(x)-0.5*abs(sin(2*x))', 'x'); x = linspace(0, pi); aret = trapz(x, are(x)) areq = quadl(are, 0, pi) Answers: aret = 0.9999

areq = 1.0000

5.23 bearing = inline('(1-(1-cos(x))/2/epsilon).^c.*cos(m*x)', 'x', 'c', 'm', 'epsilon'); m = 1; epsilon = 0.6; c = 1.5; a = acos(1-2*epsilon); Im = quadl(bearing, -a, a, [], [], c, m, epsilon)/2/pi Answer: Im = 0.2416

5.24 Ee = inline('C1./lam.^5./(exp(C2./(lam*T))-1)', 'lam', 'C1', 'C2', 'T'); C1 = 3.742e8; C2 = 1.439e4; sig = 5.667e-8; T = [300 400 500]; for k=1:length(T) App = quadl(Ee, 1, 200, [], [], C1, C2, T(k)); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a7retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

exact = sig*T(k)^4; err = 100*(exact/App-1); disp(['T = ' num2str(T(k)) ' exact = ' num2str(exact) ' Approx. = ' num2str(App) ' %error = ' num2str(err)]) end Answers: T = 300 exact = 459.027 Approx. = 458.7468 %error = 0.061087 T = 400 exact = 1450.752 Approx. = 1450.3993 %error = 0.024315 T = 500 exact = 3541.875 Approx. = 3541.4894 %error = 0.010889

5.25 arg = @(x,y) (cos(x-y).*exp(-x.*y/pi^2)); Z = dblquad(arg, 0, pi/2, pi/4, pi) Answer: Z = 0.8449

5.26 arg = @(x) (1./sqrt((6-x.^2))); Z = quadl(arg, 2, sqrt(6))-acot(sqrt(2)) Answer: Z = 1.3216e-006

5.27 function Exercise5_27 Om = pi*[0.5969, 1.4942, 2.5002, 3.5:9.5]; Cn = -T(Om)./Q(Om); N = zeros(10,1); for n=1:10 N(n) = quadl(@W, 0, 1, [], [], Om(n), Cn(n)); end disp(['N = ' num2str(N',4)]) function w = W(x, Om, Cn) w = (Cn*T(Om*x)+S(Om*x)).^2; function q = Q(x) q = 0.5*(cosh(x)+cos(x)); function s = S(x) s = 0.5*(cosh(x)-cos(x)); function t = T(x) t = 0.5*(sinh(x)+sin(x)); Answers: N = 0.6133 0.3546 0.3137 0.2955 0.2854 0.2789 0.2745 0.2712 0.2687 0.2668

5.28 function Exercise5_28 g = 9.8; alpha = 45*pi/180; v0 = 180; cd = 0.007;


[x, y] = ode45(@projectile, [0, 300], [v0*cos(alpha) v0*sin(alpha) 0 0]', [], v0, cd, g); N = length(x); indx = find(y(:,3)>0); height = @(v,x,y) (-spline(x,y,v)); opt = optimset('display','off'); xmax = fminbnd(height, 0, 1000, opt, x, y(:,3)); ymax = -height(xmax, x, y(:,3)); disp(['ymax = ' num2str(ymax) ' m at x = ' num2str(xmax) ' m']) xe = interp1(y(indx(end)-5:N, 3), x(indx(end)-5:N), 0); te = interp1(y(indx(end)-5:N, 3), y(indx(end)-5:N, 4), 0); disp(['Travel distance = ' num2str(xe) ' m and time of travel = ' num2str(te) ' s']) function p=projectile(x, z, v0, cd, g) v = sqrt(z(1)^2+z(2)^2); if v
5.29 function Exercise5_29 opt = optimset('display', 'off'); g = 9.8; co = 0.00324; L = 150; kmj = 10/70; [t, y] = ode45(@bungee, [0 50], [0 0]', [], g, co, kmj, L); splinefit = @(td, t, y) (-spline(t, y, td)); maxdisptime = fminbnd(splinefit, 0, 20, opt, t, y(:,1)); maxdisplace = -splinefit(maxdisptime, t, y(:,1)); disp(['Max distance = ' num2str(maxdisplace) ' m at ' num2str(maxdisptime) ' s.']) slack =@(td, t, y, L) (L-spline(t,y,td)); tL = fzero(slack, [0 10], opt, t, y(:,1), L); vL = -splinefit(tL, t, y(:,2)); disp(['Diver reaches ' num2str(L) ' m in ' num2str(tL) ' s at at which time the velocity is ' num2str(vL) ' m/s']) accel = g-sign(y(:,2))*co.*y(:,2).^2-kmj*(y(:,1)-L).*(y(:,1)>=L); maxacceltime = fminbnd(splinefit, 0, 20, opt, t, -accel); maxaccel = splinefit(maxacceltime, t, accel); disp(['Max acceleration = ' num2str(maxaccel) ' m/s^2 (' num2str(maxaccel/9.8) ' g) at ' num2str(maxacceltime) ' s']) function v = bungee(t, y, g, co, kmj, L) v = [y(2); g-sign(y(2))*co*y(2)^2-kmj*(y(1)-L)*(y(1)>=L)]; Answers: Max distance = 308.4649 m at 11.4679 s. Diver reaches 150 m in 5.9875 s at at which time the velocity is 43.4764 m/s Max acceleration = 12.843 m/s^2 (1.3105 g) at 11.3745 s

5.30 function Exercise5_30 a = 0.1; b = 10.0; [t, th] = ode45(@IP, linspace(0, 50, 1000), [pi/4 0]', [], b, a); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a9retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

subplot(2,1,1) plot(t, th(:,1)) subplot(2,1,2) plot(th(:,1), th(:,2)) function v = IP(t, y, b, a) v = [y(2); -a*y(2)+sin(y(1))-b*sin(y(1))*(1-1/sqrt(5-4*cos(y(1))))]; Answer: 1

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5.31 function Exercise5_31 p = 0.375; q = 7.43e-4; Z = [10 50]; zer = zeros(length(Z),1); for k = 1:2 [t, y] = ode45(@ReservoirOsc, [0 500], [Z(k) 0], [], p, q); ind = min(find(y(:,1)<0)); zer = interp1(y(ind-5:ind+4,1), t(ind-5:ind+4), 0); disp(['For Z = ' num2str(Z(k)) ': t = ' num2str(zer) ' s']) end function c = ReservoirOsc(t, y, p, q ) c = [y(2); -p*sign(y(2))*y(2)^2-q*y(1)]; Answer: For Z = 10: t = 114.2692 s For Z = 50: t = 276.1428 s

5.32 function Exercise5_32 fm = fminbnd(@fxi, 0.0, 0.99); disp(['Minimum xi = ' num2str(fm)]) function f = fxi(xi) [t, y] = ode45(@minodeoutput, [0 35], [0 0]', [], xi); f = sum((y(:,1)-1).^2); function v = minodeoutput(t, w, xi) v = [w(2); -2*xi*w(2)-w(1)+1]; © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a10 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Answer: Minimum xi = 0.71873

5.33 function Exercise5_33 e = 0.16; g = 0.4; w = 0.97; [t, y] = ode45(@tunus, [0, 60], [-1,1], [], e, w, g); plot(y(:,1), y(:,2),'k-') function d = tunus(t, z, e, w, g) c = w*t+9*pi/8; L = 1+e*sin(c)^7; Lp = 7*e*w*sin(c)^6*cos(c); d = [z(2); -((2*Lp/L+g*L)*z(2)+sin(z(1))/L)]; Answer: 1.5

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5.34 function Exercise5_34 a = 0.5; b = 0.6; k = 1; [t, z] = ode45(@Ray, [0 30],[ 0 0.1], [], a, b, k); plot(z(:,1), z(:,2), 'k-') function d = Ray(t, y, a, b, k) d = [y(2); -k*y(1)+a*y(2)-b*y(2)^3]; Answer:


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5.35 function Exercise5_35 uo = 0.05; c = 0.02; r = 1; w = 0.2; p = 2.01; pe = fzero(@Static, [0.1, 0.2], [], p); opt = odeset('RelTol', 0.00001); [t, ph] = ode45(@Mecha, [0, 800], [pe, 0], opt, uo, c, r, w, p); plot(t, ph(:,1), 'k-') hold on plot(t, uo*sin(w*t)+2*cos(ph(:,1))-2*cos(pe), 'b-') function g = Mecha(t, y, uo, c, r, w, p) den = 1/3+(1+2*r*p)*sin(y(1))^2; g = [y(2); (-(1+2*r*p)*y(2)^2*sin(y(1))*cos(y(1))-(1+r*p)*uo*w^2*sin(y(1))*sin(w*t)... -2*c*y(2)-2*y(1)+p*sin(y(1)))/den]; function d = Static(h, p) d = 2*h-p*sin(h); Answer: 0.3

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5.36 function Exercise5_36 solinit = bvpinit(linspace(0,1,5),[1, 1, 1, -10, 0.91]); sol = bvp4c(@Exode, @Exbc, solinit); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a12 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

xx = linspace(0, 1, 50); yy = deval(sol, xx); plot(xx,yy(1,:)) hold on plot(xx, yy(2,:)) plot(xx, yy(3,:)) function res = Exbc(ya,yb) % y(1) = u, y(2) = v, y(3) = w, y(4) = z, y(5) = y. res = [ ya(1) - 1; ya(2) - 1; ya(3) - 1; ya(4) + 10; yb(3) - yb(5)]; function dydx = Exode(x,y) % y(1) = u, y(2) = v, y(3) = w, y(4) = z, y(5) = y. dydx = [ 0.5*y(1)*(y(3) - y(1))/y(2) -0.5*(y(3) - y(1)) (0.9 - 1000*(y(3) - y(5)) - 0.5*y(3)*(y(3) - y(1)))/y(4) 0.5*(y(3) - y(1)) 100*(y(3) - y(5)) ]; Answer: 2.2

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5.37 function Exercise5_37 LLo = 1.2; opt = optimset('display','off'); beta = fzero(@cablebeta, 2, opt, LLo); solinit = bvpinit(linspace(0,1,5), [0.1 0.1]); sol = bvp4c(@inextcable, @BC, solinit, [], beta); z = deval(sol, 0); disp(['beta = ' num2str(beta) ' and the slope at eta = 0 is ' num2str(z(2,1))]) function a = cablebeta(beta, LLo) solinit = bvpinit(linspace(0,1,5), [0.1 0.1]); sol = bvp4c(@inextcable, @BC, solinit, [], beta); eta = linspace(0, 1, 100); zz = deval(sol, eta); a = LLo-trapz(eta, sqrt(1+zz(2,:).^2)); function c = inextcable(t, z, beta) c = [z(2); beta*sqrt(1+z(2)^2)];


function r = BC(za,zb,beta) r = [za(1); zb(1)]; Answer: beta = 2.1297 and the slope at eta = 0 is -1.2778

5.38 function Exercise5_38 qo = 1; solinit = bvpinit(linspace(0, 1, 10), [0.5, 0.5, 0.5, 0.5]); beamsol = bvp4c(@BeamODEqo, @CantBC, solinit, [], qo); eta = linspace(0, 1, 50); y = deval(beamsol, eta); subplot(2,2,1) plot(eta, y(1,:),'k-') subplot(2,2,2) plot(eta, y(2,:),'k-') subplot(2,2,3) plot(eta, y(3,:),'k-') subplot(2,2,4) plot(eta, y(4,:),'k-') function dydx = BeamODEqo(x, y, qo) dydx = [y(2); y(3); y(4); qo.*(x>0.5)]; function bc = CantBC(y0, y1, qo) bc = [y0(1); y0(2); y1(3); y1(4)]; Answer: 0.12

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5.39 function Exerxise5_39 Pbar = 8; alph = pi/2; sw = 0; tol = 0.001; gl = 157; gh = 158; qq = fzero(@Shoot, [gl, gh]*pi/180, [], Pbar, alph, sw, tol)*180/pi; sw = 1; [r, x] = NLBeam(qq*pi/180, Pbar, alph, sw, tol); disp(['Pbar = ' num2str(Pbar) ' phi(0) = ' num2str(qq) ' (deg) x(0) = '... num2str(x(1)), ' y(0) = ' num2str(x(2))]) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a14 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

function z = Shoot(q, Pbar, alph, sw, tol) z = q-NLBeam(q, Pbar, alph, sw, tol); function [res, dis]= NLBeam(y0, Pbar, alph, sw, tol) solinit = bvpinit(linspace(0,1,5), [2 1]); opt = bvpset('RelTol', tol); solu = bvp4c(@Odes, @BC, solinit, opt, Pbar, alph, y0); r = deval(solu(1,end), 1); res = r(1); dis = [999, 999]; if sw==1 xx = linspace(0,1,50); s = deval(solu,xx); dis = [trapz(xx, cos(s(1,:))), trapz(xx, sin(s(1,:)))] ; end function der = Odes(x, y, Pbar, alph, y0) der = [y(2); -Pbar*sin(y(1)+alph-y0)]; function der = BC(ya, yb, Pbar ,alph, y0) der = [ya(1); yb(2)]; Answer: Pbar = 8 phi(0) = 157.938 (deg) x(0) = -0.073844 y(0) = 0.63353

5.40 function Exercise5_40 a = 0.8; zinfi = fzero(@Eqn, [1.1 4], [], a) function slop = Eqn(zinf, a) zin = [zinf 1.05*zinf]; d(1:2) = 0; for k = 1:2 solinit = bvpinit(linspace(0, zin(k), 5), [0.1, 0.1]); exmp = bvp4c(@odeQ3, @bcQ3,solinit, [], a); et = linspace(0, zin(k), 100); wnew = deval(exmp, et); slop = wnew(1, end)-0.05; d(k) = interp1(et, wnew(1,:), 1.1); end slop = (d(2)-d(1))/0.05-0.03; function de = odeQ3(t, y, a) de = [y(2); -2*t*y(2)/sqrt(1-a*y(1))]; function b = bcQ3(ya, yb, a) b = [ya(1)-1; yb(1)]; Answer: zinfi = 1.9501

5.41 function Exercise5_41 v = 0.3; © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a15 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

solinit = bvpinit(linspace(0.2,1,5), [0.1, 0.1, 0.1, 0.1]); plat = bvp4c(@Plate, @bcPlate, solinit, [], v); rr = linspace(0.2, 1, 50); w = deval(plat, rr); subplot(2,2,1) plot(rr, w(1,:), 'k') subplot(2,2,2) plot(rr, w(2,:), 'k') subplot(2,2,3) plot(rr, w(3,:)+v*w(2,:)./rr, 'k') subplot(2,2,4) plot(rr, w(4,:)+w(3,:)./rr-w(2,:)./rr.^2, 'k') function s = Plate(r, y, v) s = [y(2); y(3); y(4); -2/r*y(4)+y(3)/r^2-y(2)/r^3+1]; function d = bcPlate(yi, yo, v) d = [yi(1); yi(2); yo(3)+v*yo(2); yo(4)+yo(3)-yo(2)]; Answer: 0.06

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5.42 function Exercise5_42 S = 0.4; solinit = bvpinit(linspace(0, 1, 10), [0.5, 0.5, 0.5, 0.5]); beamsol = bvp4c(@BeamODE, @CantileverBC, solinit, [], S); eta = linspace(0, 1, 50); y = deval(beamsol, eta); subplot(2,2,1) plot(eta, y(1,:),'k-') subplot(2,2,2) plot(eta, y(2,:),'k-') subplot(2,2,3) plot(eta, y(3,:),'k-') subplot(2,2,4) plot(eta, y(4,:),'k-') function dydx = BeamODE(x, y, S) dydx = [y(2); y(3); y(4); 1-S*y(3)]; © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a16 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

function bc = CantileverBC(y0, y1, S) bc = [y0(1); y0(2); y1(3); y1(4)+S*y1(2)]; Answer: 0.2

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5.43 function Exercise5_43 Omguess = pi/2; gbs = 3.12; Ro = 0.06; solinit = bvpinit(linspace(0,1,10), @TimoBeamInit, Omguess); beamsol = bvp4c(@TimoBeamODE, @TimoBeamBC, solinit, [], gbs, Ro); Omega = beamsol.parameters function yinit = TimoBeamInit(x) yinit = [sin(x*pi/2); cos(x*pi/2); -sin(x*pi/2); -cos(x*pi/2)]; function dydx = TimoBeamODE(x, y, Om, gbs, Ro) dydx = [y(2); y(4)-gbs*Ro^2*Om^4*y(1); y(4); -(y(2)+(gbs*Ro^4*Om^4-1)*y(3))/(gbs*Ro^2)]; function bc = TimoBeamBC(y0, y1, Om, gbs, Ro) bc = [y0(1); y0(4)-0.05; y0(3); y1(2)-y1(3); y1(4)]; Answer: Omega = 1.8444

5.44 function Exercise5_44 % Part (a) wormleadmin = @(lambda, b) (b./sin(lambda)+1./cos(lambda)); bt = [0.02, 0.05, 0.08, 0.11, 0.15, 0.18, 0.23, 0.30]; opt = optimset('display','off'); z = zeros(length(bt),1); for j = 1:length(bt) z(j )= fminbnd(wormleadmin, pi/180, 40*pi/180, opt, bt(j))*180/pi; end disp(' beta Kmin') disp([bt' z]) % Part (b) wormlead = @(lambda, b, k) (k-wormleadmin(lambda,b)); lambda1 = fzero(wormlead, [10 25]*pi/180, opt, 0.16, 1.5)*180/pi © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a17 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

lambda2 = fzero(wormlead, [25 40]*pi/180, opt, 0.16, 1.5)*180/pi Answers:

beta Kmin 0.0200 15.1866 0.0500 20.2231 0.0800 23.3098 0.1100 25.6001 0.1500 27.9830 0.1800 29.4501 0.2300 31.4950 0.3000 33.7997 lambda1 = 22.5505

lambda2 = 35.0343

5.45 gasflowtank = inline('-sqrt(k/(k-1))*sqrt(pepo.^(2/k)-pepo.^((k+1)/k))','pepo','k'); opt = optimset('display','off'); k = 1.4; maxpratio = fminbnd(gasflowtank, 0, 1, opt, k); maxexact = (2/(k+1))^(k/(k-1)); pepo = linspace(0, 1, 200); [minarray, index] = min(gasflowtank(pepo,k)); numpratio = pepo(index); disp(['max ratio exact = ' num2str(maxexact) ' max ratio w/fminbnd = ' … num2str(maxpratio) ' max ratio w/min = ' num2str(numpratio)]) Answers: max ratio exact = 0.52828 max ratio w/fminbnd = 0.5283 max ratio w/min = 0.52764

5.46 function Exercsie5_46 opt = optimset('Display', 'off'); sol = fsolve(@functExamp, [1 1 1], opt) disp(['x = ' num2str(sol(1)) ' y = ' num2str(sol(2)) ' z = ' num2str(sol(3))]) function f = functExamp(x) f = [sin(x(1))+x(2)^2+log(x(3))-7; 3*x(1)+2^x(2)+1-x(3)^3; x(1)+x(2)+x(3)-5]; Answers: x = 0.59905 y = 2.3959 z = 2.005

5.47 fs = inline('[b-z(1)*(1-cos(z(2)));a-z(1)*(z(2)-sin(z(2)))]', 'z', 'a', 'b'); a = 1; b = 3; opt = optimset('Display','off'); r = fsolve(fs, [5 pi/4], opt, a, b); disp(['k = ' num2str(r(1)) ' theta = ' num2str(r(2)*180/pi) ' degrees']) fz = inline('b*(th-sin(th))-a*(1-cos(th))', 'th', 'a', 'b'); th = fzero(fz, [pi/4], opt, a, b); disp(['k = ' num2str(a/(th-sin(th))) ' theta = ' num2str(th*180/pi) ' degrees']) Answers: k = 6.9189 theta = 55.4999 degrees k = 6.9189 theta = 55.4999 degrees © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a18 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

5.48 QQ = inline('[T1^4-T(1)^4-T(3)/sig; T(1)^4-T(2)^4-T(3)/sig; T(2)^4-T2^4-T(3)/sig]', 'T', 'T1', 'T2', 'sig'); sig = 5.667e-8; T2 = 293; T1 = 373; opt = optimset('display', 'off'); TQ = fsolve(QQ, [ 300 300 300], opt, T1, T2, sig); disp('From fsolve:') disp(['TA = ' num2str(TQ(1)) ' TB = ' num2str(TQ(2)) ' Q = ' num2str(TQ(3))]) % (b) V = [1, 0, 1/sig; 1, -1, -1/sig; 0, 1, -1/sig] \[T1^4, 0, T2^4]'; disp(' ') disp('From matrix inversion:') disp(['TA = ' num2str(V(1)^(1/4)) ' TB = ' num2str(V(2)^(1/4)) ' Q = ' num2str(V(3))]) Answers: From fsolve: TA = 352.052 TB = 326.5116

Q = 226.4312

From matrix inversion: TA = 352.052 TB = 326.5116

Q = 226.4312

5.49 syms x dfx = diff(exp(sin(x)), x); xext = solve(dfx, x) dfxx = limit(diff(dfx,x), x, xext) Answers: xext = 1/2*PI

dfxx = -exp(1)

5.50 syms x b Intg = int((2*x+5)/(x^2+4*x+5), x, 0, b) AofB = matlabFunction(Intg, 'vars', [b]) A = AofB(linspace(0, 4*pi, 10))' Answers: A = 0 1.0964 1.8252 2.3653 2.7927 3.1459 3.4467 3.7084 3.9401 4.1478

5.51 function Exercise5_51 [AlphMax fAlphaMax] = fminbnd(@(x) (-Integ(x)), 0, 1);


disp(['alphaMax = ' num2str(AlphMax) ' and f(alphaMax) = ' num2str(-fAlphaMax)]) [AlphMin fAlphaMin] = fminbnd(@Integ, 0.5, 5); disp(['alphaMin = ' num2str(AlphMin) ' and f(alphaMin) = ' num2str(fAlphaMin)]) function In = Integ(alph) In = (2+sin(10*alph)).*quadl(@Arg, 0, 1.8, [], [], alph); function A = Arg(x, alph) A = x.âlph.*sin(alph./(2-x)); Answers: alphaMax = 0.78153 and f(alphaMax) = 3.2284 alphaMin = 3.8882 and f(alphaMin) = -0.99339

5.52 function Exercise5_52 ka = pi/4; v = 0.3; b = 0.05/sqrt(12); N = 10; rot = zeros(N+1,3); for meth = 1:3 switch meth case 1 [C1 C2 C3 C4] = Coef; for n = 0:N a11 = ka^2+(1-v)/2*n.^2; a12 = (1+v)*n*ka/2; a13 = v*ka; a23 = n; a22 = (1-v)/2*ka^2+n.^2; a33 = 1+b^2*(ka^2+n.^2).^2; C11 = C1(a11,a12,a13, a22,a23,a33); C22 = C2(a11,a12,a13, a22,a23,a33); C33 = C3(a11,a12,a13, a22,a23,a33); C44 = C4(a11,a12,a13, a22,a23,a33); rot(n+1,:) = sqrt(roots([C44 C33 C22 C11])); end case 2 for n = 0:N A = MatrixElements(n, ka, b, v); rot(n+1,:) = sqrt(eig(A)); end case 3 Nroot = 3; x = logspace(-2, 1.2, 100); for n = 0:N rot(n+1,:) = FindZero(@deter, Nroot, x, [n,ka,b,v]); end end figure(meth) for m=1:3 semilogy(0:10, rot(:,m), 'ks-' ) hold on end axis tight disp(' ') disp(' n L(1,n) L(2,n) L(3,n) ') © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a20 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

disp([(0:10)' rot]) end function [C1 C2 C3 C4] = Coef syms La a11 a12 a13 a21 a22 a23 a31 a32 a33 A = [-La+a11, a12, a13; a12, -La+a22, a23; a13, a23, -La+a33]; DetA = collect(simple(det(A)), La); C = coeffs(DetA, La); C1 = matlabFunction(C(1), 'vars', [a11, a12, a13, C2 = matlabFunction(C(2), 'vars', [a11, a12, a13, C3 = matlabFunction(C(3), 'vars', [a11, a12, a13, C4 = matlabFunction(C(4), 'vars', [a11, a12, a13,

a22, a23, a33]); a22, a23, a33]); a22, a23, a33]); a22, a23, a33]);

function D = deter(La, w) n = w(1); ka = w(2); b = w(3); v = w(4); A = MatrixElements(n, ka, b, v); A(1,1) = A(1,1)-La^2; A(2,2) = A(2,2)-La^2; A(3,3) = A(3,3)-La^2; D = det(A); function Rt=FindZero(FunName, Nroot, x, w) cnt = 0; Rt = []; n = 3; D = zeros(length(x),1); % Following change required because 'deter' cannot % deal with an array of values for x for k = 1:length(x) D(k) = feval(FunName, x(k), w); end for k = 1:length(x) if sign(D(n))*sign(D(n-1))>0 n=n+1; else cnt = cnt+1; rte = fzero(FunName,[x(n-1),x(n)],[],w); Rt = [Rt rte]; n = n+2; end if n>length(x)-1||cnt==Nroot break end end function A = MatrixElements(n,ka,b,v) a11 = ka^2+(1-v)/2*n.^2; a12 = (1+v)*n*ka/2; a13 = v*ka; a23 = n; a22 = (1-v)/2*ka^2+n.^2; a33 = 1+b^2*(ka^2+n.^2).^2; A = [a11, a12, a13; a12, a22, a23; a13, a23, a33]; Answers:

n 0

L(1,n) L(2,n) L(3,n) 1.0546 0.7105 0.4646


1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 10.0000 10

10

1.5258 2.3405 3.2468 4.1920 5.1566 6.1320 7.1141 8.1004 9.0897 10.0811

0.8883 1.3203 1.8546 2.4212 2.9996 3.5831 4.1693 4.7570 5.3457 5.9351

0.2574 0.1271 0.1433 0.2348 0.3630 0.5213 0.7088 0.9253 1.1706 1.4449

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Solution to Exercises Chapter 6 6.1 % Cycloid phi = linspace(-pi, 3*pi, 200); ra = [0.5, 1, 1.5]; for k = 1:length(ra) subplot(3,1,k) plot(ra(k)*phi-sin(phi), ra(k)-cos(phi)) axis equal title(['r_a = ' num2str(ra(k))]) axis equal off end Answer: ra = 0.5

ra = 1

ra = 1.5

% Leminscate phi = linspace(-pi/4, pi/4, 101); plot(cos(phi).*sqrt(2*cos(2*phi)), sin(phi).*sqrt(2*cos(2*phi))) axis equal off Answer:

% Archimedean spiral phi = linspace(0, 6*pi, 200); plot(phi.*cos(phi), phi.*sin(phi)) axis equal off Answer:


%Logarithmic spiral phi = linspace(0, 6*pi, 200); k = 0.1; plot(exp(k*phi).*cos(phi), exp(k*phi).*sin(phi)) axis equal off Answer:

% Cardioid phi = linspace(0, 2*pi, 100); plot(2*cos(phi)-cos(2*phi), 2*sin(phi)-sin(2*phi)) axis equal off Answer:

% Astroid phi = linspace(0, 2*pi, 100); plot(4*cos(phi).^3, 4*sin(phi).^3) axis equal off Answer:


% Epicycloid en = [2*pi, 4*pi]; r = [3, 2.5]; a = [0.5 2]; for k = 1:2 subplot(1,2,k) phi = linspace(0, en(k), 200); rr = r(k); ar = a(k); plot((rr+1)*cos(phi)-ar*cos(phi*(rr+1)), (rr+1)*sin(phi)-ar*sin(phi*(rr+1))) axis equal off end Answer:

% Hypocycloid phi = linspace(0, 2*pi, 200); rr = 3; ar = [0.5, 1, 2]; for k = 1:length(ar) subplot(1,3,k) plot((rr-1)*cos(phi)+ar(k)*cos(phi*(rr-1)), (rr-1)*sin(phi)-ar(k)*sin(phi*(rr-1))) axis equal off end Answer:

% Eight curve ph = linspace(0, 2*pi, 200); a = 2; plot(a*sin(ph), a*sin(ph).*cos(ph), 'k-') axis equal off


Answer:

% Butterfly x = linspace(-1, 1, 201); a = 2; plot(x, (x.^4-x.^6).^(1/6), 'k-', x, -(x.^4-x.^6).^(1/6), 'k-') axis equal off Answer:

% Dumbbell x = linspace(-1, 1, 200); a = 2; plot(x, a*sqrt((x.^4-x.^6)), 'k-', x, -a*sqrt((x.^4-x.^6)), 'k-') axis equal off Answer:

