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Solutions Manual

Applied Numerical Methods

With MATLAB for Engineers and Scientists

Steven C. Chapra

Tufts University

CHAPTER 1

1.1 You are given the following differential equation with the initial condition, v ( t = 0) = 0,

2

v

m

c

g

dt

dv d

Multiply both sides by m / cd

2 g v c

m

dt

dv

c

m

d d

Define a = mg / cd

2 2 a v dt

dv

c

m

d

Integrate by separation of variables,

dt m

c

a v

dv (^) d ∫ ∫

2 2

A table of integrals can be consulted to find that

a

x

a x a

dx (^) 1 2 2

tanh

∫ −

Therefore, the integration yields

t C m

c

a

v

a

d = +

− 1 tanh

If v = 0 at t = 0, then because tanh

• (0) = 0, the constant of integration C = 0 and the solution is

t m

c

a

v

a

d

− 1 tanh

This result can then be rearranged to yield

= t m

gc

c

gm v

d

d

tanh

1.2 This is a transient computation. For the period from ending June 1:

v m

c g dt

dv ' = −

Applying Laplace transforms,

V

m

c

s

g sV v

Solve for

s c m

v

ss c m

g V '/

The first term to the right of the equal sign can be evaluated by a partial fraction expansion,

s c m

B

s

A

ss c m

g

( '/ ) + ' /

( '/ ) ss c m

As c m Bs

ss c m

g

Equating like terms in the numerators yields

A

m

c g

A B

Therefore,

' c '

mg B c

mg A = =−

These results can be substituted into Eq. (2), and the result can be substituted back into Eq.

(1) to give

s c m

v

s c m

mg c

s

mg c V '/

Applying inverse Laplace transforms yields

c mt cm t e v e c

mg

c

mg v

( '/ ) ('/ ) ( 0 ) ' '

− − = − +

or

( )

c mt c mt e c

mg v v e

( '/ ) ('/ ) 1 '

− − = + −

where the first term to the right of the equal sign is the general solution and the second is the

particular solution. For our case, v (0) = 0, so the final solution is

( )

c mt e c

mg v

('/ ) 1 '

− = −

(b) The numerical solution can be implemented as

v = + −

v = + −

The computation can be continued and the results summarized and plotted as:

t v dv / dt 0 0 9. 2 19.6200 6. 4 32.0374 3. 6 39.8962 2. 8 44.8700 1.

10 48.0179 0.

12 50.0102 0.

Note that the analytical solution is included on the plot for comparison.

The process can be continued to give

t y 0 0

0.5 -0. 1 -0. 1.5 -0. 2 0.

2.5 0. 3 0. 3.5 0. 4 0.

4.5 0. 5 0.

-0.

0

0.

0.

0 1 2 3 4 5

y

t m

c

e c

gm v t

⎟ ⎠

⎞ ⎜ ⎝

−⎛ = −

jumper #1: v t ( 1 e ) 44. 87 m / s

12. 5

10

68. 1

69. 5

= − =

⎟ ⎠

⎜ ⎞ ⎝

−⎛

jumper #2: ( 1 ) 14

75

14 t e

⎟ ⎠

⎜ ⎞ ⎝

−⎛ = −

t e

1. 18666
2. 87 52. 5 52. 5

− = −

t e

1. 18666
2. 14533

−

t e

1. 18666 ln 0. 14533 ln

−

t = 10.33 sec

1.8 Q in = Q out

Q 1 = Q 2 + Q 3

30 = 20 + vA 3

10 = 5 A 3

A 3 = 2 m

2

1.9 ∑ Min −∑ Mout = 0

[ 1000 + 1200 +MP+ 50 ] −[ 400 + 200 + 1400 + 200 + 350 ] = 0

Metabolic production = 300 grams

1.10 (^) ∑ %body weight= 60

4. 5 + 4. 5 + 12 + 4. 5 + 1. 5 + IW = 60

% Intracellular water body weight = 33 %

4. 5 + 4. 5 + 12 + 4. 5 + 1. 5 + IW = 60

∑ %body water=^100

7. 5 + 7. 5 + 20 + 7. 5 + 55 + TW = 100

% Transcellular water of body water = 2.5 %

t = 5 10 15 20 25 30

(b)

x = linspace(-3,3,7)

x = -3 -2 -1 0 1 2 3

2.4 (a)

v = -2:.75:

v = -2.0000 -1.2500 -0.5000 0.2500 1