% Bicuspid a = 1; x = linspace(-a, a, 200); rt = @(x,a) sqrt((a^2-x.^2).*(x-a).^2); plot(x, sqrt(a^2+rt(x,a)), 'k-', x, -sqrt(a^2+rt(x,a)), 'k-') hold on A = @(x,a) (a^2-rt(x,a)); xl = fzero(A, -a/2, [], a) x = linspace(-a, xl, 50); plot(x, sqrt(a^2-rt(x,a)), 'k-', x, -sqrt(a^2-rt(x,a)), 'k-') x= linspace(0, a, 50); plot(x, sqrt(a^2-rt(x,a)), 'k-', x, -sqrt(a^2-rt(x,a)), 'k-') axis equal off Answer: © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a4retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

% Super ellipse a = 2; b = 1; n = [2/3, 3, 7]; for k = 1:3 subplot(1,3,k) x = [linspace(-a, 0, 100) 0 linspace(0.01, a, 100)]; y = b*(1-abs(x/a).^n(k)).^(1/n(k)); plot(x, y, 'k-', x, -y, 'k-') axis equal off end Answer:

% Gear n = 12; a = 1; b = 10; t = linspace(0, 2*pi, 600); r = a+tanh(b*sin(n*t))/b; plot(r.*cos(t), r.*sin(t), 'k-') axis equal off Answer:

% Folium th = linspace(0, 2*pi, 100); a = 2; b = 3; c = cos(th); s = sin(th); r = real(sqrt(c.*(4*a*s.^2-b))); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a5retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

plot(r.*c, r.*s, 'k-') axis equal off Answer:

6.2 rts = sort(roots([.001 -.01 -0.2 1 4 -5]), 'ascend'); ind = find((rts>=-12)&(rts<=7)); for k = 1:2:length(ind) x = linspace(rts(k), rts(k+1), 25); y = polyval([.001 -.01 -0.2 1 4 -5], x); plot(x, y, 'k-' ) hold on end xlim([-12 7]) xlabel('x') ylabel('y') Answer: 70

60

50

40

30 y

20

10

0

-10 -12

-10

-8

-6

-4

-2

0

2

4

6

x

6.3 tho = pi/3; N = 17; del = 1; x1 = 0:N; x2 = x1.*del*cos(tho); y2 = x1.*del*sin(tho); N = N+1; z = zeros(1,N); plot(x1, z, 'k-s') hold on plot(x2, y2, 'k-s') plot([x1(2:N); x2(N:-1:2)], [z(N:-1:2); y2(N:-1:2)], 'k-') © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a6retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

axis([-1, N, -1, N]) axis equal off Answer:

6.4 n = input('Enter number of circles (>2): '); rb = 1.5; phi = linspace(0, 2*pi); rs = rb*sin(pi/n)/(1-sin(pi/n)); th = (1:n)*2*pi/n; plot(rb*cos(phi), rb*sin(phi), 'k', 0, 0, 'k+', (rb+rs)*cos(th), (rb+rs)*sin(th), 'k+') axis equal [x, phd] = meshgrid((rb+rs)*cos(th), phi); y = meshgrid((rb+rs)*sin(th), phi); hold on axis off plot(x+rs.*cos(phd), y+rs.*sin(phd), 'k-') Answer:

6.5 N = 15000; x = zeros(N,1); y=x; x(1) = 1; y(1) = 3.65; for n = 2:N x(n) = 1-y(n-1)+abs(x(n-1)); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a7retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

y(n) = x(n-1); end plot(x, y, 'k.') axis equal off Answer:

6.6 eta = linspace(0, 1, 10); phi = linspace(0, 2*pi, 200); c = cos(phi); s = sin(phi); a = 1.5; N = 4; for k = 1:length(eta) D = cosh(eta(k))-c; x = sinh(eta(k))./D; y = s./D; plot(a*x, a*y, 'k-', -a*x, -a*y, 'k-') hold on end phi = linspace(0, 2*pi, 30); for k = 1:length(phi) D = cosh(eta)-cos(phi(k)); x = sinh(eta)./D; y = sin(phi(k))./D; plot(a*x, a*y, 'k-', -a*x, -a*y, 'k-') hold on end axis equal off axis([-N N -N N]) Answer:


6.7 A = 2; N = 20; t = linspace(-pi, pi, 400); T = t(end)-t(1); signal = A*sin(t); dt = t(2)-t(1); x = linspace(-A, A, N); dx = x(2)-x(1); L = zeros(N-1,1); for k = 1:N-1 Index = find((signal>=x(k))&(signal
0.6

0.5 ) x ( P 0.4

0.3

0.2

0.1 -2

-1.5

-1

-0.5

0 Amplitude

0.5

1

1.5

2

6.8 h = 1:0.25:3; d = linspace(0, 5, 100); [ht, dt] = meshgrid(h, d, 1); plot(dt, 0.5*dt.^3-1.5*ht.*dt.^2+(1+ht.^2).*dt, 'k') © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a9retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

axis([0 5 0 8]) xlabel('\delta/t') ylabel('C_1') dd = 1.2; pht = 0.5*dd.^3-1.5*h.*dd.^2+(1+h.^2).*dd+.12; text(repmat(.9, 1, length(h)-1), pht(2:length(h)), num2str(h(2:end)')) text(.9, pht(end)+.4,'h/t = ') text(0.9, pht(1)-.4,num2str(h(1))) Answer: 8

7

h/t = 3

6 2.75 5

2.5 1

4

C

2.25 3 2 1.75

2

1.5 1.25 1

1

0 0

0.5

1

1.5

2

2.5 /t

3

3.5

4

4.5

5

6.9 function Exercise6_9 rb = 90.21; rt = 106; ts = 14.022; m = 8; n = 24; [t, r] = geartoothshape(rb, rt, ts, m, n); rm = max(r); subplot(1,2,1) for k = 0:n-1 polar(t+2*k*pi/n, r/rm) hold on end subplot(1,2,2) for k = 0:n-1 plot(r.*cos(t+2*k*pi/n), r.*sin(t+2*k*pi/n)) hold on axis equal off end plot(0, 0, '+b') Answer: 90

1

120

60

0.5

150

30

180

0

210

330

240

300 270


6.10 powerscreweff = inline('100*(cos(a)-mu.*tan(lamb))./(cos(a)+mu.*cot(lamb))', 'a', 'lamb', 'mu'); lam = linspace(0.01, pi/2-.01, 100); mum = [0.02, 0.05:0.05:.25]; [mu, lamb] = meshgrid(mum, lam, 1); e = powerscreweff(14.5*pi/180, lamb, mu); plot(lamb*180/pi, powerscreweff(14.5*pi/180, lamb, mu), ... '-k', lamb*180/pi, powerscreweff(7*pi/180, lamb, mu),'--k') axis([0 90 0 100]) legend('thread angle = 14.5\circ', 'thread angle = 7\circ', 'Location', 'SouthWest') xlabel('Lead angle (degrees)') ylabel('Efficiency (%)') title('Efficiency of a power screw') text(repmat(30, 1, length(mum)), powerscreweff(7*pi/180,pi/4,mum)-3,... [repmat('\mu = ', length(mum),1) num2str(mum')]) Answer: Efficiency of a power screw 100

μ = 0.02 90

) % ( y c n e i c i f f E

μ = 0.05

80

μ = 0.1

70

μ = 0.15 μ = 0.2

60

μ = 0.25 50

40

30

20

10

thread angle = 14.5 ° thread angle = 7 °

0 0

10

20

30

40 50 Lead angle (degrees)

60

70

80

90

6.11 L = 1; W = 2; d = 2; r1 = sqrt(d^2+(W/2)^2); r2 = sqrt((d+L)^2+(W/2)^2); alpha = atan(W/(2*d)); beta = atan(W/(2*(L+d))); Nv = floor(pi/alpha); Dtheta = 2*pi/Nv; theta = 0:Dtheta:2*pi-Dtheta; gama = [-alpha, alpha, beta, -beta]; rad = [r1, r1, r2, r2]; for k = 1:Nv x = rad.*cos(theta(k)+gama); y = rad.*sin(theta(k)+gama); plot([x, x(1)], [y, y(1)], 'k') hold on end plot([zeros(1,Nv); d*cos(theta)], [zeros(1,Nv); d*sin(theta)], 'k') axis equal off Answer:


6.12 N = 15; n = 1:N; Angl = cumsum(atan(1./sqrt(1:N-1))); Angl = [pi, pi-Angl]; R = sqrt(n); x = R.*cos(Angl); y = R.*sin(Angl); plot(x, y, 'k-') hold on plot([zeros(1,N);x], [zeros(1,N);y], 'k-') axis equal off ang = [pi/8, Angl(1:N-1)+diff(Angl)/2]; text(x(1)/2, 0, '1', 'VerticalAlignment','bottom') for k = 2:N text(x(k)/2, y(k)/2, ['\surd' num2str(k)]) text(R(k).*cos(ang(k)), R(k).*sin(ang(k)), '1', 'HorizontalAlignment','right') end Answer: 1

1

1 1

4

3 1

5

2

6

1

1

7 8

1

9

15 10

14 13

11

12

1

1 1 1

1 1

1

6.13 za = linspace(0.01, 3); vu1 = 0.3; sx = (1-za.*atan(1./za))*(1+vu1)-0.5./(1+za.^2); sz = 1./(1+za.^2); plot(za, sx, 'k-', za, sz, 'k--', za, -0.5*(sx-sz), 'k:') © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a12 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

legend('sigma x', 'sigma z', 'shear', 'Location', 'NorthEast') axis([0, 3, 0, 1]) xlabel('depth (z/a)') ylabel('Compressive stress and shear') title('Contact stresses in spheres as a function of depth') Answer: Contact stresses in spheres as a function of depth r a e h s d n a s s e r t s e v i s s e r p m o C

1 sigma x sigma z shear

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 0

0.5

1

1.5 depth (z/a)

2

2.5

3

6.14 zb = linspace(0.01, 3, 100); nu = 0.3; sx = 2*nu*(sqrt(1+zb.^2)-zb); sy = (2-1./(1+zb.^2)).*sqrt(1+zb.^2)-2*zb; sz = 1./sqrt((1+zb.^2)); plot(zb, sx , 'k-', zb, sz, 'k--', zb, -0.5*(sy-sz), 'k:', zb, sy, 'k-.') legend('\sigma_x' ,'\sigma_z', '\tau', '\sigma_y') axis([0, 3, 0, 1]) xlabel('depth (z/b)') ylabel('Compressive stress and shear') title('Contact stresses in cylinders as a function of depth') Answer: Contact stresses in cylinders as a function of depth r a e h s d n a s s e r t s e v i s s e r p m o C

1

0.9

0.8

x z

y

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 0

0.5

1

1.5 depth (z/b)

2

2.5

3

6.15 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a13 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

bt = [0.02, 0.05, 0.08, 0.11, 0.15, 0.18, 0.23, 0.30]; wormleadmin = inline('b./sin(lambda)+1./cos(lambda)', 'lambda', 'b'); z = zeros(1,length(bt)); for j = 1:length(bt) z(j) = fminbnd(wormleadmin, pi/180, 40*pi/180, [], bt(j)); end plot(z*180/pi, bt./sin(z)+1./cos(z), 'k--') legend('Minimum', 'Location', 'SouthWest'); hold on lamb = linspace(1, 40, 100)*pi/180; [b, lambda] = meshgrid(bt, lamb, 1); plot(lambda*180/pi, b./sin(lambda)+1./cos(lambda), 'k') axis([0, 40, 1, 2]) xlabel('\lambda') ylabel('K') lambda = 23*pi/180; pht = bt/sin(lambda)+1/cos(lambda); text(repmat(23, 1, length(bt)), pht(1:length(bt))+0.04, num2str(bt(1:end)')) text(20, pht(1)+0.04,'\beta = ') Answer: 2

1.9

0.3

1.8 0.23

1.7

1.6

0.18

1.5

0.15

1.4

0.11

K

0.08 1.3 0.05 1.2

=

0.02

1.1 Minimum 1 0

5

10

15

20

25

30

35

40

6.16 function Exercise6_16 dk = [0, 20, 50, 100, 200, 500, 1000, 2000, 5000, 10000, 20000, 50000, 100000]; options = optimset('display', 'off'); re = logspace(3.603, 7); lambda = zeros(length(re),1); label = zeros(length(dk),1); for k = 1:length(dk) for r = 1:length(re) lambda(r) = fzero(@PipeFrictionCoeff,[.008 .08], options, re(r), dk(k)); end semilogx(re, log10(lambda),'k') label(k) = lambda(end); hold on end xlabel('Reynolds number') ylabel('log_{10}(\lambda)') text(repmat(10^(7+.03), length(dk)-1,1), log10(label(2:end)'), num2str(dk(2:end)')) text(10^(7+.06), log10(label(1)), '\infty') © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a14 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

v = axis; text(10^(7+.1), v(4), 'd/k') function x = PipeFrictionCoeff(el, re, dk) if dk>100000||dk==0 x = el-(2*log10(re*sqrt(el)/2.51))^-2; else x = el-(2*log10(2.51/re/sqrt(el)+0.27/dk))^-2; end Answer: -1.1

d/k 20

-1.2

-1.3

50

-1.4 100 -1.5 )

( 0 g1 o l

200

-1.6 500 -1.7

1000 2000

-1.8 5000 -1.9

10000 20000

-2

-2.1 3 10

50000 100000

10

4

5

10 Reynolds number

10

6

10

7

6.17 th = 0:pi/90:2*pi; fill(3.+cos(th), 1.+sin(th), 'g') axis equal axis([-2, 5, -5, 4]) hold on fill([0 4.5 5 5], [-5 4 4 -5], 'r') plot([-2 5], [0 0], 'k', [0 0], [-5 4], 'k', 1, 1, 'k+') r = 0.5:.5:3; plot(1.+cos(th')*r, 1.+sin(th')*r, 'b') title('Blue circles: function minimized. Green area: feasible region') xlabel('x_1') ylabel('x_2') Answer:


Blue circles: function minimized. Green area: feasible region 4

3

2

1

0

x

2

-1

-2

-3

-4

-5 -2

-1

0

1

2 x

3

4

5

1

6.18 t = linspace(0, pi, 200); s1 = sin(t); s2 = 0.5*abs(sin(2*t)); plot(t, s1, 'k', t, s2, 'k') fill([t fliplr(t)], [s1 fliplr(s2)], 'c') axis([0, pi, 0, 1]) hold on plot([0, pi], [0.5 0.5], '--k') Answer: 1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 0

0.5

1

1.5

2

2.5

3

6.19 phi = linspace(0, 2*pi, 200); N = 2; kd = [0.25, 0.5, 1, 2]*pi; for k = 1:4 subplot(2,2,k) polar(phi, sin(2*N*kd(k)*cos(phi))./(2*N*sin(kd(k)*cos(phi))), 'k-') title(['kd/\pi = ' num2str(kd(k)/pi)]) end Answer:


kd/ = 0.25 90 1 120

kd/ = 0.5 90 1 60

0.5

150

120 30

180

0

210

330 240

0

210

300

330 240

kd/ = 1 90 1

kd/ = 2 90 1 60

0.5

120 30

180

210

330 300

60 0.5

150

0

270

300 270

120

240

30

180

270

150

60 0.5

150

30

180

0

210

330 240

300 270

6.20 th = linspace(0, pi, 100); r = 2/3; rad = [2/6, 1/6, 1/2]; c = cos(th); s = sin(th); a = [rad(1), 2*rad(1)+rad(2), rad(3)]; ch = char('w', 'w', 'g'); for k = 3:-1:1 % Displaying colors is order dependent fill(a(k)+rad(k)*c, rad(k)*s, ch(k,1)) hold on end plot([0 2*rad(3)], [0 0], 'k-') axis equal off N = 6; th = linspace(0, 2*pi, 100); c = cos(th); s = sin(th); for n = 1:N D = n^2*(1-r)^2+r; xn = r*(r+1)/2/D; yn = n*r*(1-r)/D; rn = r*(1-r)/2/D; fill(xn+rn*c, yn+rn*s, 'y') plot(xn, yn, 'k+') end % Ellipse for centers of circles. x = linspace(0, (1+r)/2, 50); v = ((4*x-(r+1))/(1+r)).^2; y = sqrt(r)*sqrt(1-v)/2; plot(x, y, 'k--') plot(rad(1), 0, 'k+', 2*rad(1)+rad(2), 0, 'k+') text(0, -0.04, '(0,0)', 'HorizontalAlignment','center') text(rad(1), -0.04, ['(' num2str(rad(1),3),',0)'], 'HorizontalAlignment','center') text(2*rad(1)+rad(2), -0.04, ['(' num2str(2*rad(1)+rad(2),3),',0)'],'HorizontalAlignment','center') Answer:


(0,0)

(0.333,0)

(0.833,0)

6.21 function Exercise6_21 xi = logspace(-1, 2, 100); Lam = 10.^(-(1:5)); for k = 1:length(Lam) y = LossFactor(xi, Lam(k)); loglog(xi, y, '-k') hold on text(0.8*interp1(y, xi, 0.1), 0.1, ['\lambda = ' num2str(Lam(k))],'Rotation', 60) end v = axis; v(3)=10^(-3); axis(v) xlabel('\xi') ylabel('\eta') function eta = LossFactor(xi, Lam) eta = Lam*xi./(1+Lam*xi); eta = eta.*(3+6*xi+4*xi.^2+2*Lam*xi.^3+Lam^2*xi.^4); eta = eta./(1+2*Lam*(2*xi+3*xi.^2+2*xi.^3)+Lam^2*xi.^4); Answer: 10

0

10

-1

=

1 0 0 .0 0

1 0 .0 0

1 .0 0

.1 0 =

=

=

5 0 0 e 1

=

10

10

-2

-3

10

-1

10

0

10

1

10

2

6.22 AB = 5; ab = linspace(0.3, AB, 50); N = inline('(m./ab+ab/m).^2', 'ab', 'm'); cusp = sqrt((1:5).*(((1:5)+1))); hold on © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a18 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

for m = 1:AB y = N(ab,m); plot(ab, y, 'k--') a = interp1(y(1:20), ab(1:20), 9); if m==AB text(a+0.1, 9, [int2str(m) ' = m']) else text(a+0.1, 9, int2str(m)) end end for m=1:AB if m==1 ab = linspace(0.3, cusp(m), 30); else ab = linspace(cusp(m-1), cusp(m), 30); end plot(ab, N(ab,m), 'k-', 'linewidth', 3) end axis([0, AB, 3, 10]) xlabel('a/b') ylabel('N_{cr}') Answer: 10

9

1

2

3

4

5= m

8

7 r c

N

6

5

4

3 0

0.5

1

1.5

2

2.5 a/b

3

3.5

4

4.5

5

6.23 wt = logspace(-2, 6, 100); for k = 1:5 a = 10^-k; y = 100*(1-1./sqrt((1+a)^2+(a*wt).^2)); loglog(wt, y, 'k-') hold on text(0.09, 1.4*y(1), ['R_g/R_i = ' num2str(a)]) end xlabel('\omega\tau_i (=\omegaR_iC_i)') ylabel('Percentage error') Answer:


10

10

2

1

R /R = 0.1

0

Rg /R i = 0.01

g

i

r o r r e e g a t n e c r e P

10

10

10

10

R /R = 0.001 g

-1

R /R = 0.0001 g

-2

i

R /R = 1e-005 g

-3

10

i

-2

10

-1

i

10

0

10

1

10

2

10

3

10

4

10

5

10

6

(= R C ) i i i

6.24 function Exercise6_24 v = 0.3; k = 10^-5; N = 100; L = logspace(-2, 1, N); Om = zeros(N, 3); for n = 1:6 for m = 1:N Om(m,:) = NatFreqShell(L(m), k, n, v)'; end loglog(1./L, Om(:,1), 'k-') hold on text(40, 1.2*Om(1,1), ['n = ' num2str(n)]) end xlabel('1/\lambda') ylabel('\Omega_1') function Om = NatFreqShell(L, k, n, v) g = n.^2+L.^2; K2 = 1+0.5*(3-v)*g+k*g.^2; K1 = 0.5*(1-v)*((3+2*v)*L.^2+n.^2+g.^2+(3-v)/(1-v)*k*g.^3); K0 = 0.5*(1-v)*((1-v^2)*L.^4+k*g.^4); Om = sort(sqrt(roots([1, -K2, K1, -K0]))); Answer: 10

0

n= 6 10

-1

n= 5 n= 4 n= 3

1

n= 2 10

-2

n= 1

10

-3

10

-1

10

0

10

1

10

2

1/


6.25 function Exercise6_25 beta = 60*pi/180; rb = 3; h = 0.5; rc = 0.5; n = 23; phi = linspace(0, beta, n); ph = [phi, beta+phi]; [Rx, Ry, Cx, Cy] = ContourFlat(phi, rb, h, beta, rc); ang = linspace(2*beta, 2*pi, 40); plot(Rx, Ry, 'k', rb*cos(ang), rb*sin(ang), 'k') hold on plot(Cx(1:5:2*n), Cy(1:5:2*n), 'k+') axis equal phd = linspace(0, 2*pi, 50); [x, phx] = meshgrid(Cx(1:5:2*n), phd); y = meshgrid(Cy(1:5:2*n), phd); hold on plot(x+rc.*cos(phx), y+rc.*sin(phx), 'k--') plot([0 Rx(4)], [0, Ry(4)], 'k-') text(Rx(4), Ry(4), '>', 'HorizontalAlignment','Right') text(-1, 0.8, 'C(\phi)', 'fontsize', 14) plot([0 Cx(end)], [0, Cy(end)], 'k-') text(1, -0.2, 'R(\phi)', 'fontsize', 14) axis off set(gca, 'fontsize', 14) function [L, dLdphi] = CamProfile(phi, rb, h, beta) arg = 2*pi*phi/beta; L = rb+h*(phi/beta-sin(arg)/2/pi); dLdphi = (h/beta)*(1-cos(arg)); L = [L fliplr(L)]; dLdphi = [dLdphi -dLdphi]; function [Rx, Ry, Cx, Cy] = ContourFlat(phi, rb, h, beta, rc) [L, dLdphi] = CamProfile(phi, rb, h, beta); theta = atan2(dLdphi, L); R = L./cos(theta); ph = [phi, beta+phi]; Ry = R.*sin(theta+ph); Rx = R.*cos(theta+ph); gama = atan(dLdphi./(L+rc)); C = (L+rc)./cos(gama); Cy = C.*sin(gama+ph); Cx = C.*cos(gama+ph); Answer:


C( )

>

R( )


Solution to Exercises Chapter 7 7.1 % Spherical helix t = linspace(0, 10*pi, 300); x = t/(2*5); plot3(sin(x).*cos(t), sin(x).*sin(t), cos(x), 'k-') axis equal Answer: 1

0.5

0

-0.5

-1

0.5 0.5

0 0 -0.5

-0.5

% Toroidal spiral a = 0.2; b = 0.8; c = 20; t = linspace(0, 2*pi, 400); plot3((b+a*sin(c*t)).*cos(t), (b+a*sin(c*t)).*sin(t), a*cos(c*t), 'k-') axis equal Answer: 0.2 0.1 0 -0.1

0.8 0.6 0.4

0.8 0.2

0.6 0.4

0

0.2

-0.2

0

-0.4

-0.2 -0.4

-0.6 -0.8

-0.6 -0.8

% Sine wave on sphere a = 10; b = 1; c = 0.3; t = linspace(0, 2*pi, 200); z = sqrt(b^2-c^2*cos(a*t).^2); plot3(z.*cos(t), z.*sin(t), c*cos(a*t), 'k-') axis equal Answer:


0.2 0.1 0 -0.1 -0.2

0.8 0.6 0.4

0.8 0.2

0.6 0.4

0

0.2

-0.2

0 -0.4

-0.2 -0.4

-0.6

-0.6

-0.8

-0.8

% Concho-spiral a = 1; b = 1.05; c = 3; u = linspace(0, 12*pi, 200); bu = b.û; plot3(a*bu.*cos(u), a*bu.*sin(u), c*bu, 'k-') axis equal Answer: 18 16 14 12 10 8 6 4

4 2

5 0 -2

0 -4 -5

% Intersection of two cylinders a = 1; b = 1.3; phi = linspace(0, 2*pi, 200); plot3(a*cos(phi), a*sin(phi), sqrt(b^2-a^2*sin(phi).^2), 'k-') axis equal view([-15, 20]) Answer: 1.2 1.1 1 0.9

0.5 0 -0.5

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

% Baseball seam a = 0.4; t = linspace(0, 4*pi, 200); g = pi/2-(pi/2-a)*cos(t); h = t/2+a*sin(2*t); plot3(sin(g).*cos(h), sin(g).*sin(h), cos(g), 'k-') © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a2retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

axis equal Answer: 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8

0.5 0.5 0 0 -0.5

-0.5

% Spherical spiral a = 0.08; phi = linspace(-12*pi, 12*pi, 400); p = atan(a*phi); plot3(cos(phi).*cos(p), sin(phi).*cos(p), -sin(p), 'k-') axis equal Answer: 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8

0.5 0.5

0 0 -0.5

-0.5

7.2 % Seashell vv = linspace(0, 2*pi, 25); uu = linspace(0, 6*pi, 45); [u, v] = meshgrid(uu, vv); E = exp(u/6/pi); C2 = cos(0.5*v).^2; x = 2*(1-E).*cos(u).*C2; y = 2*(-1+E).*sin(u).*C2; z = 1-exp(u/3/pi)-sin(v)+E.*sin(v); surf(x, y, z) axis vis3d equal Answer:


0 -1 -2 -3 -4 -5 -6 -7 -8 2 2

1 1

0

0

-1

-1 -2

-2 -3

% Figure eight torus c = 1; vv = linspace(-pi, pi, 25); uu = linspace(-pi, pi, 45); [u, v] = meshgrid(uu, vv); C = c+sin(v).*cos(u)-sin(2*v).*sin(u)/2; x = cos(u).*C; y = sin(u).*C; z = sin(u).*sin(v)+cos(u).*sin(2*v)/2; surf(x, y, z) axis vis3d equal Answer:

1 0.5

0 -0.5

-1 1.5

2 1 1

0.5 0

0 -0.5 -1

-1 -1.5

-2

% Helical spring r1 = 0.25; r2 = 0.25; T = 2.0; n = 6; [u, v] = meshgrid(linspace(0, 2*n*pi, 140), linspace(0, 2*pi, 25)); x = [1-r1.*cos(v)].*cos(u); y = [1-r1*cos(v)].*sin(u); z = r2*[sin(v)+T*u/pi]; surf(x, y, z) view([-64 0]) surf(x, y, z) axis vis3d equal off Answer:


% Cornucopia a = 0.3; b = 0.5; [u, v] = meshgrid(linspace(0, 2*pi, 22), linspace(-3, 3, 20)); x = exp(b*v).*cos(v)+exp(a*v).*cos(v).*cos(u); y = exp(b*v).*sin(v)+exp(a*v).*sin(v).*cos(u); z = exp(a*v).*sin(u); surf(x, y, z) Answer:

% Astroidal ellipsoid [u, v] = meshgrid(linspace(-pi/2, pi/2, 40), linspace(-pi, pi, 40)); a = 1; b = 1; c = 1; x = (a*cos(u).*cos(v)).^3; y = (b*sin(u).*cos(v)).^3; z = (c*sin(v)).^3; surf(x, y, z) axis vis3d equal off Answer:


% Mobius strip [s, t] = meshgrid(linspace(0, 2*pi, 30), linspace(-0.4, 0.4, 15)); x = cos(s)+t.*cos(s/2).*cos(s); y = sin(s)+t.*cos(s/2).*sin(s); z = t.*sin(s/2); surf(x, y, z) axis vis3d equal off Answer:

% Bow curve T = 0.7; [u, v] = meshgrid(linspace(0, 2*pi, 20), linspace(0, 2*pi, 40)); x = (2+T*sin(u)).*sin(2*v); y = (2+T*sin(u)).*cos(2*v); z = T*cos(u)+3*cos(v); surf(x, y, z) axis off equal vis3d shading interp

Answer:


% Hyperbolic helicoid tau = 7; [u, v] = meshgrid(linspace(-pi, pi, 100), linspace(0, 0.5, 8)); D = 1+cosh(u).*cosh(v); x = sinh(v).*cos(tau*u)./D; y = sinh(v).*sin(tau*u)./D; z = sinh(u).*cosh(v)./D; surf(x, y, z) axis vis3d equal off Answer:

% Apple surface [u, v]= meshgrid(linspace(0, 2*pi, 30), linspace(-pi, pi, 50)); x = cos(u).*(4 + 3.8*cos(v)); y = sin(u).*(4 + 3.8*cos(v)); z = (cos(v) + sin(v) - 1).*(1 + sin(v)).* log(1 - pi*v / 10) + 7.5*sin(v); h = surf(x, y, z); set(h, 'FaceAlpha', 0.4) axis off equal vis3d shading interp Answer:


7.3 function Exercise7_3 the = linspace(0, 2*pi, 30); rad = linspace(0, 1, 15)'; xc = rad*cos(the); yc = rad*sin(the); options = optimset('display','off'); nm = 2; nk = 3; n = 0; for m = 0:1:nm vs = m+2.8; for k = 1:nk v = fzero(@circularplate, [vs vs+3], options, m); mode = circularplatemode(m, v, rad, the); vs = 1.2*v; n = n+1; subplot(nm+1, nk, n) meshc(yc, xc, mode) colormap([0 0 1]) title([num2str(v) ' m=' num2str(m) ' n=' num2str(k)]) axis off end end function d = circularplate(x, n) d = besselj(n, x)*besseli(n+1, x)+besseli(n, x)*besselj(n+1, x); function [mod]= circularplatemode(m, omn, rad, the) c1mn = -besseli(m, omn)/besselj(m, omn); mod = (c1mn*besselj(m, omn*rad)+ besseli(m, omn*rad))*cos(the*m); mod = mod/max(max(abs(mod))); Answer:


3.1962 m=0 n=1

6.3064 m=0 n=2

9.4395 m=0 n=3

4.6109 m=1 n=1

7.7993 m=1 n=2

10.9581 m=1 n=3

5.9057 m=2 n=1

9.1969 m=2 n=2

12.4022 m=2 n=3

7.4 function Exercise7_4 heatslab = inline('cos(x)-x.*sin(x)/b', 'x', 'b'); tau = [linspace(0, 0.5, 9) linspace(0.6, 2, 9)]; eta = linspace(0, 1, 11); bi = 0.7; Nroot = 20; %x = linspace(0, 20*pi, 200); r = FindZeros(heatslab, Nroot, linspace(0, 20*pi, 200), bi); s = meshgrid(sin(r)./(r+sin(r).*cos(r)), eta); th = (2*cos(r*eta).*s')'*exp(-r.^2*tau); surf(tau, eta, th) xlabel('\tau') ylabel('\eta') zlabel('\theta/\theta_i') title(['Boit number = ',num2str(bi)]) axis vis3d view(-32.5, 36) function Rt = FindZeros(FunName, Nroot, x, w) f = feval(FunName, x, w); indx = find(f(1:end-1).*f(2:end)<0); L = length(indx); if L

Boit number = 0.7

1 0.9 0.8 0.7

#/ #

i

0.6 0.5 0.4 0.3

1 0.8 2

0.6 1.5

0.4 1 0.2

0.5 0

0

!

7.5 [x, y] = meshgrid(linspace(0, 1, 25), linspace(0, 1, 25)); surf(x, y, sin(2*pi*x).*sin(3*pi*y)) Answer:

1

0.5

0

-0.5

-1 1 0.8

1 0.6

0.8 0.6

0.4 0.4 0.2

0.2 0

0

7.6 [re, pr] = meshgrid(logspace(5.699, 7, 10), logspace(-0.2218, 3.301, 15)); nu = 0.037*re.^.8.*pr./(1+2.443*re.^-.1.*(pr.^(2/3)-1)); surf(log10(re), log10(pr), log10(nu)) xlabel('log_{10}(Re)') ylabel('log_{10}(Pr)') zlabel('log_{10}(Nusselt)') hold on xc = [5.699, 5.699, 7 7]; yc = [-0.2218, 3.301, -0.2218, 3.301]; zc = log10([nu(1,1), nu(end,1), nu(1,end), nu(end,end)]); plot3([xc; xc], [yc; yc], [repmat(2, 1, 4); zc], 'k') Answer:


6 5.5 ) t l e s s u N (0 g1 o l

5 4.5 4 3.5 3 2.5 2 4 3

7 2 6.5 1 6

0 log

10

-1

(Pr)

5.5 log 10 (Re)

7.7 n = linspace(6, 12, 10); rho = [0.001:0.002:0.009 linspace(0.01, 0.1, 10)]; [nn rr] = meshgrid(n, rho); rn = nn.*rr; surf(nn, rr, -rn+sqrt(rn.^2+2*rn)) xlabel('n') ylabel('\rho') zlabel('k') title('Location of neutral axis in a steel reinforced concrete beam') Answer: Location of neutral axis in a steel reinforced concrete beam

0.8 0.7 0.6 0.5 k

0.4 0.3 0.2 0.1 0.1 0.08

12 11

0.06 10 0.04

9 8

0.02

7 0

6 n

7.8 % function Exercise7_10 x = stem3data; stem3(x(:,1), x(:,2), x(:,3), 'ks', 'fill'); view(-30, 7); hold on x1 = 0:5:20; x2 = 0:200:600; mesh(x1, x2, zeros(length(x2), length(x1))); view([-30, 7]) colormap([0 0 1]) title('Deviations from mean output response') xlabel('x_1') ylabel('x_2') © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a11 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

zlabel('Deviations') box off function dat1 = stem3data dev = [1.5713, -1.1460, -2.2041, -1.5968, -2.8937, 1.1136, 1.9297, 1.1962, -3.8650, -0.4763,- 1.3223,... -0.4619, 0.4911, -0.6023, 5.8409, -0.3620, 4.3341, -2.0368, -1.5415, 0.0302, -2.1809, 1.5587,... 0.3222, 2.1478, 0.1537]; x1 = [2, 8, 11, 10, 8, 4, 2, 2, 9, 8, 4, 11, 12, 2, 4, 4, 20, 1, 10, 15, 15, 16, 17, 6, 5]; x2 = [50, 110, 120, 550, 295, 200, 375, 52, 100, 300, 412, 400, 500, 360, 205, 400, 600, 585, 540, 250,... 290, 510, 590, 100, 400]; dat1 = [x1', x2', dev']; Answer: Deviations from mean output response 6 5 4 3 s n o i t a i v e D

2 1 0 -1 -2 -3 -4 600 400 200 0 x2

0

10

5

15

20

x

1

7.9 eta = linspace(1, 3, 100); th = linspace(-pi, pi, 80); x = eta'*cos(th); y = eta'*sin(th); [t, et] = meshgrid(th, eta); sigr = 1-1./et.^2+(1+3./et.^4-4./et.^2).*cos(2*t); sigt = 1+1./et.^2-(1+3./et.^4).*cos(2*t); shear = -(1-3./et.^4+2./et.^2).*sin(2*t); z = sqrt(((sigr-sigt).^2+sigr.^2+sigt.^2)/2+3*shear.^2); [c, h] = contour(x, y, z, 15); set(h, 'LineColor','k') axis equal off Answer:


Solution to Exercises Chapter 8 8.1 %(a) disp(['probability of 8 calls = ' num2str(poisspdf(8, 5))]) disp(['probability of 2 calls = ' num2str(poisspdf(2, 5))]) %(b) disp(['probability of a busy signal = ' num2str(1-poisscdf(10, 5))]) Answers: probability of 8 calls = 0.065278 probability of 2 calls = 0.084224 probability of a busy signal = 0.013695

8.2 disp(['Probability that at least 12 withstand load = ' num2str(1-binocdf(11, 15, 0.7))]) Answer: Probability that at least 12 withstand load = 0.29687

8.3 dat = [88.4, 93.2, 87.4, 94.3, 93.0, 94.3, 89.0, 90.5, 90.8, 93.1, 92.8, 91.9;... 92.6, 93.2, 89.2, 94.8, 93.3, 94.0, 93.2, 91.7, 91.5, 92.0, 90.7, 93.8]; for k = 1:2 el = var(dat(k,:))+(mean(dat(k,:))-92)^2; disp(['L' int2str(k) ' = ' num2str(el)]) end Answers: L1 = 5.5904 L2 = 2.6936

8.4 p = 0.2; n = 24; m = 0:n; disp(['Expected value = ' num2str(n*p) ' variance = ' num2str(n*p*(1-p))]) disp(['Probability that no more than 2 are underweight = ' num2str(binocdf(2, n, p))]) disp(['Probability that none is underweight = ' num2str(binopdf(0, n, p)) ]) plot([m; m], [zeros(1, n+1); binopdf(m, n, p)], 'k', m, binocdf(m, n, p), 'k*') Answers: Expected value = 4.8 variance = 3.84 Probability that no more than 2 are underweight = 0.11452 Probability that none is underweight = 0.0047224


1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 0

5

10

15

20

25

8.5 figure(1) t = linspace(0, 1.5); r = 1-expcdf(t); plot(t, exppdf(t)./r, 'k-', t, r, 'k--') title('Hazard rate and reliability for exponential distribution') legend('Hazard rate', 'Reliability') figure(2) b = [0.5, 1, 2, 4]; for k = 1:4 subplot(2, 2, k) r = 1-weibcdf(t, 1, b(k)); plot(t, weibpdf(t, 1, b(k))./r, 'k-', t, r, 'k--') title(['Weibull: beta = ' num2str(b(k))]) legend('Hazard rate', 'Reliability') end Answers: Weibull: beta = 1

Weibull: beta = 0.5 5

1.2 Hazard rate

Hazard rate Reliability

4

Hazard rate and reliability for exponential distribution

Reliability

1

1.1 Hazard rate Reliability

3

0.8

2

0.6

1

0.4

1

0.9

0.8

0

0

0.5

1

1.5

0.2

0

0.5

1

1.5

0.7

Weibull: beta = 2

Weibull: beta = 4

3

14

0.6

Hazard rate 2.5

0.5

Hazard rate

12

Reliability

Reliability

10

2

8 1.5

0.4

6 1

4

0.3

0.5

2

0

0.2 0

0.5

1

1.5

0 0

0.5

1

1.5

0

0.5

1

1.5

8.6 dat = [1.55, 3.05, 3.65, 5.20, 7.75, 10.45, 10.85, 10.90, 12.65, 15.25 ... 15.70, 16.35, 17.70, 17.95, 19.45, 19.80, 20.05, 32.75, 35.45, 49.35]; k = 1:length(dat); ft = norminv((k-0.5)/length(dat)); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a2retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

figure(1) plot(log(dat), ft, 'k-*') beta = polyfit(log(dat), ft, 1); disp(['curve fit: mean = ' num2str(-beta(2)/beta(1)) ' std dev = ' num2str(1/beta(1))]) disp(['Eq.(a): mean = ' num2str(mean(log(dat))) ' std dev = ' num2str(std(log(dat)))]) hold on plot([log(dat(1)) log(dat(length(dat)))], [beta(2)+beta(1)*log(dat(1)) beta(2)+ ... beta(1)*log(dat(length(dat)))], 'k') title('Fit assuming data are lognormal') ylabel('F(t)') xlabel('ln(t)') text(1.25, 1.5, ['F(t) = ' num2str(beta(2)) ' + ' num2str(beta(1)) 'ln(t)']) figure(2) normplot(log(dat)-(beta(2)+beta(1)*log(dat))) Answers: curve fit: mean = 2.5072 std dev = 0.88841 Eq.(a): mean = 2.5072 std dev = 0.85441 Normal Probability Plot

Fit assuming data are lognormal 2

0.98

1.5

0.95

F(t) = -2.8221 + 1.1256ln(t)

0.90 1

y t i l i b a b o r P

0.5

) t ( F

0

-0.5

0.75

0.50

0.25 -1

-1.5

0.10 0.05

-2

0.02 -2.5 0

0.5

1

1.5

2 ln(t)

2.5

3

3.5

4

2.35

2.4

2.45

2.5

2.55 Data

2.6

2.65

2.7

2.75

8.7 dat = [2.5629, 2.5630, 2.5628, 2.5634, 2.5619, 2.5613, 2.5630, 2.5628, 2.5623, 2.5631, 2.5635, 2.5623]; [p, cp, cpk] = capable(dat, [2.560 2.565]) Answers: p = 1.5351e-004 cp = 1.3103 cpk = 1.2099

8.8 dat = [ 1115, 1310, 1540, 1502, 1258, 1315, 1085, 1567, 1883, 1203, 1270, 1015, 845, 1674; ... 1223, 375, 2265, 1910, 1018, 1452, 1890, 1782, 1522, 1792, 1000, 1820, 1940, 1120; ... 1055, 1764, 1330, 1608, 1535, 1781, 1750, 798, 1020, 865, 2130, 1421, 1109, 1481; ... 1016, 1102, 1605, 706, 2215, 785, 885, 2100, 1594, 2023, 1315, 1269, 1260, 1888; ... 910, 1730, 1102, 1578, 758, 1416, 1560, 1501, 1238, 990, 1468, 1512, 1750, 1642]; s0 = [dat(1,:), dat(2,:), dat(3,:), dat(4,:), dat(5,:)]; disp(['data set #0: geometric mean = ' num2str(geomean(s0))]) disp(['data set #0: harmonic mean = ' num2str(harmmean(s0))]) xbar0 = mean(s0); var0 = var(s0); disp(['dataset #0: mean = ' num2str(xbar0) ' std. dev. = ' num2str(sqrt(var0))]) n1 = length(s0); n2 = n1/5; nn = n1+n2-2; © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a3retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

co = sqrt(1/n1+1/n2); c1 = (n1-1)*var0; for k = 1:5 me = mean(dat(k,:)); st = std(dat(k,:)); disp(['dataset #' int2str(k) ': mean = ' num2str(me) ' std. dev. = ' num2str(st)]) [h(k), p(k), ci(k,:)] = ttest2(s0, dat(k,:), 0.025); dmean(k) = xbar0-me; end for k = 1:5 disp(['set #' num2str(k) ' lcl = ' num2str(ci(k,1)) ' diff in mean = ' num2str(dmean(k)) ... ' ucl = ' num2str(ci(k,2)) ' p_value = ' num2str(p(k)) ' hyp: ' num2str(h(k))]) end boxplot(dat') Answers: data set #0: geometric mean = 1339.9309 data set #0: harmonic mean = 1264.6733 dataset #0: mean = 1403.6571 std. dev. = 402.3852 dataset #1: mean = 1327.2857 std. dev. = 280.1281 dataset #2: mean = 1507.7857 std. dev. = 506.6656 dataset #3: mean = 1403.3571 std. dev. = 392.5288 dataset #4: mean = 1411.6429 std. dev. = 501.0445 dataset #5: mean = 1368.2143 std. dev. = 315.758 set #1 lcl = -181.3939 diff in mean = 76.3714 ucl = 334.1367 p_value = 0.50062 hyp: 0 set #2 lcl = -385.323 diff in mean = -104.1286 ucl = 177.0658 p_value = 0.40028 hyp: 0 set #3 lcl = -267.654 diff in mean = 0.3 ucl = 268.254 p_value = 0.99797 hyp: 0 set #4 lcl = -288.4656 diff in mean = -7.9857 ucl = 272.4942 p_value = 0.94832 hyp: 0 set #5 lcl = -225.2238 diff in mean = 35.4429 ucl = 296.1095 p_value = 0.757 hyp: 0 2200

2000

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1400

1200

1000

800

600

400 1

2

3

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5

8.9 dat = [88, 79, 84, 89, 81, 83, 82, 79, 82, 85, 81, 83, 90, 87, 78, 80, 87, 85, 80, 88;... 76, 83, 78, 80, 84, 86, 77, 75, 81, 78, 79, 85, 76, 80, 82, 78, 78, 77, 81, 80]; boxplot(dat') std2 = var(dat'); pvalue = 2*(1-fcdf(std2(1)/std2(2), length(dat(1,:))-1, length(dat(1,:))-1)); disp(['p-value from F-test on ratio of variances = ' num2str(pvalue)]) % Since pvalue > 0.1 use ttest2 [h, p, ci] = ttest2(dat(1,:), dat(2,:), 0.025); disp(['p-value of t-test for difference in means = ' num2str(p)]) boxplot(dat')


Answers: p-value from F-test on ratio of variances = 0.47092 p-value of t-test for difference in means = 0.0009342 90

85

80

75 1

2

8.10 function Exercise8_10 % (a) xy1 = [18, 24, 12, 30, 30, 26, 26, 22, 22, 14, 10, 10, 6, 6]'; xy2 = [52, 40, 40, 48, 32, 56, 24, 64, 16, 64, 56, 24, 48, 32]'; y = [144, 142, 124, 64 96, 74, 136, 54, 92, 96, 92, 82, 76, 68]'; x = [ones(length(y),1) xy1 xy2 xy1.^2 xy2.^2 xy1.*xy2]; [b, bi, res] = regress(y, x); disp([repmat('b(', 6, 1) int2str((1:6)') repmat(') = ', 6, 1) num2str(b)]) figure(1) normplot(res) %(b) xx1 = linspace(6, 30, 20); xx2 = linspace(16, 64, 20); [x1, x2] = meshgrid(xx1, xx2); z = fittedsurface([x1; x2], b); figure(2) surf(x1, x2, z) xlabel('x_1') ylabel('x_2') figure(3) [c h] = contour(x1, x2, z, [20:20:140 150 155]); clabel(c, h) xlabel('x_1') ylabel('x_2') %(c) opt = optimset('Display', 'off'); fitsurf = inline('-(b(1)+b(2)*x(1)+b(3)*x(2)+b(4)*x(1).^2+b(5)*x(2).^2+b(6)*x(1).*x(2))', 'x','b'); coord = fminunc(fitsurf, [10, 20]', opt, b); disp(['x1 = ' num2str(coord(1)) ' x2 = ' num2str(coord(2))]) function z = fittedsurface(x, b) [nr, nc] = size(x); x1 = x(1:nr/2,: ); x2 = x(nr/2+1:nr,: ); z = b(1)+b(2)*x1+b(3)*x2+b(4)*x1.^2+b(5)*x2.^2+b(6)*x1.*x2; Answers: b(1) = -288.2799 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a5retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

b(2) = 24.16421 b(3) = 11.51143 b(4) = -0.5072724 b(5) = -0.118116 b(6) = -0.1348869 x1 = 18.7635 x2 = 38.0155 Normal Probability Plot 0.98

0.95

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8.11 function Exercise8_11 [x, y] = regressdata1; [beta, betacl, e, ecl, stat] = regress(y, x); disp(['beta0 = ' num2str(beta(1)) ' beta1 = ' num2str(beta(2))]) figure(1) plot(1./x(:,2), beta(1)+beta(2)*x(:,2), 'k-', 1./x(:,2), y, 'k*') xlabel('x') ylabel('y') figure(2) normplot(e) r = sqrt(stat(1)); t0 = r*sqrt(length(y)-2)/sqrt(1-stat(1)); disp(['t0 = ' num2str(t0) ' p_value = ' num2str(1-tcdf(t0,length(y)-2)) ' dof = ' ... num2str(length(y)-2)]) disp('r and its confidence limits:') za2 = norminv(.975)/sqrt(length(y)-3); disp([num2str(tanh(atanh(r)-za2)) ' <= ' num2str(r ) ' <= ' num2str(tanh(atanh(r) +za2))])


function [x, y] = regressdata1 yy = [4.746, 5.466, 3.171, 1.500, 6.708, 7.158, 6.882, 1.674, 6.498, 5.598, 1.959, 5.790, 4.686... 5.211, 6.264, 3.411, 6.537, 6.336, 5.400, 4.503, 6.909, 6.930, 3.582, 3.432, 0.369]'; xx = [10.0, 12.0, 6.8, 5.4, 20.0, 19.4, 19.1, 6.1, 16.3, 12.4, 5.8, 12.7, 9.2, 11.6, 8.8, 7.2, 15.7 ... 17.6, 8.0, 10.9, 18.2, 20.4, 8.2, 7.9, 4.9]'; [xr, index] = sort(xx); y = yy(index); x = [ones(length(xx), 1), 1./xr]; Answers: beta0 = 8.9366 beta1 = -41.6073 t0 = 33.5922 p_value = 0 dof = 23 r and its confidence limits: 0.977 <= 0.98996 <= 0.99564 Normal Probability Plot

8

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8.12 function Exercise8_12 [y, x] = DataMultiRegress1; [b, bcl, e, ecl, stat] = regress(y, x, 0.05); [n, k] = size(x); sigmabar = sqrt((y'*y-b'*x'*y)/(n-k-1)); d = e/sigmabar; r = e./sqrt(1-diag(x*inv(x'*x)*x'))/sigmabar; plot(y-e, d, 'k*', y-e, r, 'ks') v = axis; hold on plot([v(1) v(2)], [0 0], 'k--') title('Residuals') ylabel('Residuals') xlabel('Average output') legend('Standardized residuals', 'Studentized residuals') function [y, X] = DataMultiRegress1 y = [0.22200, 0.39500, 0.42200, 0.43700, 0.42800, 0.46700, 0.44400, 0.37800, 0.49400, ... 0.45600, 0.45200, 0.11200, 0.43200, 0.10100, 0.23200, 0.30600, 0.09230, 0.11600, ... 0.07640, 0.43900, 0.09440, 0.11700, 0.07260, 0.04120, 0.25100, 0.00002]'; x1 = [7.3, 8.7, 8.8, 8.1, 9.0, 8.7, 9.3, 7.6, 10.0, 8.4, 9.3, 7.7, 9.8, 7.3, 8.5, 9.5, 7.4, 7.8, 7.7, 10.3, ... 7.8, 7.1, 7.7, 7.4, 7.3, 7.6]'; x2 = [0.0, 0.0, 0.7, 4.0, 0.5, 1.5, 2.1, 5.1, 0.0, 3.7, 3.6, 2.8, 4.2, 2.5, 2.0, 2.5, 2.8, 2.8, 3.0, 1.7, ... © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a7retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

3.3, 3.9, 4.3, 6.0, 2.0, 7.8]'; x3 = [0.0, 0.3, 1.0, 0.2, 1.0, 2.8, 1.0, 3.4, 0.3, 4.1, 2.0, 7.1, 2.0, 6.8, 6.6, 5.0, 7.8, 7.7, 8.0, ... 4.2, 8.5, 6.6, 9.5, 10.9, 5.2, 20.7]'; X = [ones(length(y), 1), x1, x2, x3, x1.*x2 x1.*x3, x2.*x3 x1.^2, x2.^2, x3.^2]; Answer: Residuals 3 Standardized residuals Studentized residuals

2.5

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1.5 s l a u d i s e R

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8.13 y = [85, 76, 114, 143, 164, 281, 306, 358, 437, 470, 649, 702]; x = 90:10:200; plot(x, y, 'ko') hold on [cl, sl] = polyfit(x, y, 1); [cq, sq] = polyfit(x, y, 2); plot(x, cl(2)+cl(1)*x, 'k-',x, cq(3)+cq(2)*x+cq(1)*x.^2, 'b--') figure(2) normplot(y-( cl(2)+cl(1)*x)) figure(3) normplot(y- (cq(3)+cq(2)*x+cq(1)*x.^2)) Answers: 800

700

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Normal Probability Plot

Normal Probability Plot

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Solutions to Exercises Chapter 9 9.1 function Exercise9_1 initcond = [2.0, 0.0, 0.0, 0.5]; ti = 0.0; tf = 20.0; spacing = [400; 4000]; options(1)=odeset('RelTol', 1e-3, 'AbsTol', [1e-3, 1e-3, 1e-3, 1e-3]); options(2)=odeset('RelTol', 1e-6, 'AbsTol', [1e-6, 1e-6, 1e-6, 1e-6]); for i = 1:2 tspan = linspace(ti, tf, spacing(i)); [t, x] = ode45(@orbit, tspan, [initcond]', options(i)); angmom = x(:,1).^2.*x(:,4); subplot(2, 2, 2*i-1) polar(x(:,3), x(:,1), 'k-') subplot(2, 2, 2*i) plot(t, angmom, 'k-') axis([0 20 1.6 2.2]); xlabel('Time') ylabel('Angular Momentum/Unit Mass'); end function xdot = orbit(t, x) xdot = [x(2); x(1)*x(4)^2-4.0*pi^2/x(1)^2; x(4); -2.0*x(2)*x(4)/x(1)]; Answers: 90

2

120

60 1

150

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0

210

r a l u s g s n a A M

330 240

300 270

90

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n U / m u t n e m o M

t i n U / m u t n e m o M r a l u g n A

2.2 2.1 2 1.9 1.8 1.7 1.6

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Time

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10 Time


9.2 ro = [2.0, 2.0]; tho = [ 2, 0.2]; for i = 1:2 r = roots([tho(i)^2-8*pi^2/ro(i), 8*pi^2, -ro(i)^2*tho(i)^2]) end Answers: r = 2.0000 0.2255 r= 2.0000 0.0020

9.3 % (a) Om = [linspace(0.0, 0.8, 10), linspace(0.8, 1.2, 25), linspace(1.2, 2, 10)]; zet = linspace (0.05, 0.99, 15); [Omm, zeta] = meshgrid(Om, zet); AmAt = 1./sqrt((1-Omm.^2).^2+(2*zeta.*Omm).^2); mesh(Omm, zeta, AmAt); colormap([0 0 0]) xlabel('\Omega'); ylabel('\zeta'); zlabel('A_m/A_t') % (b) e = [-2, 2]; err = 1-e/100; Omega = (6000/60)/150; for i = 1:2 zetta(i) = sqrt((-(1-Omega^2)^2+1/err(i)^2))/2/Omega; end disp(['At -2%, zeta = ' num2str(zetta(1))]) disp(['At +2%, zeta = ' num2str(zetta(2))]) Answers: At -2%, zeta = 0.60584 At +2%, zeta = 0.64194

10

8

6 At /m A

4

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0 1 0.8

2 0.6

1.5 0.4

1 0.2

0.5 0

0


9.4 % (a) Om = [linspace(0.0, 0.8, 10), linspace(0.8, 1.2, 25), linspace(1.2, 2, 10)]; zet = linspace (0.05, 0.99, 15); [Omm, zeta] = meshgrid(Om, zet); dmdt = Omm.^2./sqrt((1-Omm.^2).^2+(2*zeta.*Omm).^2); mesh(Omm, zeta, dmdt); colormap([0 0 0]) xlabel('\Omega'); ylabel('\zeta'); zlabel('d_m/d_t') % (b) e = [-2, 2]; d = 1-e/100; z = 0.1; w = 1500/60; for k = 1:2 Omega = abs((roots([(1-1/d(k)^2) (-2+4*z^2) 1]))); f(k) = w./sqrt(min(Omega)); end disp(['At -2%, f = ' num2str(f(1)) ' Hz']) disp(['At +2%, f = ' num2str(f(2)) ' Hz']) disp(['Maximum frequency is ' num2str(max(f)) ' Hz']) Answers: At -2%, f = 34.8208 Hz At +2%, f = 35.1854 Hz Maximum frequency is 35.1854 Hz

10

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6 dt / m d

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9.5 function Exercise9_5 tspan = linspace(0, 20, 200); xo = 1; vo = 1; zet = [0.1, 1, 2]; for k = 1:3 [t, x] = ode45(@FreeOscill, tspan, [xo, vo]', [], zet(k)); subplot(3, 2, 2*k-1), plot(t , x(:,1), 'k-'); xlabel('\tau') ylabel('x(\tau)') © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a3retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

title(['ode45 \zeta = ' num2str(zet(k))]); subplot(3, 2, 2*k) plot(t, Analytical(t, zet(k), xo, vo),'k-'); xlabel('\tau') ylabel('x(\tau)') title(['Analytical \zeta = ' num2str(zet(k))]); end function y = Analytical(t, z, xo, vo) if z < 1 r = sqrt(1-z^2); y = xo*exp(-z*t).*cos(r*t)+(vo+z*xo)/r*exp(-z*t).*sin(r*t); elseif z == 1 y = (xo+(xo+vo)*t).*exp(-t); else r = sqrt(z^2-1); y = xo*exp(-z*t).*cosh(r*t)+(vo+z*xo)/r*exp(-z*t).*sinh(r*t); end function xdot = FreeOscill(t, x, z) xdot = [x(2); -2*z*x(2)-x(1)]; Answer: ode45 = 0.1

Analytical:

2 1

)( x

1

0

)( x

-1 -2

0 -1 -2

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5

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20

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5

ode45 = 1

Analytical:

0.5

x

20

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ode45 = 2

Analytical:

1.5

= 2

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1

1

)(

0.5 0

15

= 1

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x

20

1

)(

0

)(

15

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1 x

10

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)(

= 0.1

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9.6 function Exercise9_6 [t, amp] = Dat; plot(t, amp, 'ks'); hold on xo = [0.1, 1.1, 0.9]; opt = optimset('display', 'off'); x = lsqcurvefit(@xdamped, xo, t, amp, [], [], opt); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a4retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

t = linspace(0, 30, 200); d = 0.1; v = axis; plot(t, xdamped(x, t), 'k-'); xlabel('Time'); ylabel('Amplitude'); text(20, v(4)-d, 'Fitted values') text(20, v(4)-2*d, ['\zeta = ' num2str(x(1), 3)]) text(20, v(4)-3*d, ['\omega_n = ' num2str(x(2), 3) ' rad/s']) text(20, v(4)-4*d, ['X_o = ' num2str(x(3), 3) ' units']) function amp = xdamped(x, t) r = sqrt(1-x(1)^2); amp = x(3)*exp(-x(1)*x(2)*t).*sin(x(2)*t*r+atan(r/x(1)))/r; function [t, amp] = Dat t = [0.000, 0.692, 1.538, 2.308, 3.077, 3.846, 4.615, 5.385, 6.154, ... 6.923, 7.692, 8.461, 9.231, 10.00, 10.77, 11.54, 12.31, 13.07, ... 13.85, 14.15, 15.85, 16.23, 17.92, 18.46, 19.23, 20.00, 20.77, ... 21.54, 22.31, 23.08, 23.85, 24.61, 25.38, 26.15, 26.92, 27.69, ... 28.46, 29.23, 30.00]; amp = [0.801, 0.365, -0.386, -0.562, -0.114, 0.349, 0.338, -0.301, ... -0.204, 0.104, 0.228, 0.008, -0.010, -0.0151, -0.00688, 0.118, ... 0.0882, -0.028, -0.0871, -0.0551, -0.0220, 0.0687, 0.0376, -0.040, ... -0.0514, -0.00641, 0.0379, 0.0167, -0.0184, -0.0259, -0.0141, 0.0149, ... 0.0115, 0.00367, -0.0148, -0.0125, 0.0157, 0.00263, -0.00727]; Answer: 1 Fitted values 0.8

= 0.122

0.6

X = 0.824 units o

= 1.6 rad/s n

e d u t i l p m A

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9.7 function Exercise9_7 initcond = [1.0, 0.0, 0.0]; tspan = linspace(0, 1, 100); xmu = 0.2; omega = 6.0; L = 3.0; [t, x] = ode45(@rod, tspan, [initcond]', [], xmu, omega); indx = find(x(:,1) <= L); polar(x(indx, 3), x(indx, 1)); t3 = interp1(x(:,1), t, L); disp(['Time at which mass leaves rod is ' num2str(t3, 3) ' s']) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a5retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

function xdot = rod(t, x, xmu, omega) xdot = [x(2); omega^2*x(1)-2.*xmu*omega*x(2); omega]; Answer: Time at which mass leaves rod is 0.325 s 90

3

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9.8 function Exercise9_8 zet = 0.1; omegan = 4; omega = linspace(0, 10, 100); [mag2, phas2] = Anal(omega, omegan, zet); sys = freqresp(zet, omegan); [mag1, phas1] = bode(sys, omega); subplot(2, 2, 1) plot(omega,mag1(1,:), '-k') xlabel('Excitation frequency (rad/s)'); ylabel('Magnitude'); title(['Bode']); subplot(2, 2, 3), plot(omega, phas1(1,:),'k-'); xlabel('Excitation frequency (rad/s)'); ylabel('Phase (Degrees)') subplot(2, 2, 2) plot(omega, mag2, '-k') xlabel('Excitation frequency (rad/s)'); ylabel('Magnitude') title('Analytical'); subplot(2,2,4) plot(omega, phas2, 'k-') xlabel('Excitation frequency (rad/s)'); ylabel('Phase (Degrees)') function sys = freqresp(zet, omegan) N = [2*zet*omegan, omegan^2]; D = [1, 2*zet*omegan, omegan^2]; sys = tf(N, D); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a6retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

function [mag, phas] = Anal(omega, omegan, zet) Om = omega/omegan; O1 = 1-Om.^2; Z1 = 2*zet*Om; mag = sqrt((1+Z1.^2)./(O1.^2+Z1.^2)); phas = (atan(Z1)-atan2(Z1, O1))*180/pi; Answer: Bode

Analytical

6

e d u t i n g a M

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e d u t i n g a M

4 3 2 1 0

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Excitation frequency (rad/s) ) s e e r g e D ( e s a h P

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2 4 6 Excitation frequency (rad/s)

9.9 function Exercise9_9 z = 0.1; to = [15, 6, 0.7]; for k=1:length(to) figure(k) [t, y] = ode45(@Ramp, [0 30], [0 0]', [], z, to(k)); plot(t, y(:,1), 'k-', [0, to(k), 30], [0, 1, 1], 'k--') xlabel('\tau') ylabel('y(\tau)') title(['\tau_o = ' num2str(to(k)) ' \zeta = ' num2str(z)]) end function xdot = Ramp(t, x, z, to) h = (t - (t-to).*(t>to))/to; xdot = [x(2); -2*z*x(2)-x(1)+h]; Answer:


= 6 = 0.1 o

= 15 = 0.1 o 1.4

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1

1

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)(

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y 0.6

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25

30

9.10 function Exercise9_10 m = 10.0; k = 150*10^3; c = 50; b = [1*10^-6, 1000*10^-6]; forceampl = 0; forcefreq = 0; figure(1) for i = 1:2 [t, x]=ode45(@sdofstops, [0 1.4], [0, 2]', [], m, k, c, b(i), forceampl, forcefreq); subplot(2,1,i) plot(t, x(:,1), 'k-') xlabel('Time t'); ylabel('x(t)'); title(['Dead zone b =' num2str(b(i)*10^6) ' um']) end figure(2) forceampl = 20; forcefreq = 12; b = 5e-6; [t, x] = ode45(@sdofstops, [0 2], [0 0]', [], m, k, c, b, forceampl, forcefreq); plot(t, x(:,1), 'k-') xlabel('Time t') ylabel('x(t)') title('Forced Response') © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a8retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

function z = sdofstops(t, x, m, k, c, b, forceampl, forcefreq) if abs(x(1))<=b h = 0; else h = k*(x(1)-b*sign(x(1))); end z = [x(2); -2*c*x(2)/m-h/m+forceampl*cos(forcefreq*t)/m]; Answer: Dead zone b =1 um 0.02

4

x 10

-4

Forced Response

0.015 0.01 ) t ( x

3

0.005 0 -0.005

2

-0.01 -0.015

0

0.2

0.4

0.6

0.8

1

1.2

1.4 ) t ( x

Time t Dead zone b =1000 um

1

0.02 0.015 0

0.01 ) t ( x

0.005 0

-1

-0.005 -0.01 -0.015 0

0.2

0.4

0.6

0.8

1

1.2

1.4

Time t

-2 0

0.2

0.4

0.6

0.8

1 Time t

1.2

1.4

1.6

1.8

2

9.11 function Exercise9_11 L = 2.0; b = [10, 0.2]; tspan=[0.0:0.1:10]; for i = 1:2 [t, x] = ode45(@Freeosc, [0, 10], [0.2, 0]', [], b(i), L); subplot(2,1,i) plot(t, x(:,1), 'k-'); xlabel('Time (s)') ylabel('\theta'); title(['b = ' num2str(b(i)) ' m/s^2']); end function xdot = Freeosc(t, x, b, L) xdot = [x(2); -b/L*cos(x(1))-9.81/L*sin(x(1))]; Answer:


b = 10 m/s 2 0.5 0 -0.5

-1 -1.5 -2

0

1

2

3

4

5 Time (s) b = 0.2 m/s

6

7

8

9

10

6

7

8

9

10

2

0.3 0.2 0.1 0

-0.1 -0.2 -0.3 -0.4 0

1

2

3

4

5 Time (s)

9.12 function exercise9_12 gam = 1; zet = 0.15; a = 30; Dst = 0.3; [t, zn] = ode45(@NonLinearSpring, [0 25], [0 0], [], zet, a, gam, Dst); plot(t, (zn(:,1)+Y(t, gam)), 'k-') hold on plot(t, Y(t, gam), 'k-') xlabel('\tau') ylabel('x(\tau)/y_{max}') function Out = NonLinearSpring(t, z, zet, a, g, Dst) yinacc = g^2*(1-g*t).*exp(-g*t); Out = [z(2); -2*zet*z(2)-z(1)-a*(z(1)-Dst)^3-yinacc-a*Dst^3]; function yin = Y(t, g) yin = 1-(1+g*t).*exp(-g*t); Answer: 1.4

1.2

1

0.8 x

ya /m

)(

x

0.6

0.4

0.2

0 0

5

10

15

20

25

9.13 function Exercise9_13 wr = 1; mr = 0.05; z1 = 0; z2 = 0.05:0.05:0.3; Om = linspace(0.6, 1.4, 200); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a10 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

for k = 1:length(z2) plot(Om, H11(Om, wr, mr, z1, z2(k)), 'k-') hold on end xlabel('\Omega') ylabel('H_{11}(\Omega)') xlim([0.6, 1.4]) A = sqrt((wr^2-1)^2+mr*(2+mr)); B = 1+wr^2*(1+mr); Om1 = sqrt((B-A)/(2+mr)) Om2 = sqrt((B+A)/(2+mr)) function h = H11(Om, wr, mr, z1, z2) D = Om.^4-j*2*(z1+z2*wr*mr+z2*wr)*Om.^3 ... -(1+mr*wr^2+wr^2+4*z1*z2*wr)*Om.^2+j*2*(z2*wr+z1*wr^2)*Om+wr^2; E = wr^2+j* 2*z2*wr*Om-Om.^2; h = abs(E./D); Answers: Om1 = 0.9186

Om2 = 1.0753

16

14

12

10

) (1

8

1

H 6

4

2

0 0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

9.14 k1 = 14600; k2 = 21900; L1 = 1.520; L2 = 1.22; m = 730; Ic = 1360; K = [k1+k2, (L2*k2-L1*k1); (L2*k2-L1*k1), (k1*L1^2+k2*L2^2)]; M = [m, 0; 0, Ic]; [Modes, Eval] = eig(K, M); NatFreq = sqrt(diag(Eval)); disp(['omega1 = ' num2str(NatFreq(1)) ' rad/s omega2 = ' num2str(NatFreq(2)) ' rad/s']) disp(' ') disp(['Mode shapes are:']) disp(Modes) Answers: omega1 = 6.6934 rad/s omega2 = 7.3464 rad/s Mode shapes are: 0.0244 -0.0279 -0.0204 -0.0178

9.15 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a11 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

z1 = 0.1; z2 = 0.1; mr = 1; wrr = [0.05, 0.25]; t = linspace(0, 100, 500); for k = 1:2 subplot(2,1,k) wr = wrr(k); N = [4*wr*z1*z2, 2*wr*(z1*wr+z2), wr^2]; D = [1, 2*(z1+z2*wr*mr+z2*wr), (1+mr*wr^2+wr^2+4*z1*z2*wr), 2*(z2*wr+z1*wr^2), wr^2]; sys = tf(N, D); [TR, t] = impulse(sys, t); plot(t,TR, 'k-', [0,100], [0 0], 'k--') xlabel('Time') ylabel('T_R') title(['\omega_r = ' num2str(wr)]) end Answer: = 0.05 r 0.06 0.04 0.02 T

R

0 -0.02 -0.04

0

10

20

30

40

50 Time

60

70

80

90

100

60

70

80

90

100

= 0.25 r 0.3 0.2 0.1 T

R

0 -0.1 -0.2 0

10

20

30

40

50 Time

9.16 m1 = 10; m2 = 40; JG = 8; k1 = 10000; k2 = 35000; k3 = 8100; L1 = 1; L2 = 0.3; g = 9.81; L3 = (L1-L2)/2; A(1,1) = k1+k2+k3; A(1,2) = -k3; A(1,3) = k1*L1-k2*L2+k3*L3; A(2,1) = A(1,2); A(2,2) = k3; A(2,3) = -k3*L3; A(3,1) = A(1,3); A(3,2) = A(2,3); A(3,3) = k1*L1^2+k2*L2^2+k3*L3^2; xt = A\[m1*g, m2*g, 0]'; tho = asin(xt(3)); M = diag([m1, m2, JG]); A(3,3) = -A(1,3)*xt(1)*sin(tho)+k3*L3*xt(2)*sin(tho)+A(3,3)*(cos(tho)^2-sin(tho)^2); A(1,3) = A(1,3)*cos(tho); A(3,1) = A(1,3); A(2,3) = A(2,3)*cos(tho); A(3,2) = A(2,3); K = A; [Modes, EVal] = eig(K,M); NF = sqrt(diag(EVal)) Modes Answers: NF = 12.5989 41.9371 73.2299 Modes = 0.0233 -0.0260 0.3143 0.1571 -0.0126 -0.0127 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a12 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

0.0304

0.3512

0.0268

9.17 M = diag([21 21 21 21 21 21 98 49]); K = [51 -51 0 0 0 0 0 0 -51 102 -51 0 0 0 0 0 0 -51 102 -51 0 0 0 0 0 0 -51 102 -51 0 0 0 0 0 0 -51 102 -51 0 0 0 0 0 0 -51 117 -66 0 0 0 0 0 0 -66 81 -15 0 0 0 0 0 0 -15 15]*10^6; rnk = rank(K) [modes E] = eig(K,M); w = sqrt(E); hold on w(1,1) = 0; for n = 1:8 plot([1 8], [2*n-1, 2*n-1], 'k--') plot(1:8, 2*n-1+modes(:,n)/max(abs(modes(:,n))), 'sk-') text(-.4, 2*n-1, ['\omega_{' num2str(n) '}=' num2str(w(n,n),4)], 'fontsize', 14, 'HorizontalAlignment', 'left') end v = axis; v(1) = 0.5; axis(v); axis off Answer: 8=3041 7=2805 6=2406 5=1862 4=1219 3=704.8 2=458.3 1=0

9.18 Jp = 2; Jt = 1.2; k1 = 2.5e6; k2 = k1; m = 10; M = diag([m, Jt, m, Jt]); IM = inv(M); K = diag([k1, k2, k1, k2]); G = zeros(4,4); Mstar = [K, G; G, M]; Ks11 = K*IM*K; omega = linspace(0, 1500, 10); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a13 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

for n = 1:length(omega) G(2,4) = -Jp*omega(n); G(4,2) = Jp*omega(n); Kstar = [Ks11, K*IM*G; G'*IM*K, K+G'*IM*G]; Lambda(n,:) = sort(sqrt(eig(Kstar, Mstar)))'; end plot(omega', Lambda, 'k-s') v = axis; axis([v(1) v(2) 0 v(4)]) xlabel('\omega') ylabel('\Omega') Answers: 3500

3000

2500

2000

1500

1000

500

0 0

500

1000

1500


Solutions to Exercises Chapter 10 10.1 % Parts a and b theta = linspace(0, 2*pi, 11); k = 1; p = 2*pi*k*[cos(theta(6)) cos(theta(4))+sqrt(-1)*sin(theta(4)) cos(theta(4))-sqrt(-1)*sin(theta(4)) cos(theta(5))+sqrt(-1)*sin(theta(5)) cos(theta(5))-sqrt(-1)*sin(theta(5))]; z = -p; figure(1); plot(real(p), imag(p), 'bx', real(z), imag(z), 'ro'); title('Map of delay approximation'); dplant = zpk(z,p,1); dplant = 1/(dcgain(dplant))*dplant; gplant = tf(1, [1 1]); [y1, t1] = step(dplant*gplant); [y0] = step(gplant, t1); figure(2); plot(t1, y0, 'k-', t1, y1, 'k--'); legend('Orginal', 'Delayed'); title('Step response, original and delayed system'); % Part c p2 = 0.5*p; z2 = -p2; dplant2 = zpk(z2,p2,1); dplant2 = 1/(dcgain(dplant2))*dplant2; [y2,t2] = step(dplant2*gplant); [y0] = step(gplant,t2); % Increase k to 2 p3 = 2*p; z3 = -p3; dplant3 = zpk(z3, p3, 1); dplant3 = 1/(dcgain(dplant3))*dplant3; [y3,t3] = step(dplant3*gplant); [y0] = step(gplant,t3); figure(3); plot(t3, y0,t1, y1, t2, y2, t3, y3); legend('Org', 'k=1', 'k=0.5', 'k=2', 'Location', 'SouthEast'); title('Step response for variety of delays'); Answers:


Step response, original and delayed system

Map of delay approximation

1.2

6

Orginal Delayed 1 4

0.8 2

0.6 0

0.4

-2

0.2

-4

0

-6 -8

-0.2 -6

-4

-2

0

2

4

6

0

8

1

2

3

4

5

6

7

Step response for variety of delays 1.2

Step response with variety of poles 1.2

1 1

0.8 0.8

0.6 0.6

0.4

0.4

0.2

0.2

0

Org k=1 k=0.5

0

0 5 3

k=2

7

-0.2 0

1

2

3

4

5

6

7

8

9

-0.2 0

1

2

3

4

5

6

7

8

10.2 m1 = 500; m2 = 100; b = 1000; k1 = 2000; k2 = 1e4; num = [b*k2 k1*k2]; den = [m1*m2, b*(m1+m2), k1*(m1+m2)+k2*m1, k2*b, k1*k2]; [mag, phs, w] = bode(num, den); logW = log10(w/(2*pi)); logmag = 20*log10(mag); [mx, inmx] = max(20*log10(mag)); fa = 10^(interp1(logmag(inmx:end), logW(inmx:end), 20*log10(0.9))); fc = 10^(interp1(logmag(inmx:end), logW(inmx:end), 20*log10(0.1))); disp(['frequency @ 0.1e = ' num2str(fa) ' Hz frequency @ 0.9e = ' num2str(fc) ' Hz']) mload = linspace(500, 1000, 20); z = zeros(length(w), 20); for i = 1:length(mload) m1 = mload(i); den = [m1*m2, b*(m1+m2), k1*(m1+m2), k2*m1, k2*b, k1*k2]; [mag] = bode(num, den, w); z(:,i) = 20*log10(mag); end mesh(mload, logW, z) xlabel('Load (kg)') ylabel('log_{10}(freqency in Hz)') zlabel('Response (dB)')


Answers: frequency @ 0.1e = 0.5237 Hz

frequency @ 0.9e = 2.0994 Hz

50

) B d ( e s n o p s e R

0

-50

-100

-150

-200

-250 3 2

1000 1

900 800

0 700 -1 log

10

(freqency in Hz)

600 -2

500 Load (kg)

10.3 g0 = tf(1, [1, 1]); g1 = tf(3.7, [0.75, 0.6]); g2 = tf(1.63, [0.94, 0.92]); g3 = tf([0.7, 7, 17], [1, 2, 5.2, 4]); t = linspace(0, 2, 200); y0 = step(g0, t); y1 = step(g1, t); y2 = step(g2, t); y3 = step(g3, t); figure(1) plot(t, y0, t, y1, t, y2, t, y3) legend('g0', 'g1', 'g2', 'g3') title('Open loop response') xlabel('Time (s)') r0 = step(feedback(g0, 20), t); r1 = step(feedback(g1, 20), t); r2 = step(feedback(g2, 20), t); r3 = step(feedback(g3, 20), t); figure(2) plot(t, r0, t, r1, t, r2, t, r3) legend('g0', 'g1', 'g2', 'g3') title('Closed loop response') xlabel('Time (s)') Answers:


Open loop response

Closed loop response

5

0.07

g0 g1 g2 g3

4.5

g0 g1 g2 g3

0.06

4

0.05

3.5

3 0.04

2.5 0.03

2

1.5

0.02

1 0.01

0.5

0 0

0.2

0.4

0.6

0.8

1 Time (s)

1.2

1.4

1.6

1.8

2

0 0

0.2

0.4

0.6

0.8

1 Time (s)

1.2

1.4

1.6

1.8

2

10.4 function Exercise10_4 for m = 1:3 [t0, y0] = ode45(@f, [0 8], [0], [], 10^(m-1), 0, 1); [t1, y1] = ode45(@f, [0 8], [0], [], 10^(m-1), 1, 1); [t2, y2] = ode45(@f, [0 8], [0], [], 10^(m-1), 2, 1); [t3, y3] = ode45(@f, [0 8], [0], [], 10^(m-1), 3, 1); figure(m) plot(t0, y0, t1, y1, t2, y2, t3, y3); legend('none', '(a)', '(b)', '(c)'); title(['Step response with gain = ' num2str(10^(m-1))]); end function ydot = f(t, y, p1, p2, p3) % p1 = gain k % p2 = nonlinearity type % p3 = time at step response e = -y; if(t > p3) e = 3-y; end switch p2 case 1 v = 0.2*(e^3 - e); case 2 if(abs(e) > 1) v = e + sin(e); else v = 0; end case 3 v = atan(e); otherwise v = e; end ydot = -y + p1*v; Answers:


Step response with gain = 1 2

Step response with gain = 10 3

none

none (a) (b)

(a) (b) (c)

2.5

(c)

1.5

2

1 1.5

1

0.5

0.5

0 0

-0.5 0

1

2

3

4

5

6

7

8

-0.5 0

1

2

3

4

5

6

7

8

Step response with gain = 100 3 none (a) (b) (c)

2.5

2

1.5

1

0.5

0 0

1

2

3

4

5

6

7

8

10.5 Lead = tf([1, 6], [1, 50]); Notch0 = zpk([-3+30i, -3-30i], [-60, -60], 1); Notch0 = 1/(dcgain(Notch0))*Notch0; Notch1 = zpk([-3+28i, -3-28i], [-60, -60], 1); Notch1 = 1/(dcgain(Notch1))*Notch1; Notch2 = zpk([-3+34i, -3-34i], [-60, -60], 1); Notch2 = 1/(dcgain(Notch2))*Notch2; t = linspace(0, 3, 200); y0 = step(feedback(10.0*Notch0*Lead*Pointer,1), t); y1 = step (feedback(10.0*Notch1*Lead*Pointer,1), t); y2 = step (feedback(10.0*Notch2*Lead*Pointer,1), t); plot(t, y0, t, y1, t, y2) legend('No notch', 'Notch 1', 'Notch 2') xlabel('Time (s)') ylabel('Step response') Answers:


1.4 No notch Notch 1 Notch 2 1.2

e s n o p s e r p e t S

1

0.8

0.6

0.4

0.2

0 0

0.5

1

1.5 Time (s)

2

2.5

3

10.6 g0 = tf(9, [1, 0, -9]); lead = tf([1, 2], [1, 6]); figure(1) rlocus(lead*g0) % Use % [k, p] = rlocfind(lead*g0) % to determine suitable gain k k = 5; % Now find a range of kd off nominal kd = linspace(-1, -20, 50); pMax = zeros(length(kd),1); for i = 1:length(kd) g0 = tf(9, [1 0 kd(i)]); clSys = feedback(k*lead*g0, 1); pMax(i) = max(real(pole(clSys))); end figure(2) plot(sqrt(-kd), pMax, sqrt(-kd), zeros(length(kd), 1), 'r:') xlabel('Pole location') ylabel('Maximum pole') title('Stability of rocket') Answers: Root Loc us

Stability of rocket 1

25

20

0.5 15

s i x A y r a n i g a m I

10

e l o p

5

m u m i x a M

0

-5

0

-0.5

-1 -10

-15

-1.5 -20

-25 -8

-6

-4

-2 Real Ax is

0

2

4

-2 1

1.5

2

2.5 Pole location

3

3.5

4

4.5


10.7 m = 1200; b = 70; kp = 100; g = tf(1, [m, b]); % (a) [y1, t1] = step(feedback(g, kp)); figure(1) plot(t1, y1) title('(a) Step response disturbance to ride'); % (b) [y2, t2] = step(feedback(kp*g, 1)); figure(2) plot(t2, y2) title('(b) Step response command to ride'); %(c) - disturbance figure(3) hold on kp = linspace(50, 150, 10); t = linspace(0, 50, 200); for i = 1:length(kp) [y] = step(feedback(g, kp(i)), t); plot(t, y) end title('(c) Step response disturbance to ride'); % (c)- command figure(4) hold on for i = 1:length(kp) y = step(feedback(kp(i)*g, 1), t); plot(t, y); end title('(c) Step response command to ride'); Answers: 6

x 10

-3

(b) Step response command to ride

(a) Step response disturbance to ride

0.7

0.6

5

0.5 4

0.4 3

0.3

2

0.2

1

0.1

0

0 0

5

10

15

20

25

30

35

40

45

0

5

10

15

20

25

30

35

40

45


9

x 10

-3

(c) Step response disturbance to ride

(c) Step response command to ride 0.7

8 0.6

7 0.5

6

0.4

5

4

0.3

3 0.2

2 0.1

1

0 0

5

10

15

20

25

30

35

40

45

50

0 0

5

10

15

20

25

30

35

40

45

50

10.8 m = 5000; k = 8e5; b = 1.2e4; % (a) num = [1, 0]; den = [m, b, k]; rlocus(tf(num, den)) % Use % k = rlocfind(tf(num, den)) % to find stable value of k k = 114.5e3; % (b) plant = tf([b, k], [m, b, k]); figure(1) step(plant); title('Step response of uncontrolled suspension') figure(2) bode(plant) title('Bode plot of uncontrolled suspension') % (c) kval = linspace(4e5, 10e5, 20); t = linspace(0 ,5, 100); z = zeros(length(kval), length(t)); for i = 1:length(kval) plant = tf([b, kval(i)], [m, b, kval(i)]); y = step(plant, t); z(i,:) = y'; end figure(3) mesh(t, kval, z) xlabel('Time (s)') ylabel('k (N/m)') title('Response for various spring constants') colormap([0 0 0 ]) Answers:


Bode plot of unc ontrolled s us pens ion Step res pons e of unc ontrolled s us pens ion

10

1.6

) B d ( 1.4

e d u t i n g a M

1.2

0

-10

-20

-30

1

e d u t i l p m A

-40

-50

0.8

0

0.6

) g e d (

0.4

e s a h P

0.2

-45

-90

-135

0 0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

10

-1

Time (s ec )

10

0

10

1

10

2

10

3

Frequenc y (rad/s ec )

Response for various spring constants

1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 10 9

5

8 x 10

5

4 7

3 6

2 5

k (N/m)

1 4

0 Time (s)

10.9 num = [1, 0.03]; den = conv([1, 0], conv([1, 0.09], conv([1, 0.04], [1, -0.0004]))); g = tf(num, den); figure(1) rlocus(g) title('Root locus of ship system'); % (b) z = 0.003; p = 10*z; lead = tf([1 z], [1 p]); figure(2) rlocus(lead*g) % k = rlocfind(lead*g) k = 3.5e-5; % From the above result title('Finding a lead controller') figure(3) step(feedback(k*lead*g, 1)) title('Step response of lead controller'); % (c) p = 0.0001; numPD = conv([1 p], num); gPD = tf(numPD, den); figure(4) rlocus(gPD) title('Root locus of PD control system'); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a9retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

% k = rlocfind(gPD) k = 0.0075; % From the above result kp = p*k; kd = k; figure(5) t = linspace(1, 500, 500); step(feedback(k*gPD,1),t) title(['Step response with kp = ' num2str(kp) ' kd = ' num2str(kd)]) Answers: Finding a lead c ontroller

Root loc us of s hip s y s tem 0.4

0.4

0.3

0.3

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0.2 s i x A y r a n i g a m I

s i x A

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Step res pons e of lead c ontroller

Root loc us of PD c ontrol s y s tem

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Step res pons e with k p = 7.5e-007 k d = 0.0075 1.4

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10.10 G = tf(4, [1 4]); [nd, dd] = pade(0.2, 10); % Padé approximation of models with time delays plant = tf(nd, dd)*G; % (a) PI control control = tf([1, 2], [1, 0]); figure(1) rlocus(control*plant) % kp = rlocfind(control*plant); title('Root locus of A/F ratio system'); kp = 1.384; figure(2); step(feedback(kp*control*plant, 1)) title(['Step response of PI controller with kp = ' num2str(kp)]) % (b) Smith predictor % We pick a simple PE control with gain = 10 t = linspace(0, 1, 200); smith = feedback(10, G*(1-tf(nd, dd))); figure(3) step(feedback(smith*plant, 1), t); title('Step response of ideal smith predictor with kp = 10'); % (c) Smith predictor with incorrect time delays and system model % First compute the effect of system delay mismatch figure(4) hold on dels = linspace(0.1, 0.3, 10); for i = 1:length(dels); [nd, dd] = pade(dels(i),10); plant1 = tf(nd,dd)*G; step(feedback(smith*plant1, 1), t) end title('Step response with system delay mismatch'); % Now show the effect of system mismatch G2 = tf(6, [1, 5]); [nd, dd] = pade(0.2, 10); plant2 = tf(nd, dd)*G2; figure(5) step(feedback(smith*plant2, 1), t) title('Step response with system mismatch') Answers: Step res pons e of PI c ontroller with k p = 1.384

Root loc us of A/F ratio s y s tem

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300

1.2 200

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Step res pons e of ideal s mith predic tor with k p = 10

Step res pons e with s y s tem delay mis matc h

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10.11 w = 250*pi; xi = 20/w; g = tf(9, [1, 0, 9]); f = tf(w^2, [1, 2*xi*w, w^2]); plant = f*g; z = 8; p = 5*z; control = tf([1, z], [1, p]); figure(1) rlocus(control*plant) % pause; % readjust scale % k = rlocfind(control*plant) k = 75; figure(2) step(feedback(k*control*plant,1)) title('Step response of motor cycle system under lead control') % The effect of the inclinometer's bandwidth limitations % is negligible on the system right now. If it is shifted % to a much lower frequency, the control design is much harder % and the performance unacceptably low w = 14.8*pi; xi = 0.4; fp = tf(w^2, [1, 2*xi*w, w^2]); plantP = fp*g; figure(3) rlocus(control*plantP) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a12 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

% k = rlocfind(control*plantP) k = 42; figure(4); step(feedback(k*control*plantP, 1)) title('Step response with low bandwidth sensor') Answers: Root Loc us

Step res pons e of motor c y c le s y s tem under lead c ontrol

4000

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200 1.6

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Solution to Exercises Chapter 11 11.1 function Exercise11_1 g = 9.81; nu = 1.3e-6; k = 0.00026; L = 25.0; Q = 0.003; h = 3.0; SumKml = 0.5+0.3+0.3+0.15+1.0; D = fzero(@ColebrookFriction, [0.05], [], g, nu, k, L, Q, h, SumKml); disp(['D = ' num2str(D)]) function value = ColebrookFriction(D, g, nu, k, L, Q, h, SumKml) lambda = D^5*h*g*pi^2/(8.0*Q^2*L)-SumKml*D/L; Re = 4.0*Q/(pi*D*nu); value = 1/sqrt(lambda)+2*log10((k/D)/3.7+2.51/(sqrt(lambda)*Re)); Answer: D = 0.046979

11.2 function Exercise11_2 p = 0.375; q = 7.43e-4; Z = 5:5:50; zer = zeros(length(Z),1); for k = 1:length(Z) [t, y] = ode45(@ReservoirOsc, [0 500], [Z(k) 0], [], p, q); ind = min(find(y(:,1)<0)); zer(k) = interp1(y(ind-5:ind+4,1), t(ind-5:ind+4), 0); end plot(Z, zer, 'k-') xlabel('Initial height [Z(0)]') ylabel('Time to first zero crossing (s)') function c = ReservoirOsc(t, y, p, q ) c = [y(2); -p*sign(y(2))*y(2)^2-q*y(1)]; Answer: ) s ( g n i s s o r c o r e z t s r i f

280

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o t 160 e m i T

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35

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11.3 function Exercise11_3 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a1retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

mu = 0.45; rho = 900.0; nu = mu/rho; h = 0.01; g = 9.81; Ubelt = 1.5; nx = 100; xmax = h; x = linspace(0, xmax, nx); nt = 41; tmax = 0.4; t = tmax/(nt^2)*(1:nt).^2; u = pdepe(0, @pdfslpde2,@pdfslic2,@pdfslbc2,x , t, [], nu, g, rho, Ubelt); hold on for ijd = 2:nt plot(x, u(ijd,:), 'k') end ylabel('Vertical Velocity (m/s)') xlabel('y (m)') xp = x(nx/2); yp = u(nt,nx/2); text(xp+0.0021, yp+0.1, ['t = ' num2str(tmax, 4) ' s']) plot([xp, xp+0.002], [yp, yp+0.1], 'k') v = rho*g/2/mu*x.*x-rho*g*h/mu*x+Ubelt; plot(x, v, '.k') box on function [c,f,s] = pdfslpde2(x, t, u, DuDx, nu, g, rho, Ubelt) c = 1.0; f = nu*DuDx; s = -g; function u0 = pdfslic2(x, nu, g, rho, Ubelt) u0 = 0; function [pl,ql,pr,qr] = pdfslbc2(xl, ul, xr, ur, t, nu, g, rho, Ubelt) pl = ul-Ubelt; ql = 0; pr = 0; qr = 1; Answer: 1.6

) s / m ( y t i c o l e V l a c i t r e V

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11.4 function Exercise11_4 mu = 0.02; rho = 800.0; nu = mu/rho; h = 0.02; tau = 0.01; ny = 100; ymax = h; © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a2retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

y = linspace(0, ymax, ny); nt = 41; tmax = 1.0; t = tmax/(nt^2)*(1:nt).^2; u = pdepe(0, @pdfslpde3, @pdfslic3, @pdfslbc3, y, t, [], nu, tau, rho); hold on for ijd = 2:nt plot(u(ijd,:), y, 'k') end xlabel('Horizontal Velocity (m/s)') ylabel('y (m)') text(u(nt,ny/4)+0.25, y(ny/4), ['t = ' num2str(tmax, 4) ' s']) box on function [c,f,s] = pdfslpde3(x, t, u, DuDy, nu ,tau, rho) c = 1.0; f = nu*DuDy; s = 0.0; function u0 = pdfslic3(y, nu, tau, rho) u0 = 0; function [pl, ql, pr, qr] = pdfslbc3(yl, ul, yr, ur, t, nu, tau, rho) pl = tau; ql = 1; pr = 0; qr = 1; Answer: 0.02 0.018 0.016 0.014 ) 0.012 m ( 0.01 y

0.008 0.006 t=1 s 0.004 0.002 0 0

0.5

1 1.5 Horizontal Velocity (m/s)

2

2.5

11.5 function Exercise11_5 mu = 0.02; rho = 800.0; nu = mu/rho; h = 0.003; dPdz = 10000; Utop = 1.5; ny = 100; ymax = h; y = linspace(0, ymax, ny); nt = 50; tmax = 1.0; t = tmax/(nt^2)*(1:nt).^2; u = pdepe(0, @pdfslpde, @pdfslic, @pdfslbc, y, t, [], nu, dPdz, rho, Utop); hold on for ijd = 2:nt plot(u(ijd,:),y*1000,'k') end © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a3retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

xlabel('Horizontal Velocity (m/s)') ylabel('y (mm)') box on text(u(nt,ny/2)+0.05, y(ny/2)*1000, ['t = ' num2str(tmax, 4) ' s']) function [c,f,s] = pdfslpde(x, t, u, DuDy, nu, dPdz, rho, Utop) c = 1.0; f = nu*DuDy; s = dPdz/rho; function u0 = pdfslic(y, nu, dPdz, rho, Utop) u0 = 0; function [pl,ql,pr,qr] = pdfslbc(yl, ul, yr, ur, t, nu, dPdz, rho, Utop) pl = ul; ql = 0; pr = ur-Utop; qr = 1; Answer: 3

2.5

2

) m m (

1.5

t = 1s

y

1

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0 0

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0.6 0.8 Horizontal Velocity (m/s)

1

1.2

1.4

1.6

11.6 function Exercise11_6 c = 2; Vinfty = 100; a = 10; alpha = a*pi/180; [x, y ]= meshgrid(linspace(-1.0, 3.0, 100), linspace(-1.5, 1.5, 100)); m = 100; xf = linspace(0.01, c, m); thetaf = 1.0 - 2.0 .*xf./c; thetaf = acos(thetaf); gammaf = 2.0*alpha*Vinfty.*(1.0 + cos(thetaf))./sin(thetaf); psi = Vinfty*cos(alpha).*y-Vinfty*sin(alpha).*x; for i = 1:m r = sqrt((x-xf(i)).^2 + y.^2); psi = psi + (c/m)*(gammaf(i)/2.0/pi).*log(r); end contour(x,y, psi,60 , 'k') axis equal


Answer: 1.5

1

0.5

0

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-1

-1.5 -1

-0.5

0

0.5

1

1.5

2

2.5

3

11.7 function Exercise11_7 nx = 100; xmin = -2.5; xmax = 2.5; ny = 100; ymin = -2.0; ymax = 2.0; x = linspace(xmin, xmax, nx); y = linspace(ymin, ymax, ny)'; [x, y]=meshgrid(x ,y); U0 = 1.0; D = 1.0; K1 = U0*D^2/4; xK1 = 0.0; yK1 = 0.75; K2 = U0*D^2/4; xK2 = 0.0; yK2 = -0.75; thetaK1 = atan2(y-yK1, x-xK1); rK1 = sqrt((x-xK1).*(x-xK1)+(y-yK1).*(y-yK1) ); thetaK2 = atan2(y-yK2,x-xK2); rK2 = sqrt((x-xK2) .* (x-xK2) + (y-yK2) .* (y-yK2) ); StreamFunction = U0.*y - K1.*sin(thetaK1)./ rK1-K2.*sin(thetaK2)./ rK2 ; levmin = StreamFunction(1,nx); levmax = StreamFunction(ny,nx/2); levels = linspace(levmin,levmax,51)'; figure(1) contour(x,y,StreamFunction,levels) axis equal axis([xmin,xmax,ymin,ymax]) ylabel('y') xlabel('x') colormap([0 0 0]) x = linspace(xmin,xmax,nx); thetaK2 = atan2(0.0-yK2,x-xK2); rK2 = sqrt((x-xK2).*(x-xK2)+(0.0-yK2).*(0.0-yK2)); thetaK1 = atan2(0.0-yK1,x-xK1); rK1 = sqrt((x-xK1).*(x-xK1)+(0.0-yK1).*(0.0-yK1)); phi=U0*x+K1*cos(thetaK1)./rK1+K2*cos(thetaK2)./rK2; delx=x(2)-x(1); figure(2) plot(x(2:nx), diff(phi)/delx, 'k-') ylabel('horizontal velocity along the x axis') xlabel('x') © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a5retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Answers: a x

2

e h t

1.5

g n o l a

1

y t i c o l e v

0.5

0

l a t n o z i r o h

y

-0.5

-1

2

1.8

1.6

1.4

1.2

1

-1.5

-2 -2.5

-2

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-1

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0 x

0.5

1

1.5

2

2.5

0.8 -2.5

-2

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-1

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1

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2.5

11.8 function Exercise11_8 modes = 20; alpha = 4.*pi/180.0; Vinfty = 100; chord = 2.0; a = zeros(modes,1); m = 100; n = 200; tb = acos(1.0-0.2025*2); a0 = alpha-quadl(@camber1, 0, tb, 1.e-5, [], 0)/pi... -quadl(@camber2, tb, pi, 1.e-5, [], 0)/pi; for i = 1:modes a(i) = 2/pi*quadl(@camber1, 0, tb, 1.e-5, [], i)... +2/pi*quadl(@camber2, tb, pi, 1.e-5, [], i); end q = 1:modes; disp(['A(0) = ' num2str(a0, '%8.6f')]) fprintf(1, 'A(%2.0f) = %8.6f\n', [q; a']) xf = linspace(0.001, chord, m); thetaf = acos(1-2.*xf/chord); summ = a'*sin(q'*thetaf); gammaf = 2.0*Vinfty*(a0*(1.0 + cos(thetaf))./sin(thetaf)+summ); figure(1) plot(xf, gammaf, 'k-') hold on plot(xf, 2.0*Vinfty*summ, 'r--') [x, y] = meshgrid(linspace(-1.0, 3, n), linspace(-1.5, 1.5, n)); psi = Vinfty*cos(alpha).*y-Vinfty*sin(alpha).*x; for i = 1:m r = sqrt((x-xf(i)).^2 + y.^2); psi = psi + (chord/m)*(gammaf(i)/2.0/pi).*log(r); end figure(2) contour(x, y, psi, 50, 'k') axis([-1, 2.5, -0.4, 0.4]); hold on x = linspace(0.0, 0.2025, m); z = 2.6595*x.*(x.*x-0.6075*x+0.1147); h = plot(chord*x, chord*z, 'k-'); set(h, 'LineWidth', 2) x= linspace(0.2025, 1.0, m); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a6retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

z= 0.02208*(1.0-x); h = plot(chord*x, chord*z, 'k-'); set(h, 'LineWidth', 2) function func = camber1(theta, n) x = 0.5*(1-cos(theta)); dzdx = 2.6595*(3*x.^2-0.6075*2*x+0.1147); func = dzdx.*cos(n*theta); function func = camber2(theta, n) func = -0.02208*cos(n*theta);

Answers: A(0) = 0.041156 A( 1) = 0.095484 A( 2) = 0.079150 A( 3) = 0.056780 A( 4) = 0.033789 A( 5) = 0.014870 A( 6) = 0.002613 A( 7) = -0.002884 A( 8) = -0.003450 A( 9) = -0.001674 A(10) = 0.000231 A(11) = 0.001114 A(12) = 0.000910 A(13) = 0.000197 A(14) = -0.000378 A(15) = -0.000499 A(16) = -0.000242 A(17) = 0.000098 A(18) = 0.000269 A(19) = 0.000205 A(20) = 0.000014 400

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2.5

11.9 function Exercise11_9 R = 1.0; Q = 1.0; nn = 5; © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a7retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

etamax = 0.04; ksimin = -0.2; ksimax = -0.0; etaoffarray = linspace(0, etamax, nn); ksioffarray = linspace(ksimin, ksimax, nn); maxcamber = zeros(nn,nn); maxthick = zeros(nn,nn); nplot = 200; for in = 1:nn ksioff = ksioffarray(in); for jn = 1:nn etaoff = etaoffarray(jn); zetaoff = complex(ksioff,etaoff); thetaTE = -asin(etaoff/R); ksiTE = ksioff+sqrt(R^2-etaoff^2); ksiLE=ksioff-sqrt(R^2-etaoff^2); lambda = ksioff+sqrt(R^2-etaoff^2); zetaTE = complex(ksiTE, 0); zetaLE = complex(ksiLE, 0); zTE = zetaTE+lambda^2/zetaTE; zLE = zetaLE+lambda^2/zetaLE; chord = real(zTE-zLE); thetaLE = pi-thetaTE; thetatop = fliplr(linspace(thetaTE, thetaLE, nplot)); thetabottom = linspace(thetaLE, 2*pi+thetaTE, nplot); zetatop = R*exp(1i*thetatop)+zetaoff; zetabottom = R*exp(1i*thetabottom)+zetaoff; ztop = zetatop+lambda^2./zetatop; zbottom = zetabottom+lambda^2./zetabottom; ytop = imag(ztop); xtop = real(ztop); xtopeven = linspace(xtop(1), xtop(nplot), nplot); ytopeven = interp1(xtop, ytop, xtopeven); ybottom = imag(zbottom); xbottom = real(zbottom); xbottomeven = linspace(xbottom(1), xbottom(nplot), nplot); ybottomeven = interp1(xbottom, ybottom, xbottomeven); ythickness = (ytopeven-ybottomeven); ymidline = (ytopeven+ybottomeven)/2; maxthick(in,jn) = max(ythickness)/chord; maxcamber(in,jn) = max(ymidline)/chord; end end subplot(1,2,1) for in = 1:nn plot(ksioffarray, maxthick(:,in), 'k-') hold on end axis([ksimin 0 0 0.3]) xlabel('\xi_{off}') ylabel('t/L') subplot(1,2,2) for jn = 1:nn plot(etaoffarray, maxcamber(jn,:), 'k-') hold on end axis([0 etamax 0 0.02]) xlabel('\eta_{off}') © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a8retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

ylabel('b/L') Answers: 0.3

0.02

0.018 0.25 0.016

0.014 0.2 0.012 L / t

L / b

0.15

0.01

0.008 0.1 0.006

0.004 0.05 0.002

0 -0.2

0 -0.15

-0.1

off

-0.05

0

0

0.01

0.02

0.03

0.04

off

11.10 function Exercise11_10 R = 1.0; Q = 1.0; alpha = 6.0*pi/180; ksioff = -0.093*R; etaoff = 0.08*R; zetaoff = complex(ksioff,etaoff); thetaTE = -asin(etaoff/R); Gamma = 4*pi*Q*R*sin(alpha-thetaTE); theta = linspace(0, 2*pi, 1000); zetac = R*exp(1i*theta)+zetaoff; ksiTE = ksioff + sqrt(R^2-etaoff^2); ksiLE = ksioff - sqrt(R^2-etaoff^2); lambda = ksioff + sqrt(R^2-etaoff^2); thetaLE = pi-thetaTE; nplot = 200; thetatop = linspace(thetaTE,thetaLE,nplot); thetatemp = zeros(nplot,1); for ijd = 1:nplot thetatemp(ijd) = thetatop(nplot+1-ijd); end thetatop = thetatemp; thetabottom = linspace(thetaLE, 2*pi+thetaTE, nplot); zetatop = R*exp(1i*thetatop)+zetaoff; zetabottom = R*exp(1i*thetabottom)+zetaoff; ztop = zetatop+lambda^2./zetatop; zbottom = zetabottom+lambda^2./zetabottom; ytop = imag(ztop); xtop = real(ztop); plot(ztop, 'k-') hold on plot(zbottom, 'k-') hold on axis([xtop(1)-0.5, xtop(nplot)+0.5, -3, 2]) xlabel('x') ylabel('y or C_p') © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a9retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

w = (Q*exp(-1i*alpha)-Q*exp(1i*alpha)*R^2./(zetatop-zetaoff).^2+ ... 1i*Gamma/(2*pi)./(zetatop-zetaoff))./(1.0-lambda^2./zetatop.^2); qsquared = w.*conj(w); Cptop = 1.0-qsquared/Q^2; plot(real(ztop(1:nplot-1)), Cptop(1:nplot-1), 'k:') hold on w = (Q*exp(-1i*alpha)-Q*exp(1i*alpha)*R^2./(zetabottom-zetaoff).^2+ ... 1i*Gamma/(2*pi)./(zetabottom-zetaoff))./(1.0-lambda^2./zetabottom.^2); qsquared = w.*conj(w); Cpbottom = 1.0-qsquared/Q^2; plot(real(zbottom(1:nplot-1)), Cpbottom(1:nplot-1), 'k--') Answer: 2

1.5

1

0.5

0 C rp o

-0.5

y -1

-1.5

-2

-2.5

-3 -2

-1.5

-1

-0.5

0 x

0.5

1

1.5

2

11.11 function Exercise11_11 R = 1.0; Q=1.0; nalpha = 21; alphamin = -10; alphamax = 10; alpha = linspace(alphamin, alphamax, nalpha)*pi/180; ksioff = -0.093*R; neta = 5; etaoffarray = linspace(0.0,0.08,neta)*R; maxthick = zeros(neta,1); maxcamber = zeros(neta,1); hold on for in = 1:neta %foil geometry etaoff = etaoffarray(in); thetaTE = -asin(etaoff/R); Gamma = 4*pi*Q*R*sin(alpha-thetaTE); lambda = ksioff+sqrt(R^2-etaoff^2); ksiTE = ksioff+sqrt(R^2-etaoff^2); ksiLE = ksioff-sqrt(R^2-etaoff^2); zetaTE = complex(ksiTE,0); zetaLE = complex(ksiLE,0); zTE = zetaTE+lambda^2/zetaTE; zLE = zetaLE+lambda^2/zetaLE; chord = real(zTE-zLE); thetaLE = pi-thetaTE; nplot = 200; © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a10 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

thetatop = linspace(thetaTE,thetaLE,nplot); thetatemp = zeros(1,nplot); for ijd = 1:nplot thetatemp(ijd) = thetatop(nplot+1-ijd); end thetatop = thetatemp; thetabottom = linspace(thetaLE, 2*pi+thetaTE, nplot); zetaoff = complex(ksioff,etaoff); zetatop = R*exp(1i*thetatop)+zetaoff; zetabottom = R*exp(1i*thetabottom)+zetaoff; ztop = zetatop+lambda^2./zetatop; zbottom = zetabottom+lambda^2./zetabottom; ytop = imag(ztop); xtop = real(ztop); xtopeven = linspace(xtop(1),xtop(nplot),nplot); ytopeven = interp1(xtop,ytop,xtopeven); ybottom = imag(zbottom); xbottom = real(zbottom); xbottomeven = linspace(xbottom(1),xbottom(nplot),nplot); ybottomeven = interp1(xbottom,ybottom,xbottomeven); ythickness = (ytopeven-ybottomeven); ymidline = (ytopeven+ybottomeven)/2; maxthick(in) = max(ythickness)/chord; maxcamber(in) = max(ymidline)/chord; %Lift calculation LiftCoeff = Q*Gamma/(0.5*Q^2*chord); plot(alpha*180/pi, LiftCoeff, 'k-') end axis([alphamin, alphamax, -3, 3]) grid on box on xlabel('\alpha (degrees)') ylabel('C_L') text(0.4, -0.8, ['t/L=' num2str(maxthick(1),4) ... ' b/L=' num2str(maxcamber(1), '%2.1f')]) plot([0.3,-2.8], [-0.8,-0.375], '-k') hold on text(-7.6, 1.2, ['t/L=' num2str(maxthick(5),4) ... ' b/L=' num2str(maxcamber(5),4)]) plot([-7.3,-3.0], [1.05,0.3], '-k') Answer: 3

2

t/L=0.1215 b/L=0.04013 1

L

0

C

t/L=0.1205 b/L=0.0 -1

-2

-3 -10

-8

-6

-4

-2

0

2

4

6

8

10

(degrees)


11.12 function Exercise11_12 n = 0.014; yr = 3.0; g = 9.81; b = 1.5; m = 1.0; ywcrit = fzero(@Channel1, 6, [], b, m, yr); Awcrit = b*ywcrit*(2+ywcrit/(b*m)); Qcrit = Awcrit*sqrt(2*g*(yr-ywcrit)); Pwcrit = 2*(b+ywcrit*sqrt(1+1/m^2)); Rhwcrit = Awcrit/Pwcrit; S0crit = (Qcrit*n/(1.0*Awcrit*Rhwcrit^0.666))^2; nslope = 100; Slope = linspace(0.00001, S0crit-0.0001, nslope); ywflow = zeros(nslope,1); for is = 1:nslope S0 = Slope(is); ywflow(is) = fzero(@Channel2, [ywcrit, yr], [], b, m, S0, n, g, yr); end Awflow = b*ywflow.*(2+ywflow/(b*m)); Qflow = Awflow.*sqrt(2*g*(yr-ywflow)); Smax = 0.0025; plot(Slope*1000, Qflow, 'k') axis([0, Smax*1000, 0, 60]) xlabel('Channel Slope \times 10^{-3}') ylabel('Flow rate (m^3/s)') hold on plot(S0crit*1000, Qcrit, 'ks') hold on plot([Slope(nslope)*1000, S0crit*1000], [Qflow(nslope), Qcrit]) hold on plot([S0crit*1000, Smax*1000], [Qcrit, Qcrit], 'k') hold on text(0.75, 30, ['Critical Flow Rate = ' num2str(Qcrit,5) ' m^3/s']) text(0.75, 27, ['Critical Slope = ' num2str(S0crit,4)]) plot([1.65, (S0crit)*1000], [32, Qcrit], 'k') function y = Channel1(ywc,b,m,yr) y = ywc+0.5*b*ywc*(2+ywc/(b*m))/(2*b+2*ywc/m)-yr; function y = Channel2(y,b,m,S0,n,g,yr) y = (b*y*(2+y/(b*m)))*(((b*y*(2+y/(b*m)))/(2*(b+y*sqrt(1+1/m^2))))^0.6666* ... sqrt(S0))/n-sqrt(2*g*b^2*(yr-y)*(2*y+y^2/(b*m))^2); Answer:


60

50

m ( )e s t 3/a r

40

30

Critical Flow Rate = 45.334 m

w o l F

3

/s

Critical Slope = 0.002227

20

10

0 0

0.5

1 Channel Slope

1.5

2

2.5

10 -3

11.13 function Exercise11_13 g = 9.81; m = 1.0; slope = 0.003; b = 1.5; Q = 45.3336; n = 0.014; y0 = fzero(@flow1, 6, [], Q, n, b, m, slope); A0 = b*y0*(2+y0/(b*m)); P0 = 2*(b+y0*sqrt(1+1/m^2)); yc = fzero(@flow2, 3, [], Q, g, b, m); [y, x] = ode45(@dchannel, [yc, 0.94*y0], 0, [], yc, y0, slope); xmax = 1000; xmin = 0.0; plot(x, y, 'k') axis([0, 1000, 1.5, 2.5]) xlabel('x (m)') ylabel('Water depth, y (m)') hold on plot([xmin,xmax], [y0,y0],' --k') text((xmax-xmin)/2, y0-0.05, 'y_0') plot([xmin,xmax], [yc,yc], '--k') text((xmax-xmin)/2, yc+0.04, 'y_c') function f = flow1(y, Q, n, b, m, slope) f = Q-1.0/n*b*y*(2+y/(b*m))*(b*y*(2+y/(b*m))/(2*(b+y*sqrt(1+1/m^2))))^0.667*sqrt(slope); function f = flow2 (y, Q, g, b, m) f = Q^2-g*(b*y*(2+y/(b*m)))^3/(2*b+2*y/m); function dydx = dchannel(y, x, yc, y0, slope) dydx(1) = 1/slope*(1-(yc/y(1))^3)/(1-(y0/y(1))^(10/3)); Answer:


2.5

2.4

) m (

2.3

y

2.2

, h t p e d

y

c

2.1

y0

2 r e t 1.9 a W 1.8

1.7

1.6

1.5 0

100

200

300

400

500 x (m)

600

700

800

900

1000

11.14 function Exercise1114 omega = 2*pi; nu = 5.0*10^(-6); rho = 1000; dPdz = -1e6; nr = 200; nt = 16; tmax = 2*2*pi/omega; t = linspace(0, tmax, nt); rmax = [0.001, 0.005, 0.01]; for kk = 1:3 subplot(3,1,kk) r = linspace(0, rmax(kk),nr); u = pdepe(1, @pulse14, @pulseic14, @pulsebc14, r, t, [], omega, nu, rho, dPdz); hold on for ijd = 1:nt plot(u(ijd,:)/(dPdz/rho/omega)+(ijd-1),r/rmax(kk),'k') plot(u(ijd,:)/(dPdz/rho/omega)+(ijd-1),-r/rmax(kk),'k') plot([(ijd-1),(ijd-1)],[-1,1],':k') end xlabel('u_z/[\partialP/\partialz/(\rho\omega)]') ylabel('r/R') title(['R = ', num2str(rmax(kk)*1000'), ' mm']) box on end function [c, f, s] = pulse14(r, t, u, DuDr, omega, nu, rho, dPdz) c = 1.0; f = nu*DuDr; s = (dPdz/rho)*sin(omega*t); function u0 = pulseic14(r, omega, nu, rho, dPd) u0 = 0; function [pl, ql, pr, qr] = pulsebc14(rl, ul, rr, ur, t, omega, nu, rho, dPdz) pl = 0; ql = 1; pr = 0; qr = ur; Answer:


R = 1 mm 1 0.5 R / r

0 -0.5 -1

0

5

10

15

10

15

10

15

u /[ P/ z/( )] z

R = 5 mm 1 0.5 R / r

0 -0.5 -1

0

5 u /[ P/ z/( )] z

R = 10 mm 1 0.5 R / r

0 -0.5 -1 0

5 u /[ P/ z/( )] z


Solution to Exercises Chapter 12 12.1 syms eta t th = 1-erf(eta/(2*t))-exp(eta+t^2)*(1-erf(eta/(2*t)+t)); dt = simple(diff(th,t,1)); deta2 = simple(diff(th,eta,2)); Eqn = simple(deta2-dt/(2*t)) BC = simple(subs(th, eta, 0)-1-subs(diff(th,eta,1), eta, 0))

12.2 tau = @(x) (exp(x.^2).*erfc(x)-1/2); tauHalf = fzero(tau, 0.5) Answer: tauHalf = 0.7691

12.3 function Exercise12_3 tspan = [0, 10^5]; y0 = 5; Tamb = 30; h = 2; A = 0.3; Tinf = 30; m = 3.5; c = 4.2; emm = 0.5; sigma = 5.667e-8; [t, T] = ode45(@odefun1, tspan, y0, [], h , A , Tinf, m, c); % Convection only [tr, Tr] = ode45(@odefun2, tspan, y0, [], h , A , Tinf, m, c, emm, sigma); % Convection and radiation plot(t/3600, T, 'k-', tr/3600, Tr, 'k--') xlabel('Time (hr)') ylabel('Temperature (\circC)') legend('Convection only','Convection + radiation', 4) theta = (Tamb-T)/(Tamb-y0); tau = interp1(theta, t, 0.368)/3600 thetar = (Tamb-Tr)/(Tamb-y0); taur = interp1(thetar, tr, 0.368)/3600 function dy = odefun1(t, T, h , A , Tinf, m, c) dy = -h*A*(T-Tinf)/(m*c*1000);

% Convection only

function dy = odefun2(t, T, h , A , Tinf, m, c, emm, sigma) % Convection + radiation dy = -(h*A*(T-Tinf)+A*emm*sigma*((T+273.15)^4-(Tinf+273.15)^4))/(m*c*1000); Answers: tau = 6.8121

taur = 2.7700


30

25

( e r )u C t

20

°

a r e p m e T

15

10

Convection only Convection + radiation 5 0

5

10

15 Time (hr)

20

25

30

12.4 function Exercise12_4 L = 0.1; k = 200; h = 100; P = 0.1; Ac = 0.001; t = L/100; Msqu = 2 *h*L^2/(k*t)*sqrt(1+(t/(2*L))^2); solinit = bvpinit(linspace(0, 0.9999, 8), [0,0]); sol = bvp4c(@Fin, @Finbc, solinit, [], Msqu); x = linspace(0, 0.99); y = deval(sol, x); etaf = -1/Msqu*y(2,1) M = sqrt(Msqu); etafAnal = 1/M*besseli(1, 2*M)/besseli(0, 2*M) subplot(2, 1, 1) plot(x, y(1,:), 'k-') title('1-D Triangular Fin') xlabel('\eta') ylabel('\theta') subplot(2, 1, 2) q = -k*y(2,:); plot(x, q(:), 'k-') xlabel('x (m)') ylabel('q" (W/m^2)') function dydx = Fin(x, y, Msqu) dydx = [y(2); (y(2)+Msqu*y(1))/(1-x)]; function res = Finbc(ya, yb, Msqu) res = [ya(1)-1; yb(2)]; Answers: etaf = 0.2900

etafAnal = 0.2900


1-D Triangular Fin 1 0.8 0.6

0.4 0.2 0

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.6

0.7

0.8

0.9

1

600 500 m 400 / 2)W 300 ( " q

200 100 0 0

0.1

0.2

0.3

0.4

0.5 x (m)

12.5 h = 5; k = 0.1; ri = 0.01; L = 1; Ti = 100; Tinf = 20; r0 = linspace(ri, 5*k/h, 100); q = 2*pi*L*(Ti-Tinf)./(1/k.*log(r0./ri)+1./(r0*h)); plot(r0, q, 'k-') xlabel('r_0 (m)') ylabel('Heat transfer rate (W)') Answer: 30

) W ( e t a r r e f s n a r t t a e H

29

28

27

26

25

24

23

22

21

20 0.01

0.02

0.03

0.04

0.05

0.06 r0 (m)

0.07

0.08

0.09

0.1

0.11

12.6 function Exercise12_6 Bi = .5; Nroot = 15; CylinderRoots = inline('x.*besselj(1, x)-Bi*besselj(0, x)', 'x', 'Bi'); d = FindZeros(CylinderRoots, Nroot, linspace(0, 50, 200), Bi); Cn = 2*besselj(1, d)./(d.*(besselj(0, d).^2+besselj(1, d).^2)); tau = fzero(@Flip, 0.5, [], d, Cn) function gg = Flip(t, d, Cn) gg = 0.5-sum(Cn.*exp(-t*d.^2).*besselj(0,d)); function Rt = FindZeros(FunName, Nroot, x, w) % See Section 5.5.1 f = feval(FunName, x, w); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a3retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

indx = find(f(1:end-1).*f(2:end)<0); L = length(indx); if L
12.7 Bi = 0.5; Nroot = 20; ntime = 30; timeMax = 0.5; nradial = 25; tau = linspace(0, timeMax, ntime); xi = linspace(0.001, 1, nradial); SphereRoots = inline('sin(x)-Bi*sin(x)-x.*cos(x)', 'x', 'Bi'); zet = FindZeros(SphereRoots, Nroot, linspace(0.1, 70, 120), Bi); % See Exercise 12.6 Cn = 4.*(sin(zet)-zet.*cos(zet))./(2*zet-sin(2*zet)); [dum, Cnn] = meshgrid(xi, Cn); theta = exp(-zet.^2*tau)'*(Cnn.*sin(zet*xi)./(zet*xi)); mesh(xi, tau, theta) colormap([0 0 0]) xlabel('\zeta') ylabel('\tau') zlabel('\theta') view(49.5,-34) Answer:

1.1 1 0.9 0.8

0.7 0.6 0.5 1

0.4 0

0.8 0.1

0.6 0.2 0.4

0.3 0.2

0.4

0.5

0

12.8 function Exercise12_8 CylinderRoots = inline('x.*besselj(1, x)-Bi*besselj(0, x)', 'x', 'Bi'); Bi = .5; Nroots = 15; tau = [0, 0.001, 0.005]; xi = linspace(0, 1, 100); r = FindZeros(CylinderRoots, Nroots, linspace(0.1, 60, 100), Bi); Cnn = Cn(r); co = char('k-', 'k--', 'k:.'); T = zeros(length(xi),1); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a4retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

for k = 1:length(tau) for n = 1:length(xi) T(n) = sum(Cnn.*besselj(0, r*xi(n)).*exp(-tau(k)*r.^2)); end plot(xi, T, co(k,:)) hold on end legend(['\tau = ' num2str(tau(1))], ['\tau = ' num2str(tau(2))], ['\tau = ' num2str(tau(3))], 'Location','SouthWest') ylim([0.98, 1.005]) function res = Cn(r) res = 2*besselj(1, r)./(r.*(besselj(0, r).^2+besselj(1, r).^2)); Answer: 1.005

1

0.995

), ( 0.99

0.985

=0 = 0.001 = 0.005 0.98 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

12.9 function Exercise12_9 CylinderRoots = inline('x.*besselj(1, x)-Bi*besselj(0, x)', 'x', 'Bi'); Bi = .5; Nroots = 15; r = FindZeros(CylinderRoots, Nroots, linspace(0.1, 60, 100), Bi); tau = fzero(@Flux, 1, [], r) function res = Flux(t, r) res = 1/exp(1)-sum(2*besselj(1, r)./(r.*(besselj(0, r).^2+besselj(1, r).^2)).*exp(-t*r.^2)); Answer: tau = 1.2521

12.10 function Exercise12_10 L = 0.1; flux = 12.5; TL = 300; k0 = 0.048; ks = 0.00032; solinit = bvpinit(linspace(0, L, 10), [100]); sol = bvp4c(@heatVark, @heatbc, solinit, [], flux, k0, ks, TL); x = linspace(0, L, 10); T = deval(sol, x); AnalSol = inline('k0*(T-TL)-ks/2*(T.^2-TL^2)-q*(x-L)', 'T', 'k0', 'ks', 'TL', 'x', 'q', 'L'); T0anal = fzero(AnalSol, [250 350], [], k0, ks, TL, 0, flux, L); k0 = -k0+ks*mean(T(1,:)); ks = 0.0; solavg = bvp4c(@heatVark, @heatbc, solinit, [], flux, k0, ks, TL); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a5retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Tavg = deval(solavg, x); T0analA = fzero(AnalSol, [250 350], [], k0, ks, TL, 0, flux, L); disp(['When k = k(T) and T determined from bvp4c, T(0) = ' num2str(T(1,1)) ' K']) disp(['When k = k(T) and T determined from analytical solution, T(0) = ' num2str(T0anal) ' K']) disp(['When k = k(Tavg) and T determined from bvp4c, T(0) = ' num2str(Tavg(1,1)) ' K']) disp(['When k = k(Tavg) and T determined from analytical solution, T(0) = ' num2str(T0analA) ' K']) function dydx = heatVark(x, y, flux, k0, ks, T2) dydx = -flux/(-k0+ks*y(1)); function res = heatbc(ya, yb, flux, k0, ks, T2) res = [yb(1)-T2]; Answers: When k = k(T) and T determined from bvp4c, T(0) = 324.1048 K When k = k(T) and T determined from analytical solution, T(0) = 324.1049 K When k = k(Tavg) and T determined from bvp4c, T(0) = 275.9346 K When k = k(Tavg) and T determined from analytical solution, T(0) = 275.9346 K

12.11 Tb = 80; Tinf = 100; h = 25; d = 0.005; P = pi*d; Ac = pi*d^2/4; Np = 20; k = [20, 100, 200, 300, 400]; L = linspace(0.005, 0.1, Np); col = char('k-', 'b-', 'r-', 'g-', 'c-'); for i = 1:length(k) y = (Tb-Tinf)./cosh(sqrt(h*P/(k(i)*Ac))*L); plot(L, y, col(i,:)) hold on end % plot(L, y(1,:), L, y(2,:), L, y(3,:), L, y(4,:), L, y(5,:)) legend('k = 20 W/m-K', num2str(k(2)), num2str(k(3)), num2str(k(4)), num2str(k(5)), 'Location', 'NorthWest') xlabel('L (m)') ylabel('Temperature Error (K)') Answer: 0

) K ( r o r r E e r u t a r e p m e T

-2

k = 20 W/m-K 100 200 300

-4

400

-6

-8

-10

-12

-14

-16

-18

-20 0

0.01

0.02

0.03

0.04

0.05 L (m)

0.06

0.07

0.08

0.09

0.1

12.12 function Exercise12_12 Tw = 40; qw = 10; Re = 40; Pr = 5; © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a6retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

R = 0.01; L = 0.75; k = 0.6; Rt = 101; zt = 50; dxi = 1/(Rt-1); m = 1; xi = linspace(0, 1, Rt); zeta = linspace(0, 1, zt); TT = pdepe(m, @pdex1pde, @pdex1ic, @pdex1bcT, xi, zeta, [], Tw, qw, Re, Pr, R, L, k); TF = pdepe(m, @pdex1pde, @pdex1ic, @pdex1bcF, xi, zeta, [], Tw, qw, Re, Pr, R, L, k); TmT = zeros(zt,1); NuT = TmT; TmF = TmT; NuF = TmT; for i = 1:zt TmT(i) = 2*trapz(xi, xi.*TT(i,:)); dThdxiT = (TT(i,Rt)-TT(i,Rt-1))/(dxi*(TmT(i)-TT(i,Rt))); NuT(i) = -2*dThdxiT; TmF(i) = 2*trapz(xi,xi.*TF(i,:)); dThdxiF = (TF(i,Rt)-TF(i,Rt-1))/(dxi*(TmF(i)-TF(i,Rt))); NuF(i) = -2*dThdxiF; end ThT =( TT(end,:)-ones(1,Rt)*Tw)/(TmT(end)-Tw); ThF = (TF(end,:)-ones(1,Rt)*TF(end,Rt))/(TmF(end)-TF(end,Rt)); figure(1) plot(xi, ThT, '-', xi, ThF, '--') xlabel('\xi') ylabel('\theta') legend('Constant wall temperature', 'Constant wall heat flux') figure(2) plot(zeta, NuT, '-', zeta, NuF, '--') xlabel('\zeta') ylabel('Nu') ylim([4 12]) legend('Constant wall temperature', 'Constant wall heat flux') figure(3) surf(xi(1:3:Rt), zeta(1:3:zt), TT(1:3:zt,1:3:Rt)) zlabel('\theta') xlabel('Radial Distance \xi') ylabel('Axial distance \zeta') function [c, f, s] = pdex1pde(xi, zeta, T, DTDxi, Tw, qw, Re, Pr, R, L, k) c = Re*Pr*R/L; f = DTDxi; s = 0; function T0 = pdex1ic(xi, Tw, qw, Re, Pr, R, L, k) T0 = 20; %deg C function [pl,ql,pr,qr] = pdex1bcT(xil, Tl, xir, Tr, zeta, Tw, qw, Re, Pr, R, L, k) pl = 0; ql = 1; pr = Tr-Tw; qr = 0; function [pl, ql, pr, qr] = pdex1bcF(xil, Tl, xir, Tr, zeta, Tw, qw, Re, Pr, R, L, k) pl = 0; ql = 1; pr = -qw; qr = k/R; Answers:


2.5 12

Constant wall temperature

Constant wall temperature

Constant wall heat flux

Constant wall heat flux 11

2 10

9

1.5

u N

8

1 7

6

0.5 5

0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

4

1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

40

35

30

25

1

20 1

0.8 0.8

0.6

0.6 0.4

0.4 0.2

0.2 Axial distance

0

0

Radial Distance

12.13 function Exercise12_13 Pr = [0.07, 0.7, 7.0]; etaMax = [15, 8, 8]; xm = [15, 5, 5]; disp([' Pr d_u d_T d_u/d_T Pr^(1/3)']) for k = 1:3 figure(k) solinit = bvpinit(linspace(0, etaMax(k), 8), @BlasiusTguess); sol = bvp4c(@BlasiusT, @BlasiusTbc, solinit, [], Pr(k)); eta = linspace(0, etaMax(k), 1000); y = deval(sol, eta); subplot(2, 1, 1) plot(eta, y(1,:), '-.k', eta, y(2,:), '-k', eta,y(3,:), '--k') xlabel('\eta') ylabel('y_i (i = 1, 2, 3)') legend('Stream function', 'Velocity', 'Shear', 'Location', 'NorthWest') title(['Pr = ' num2str(Pr(k))]) axis([0, xm(k), 0, 2]) subplot(2, 1, 2) plot(eta, y(4,:), '-k', eta, y(5,:), '--k') axis([0, xm(k), 0, 2]) legend('Temperature', 'Heat flux', 'Location', 'NorthWest') xlabel('\eta') ylabel('y_i (i = 4, 5') © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a8retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

eta_ubl = eta(find(y(2,:)>0.99,1,'first')); % Velocity boundary layer thickness eta_Tbl = eta(find(y(4,:)>0.99, 1, 'first')); % Temperature boundary layer thick disp([Pr(k), eta_ubl, eta_Tbl, eta_ubl/eta_Tbl, Pr(k)^(1/3)]) end function F = BlasiusT(eta, y, Pr) F=[y(2); y(3); -0.5*y(1)*y(3); y(5); -Pr*0.5*y(1)*y(5)]; function res=BlasiusTbc(ya, yb, Pr) res = [ya(1); ya(2); ya(4); yb(2)-1; yb(4)-1]; function y = BlasiusTguess(eta) y = [eta, eta^0.5, 5-eta, eta/10, 0.5*(1-eta/10)]; Answers:

) 3

Pr d_u d_T d_u/d_T Pr^(1/3) 0.0700 4.9249 13.6637 0.3604 0.4121 0.7000 4.9169 5.6296 0.8734 0.8879 7.0000 4.9169 2.4505 2.0065 1.9129 Pr = 0.07

) 3

2

, 2

Stream function Velocity Shear

1.5

, 1 =

Velocity 1.5

=

1 i

Stream function

, 1

i ( y

Pr = 0.7 2

, 2

Shear

1

i (

0.5

0

y

0

5

10

i

0.5

0

15

0

0.5

1

1.5

2

3

3.5

4

4.5

5

3

3.5

4

4.5

5

2

2

5

Temperature

, 4

Heat flux

1.5

5

Temperature

, 4

=

Heat flux

1.5

=

1

i ( y

2.5

1

i (

i

y

0.5

0 0

5

10

15

i

0.5

0

) 3

0

0.5

1

1.5

2

2.5

Pr = 7 2

, 2

Stream function Velocity Shear

1.5

, 1 =

1

i ( y

i

0.5

0

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

3

3.5

4

4.5

5

2 5

Temperature

, 4

Heat flux

1.5

= 1

i ( y

i

0.5

0 0

0.5

1

1.5

2

2.5

12.14 function Exercise12_14 Pr = 0.7; etaMax = 8; xm = 5; solinit = bvpinit(linspace(0, etaMax, 8), @BlasiusTguess); sol = bvp4c(@BlasiusT, @BlasiusTbc, solinit, [], Pr); eta = linspace(0, etaMax); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a9retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

y = deval(sol, eta); uinf = 1; nu = 1.5e-5; Rec = 5e5; xc = Rec*nu/uinf; yy = 10*xc/sqrt(Rec); x = linspace(.001, xc, 50); yx = linspace(0.001, yy, 30); [X, Y] = meshgrid(x, yx); et = Y.*(uinf./sqrt(nu*X)); T = interp1(eta, y(4,:), et); [c, h] =contour(X, Y, T); clabel(c,h) xlabel('x (m)') ylabel('y (m)') title('Isotherms') ylim([0 .07]) function F = BlasiusT(eta, y, Pr) F=[y(2); y(3); -0.5*y(1)*y(3); y(5); -Pr*0.5*y(1)*y(5)]; function res=BlasiusTbc(ya, yb, Pr) res = [ya(1); ya(2); ya(4); yb(2)-1; yb(4)-1]; function y = BlasiusTguess(eta) y = [eta, eta^0.5, 5-eta, eta/10, 0.5*(1-eta/10)]; Answer: Isotherms 0.05

0.045

0.04 0.9 0.035

) m (

0.03

0.9

0.025

0.8

0.8

0.7

y 0.02

0.015

0.01

0.7

0. 8

0.6

0.5

0.7

0.5

0.4

0.6

0.4

0.3

0.5 0.4 0.3 0.2 0.1

0.005

0.6

0. 9

0.3 0.2 0.2 0.1

0.1

0 1

2

3

4

5

6

7

x (m)

12.15 function Exercise12.15 Pr = 0.7; etaMax = 8; xm = 5; solinit = bvpinit(linspace(0,etaMax,8), @BlasiusTguess); sol = bvp4c(@BlasiusT, @BlasiusTbc, solinit, [], Pr); eta = linspace(0, etaMax); y = deval(sol, eta); uinf = 1; nu = 1.5e-5; Rec = 5e5; xc = Rec*nu/uinf; yy = 10*xc/sqrt(Rec); x = linspace(.001, xc, 50); yx = linspace(0.001, yy, 30); [X, Y] = meshgrid(x, yx); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a10 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

et = Y.*sqrt(uinf./(nu*X)); SF = interp1(eta, y(1,:), et).*uinf.*sqrt(nu*X/uinf); [c, h] =contour(X, Y, SF, 10); clabel(c, h) xlabel('x (m)') ylabel('y (m)') title('Isotherms') ylim([0.01, 0.09]) function F = BlasiusT(eta, y, Pr) F=[y(2); y(3); -0.5*y(1)*y(3); y(5); -Pr*0.5*y(1)*y(5)]; function res=BlasiusTbc(ya, yb, Pr) res = [ya(1); ya(2); ya(4); yb(2)-1; yb(4)-1]; function y = BlasiusTguess(eta) y = [eta, eta^0.5, 5-eta, eta/10, 0.5*(1-eta/10)]; Answer: Isotherms 0.09

0.08

0.054066

0.07

0.048061 0.06 ) m (

0.042055 0.042055 0.03605

0.05

0.03605 0.030044

y 0.030044

0.024039

0.04 0.024039

0.018033

0.024039

0.018033

0.03

0.012028

0.018033 0.012028 0.012028

0.02

0.0060227 0.0060227

0.0060227 0.01 1

2

3

4

5

6

7

x (m)

12.16 function Exercise12_16 Pr = [0.07, 0.7, 7]; etaMax = [11, 8, 8]; xm = [10, 5, 5]; ym = [2, 0.8, 0.5]; for k = 1:3 figure(k) solinit = bvpinit(linspace(0, etaMax(k), 5), @NaturalConvGuess); sol = bvp4c(@NaturalConv, @NaturalConvbc, solinit, [], Pr(k)); eta = linspace(0, etaMax(k), 300); y = deval(sol, eta); subplot(2, 1, 1) plot(eta, y(1,:), '-.k', eta,y(2,:), '-k', eta, y(3,:), '--k') legend('Stream function', 'Velocity', 'Shear') axis([0, xm(k), -0.2, ym(k)]) xlabel('\eta') ylabel('y_i (i = 1, 2, 3)') title(['Pr = ' num2str(Pr(k))]) subplot(2, 1, 2) plot(eta, y(4,:), '-k', eta, y(5,:), '--k') legend('Temperature', 'Heat flux') axis([0 xm(k) -1.2 1]) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a11 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

xlabel('\eta') ylabel('y_i (i = 4, 5)') [etaMaxv, Vmax] = fminbnd(@Vel, 0,5, [], eta, y(2,:)); disp(['Pr = ' num2str(Pr(k)) ' etaMax = ' num2str(etaMaxv) ' Vmax =' num2str(-Vmax)]) end function ff = NaturalConv(eta,y,Pr) ff = [y(2); y(3); -3*y(1)*y(3)+2*y(2)^2-y(4); y(5); -3*Pr*y(1)*y(5)]; function res = NaturalConvbc(ya,yb,Pr) res=[ya(1); ya(2); ya(4)-1; yb(2); yb(4)]; function y = NaturalConvGuess(eta) y = [eta, eta-eta^2/10, 1-eta, 1-eta/10, -.5*(1-eta/10)]; function V = Vel(x, eta, y2) V=-spline(eta, y2, x); Answers: Pr = 0.07 etaMax = 1.2307 Vmax =0.4551 Pr = 0.7 etaMax = 0.96452 Vmax =0.27838 Pr = 7 etaMax = 0.72832 Vmax =0.13138 ) 3

) 3

Pr = 0.07 2

, 2

Stream function Velocity

0.6

, 1

Shear 0.4

=

=

1

i ( y

Velocity Shear

1.5

, 1

Pr = 0.7 0.8

, 2

Stream function

i

i (

0.5

y

0.2 i

0 0 0

1

2

3

4

5

6

7

8

9

-0.2

10

0

0.5

1

1.5

2

Heat flux

0.5

=

4

4.5

5

Temperature

, 4

Heat flux

0.5

=

0

i ( y

3.5

1

) 5

Temperature

, 4

3

1

) 5

2.5

0

i ( i

-0.5

y

-1

i

-0.5

-1 0

1

2

3

4

5

6

7

8

9

10

0

0.5

1

1.5

2

) 3

0.5

, 2

0.4

, 1

0.3

3

3.5

4

4.5

5

Pr = 7 Stream function Velocity Shear

0.2

=

0.1

i ( y

2.5

i

0 -0.1 -0.2

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

1

) 5

Temperature

, 4

Heat flux

0.5

=

0

i ( y

i

-0.5

-1 0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

12.17 function Exercise12_17 Pr = [0.07, 0.7, 7]; etaMax = [11, 8, 8]; xm = [10, 5, 5]; ym = [2, 0.8, 0.5]; © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a12 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

for k = 1:3 solinit = bvpinit(linspace(0, etaMax(k), 5), @NaturalConvGuess); sol = bvp4c(@NaturalConv,@NaturalConvbc, solinit, [], Pr(k)); eta = linspace(0, etaMax(k), 300); y = deval(sol, eta); [etaMaxv Vmax] = fminbnd(@Vel, 0, 5, [], eta, y(2,:)); Vmax = -Vmax; delV = spline(y(2, 2:end)/Vmax, eta(2:end), 0.01); delT = spline(y(4, 2:end), eta(2:end), 0.01); disp(['Pr = ' num2str(Pr(k)) ' deltaT = ' num2str(delT) ' deltaV = ' num2str(delV)]) end function ff = NaturalConv(eta, y, Pr) ff = [y(2); y(3); -3*y(1)*y(3)+2*y(2)^2-y(4); y(5); -3*Pr*y(1)*y(5)]; function res = NaturalConvbc(ya, yb, Pr) res=[ya(1); ya(2); ya(4)-1; yb(2); yb(4)]; function y = NaturalConvGuess(eta) y = [eta, eta-eta^2/10, 1-eta, 1-eta/10, -.5*(1-eta/10)]; function V = Vel(x, eta, y2) V=-spline(eta, y2, x); Answer: Pr = 0.07 deltaT = 10.2661 deltaV = 9.4841 Pr = 0.7 deltaT = 4.4729 deltaV = 5.6483 Pr = 7 deltaT = 1.7772 deltaV = 6.4123

12.18 function Exercise12_18 Pr = 0.7; etaMax = 8; xm = 5 ;ym = 0.8; g = 9.98; Tinf = 300; Ts = 400; nu = 1.6e-5; beta = 1/Tinf; x = 0.1:0.1:0.5; solinit = bvpinit(linspace(0,etaMax,5), @NaturalConvGuess); sol = bvp4c(@NaturalConv, @NaturalConvbc, solinit, [], Pr); eta = linspace(0, etaMax, 100); y = deval(sol, eta); y1 = linspace(1e-5, 0.01, 100); co = g*beta*(Ts-Tinf)/nu^2; u1 = zeros(length(x), length(y1)); v1 = u1; for i = 1:length(x) Gr = co*x(i)^3; eta1 = y1/x(i)*(Gr/4)^0.25; u1(i,:) = spline(eta, y(2,:), eta1)*sqrt(Gr)*2*nu/x(i); fp = spline(eta, y(2,:), eta1); f = spline(eta, y(1,:), eta1); v1(i,:) = nu./x(i).*(Gr/4)^0.25.*(eta1.*fp-3*f); end subplot(2, 1, 1) plot(y1, u1(1,:), y1,u1(2,:), y1, u1(3,:), y1, u1(4,:), y1, u1(5,:)) xlabel('y (m)') ylabel('u (m/s)') subplot(2, 1, 2) plot(y1, v1(1,:), y1, v1(2,:), y1, v1(3,:), y1, v1(4,:), y1, v1(5,:)) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a13 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

xlabel('y (m)') ylabel('v (m/s)') legend('x = 0.1 m', 'x = 0.2', 'x = 0.3', 'x = 0.4.', 'x = 0.5', 'Location', 'SouthWest') function ff = NaturalConv(eta, y, Pr) ff = [y(2); y(3); -3*y(1)*y(3)+2*y(2)^2-y(4); y(5); -3*Pr*y(1)*y(5)]; function res = NaturalConvbc(ya, yb, Pr) res=[ya(1); ya(2); ya(4)-1; yb(2); yb(4)]; function y = NaturalConvGuess(eta) y = [eta, eta-eta^2/10, 1-eta, 1-eta/10, -.5*(1-eta/10)]; function V = Vel(x, eta, y2) V=-spline(eta, y2, x); Answer: 0.8

) s / m (

0.6

0.4

u 0.2

0

0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

0.006

0.007

0.008

0.009

0.01

y (m)

0 -0.002 ) s / m ( v

-0.004 -0.006 x = 0.1 m x = 0.2

-0.008

x = 0.3 -0.01

x = 0.4. x = 0.5

-0.012 0

0.001

0.002

0.003

0.004

0.005 y (m)

12.19 function Exercise12_19 Pr = 0.7; etaMax = 8; xm = 5 ;ym = 0.8; solinit = bvpinit(linspace(0,etaMax,5), @NaturalConvGuess); sol = bvp4c(@NaturalConv, @NaturalConvbc, solinit, [], Pr); eta = linspace(0, etaMax, 300); y = deval(sol, eta); g = 9.98; Tinf = 300; Ts = 400; nu = 1.6e-5; beta = 1/Tinf; co =g*beta*(Ts-Tinf)/nu^2; y1 = linspace(.001, 0.035, 30); x1 = linspace(.001, 1, 50); [X Y] = meshgrid(x1, y1); Gr = co*X.^3; eta1 = Y./X.*(Gr/4).^0.25; f = spline(eta, y(1,:), eta1); SF = 4*nu*f.*(Gr/4).^0.25; [c, h] = contour(X, Y, SF); clabel(c, h) xlabel('x (m)') ylabel('y (m)')


function ff = NaturalConv(eta, y, Pr) ff = [y(2); y(3); -3*y(1)*y(3)+2*y(2)^2-y(4); y(5); -3*Pr*y(1)*y(5)]; function res = NaturalConvbc(ya, yb, Pr) res=[ya(1); ya(2); ya(4)-1; yb(2); yb(4)]; function y = NaturalConvGuess(eta) y = [eta, eta-eta^2/10, 1-eta, 1-eta/10, -.5*(1-eta/10)]; function V = Vel(x, eta, y2) V=-spline(eta, y2, x); Answer: 0.035

0.03

0 . 0 0 1

0 . 0 0 2

0 . 0 0 3

0 . 0 0 4

0 . 0 0 5

0 . 0 0 6

0 . 0 0 1

0 . 0 0 2

0 . 0 0 3

0 . 0 0 4

0 . 0 0 5

0 . 0 0 6

7 0 0 . 0

8 0 0 . 0

0 . 0 0 9

0.025

0.02

) m ( y

0 . 0 0 7

0 . 0 0

0 . 0 0 8

0.015

0.01

0.005

0 .0 0 1

0. 0 02

0. 0 04

0. 0 03

0.004

0.2

0.005

0.003 0.002

0.002

0.001 0.1

0. 0 07

0. 0 06

0. 0 05

0.3

0.4

0.001 0.5 x (m)

0.6

0.7

0.8

0.9

1

12.20 function Exercise12_20 F12 = dblquad(@kernel2, -1, 0, -1, 0, [], [], 5)/pi function f = kernel2(x, y, dist) L = length(x); S = [x; repmat(y, 1, L); dist*ones(1, L)]; nd = repmat([0 1 0]', 1, L); nf = repmat([0 0 -1]', 1, L); f = dot(nd, S).*dot(nf, S)./dot(S, S).^2; Answer: F12 = 0.0012

12.21 function Exercise12_21 sigma = 5.667e-8; T1 = 100+273.15; T2 = 20+273.15; q1 = sigma*(T1^4-T2^4) Tm1 = ((T1^4+T2^4)/2)^0.25 % Algebraic solution to one shield problem qm1 = sigma*(T1^4-Tm1^4) % Flux for one shield from algebraic solution opt = optimset('display', 'off'); x1 = fsolve(@OneShield, [q1 (T1+T2)/2], opt, T1, T2, sigma) x2 = fsolve(@TwoShield, [q1, (T1+T2)/2, (T1+T2)/2-10], opt, T1, T2, sigma)


function F = OneShield(x, T1, T2, sigma) F = [x(1)-sigma*(T1^4-x(2)^4); x(1)-sigma*(x(2)^4-T2^4)]; function F = TwoShield(x, T1, T2, sigma) F = [x(1)-sigma*(T1^4-x(2)^4); x(1)-sigma*(x(2)^4-x(3)^4); x(1)-sigma*(x(3)^4-T2^4)]; Answers: q1 = 680.2032 Tm1 = 340.1477 qm1 = 340.1016 x1 = 340.1016 340.1477 x2 = 226.7344 352.1997 326.6584

12.22 function Exercise12_22 sigma = 5.667e-8; llow = 0.5; lhigh = 300; T = [300, 400, 500]; C1 = 3.742e8; C2 = 1.439e4; format short e disp([' Temp E(int) E(exact)']) for i = 1:3 Etot = quadl(@intg, llow, lhigh, [], [], T(i), C1, C2); Etotchk = sigma*T(i)^4; disp([T(i), Etot, Etotchk]) end format short function E = intg(lambda, T, C1, C2) E = C1./(lambda.^5.*(exp(C2./(lambda.*T))-1)); Answers:

Temp E(int) E(exact) 3.0000e+002 4.5895e+002 4.5903e+002 4.0000e+002 1.4507e+003 1.4508e+003 5.0000e+002 3.5418e+003 3.5419e+003

12.23 sigma = 5.6693e-8; N = 5; A = [3, 5, 3, 2.5, 2.5]; epsilon = [0.7, 0.3, 0.85, 0.45, 0.45]; T = [550 700 650 600 600]; F = -[0, 0.3615, 0.277, 0.0957, 0.2658;... 0.2169, 0, 0.2169, 0.2831, 0.2831;... 0.277, 0.3615, 0, 0.2658, 0.0957;... 0.1148, 0.5662, 0.3190, 0, 0;... 0.3190, 0.5662, 0.1148, 0, 0]; Q = [0, 0, 0, 0, 0]; c = [0, 0, 0, 0, 0]; b = sigma*epsilon./(1-epsilon).*(1-c).*T.^4+c.*Q./A; d = (1-c).*1./(1-epsilon)+c; F = F+diag(d); q0 = F\b'; Q = A.*epsilon./(1-epsilon).*(1-c).*(sigma*T.^4-q0'); q = Q./A; disp('Enclosure Results') disp('Heat(W), Temp(K), Heat Flux(W/m^2)') format short g disp([Q' T' (Q./A)']) format short © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a16 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Answers: Enclosure Results Heat(W) Temp(K) -8559.6 550 8063.7 700 4450.9 650 -2373.5 600 -1581.5 600

Heat Flux(W/m^2) -2853.2 1612.7 1483.6 -949.39 -632.59


Solution to Exercises Chapter 13 13.1 function Exercise13_1 L1 = 0.5; L2 = 0.3; P1 = 3; P2 = 1; P = 3; x1 = 0.0:0.1:pi/2; x2 = 0.0:0.1:pi/2; [x1, x2] = meshgrid(x1,x2); PE= -P1*L1*sin(x1)-P2*L2*sin(x2)+2*P*(L1*(1-cos(x1))+L2*(1-cos(x2))); subplot(1,2,1) c = contour(x1, x2, PE, [-.4:.1:-.1 .5:1:20], 'k'); clabel(c) axis([0, 1.5, 0, 1.5]); xlabel('\alpha') ylabel('\beta') subplot(1,2,2) meshc(x1, x2, PE) axis([0, 1.5, 0, 1.5, -0.5, 2.5]); colormap([0 0 0]) xlabel('\alpha') ylabel('\beta') zlabel('PE'); view (150,10); options = optimset('LargeScale', 'off'); [xopt, fopt] = fminunc(@func13_1, [1,1], options, L1, L2, P1, P2, P); disp(['alpha = ' num2str(xopt(1)) ' beta = ' num2str(xopt(2)) ' PE = ' num2str(fopt) ]) function PE = func13_1(x, L1, L2, P1, P2, P) PE=-P1*L1*sin(x(1))-P2*L2*sin(x(2))+2*P*(L1*(1-cos(x(1)))+L2*(1-cos(x(2)))); Answers: alpha = 0.46365 beta = 0.16515 PE = -0.37893 1.5

2. 1.5

2.5

2 0.5 1.5

1

1 E P

0.5 0.5 0

-0.1

-0.2 -0.5 0

-0.3

0.5 1

0 0

0.5

1

0.5 1.5

1.5

1.5

0

1

13.2 function Exercise13_2 A = 9.3; n = 15; theta = linspace(0.5, 2, n); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a1retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

d = linspace(1.5, 4, n); [dd pphi] = meshgrid(d, theta); f = 1./(A./dd-dd.*cot(pphi)+2*dd./sin(pphi)); subplot(1,2,1) h = contour(dd, pphi*180/pi, f, (0.09:0.005:0.125), 'k'); clabel(h) xlabel('h') ylabel('\theta (\circ)') subplot(1,2,2); mesh(dd, pphi*180/pi, f); colormap([0 0 0]) xlabel('h') ylabel('\theta (\circ)') zlabel('1/p') options = optimset('LargeScale', 'off'); [xopt, fopt] = fminsearch(@flowrate, [1 1], options, A); disp(['h = ' num2str(xopt(1)) ' theta = ' num2str(xopt(2)*180/pi) ' deg']) function f = flowrate(x, A) f = -1/(A/x(1)-x(1)*cot(x(2))+2*x(1)/sin(x(2))); Answers: h = 2.3172 theta = 60 deg 0.1 110

0.09 0.1

0.095

0.105

100

0.13

0.12

90

0.11 80

°)(

p / 1

70

0.1

0.09

60

0.115 0.11

50

0.12

150 4

100

40

0.105 0.1

30 0.105 1.5

0.08

3.5 3

0.095

50

2.5

0.09 2

2.5

3

3.5

4

(°)

2 0

1.5

h

h

13.3 function Exercise13_3 k1 = 8; k2 = 8; L1 = 10; L2 = 10; F = 5; [x1, x2] = meshgrid(linspace(-5, 15, 15), linspace(-5, 15, 15)); subplot(1,2,1); h = contour(x1, x2, spring([x1, x2], k1, k2, L1, L2, F), [-25:10:25 100:50:500],'k'); xlabel('x_1') ylabel('x_2') axis([-5 15 -5 10]) clabel(h) subplot(1,2,2) mesh(x1, x2, spring([x1, x2], k1, k2, L1, L2, F)) colormap([0 0 0]) axis([-5, 15, -5, 10, -100, 1000]) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a2retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

view(50,15) xlabel('x_1') ylabel('x_2') zlabel('PE') [x, fx] = fminsearch(@spring, [1,1], [], k1, k2, L1, L2, F); disp(['x1 = ' num2str(x(1)) ' x2 = ' num2str(x(2)) ' PE = ' num2str(fx)]) function PE = spring(x, k1, k2, L1, L2, F) [r, c]= size(x); x1 = x(1:r,1:c/2); x2 = x(1:r,(c/2+1):c); PE1=1/2*k1*(sqrt(x1.^2+(L1-x2).^2)-L1).^2; PE2=1/2*k2*(sqrt(x1.^2+(L2+x2).^2)-L2).^2; PE=(PE1+PE2-F*x1); Answers: x1 = 4.1289 x2 = 3.1281e-005

PE = -15.2802

10

500 450

1000 900 350 800

250 700

5

600 100 500

x

2

E P

25 5

300 200

-5

0

100

15 30040 4

0

5

-100 -5

200 150

10

0

150 -5 -5

400

5

5 200 0

0

10 5 x1

10

15

15 x1

-5

x

2

13.4 function Exercise13_4 n = 0.17; a = 0.77; b = 0.37; Kt = 200; fmax = 2; tc = 7; taux = 3; L = 500; D=100; d = 0.3; [v, f] = meshgrid(linspace(20, 150, 10), linspace(0.5, 10, 10)); F1 = MachineObj([v, f], d, a, b, n, Kt, tc, taux, L, D); subplot(1,2,1); g = contour(v, f, F1, [0:0.1:3.94, 4:0.5:5, 10:15:50]); clabel(g) hold on; plot([20, 200], [fmax, fmax], 'r') xlabel('V') ylabel('f'); subplot(1,2,2); mesh(v, f, F1); colormap([0 0 0]) xlabel('V') ylabel('f') zlabel('T'); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a3retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

options = optimset('Algorithm', 'active-set'); [x, f] = fmincon(@MachineObj, [10, 1], [], [], [], [], [], [inf, fmax], @MachineCon,... options, d, a, b, n, Kt, tc, taux, L, D); disp(['V = ' num2str(x(1)) ' f = ' num2str(x(2)) ' T = ' num2str(f)]) function f = MachineObj(x, d, a, b, n, Kt, tc, taux, L, D) [r, c] = size(x); x1 = x(1:r,1:c/2); x2 = x(1:r,(c/2+1):c); Tm = (L*D*pi)./(1000*x1.*x2); Tl = (Kt./(x1.*x2.â*d^b)).^(1/n); f = Tm + Tm./Tl.*tc + taux; function [C, Ceq] = MachineCon(x, d, a, b, n, Kt, tc, taux, L, D) C(1) = x(2)-2; Ceq = []; Answers: V = 100.4641 f = 2

T = 3.942

10

9 350 8

3.7 40

300

7

250

25 10

200

6 T f

5

3.8

5 4.5

150 100

4

50 3.9 0

3 3.9 4

10

4.5

2

150 4

5

5

10

100

1 20

50 40

60

80 V

100

120

140

f

0

0

V

13.5 function Exercise13_5 k = [5, 10, 8]; p = 100; [x1, x2] = meshgrid(linspace(0, 10, 20), linspace(5, 15, 20)); subplot(1,2,1) con = contour(x1, x2, PE5([x1, x2], k, p), [-550:22:-400]); clabel(con) xlabel('x_1') ylabel('x_2') subplot(1,2,2) meshc(x1, x2, PE5([x1, x2], k, p)) colormap([0 0 0]) xlabel('x_1') ylabel('x_2') zlabel('PE') view(50,15) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a4retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

PE=inline('0.5*k(2)*x(1)^2+0.5*k(3)*(x(2)-x(1))^2+0.5*k(1)*x(2)^2-p*x(2)','x',... 'k','p'); options = optimset('LargeScale', 'off'); [x, f] = fminunc(@PE5, [5,5], options, k, p); disp(['x1 = ' num2str(x(1)) ' x2 = ' num2str(x(2)) ' PE = ' num2str(f)]) function PE = PE5(x, k, p) [r, c]= size(x); x1 = x(1:r,1:c/2); x2 = x(1:r,(c/2+1):c); PE = 0.5*k(2)*x1.^2+0.5*k(3)*(x2-x1).^2+0.5*k(1)*x2.^2-p*x2; Answers: x1 = 4.7059 x2 = 10.5882

PE = -529.4118

15

14

200

13

100

12

0

-462 -506

-418 -100

-528

11

-200 x

2

10

E P -300

9 -400 8

-484 -500

7 -440

-600 0

6

15 5

10

5 0

2

4

6

8

10

x1

10

5

x2

x1

13.6 function Exercise13_6 k1 = 4500; k2 = 1650; k3 = 1100; k4 = 2250; k5 = 550; k6 = 9300; K = [k1+k3+k4 -k3 -k4;... -k3 k2+k3+k5 -k5;... -k4 -k5 k4+k5+k6]; P = [1100, 1800, 3300]'; options = optimset('LargeScale', 'off'); [x, f]=fminunc(@Func6, [0, 0, 0]', options, K, P) disp(['x1 = ' num2str(x(1)) ' x2 = ' num2str(x(2)) '

x3 = ' num2str(x(3))])

function PE = Func6(x, K, P) PE = 0.5*x'*K*x-x'*P; Answers: x1 = 0.34755 x2 = 0.72301

x3 = 0.37022

13.7 function Exercise13_7 I = 6; J = 5; Li = 7.5; W = 60*(1:J); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a5retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

k = 450+225*(J/3-(1:I)).^2; Lspr(1:J) = Li; xo = [cumsum(Lspr) zeros(1,J)]; options = optimset('LargeScale', 'off'); [x, f] = fminunc(@Prob7, xo, options, Li, k, W); n = (2:6)'; X = x(1:5)'; Y = x(6:10)'; disp([repmat('X(', 5,1) num2str(n) repmat(') = ', 5,1) num2str(X) ... repmat(' Y(', 5,1) num2str(n) repmat(') = ', 5,1) num2str(Y)]) function PE = Prob7(x, Li, k, w) X = [0, x(1:5), 6*Li]; Y = [0, x(6:10), 0]; dL = sqrt(diff(X).^2+diff(Y).^2)-Li; PE = 0.5*sum(k.*dL.^2)+sum(w.*x(6:10)); Answers: X(2) = 7.91676 X(3) = 16.2538 X(4) = 24.3557 X(5) = 32.2231 X(6) = 39.3033

Y(2) = -3.8577 Y(3) = -7.1459 Y(4) = -8.8364 Y(5) = -8.2859 Y(6) = -5.1603

13.8 function Exercise13_8 E = 1724; L = 127; t = 25.4; R = [27.9, 43.2, 53.3, 73.7, 99.1]; F = [385, 536, 2317, 7820, 18230]; x = lsqcurvefit(@func8, [1 1], R, F, [], [], [], E, L, t); disp(['a = ' num2str(x(1)) ' b = ' num2str(x(2))]) plot(R, F*1e-3, 'ko'); hold on; R = linspace(25, 100, 50); plot(R, func8(x, R, E, L, t)*1e-3, 'k'); xlabel('R (mm)') ylabel('F (MN)') legend('Experimental values', 'Fitted curve', 'Location', 'NorthWest'); function F = func8(x, R, E, L, t) F = pi^x(1)*E*R.^x(2)*t^(4-x(2))/(4*L^2); Answers: a = -3.3182 b = 3.1588 20

18

Experimental values Fitted curve

16

14

12 ) N M (

10

F 8

6

4

2

0 20

30

40

50

60 R (mm)

70

80

90

100


13.9 function Exercise13_9 t = 7:2:41; ingred = [0.488, 0.473, 0.448, 0.435, 0.431, 0.453, 0.421,... 0.405, 0.405, 0.405, 0.403, 0.391, 0.403, 0.398, 0.393,... 0.398, 0.388, 0.388]; x = lsqcurvefit(@fit8, [1,1], t, ingred); disp(['a = ' num2str(x(1)) ' b = ' num2str(x(2))]) function ingred = fit8(x, t, ingred) ingred = x(1)+(0.51-x(1))*exp(-x(2)*(t-8)); Answers: a = 0.39177 b = 0.13944

13.10 function Exercise13_10 x0 = [0, 0]; pos = [0, 2.74, 6.1, 18.3, 27.4, 36.6]; t = [0, 2.05, 3.1, 4.8, 5.6, 6.8]; x = lsqnonlin(@func10,[0, 0], [], [], [], pos, t); disp(['v(t) = ' num2str(x(1)*4) 't^3 + ' num2str(x(2)*3) 't^2 + 1.22t']) function velfunc = func10(x, pos, t) velfunc = pos - (x(1).*t.^4 + x(2).*t.^3 + 0.61*t.^2); Answers: v(t) = -0.032693t^3 + 0.2515t^2 + 1.22t

13.11 function Exercise13_11 N = 40; xc = 1.25; yc = 1.5; th = linspace(0, 2*pi*(1-1/N), N); r = 0.75+0.02*randn(N,1); x = xc+r'.*cos(th); y = yc+r'.*sin(th); plot(x, y, 'k*') hold on plot([0, 0, 2.5, 2.5, 0], [0, 2.5, 2.5, 0, 0], 'k-') axis equal off opt = optimset('display', 'off'); x = lsqnonlin(@circl, [1, 2, 1], [], [], opt, x, y); th = linspace(0, 2*pi, 100); plot(x(1)+x(3)*cos(th), x(2)+x(3)*sin(th),'k-') plot(x(1), x(2), '+k') text(0.1, 0.7, ['r = ' num2str(x(3),5)]) text(0.1, 0.55, ['x_c = ' num2str(x(1),5)]) text(0.1, 0.40, ['y_c = ' num2str(x(2),5)]) function f = circl(x, xd, yd) f = x(3)-sqrt((x(1)-xd).^2+(x(2)-yd).^2);


Answer:

r = 0.75083 x c = 1.2491 y

c

= 1.5011

13.12 function Exercise13_12 p = [1, 1, 1; -1, 2, 4; 2, 3, 4; -3, -4, 1]; SphereDiamObj = @(x, p) (2*x(4)); options = optimset('Algorithm', 'active-set'); x = fmincon(SphereDiamObj, [4, 4, 4, 4], [], [], [], [], ... [-inf, -inf, -inf, 0], [], @SphereConstraints, options, p); disp(['xc = ' num2str(x(1)) ' yc = ' num2str(x(2)) ' zc = ' num2str(x(3)) ' R = ' num2str(x(4))]) function [C,Ceq] = SphereConstraints(x, p) r = size(p); for k = 1:r C(k) = (x(1)-p(k,1))^2+(x(2)-p(k,2))^2+(x(3)-p(k,3))^2-x(4)^2; end Ceq = []; Answers: xc = -0.5 yc = -0.5 zc = 2.5 R = 4.5552

13.13 function Exercise13_13 x0 = [0, 0]; p = 1; e = 2; h = 2; vlb = [0.1, 0.1]; vub = [inf, inf]; options = optimset('Algorithm', 'active-set'); [x, f] = fmincon(@TrussObj,x0,[],[],[],[],vlb,vub,... @TrussCon, options, p, h, e); disp(['x1 = ' num2str(x(1)) ' x2 = ' num2str(x(2)) ' f = ' num2str(f)]) function f = TrussObj(x, p, h, e) f = (p*h/e)*(1/(x(1)+sqrt(2)*x(2))); function [C, Ceq] = TrussCon(x, p, h, e) s1x = p*(x(2)+sqrt(2)*x(1))/(sqrt(2)*x(1)^2+2*x(1)*x(2)); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a8retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

s2x = p*(1/(x(1)+ sqrt(2)*x(2))); s3x = -p*(x(2)/(sqrt(2)*x(1)^2+2*x(1)*x(2))); C(1) = s1x-17.5; C(2) = s2x-17.5; C(3) = s3x-12; Ceq = []; Answers: x1 = 8.3177 x2 = 26.924 f = 0.021554

13.14 function Exercise13_14 x0(1:5) =1; lb(1:5) = 0; Imin(1:5) =1; Imax(1:5) = 2; v = [3, 2, 1, 2, 1]; options = optimset('Algorithm', 'active-set'); [x, P] = fmincon(@CircuitObj, x0, [], [], [], [], lb, [],... @CircuitCon, options, v, Imin, Imax); X = x'; n = (1:5)'; disp([repmat('R(', 5,1) num2str(n) repmat(') = ', 5,1) num2str(X)]) disp(['Power = ' num2str(P) ' W']) function f = CircuitObj(x, v, Imin, Imax) f = sum(v.^2./x); function [C, Ceq] = CircuitCon(x, v, Imin, Imax) C = zeros(10,1); for n = 1:5 C(n) = -v(n)/x(n)+Imin(n); C(n+5) = v(n)/x(n)-Imax(n); end Ceq(1) = v(1)/x(1)+v(3)/x(3)-v(5)/x(5); Ceq(2) = v(2)/x(2)-v(3)/x(3)-v(4)/x(4); Answers: R(1) = 3 R(2) = 1 R(3) = 1 R(4) = 2 R(5) = 0.5 Power = 12 W

13.15 function Exercise13_15 d = -[7, 12, 13, 7, 4, 17, 4, 9, 12, 9, 10, 9,... 6, 14, 8, 16, 21, 8, 8, 7, 4, 8, 13, 4]; c = [300, 740, 300, 430, 210, 360,... 270, 600, 490, 250, 830, 290,... 460, 540, 380, 390, 470, 400,... 800, 380, 760, 600, 680, 310]; H = 2*diag(d); A = zeros(10,24); k = 0; for i = 1:4 for j = 1:6 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a9retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

A(i,j+k*6)=-1; end k = k+1; end k = 0; for i = 1:6 A(i+4,1+k) = 1; A(i+4,7+k) = 1; A(i+4,13+k) = 1; A(i+4,19+k) = 1; k = k+1; end a = [8, 24, 20, 24, 16, 12]; b = -[29, 41, 13, 21]; lb = zeros(24,1); options = optimset('LargeScale','off'); x0 = [5, 5, 10, 5, 5, 5, 5, 5, 0, 20, 0, 5, 0, 0, 0, 0, 15, 0, 0, 10, 0, 0, 0, 0]; xopt = quadprog(H, c, A, [b, a]', [], [], lb, [], x0, options); x = zeros(6,4); for k = 1:4 rs =6*(k-1)+1; x(1:6,k) = xopt(rs:rs+5); end disp(int2str(x)) Answers: -0 8 -0 0 0 3 0 21 20 0 -0 0 0 24 -0 0 3 -0 13 0 6 6 -0 0

13.16 function Exercise13_16 x0 = [0.1, 0.1]; LBnd = [0.15, 0.15]; UBnd = [3.0, 3.5]; [x, fopt] = fseminf(@TwobarsObj, x0, 1, @TwobarsCon, [], [], [], [], LBnd, UBnd); disp(['x(1) = ' num2str(x(1)) ' x(2) = ' num2str(x(2)) ' w = ' num2str(fopt)]) subplot(1,2,1) ylabel('stress \sigma_1 (GPa)') xlabel('\theta (degrees)') axis([-90, 90, -1.5, 1.5]) hold off subplot(1,2,2) ylabel('stress \sigma_2 (GPa)') xlabel('\theta (degrees)') axis([-90, 90, -1.5, 1.5]) function f = TwobarsObj(x) ro = 7833.4117; h = 2.159; Aref = 6.4516e-04; f = 2*ro*h*x(2)*sqrt(1+x(1)^2)*Aref; function [C, Ceq, K1, s] =TwobarsCon(x, s) P = 35585.776; sgm0 = 127.553e06; Aref = 6.4516e-04; © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a10 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

if isnan(s(1,1)) s = [5, 0]; end theta = -90:s(1,1):90; b1 = x(1)*cos(theta/180*pi); b2 = P*sqrt(1+x(1)^2)/(2*x(1)*x(2)*Aref); sgm1 = (b1+sin(theta/180*pi))*b2; sgm2 = (b1-sin(theta/180*pi))*b2; K1(1,:) = sgm1-sgm0; K1(2,:) = sgm2-sgm0; K1(3,:) = -sgm1-sgm0; K1(4,:) = -sgm2-sgm0; C = []; Ceq = []; subplot(1,2,1) plot(theta, (K1(1,:)+sgm0)*1e-9, 'k') hold on subplot(1,2,2) plot(theta, (K1(2,:)+sgm0)*1e-9, 'k') hold on Answers: x(1) = 0.80883 x(2) = 0.45046 w = 12.6431 1.5

1.5

1

1

0.5 ) a P G ( s1 s e r t s

0.5 ) a P G ( s2 s e r t s

0

0

-0.5

-0.5

-1

-1

-1.5

-1.5 -50

0

(degrees)

50

-50

0

50

(degrees)

13.17 function Exercise13_17 global K P k1 = 4500; k2 = 1650; k3 = 1100; k4 = 2250; k5 = 550; k6 = 9300; K = [k1+k3+k4, -k3, -k4;... -k3, k2+k3+k5, -k5;... -k4, -k5, k4+k5+k6]; P = [1100, 1800, 3300]'; nvars = 3; lb = [0 0 0]; ub = [1 1 1]; options = gaoptimset; options = gaoptimset(options, 'HybridFcn', {@fmincon []}); options = gaoptimset(options, 'Display', 'off'); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a11 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

options = gaoptimset(options, 'PlotFcns', {@gaplotbestindiv}); x = ga(@ObjExercise13_17,nvars,[],[],[],[],lb,ub,[],options); disp(['x1 = ' num2str(x(1)) ' x2 = ' num2str(x(2)) ' x3 = ' num2str(x(3))]); function PE = ObjExercise13_17(x) global K P PE = 0.5*x*K*x'-x*P; Answers: x1 = 0.34755 x2 = 0.72301 x3 = 0.37022 Current Best Individual 0.8

l a u d i v i d n i

0.7

0.6

t s e b

0.5

t n e r r u C

0.4

0.3

0.2

0.1

0 1

2 Number of variables (3)

3

13.18 function Exercise13_18 global p e h p = 1; e = 2; h = 2; nvars = 2; lb = [0.1, 0.1]; ub = [inf, inf]; options = gaoptimset; options = gaoptimset(options, 'Display', 'off'); options = gaoptimset(options, 'PlotFcns', {@gaplotbestindiv}); x = ga(@obj_truss_Ex13_18, nvars, [], [], [], [], lb, ub, @con_truss_Ex13_18, options); disp(['x1 = ' num2str(x(1)) ' x2 = ' num2str(x(2))]); function f = obj_truss_Ex13_18(x) global p e h f = (p*h/e)*(1/(x(1)+sqrt(2)*x(2))); function [C, Ceq] = con_truss_Ex13_18(x) global p s1x = p*(x(2)+sqrt(2)*x(1))/(sqrt(2)*x(1)^2+2*x(1)*x(2)); s2x = p*(1/(x(1)+ sqrt(2)*x(2))); s3x = -p*(x(2)/(sqrt(2)*x(1)^2+2*x(1)*x(2))); C(1) = s1x-17.5; C(2) = s2x-17.5; C(3) = s3x-12; Ceq = []; Answers: x1 = 67.8144 x2 = 79.4752


Current Best Individual 80

l a u d i v i d n i

70

60

t s e b

50

t n e r r u C

40

30

20

10

0 1

2 Number of variables (2)

13.19 function Exercise13_19 global E Rho c P nvars = 5; P = 100; lb = [0.05, 0.2, 0.2, 0.35, 3]; ub = [0.5, 0.5, 0.6, 0.5, 6]; E = [1.6, 70, 200]*10^9; Rho = [100, 2770, 7780]; c = [500, 1500, 800]; A = [1, -1, 0, 0, 0; -1, 1, 0, 0, 0; 0, 1, -1, 0, 0 ; 0, -1, 1, 0, 0]; b = [0, 0.15, 0, 0.01]'; options = gaoptimset; options = gaoptimset(options,'PlotFcns', { @gaplotpareto }); gamultiobj(@obj_Ex13_19,nvars,A,b,[],[],lb,ub,options); function f = obj_Ex13_19(x) global E Rho c P EI = (2*x(4)/3)*(E(1)*x(1)^3+E(2)*(x(2)^3-x(1)^3)+E(3)*(x(3)^3-x(2)^3)); mu = 2*x(4)*(Rho(1)*x(1)+Rho(2)*(x(2)-x(1))+Rho(3)*(x(3)-x(2))); c1 = mu*x(5)-2800; f(1) = (-pi/(2*x(5)^2)*sqrt(EI/mu))+P*max(0,c1); f(2) = 2*x(4)*(c(1)*x(1)+c(2)*(x(2)-x(1))+c(3)*(x(3)-x(2)))+P*max(0,c1); Answers:


Pareto front 300

250

2 e v i t c e j b O

200

150

100

50 -400

-350

-300

-250

-200 Objective 1

-150

-100

-50

0

13.20 function Exercise13_20 % Part (a) global P nvars = 7; P = 100; lb = [2.6, 0.7, 17, 7.3, 7.3, 2.9, 5]; ub = [3.6, 0.8, 28, 8.3, 8.3, 3.9, 5.5]; A = zeros(4, 7); A(1,1) = -1; A(1,2 )= 5; A(2,1) = 1; A(2,2) = -12; A(3,4) = -1; A(3,6) = 1.5; A(4,5) = -1; A(4,7) = 1.1; b = [0, 0, -1.9, -1.9]; options = gaoptimset; [x, V] = ga(@obj_Ex13_20a, nvars, A, b, [], [], lb, ub, @con_Ex13_20a, options); for n = 1:length(x) if n == 2 || n == 3 a = []; else a = ' cm'; end disp(['x(' int2str(n) ') = ' num2str(x(n)) a]) end disp(['Volume = ' num2str(V) ' cm^3']) % Part (b) options = gaoptimset; options = gaoptimset(options,'PlotFcns', { @gaplotpareto }); gamultiobj(@obj_Ex13_20b,nvars,A,b,[],[],lb,ub,options); function f = obj_Ex13_20a(x) f = 0.7854*x(1)*x(2)^2*(3.3333*x(3)^2+14.9334*x(3)-43.0934) ... -1.508*x(1)*(x(6)^2+x(7)^2)+7.477*(x(6)^3+x(7)^3) ... +0.7854*(x(4)*x(6)^2+x(5)*x(7)^2); function [C, Ceq] = con_Ex13_20a(x) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a14 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

C(1) = 1/(x(1)*x(2)^2*x(3))-1/27; C(2) = 1/(x(1)*x(2)^2*x(3)^2)-1/397.5; C(3) = x(4)^3/(x(2)*x(3)*x(6)^4)-1/1.93; C(4) = x(5)^3/(x(2)*x(3)*x(7)^4)-1/1.93; C(5) = sqrt((745*x(4)/(x(2)*x(3)))^2+16.9*10^6)/(0.1*x(6)^3)-1100; C(6) = sqrt((745*x(5)/(x(2)*x(3)))^2+157.5*10^6)/(0.1*x(7)^3)-850; C(7) = x(2)*x(3)-40; Ceq = []; function f = obj_Ex13_20b(x) global P [C, Ceq] = con_Ex13_20a(x); penalty = P*(max(0,C(1))+max(0,C(2))+max(0,C(3))+max(0,C(4))+max(0,C(5))+max(0,C(6))+max(0,C(7))); f(1) = obj_Ex13_20a(x)+penalty; f(2) = sqrt((745*x(4)/(x(2)*x(3)))^2+16.9*10^6)/(0.1*x(6)^3)+penalty; Answers: Part (a) x(1) = 3.4991 cm x(2) = 0.7 x(3) = 17.0002 x(4) = 7.3015 cm x(5) = 7.7146 cm x(6) = 3.3503 cm x(7) = 5.2867 cm Volume = 2994.0775 cm^3 Part (b) Pareto front 1100

1050

1000

2 e v i t c e j b O

950

900

850

800

750

700

650 3000

3050

3100

3150

3200 3250 Objective 1

3300

3350

3400

3450


Solution to Exercises Chapter 14 14.1 function Exercise14_1 rho = 850; cp = 3800; Sm = 145; mb = 0.18; cpb = 3300; Tb = 37; alpha = 1e-7; S = 8e5; R = 0.05; r = linspace(0, R, 51); t = linspace(0, 3000, 1001); sol = pdepe(2, @tumorPDE, @tumorIC, @tumorBC, r, t, [], alpha, S, rho, cp, mb, cpb, Tb, Sm); T = sol(:, :); r = r*100; plot(r, T(61,:), 'k-.', r, T(134,:), 'k-', r, T(201,:), 'k--', r, T(501,:), 'k:', [1, 1], [30, 70], 'k--') xlabel('r (cm)') ylabel('Temperature (\circC)') legend('t = 300 s', 't = 400 s', 't = 600 s', 't = 1500 s') ylim([30 70]) function [c, f, s] = tumorPDE(r, t, T, DTDr, alpha, S, rho, cp, mb, cpb, Tb, Sm) c = 1; f = DTDr*alpha; s = (S*(r<0.008)*(t<200)+Sm+mb*cpb*(Tb-T))/(rho*cp); function T0 = tumorIC(r, alpha, S, rho, cp, mb, cpb, Tb, Sm) T0 = 37; function [pl, ql, pr, qr] = tumorBC(rl, Tl, rr, Tr, t, alpha, S, rho, cp, mb, cpb, Tb, Sm) pl = 0; ql = 1; pr = 0; qr = 1; Answer: 70 t = 300 s t = 400 s t = 600 s t = 1500 s

65

60 ( e r )u

55

t °C

a r e p m e T

50

45

40

35

30 0

0.5

1

1.5

2

2.5 r (cm)

3

3.5

4

4.5

5

14.2 function Exercise14_2 rho = 850; cp = 3800; Sm = 145; mb = 0.18; cpb = 3300; Tb = 37; alpha = 1e-7; S = 8e4; R = 0.05; r = linspace(0, R, 501); t = linspace(0, 80000, 1001); sol = pdepe(2, @tumorPDE2, @tumorIC, @tumorBC, r, t, [], alpha, S, rho, cp, mb, cpb, Tb, Sm); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a1retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

T = sol(:, :); r1 = r(1:101); Q1= 4*pi*S/3*r1(end)^3; Q2 = trapz(r1,4*pi*r1.^2.*(mb*cpb*(Tb-T(end,1:101)))); Q4 = 4*pi*Sm/3*r1(end)^3; [uout, duoutdx] = pdeval(2, r, T(end,:), r1(end)); Q3 = 4*pi*r1(end)^2*alpha*rho*cp*duoutdx; Q = Q1+Q2+Q3+Q4; disp(['Q1 = ' num2str(Q1) 'Q2 = ' num2str(Q2) ' Q3 = ' num2str(Q3) ' Q4 = ' num2str(Q4)]) disp(['Sum Qj = ' num2str(Q)]) function [c, f, s] = tumorPDE2(r, t, T, DTDr, alpha, S, rho, cp, mb, cpb, Tb, Sm) c = 1; f = DTDr*alpha; s = (S*(r<0.01)*(t<80000)+Sm+mb*cpb*(Tb-T))/(rho*cp); function T0 = tumorIC(r, alpha, S, rho, cp, mb, cpb, Tb, Sm) T0 = 37; function [pl, ql, pr, qr] = tumorBC(rl, Tl, rr, Tr, t, alpha, S, rho, cp, mb, cpb, Tb, Sm) pl = 0; ql = 1; pr = 0; qr = 1; Answers: Q1 = 0.3351Q2 = -0.018606 Q3 = -0.2919 Q4 = 0.00060737 Sum Qj = 0.025208

14.3 T0 = 20; alpha = 1.5e-7; k=0.6; euler = 0.577216; Q=1; t = linspace(0, 10, 100); r = 0.001/2; z = r^2./(4*alpha*t); Texact = T0+Q/(4*pi*k)*expint(z); Tapprox = T0+Q/(4*pi*k)*(-euler-log(r^2./(4*alpha*t))); plot(t, Texact, 'k-', t, Tapprox, 'k--') xlabel('Time (s)') ylabel('Temperature (\circC)') legend('Exact', 'Approximate', 'Location', 'SouthEast') Answer: 20.4

20.3

20.2 ( e r )u t °C a r e p m e T

20.1

20

19.9

19.8 Exact Approximate 0

1

2

3

4

5 Time (s)

6

7

8

9

10


14.4 function Exercise14_4 Q = 1; T0 = 20; R = 0.0005; Rs = R/10; k = 0.6; alpha = 1.5e-7; t = linspace(0, 10, 500); r = logspace(0, 3, 100)*Rs; % r variable for pdepe, Note: r1(34) = R sol = pdepe(1, @LinePDE, @LineIC, @LineBC, r, t, [], Rs, k, alpha, T0, Q); T = sol(:,:); z = R^2./(4*alpha*t); Texact = T0+Q/(4*pi*k)*expint(z); plot(t, T(:,34), 'k-', t, Texact, 'k--') legend('T_{pdepe}', 'T_{exact}', 'Location', 'SouthEast') xlabel('Time (sec)') ylabel('Temperature (\circC)') function [c, f, s] = LinePDE(r, t, T ,DTDr, R, k, alpha, T0, Q) c = 1/alpha; f = DTDr; s = 0; function T0 = LineIC(r, R, k, alpha, T0, Q) T0 = 20; function [pl,ql,pr,qr] = LineBC(rl, Tl, rr, Tr, t, R, k, alpha, T0, Q) pl = Q/(2*R*pi); ql = k; pr = Tr-T0; qr = 0; Answer: 20.4

20.35

20.3 ( e r )u

20.25

t °C

a r e p m e T

20.2

20.15

20.1

20.05 T pdepe T ex ac t 20 0

1

2

3

4

5 Time (sec)

6

7

8

9

10

14.5 syms K1 K2 Kc H x1 x2 x3 y D X = solve('Kc=(x2-x1)/x1',... 'K1 =H*(x2-x3)/(x1-x2)',... 'K2=(1+x3)*H/(x2-x3)', 'x1', 'x2', 'x3'); x1 = X.x1 x2 = X.x2 x3 = X.x3 Answers: x1 = -H^2/(H^2*Kc + H^2 + H*K1*Kc + K1*K2*Kc) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a3retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

x2 = -(H^2*Kc + H^2)/(H^2*Kc + H^2 + H*K1*Kc + K1*K2*Kc) x3 = -(H^2*Kc + H^2 + H*K1*Kc)/(H^2*Kc + H^2 + H*K1*Kc + K1*K2*Kc)

14.6 Kc1 = 4.25e-7; PCO2 = 45; HCO2 = 22618; CO2 = PCO2/HCO2; Kw =1e-14; Hi = 10^(-7.4); OHi = 1e-14/Hi; ybase = linspace(0, 1e-6, 25); HCO3base = Kc1*CO2/Kw*(OHi+ybase); Hbase = Kw./(OHi+ybase); yacid = linspace(0, 1e-7, 10); HCO3acid = Kc1*CO2./(Hi+yacid); Hacid = Hi+yacid; plot( HCO3acid, -log10(Hacid), 'k-') hold on plot( HCO3base, -log10(Hbase), 'k-') xlabel('Bicarbonate concentration (M)') ylabel('pH') Answer: 8.4

8.2

8

7.8

H p

7.6

7.4

7.2

7

6.8 0

0.02

0.04 0.06 Bicarbonate concentration (M)

0.08

0.1

0.12

14.7 function Exercise14_7 C = linspace(0,150e-6,100); CHb = 0.0022; [Y, dDtdC] = OxyHemo(C, CHb); figure(1) plot(C*1e6, Y, 'k-') xlabel('[C_{O2}]_d (\muM)') ylabel('Fractional Hb Saturation (Y)') figure(2) plot(C*1e6, dDtdC, 'k-') xlabel('[C_{O2}]_d (\muM)') ylabel('\partialC_{T,O2}/\partialC_{O2}') function [Y, dCtdC] = OxyHemo(C, CHb) n = 2.34; C50 = 35.14e-6; Y = C.^n./(C50.^n+C.^n); dCtdC = 1+4*n*CHb.*Y.^2.*C50.^n./C.^(n+1); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a4retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Answers: ) Y (

1

n o i t a r u t a S

0.9

b H

0.6

l a n o i t c a r F

180

160

0.8 140

0.7 120

2 O

100

C

0.5

/ 2O , T

80

C

0.4

60

0.3 40

0.2

20

0.1

0 0

50

100

150

[C O2 ] d (μM)

0 0

50

100

150

[C O2 ] d (μM)

14.8 pCO2 = 45; pH = 7.4; Hco2 = 22618; Kc1 = 4.25e-7; H = 10^(-pH); CO2 = pCO2/Hco2; HCO3 = Kc1*CO2/H; disp(['Bicarbonate concentration = ' num2str(HCO3) ' M']) Answer: Bicarbonate concentration = 0.02124 M

14.9 function Exercise14_9 L = 5.5; h = 0.01; Vavg = 0.3; D = 2e-5; Da = 0.45; Cin = 150e-6; Cmm = 10e-6; Lh = L/h; hD = h/D; y = linspace(0, 1, 101); x = linspace(0, 1, 101); sol = pdepe(0, @pdepbx, @pbICx, @pbBCx, y, x, [], Lh, hD, Vavg, Da, Cin, Cmm); C = sol(:,:)*1e6; Pe = Vavg*hD; Cf = zeros(101,2); for j = [1,101] for i = 1:101 Cf(i,j) = Cin*(1+Da*((1-3*(h-y(j))^2)/6-Lh/Pe*x(i)))*1E6; end end plot(x, C(:,1), 'k-', x, Cf(:,1), 'k--') hold on plot(x, C(:,end), 'k-', x, Cf(:,end), 'k--') xlabel('Distance along reactor (x/L)') ylabel('Concentration (\muM)') text(0.14, 60, 'y/h = 0') text(0.39, 84, 'y/h = 1') axis([0 1 0 160]) legend('pdepe solution', 'Approximate solution', 'Location', 'NorthEast')


function [c, f, s] = pdepbx(y, x, C, DCDy, Lh, hD, Vavg, Da, Cin, Cmm) c = Vavg*hD; f = DCDy*Lh; s = 0; function Czero = pbICx(y, Lh, hD, Vavg, Da, Cin, Cmm) Czero = Cin; function [pa, qa, pb, qb] = pbBCx(yl, Cl, yr, Cr, x, Lh, hD, Vavg, Da, Cin, Cmm) pb = 0; qb = 1; pa = -Da*Cin*Cl/(Cl+Cmm); qa = 1/(Lh); Answer: 160 pdepe solution Approximate solution 140

120 ( n o i t) a M μr t n e c n o C

100

y/h = 1 80

60

y/h = 0

40

20

0 0

0.1

0.2

0.3

0.4 0.5 0.6 Distance along reactor (x/L)

0.7

0.8

0.9

1

14.10 function Exercise14_10 L = 5.5; h = 0.01; Vavg = 0.3; D = 2e-5; Da = 0.45; Cin = 150e-6; Cmm = 10e-6; Lh = L/h; hD = h/D; y = linspace(0, 1, 101); x = linspace(0, 1, 101); sol = pdepe(0, @pdepbyy, @pbICyy, @pbBCyy, y, x, [], Lh, hD, Vavg, Da, Cin, Cmm, 1); C = sol(:,:)*1e6; plot(x, C(:,1), 'k-') hold on plot(x, C(:,end), 'k-') xlabel('Distance along reactor (x/L)') ylabel('Concentration (\muM)') text(0.16, 120, 'y/h = 0') text(0.39, 145, 'y/h = 1') ylim([90 160]) function [c, f, s] = pdepbyy(y, x, C, DCDy, Lh, hD, Vavg, Da, Cin, Cmm, plug) C50 = 35e-6; n = 2.54; CHb = 2.2e-3; Y = C^n/(C^n+C50^n); dCdC = 1+4*n*CHb*Y^2*C50^n/C^(n+1); c = Vavg*hD*dCdC; f = DCDy*Lh; © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a6retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

s = 0; function Czero = pbICyy(y, Lh, hD, Vavg, Da, Cin, Cmm, plug) Czero = Cin; function [pa, qa, pb, qb] = pbBCyy(yl, Cl, yr, Cr, x, Lh, hD, Vavg, Da, Cin, Cmm, plug) pb = 0; qb = 1; pa = -Da*Cin*Cl/(Cl+Cmm); qa = 1/(Lh); Answer: 160

150 y/h = 1 ( n o i t) M a μr t n e c n o C

140

130

120

y/h = 0

110

100

90 0

0.1

0.2

0.3

0.4 0.5 0.6 Distance along reactor (x/L)

0.7

0.8

0.9

1

14.11 function Exercise14_11 R = 0.02; Xb = 5.234e-8; Smax = 15e-9; D = 2e-5; rho = 1; K = 25e-10; S = Smax*R^2/(D*Xb*rho); KKb = K/Xb; c = char('k--', 'k-'); chiR = 1.0; chi0 = 0.0005; solinit = bvpinit(linspace(chi0, chiR, 5), [0 0]); chi = linspace(chi0, chiR, 50); for MM = 0:1 sol = bvp4c(@TumorODE, @TumorBC, solinit, [], S, KKb, MM); y = deval(sol, chi); subplot(2,1,1) plot(chi, y(1,:), c(MM+1,:)) hold on ylabel('\psi') xlabel('\chi') if MM == 1 plot(chi, 1+S/6*(chi.^2-1), 'ko') legend('K = 0', ['K = ' num2str(K)], 'Exact solution (K = 0)', 'Location', 'NorthWest') end subplot(2,1,2) plot(chi, y(2,:), c(MM+1,:)) hold on ylabel('d\psi/d\chi') xlabel('\chi') if MM == 1 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a7retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

plot(chi, S/3*chi, 'ko') legend('K = 0', ['K = ' num2str(K)], 'Exact solution (K = 0)', 'Location', 'NorthWest') end end function dy = TumorODE(chi, y, S, KKb, MM) if MM == 1 dy = [y(2); S*y(1)/(KKb+y(1))-2/chi*y(2)]; else dy = [y(2); S-2/chi*y(2)]; end function res = TumorBC(ya, yb, S, KKb, MM) res = [ya(2); yb(1)-1]; Answer: 1 K= 0 K = 2.5e-009 Exact solution (K = 0)

0.8 0.6

0.4 0.2 0

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.6

0.7

0.8

0.9

1

2 K= 0 K = 2.5e-009

1.5

d /

Exact solution (K = 0)

1

d 0.5

0 0

0.1

0.2

0.3

0.4

0.5

14.12 function Exercise14_12 Rc = 10e-6; Rt=100e-6; Vmax = 0.0005; CHb = 0.0022; Cin = 120e-6; Smax = 20e-6; Cmm = 10e-6; z = linspace(0, 0.001, 101); r = linspace(0, 10e-6, 101); options = odeset('RelTol', 1e-6, 'AbsTol', 1e-9); Dc = [2e-9, 2.2e-9, 1.8e-9, 2e-9, 2e-9]; Dt = [1e-9, 1e-9, 1e-9, 1.1e-9, 0.9e-9,]; for n = 1:length(Dc) sol1 = pdepe(1, @CapillPDE, @CapillIC, @CapillBC, r, z, options, ... Rc, Rt, Dc(n), Dt(n), Vmax, CHb, Cin, Smax, Cmm); C = sol1(:,:); Cm = 4/Rc^2*trapz(r, (1-(r/Rc).^2).*C(end,:).*r)*1e6; disp(['Dc = ' num2str(Dc(n)) ' m^2/s Dt = ' num2str(Dt(n)) ' m^2/s Cm = ' num2str(Cm) ' uM']) end function [c, f, s] = CapillPDE(r, z, C, DCDr, Rc, Rt, Dc, Dt, Vmax, CHb, Cin, Smax, Cmm) n = 2.34; C50 = 35.14e-6; Y = C^n/(C50^n+C^n); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a8retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

dCtdC = 1+4*n*CHb*Y^2*C50^n/C^(n+1); c = dCtdC*Vmax*(1-(r/Rc)^2); f = Dc*DCDr; s = 0; function C0 = CapillIC(r, Rc, Rt, Db, Dt, Vmax, CHb, Cin, Smax, Cmm) C0 = Cin; function [pa, qa, pb, qb] = CapillBC(rl, Cl, rr, Cr, z, Rc, Rt, Dc, Dt, Vmax, CHb, Cin, Smax, Cmm) solinit = bvpinit(linspace(Rc, Rt, 10), [0 0]); sol = bvp4c(@TissueODE, @TissueBC, solinit, [], Cr, Dt, Smax, Cmm); pb = -Dt*sol.y(2,1); qb = 1; pa = 0; qa = 1; function res = TissueBC(ya, yb, Cr, Dt, Smax, Cmm) res = [ya(1)-Cr; yb(2)]; function dydx = TissueODE(r, y, Cr, Dt, Smax, Cmm) dydx = [y(2); Smax/Dt*y(1)/(y(1)+Cmm)-y(2)/r]; Answers: Dc = 2e-009 m^2/s Dt = 1e-009 m^2/s Cm = 45.2942 uM Dc = 2.2e-009 m^2/s Dt = 1e-009 m^2/s Cm = 45.1111 uM Dc = 1.8e-009 m^2/s Dt = 1e-009 m^2/s Cm = 45.517 uM Dc = 2e-009 m^2/s Dt = 1.1e-009 m^2/s Cm = 44.0989 uM Dc = 2e-009 m^2/s Dt = 9e-010 m^2/s Cm = 46.645 uM

14.13 function Exercise14_13 gate = char('n', 'm', 'h'); tspan = linspace(0, 100, 300); for k = 1:3 [t, y] = ode45(@HH, tspan, [0], [], gate(k)); disp([gate(k) '(100) = ' num2str(y(end))]) end function dy = HH(t, y, g) V = -65+65*((t>2)&(t<5)); dy = alph(V+65, g)-(alph(V+65, g)+bet(V+65, g))*y(1); Answers: n(100) = 0.31768 m(100) = 0.05294 h(100) = 0.59611

14.14 function Exercise14_14 gKbar = 0.036; gNabar = 0.120; gL = 0.0003; EL = -49; EK = -77; gate = char('n', 'm', 'h'); tspan = linspace(0, 20, 500); IC = [0.3177, 0.0529, 0.5961]; Vs = [-50, -25, 0, 25]+65; for nn = 1:4 subplot(4,1,nn)

ENa = 50;


for k = 1:3 [t, y] = ode45(@HH2, tspan, [IC(k)], [], gate(k), Vs(nn)); switch gate(k) case 'n' n = y(:,1); case 'm' m = y(:,1); case 'h' h = y(:,1); end end V = -65+Vs(nn)*((t>2) & (t<18)); Iion = gKbar.*n.^4.*(V-EK)+gNabar.*m.^3.*h.*(V-ENa)+gL.*(V-EL); plot(t, Iion, 'k-') ylabel('I_{ion} (mA/cm^2)') text(9, min(Iion), ['V_{pot} = ' num2str(Vs(nn)-65) ' mV']) if nn == 4 xlabel('t (ms)') end end function dy = HH2(t, y, g, Vs) V = -65+Vs*((t>2)&(t<18)); dy = alph(V+65, g)-(alph(V+65, g)+bet(V+65, g))*y(1); Answers: 0.1

m c

2)/

A m (n I

0

o i

-0.1

m c 2)/ A m (n I

o i

m c 2)/ A m (n I

o i

m c 2)/ A m (n

V pot = -50 mV 0

2

4

6

8

0

2

4

6

8

0

2

4

6

8

0

2

4

6

8

12

14

16

18

20

12

14

16

18

20

12

14

16

18

20

12

14

16

18

20

0 V

-1 -2

pot

= -25 mV

10

2

0 V -2

pot

= 0 mV

10

4 2 0

V pot = 25 mV

o

Ii

10

1

-2 10 t (ms)

14.15 function Exercise14_15 Vr = -62.607; gKbar = 0.036; gNabar = 0.120; gL = 0.0003; EL = -49; EK = -77; ENa = 50; Cm = 0.001; IC = [0.3177, 0.0529, 0.5961, -62.607]; Vs = [100, 150, 200, 250]; c = char('k-', 'k--', 'k:', 'k-.'); options = odeset('MaxStep', 0.1); for nn=1:4 [t, y] = ode45(@HodHux2, [0, 10], IC, options, gKbar, gNabar, gL, EL, EK, ENa, Cm, Vr, Vs(nn)); plot(t, y(:,4), c(nn,:)) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a10 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

hold on end ylabel('Membrane Potential (mV)') xlabel('Time (ms)') A = 'V_s = '; B = ' mV'; legend([A int2str(Vs(1)) B], [A int2str(Vs(2)) B], [A int2str(Vs(3)) B], [A int2str(Vs(4)) B]) function hh = HodHux2(t, y, gKbar, gNabar, gL, EL, EK, ENa, Cm, Vr, Vs) % y(1) = n, y(2) = m, y(3) = h, y(4) = V Vd = y(4)-Vr; dn = alph(Vd, 'n')-(alph(Vd, 'n')+bet(Vd, 'n'))*y(1); dm = alph(Vd, 'm')-(alph(Vd, 'm')+bet(Vd, 'm'))*y(2); dh = alph(Vd, 'h')-(alph(Vd, 'h')+bet(Vd, 'h'))*y(3); dV = -(gKbar*y(1)^4*(y(4)-EK)+gNabar*y(2)^3*y(3)*(y(4)-ENa)+gL*(y(4)-EL))/Cm; dV = dV+Vs*((t>2) & (t<2.1)); hh = [dn; dm; dh; dV]; Answers: 60 ) V m (

V s = 100 mV V s = 150 mV V

40

V , l a i t n e t o P e n a r b m e M

s

= 200 mV

V s = 250 mV 20

0

-20

-40

-60

-80 0

1

2

3

4

5 Time (ms)

6

7

8

9

10

14.16 function Exercise14_16 Vr = -62.607; gKbar = 0.036; gNabar = 0.120; gL = 0.0003; EL = -49; EK = -77; ENa = 50; Cm = 0.001; IC = [0.3177, 0.0529, 0.5961, -62.607]; td = [11, 12]; options = odeset('MaxStep', 0.1); for nn=1:2 subplot(2,1,nn) [t, y] = ode45(@HodHux3, [0, 20], IC, options, gKbar, gNabar, gL, EL, EK, ENa, Cm, Vr, td(nn)); plot(t, y(:,4), 'k-') ylabel('Membrane Potential (mV)') xlabel('Time (ms)') end function hh = HodHux3(t, y, gKbar, gNabar, gL, EL, EK, ENa, Cm, Vr, td) % y(1) = n, y(2) = m, y(3) = h, y(4) = V Vd = y(4)-Vr; dn = alph(Vd, 'n')-(alph(Vd, 'n')+bet(Vd, 'n'))*y(1); © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a11 retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

dm = alph(Vd, 'm')-(alph(Vd, 'm')+bet(Vd, 'm'))*y(2); dh = alph(Vd, 'h')-(alph(Vd, 'h')+bet(Vd, 'h'))*y(3); dV = -(gKbar*y(1)^4*(y(4)-EK)+gNabar*y(2)^3*y(3)*(y(4)-ENa)+gL*(y(4)-EL))/Cm; dV = dV+200*((t>2) & (t<2.1))+200*((t>td) & (t
l a i t n e t o P e n a r b m e M ) V m ( l a i t n e t o P

60 40 20 0 -20 -40 -60 -80

0

2

4

6

8

10

12

14

16

18

20

12

14

16

18

20

Time (ms)

60 40 20 0

e n a r b m e M

-20 -40 -60 -80 0

2

4

6

8

10 Time (ms)

14.17 function Exercise14_17 Vr = -62.607; gKbar = 0.018; gNabar = 0.120; gL = 0.0003; EL = -49; EK = -77; ENa = 50; Cm = 0.001; IC = [0.3177, 0.0529, 0.5961, -62.607]; options = odeset('MaxStep', 0.1); [t, y] = ode45(@HodHux, [0, 100], IC, options, gKbar, gNabar, gL, EL, EK, ENa, Cm, Vr); plot(t, y(:,4), 'k-') ylabel('Membrane Potential (mV)') xlabel('Time (ms)') function hh = HodHux(t, y, gKbar, gNabar, gL, EL, EK, ENa, Cm, Vr) % y(1) = n, y(2) = m, y(3) = h, y(4) = V Vd = y(4)-Vr; dn = alph(Vd, 'n')-(alph(Vd, 'n')+bet(Vd, 'n'))*y(1); dm = alph(Vd, 'm')-(alph(Vd, 'm')+bet(Vd, 'm'))*y(2); dh = alph(Vd, 'h')-(alph(Vd, 'h')+bet(Vd, 'h'))*y(3); dV = -(gKbar*y(1)^4*(y(4)-EK)+gNabar*y(2)^3*y(3)*(y(4)-ENa)+gL*(y(4)-EL))/Cm; dV = dV+200*((t>2) & (t<2.1)); hh = [dn; dm; dh; dV]; Answer:


60

) V m ( l a i t n e t o P e n a r b m e M

40

20

0

-20

-40

-60

-80 0

10

20

30

40

50 Time (ms)

60

70

80

90

100


An Engineers Guide to Matlab Solutions

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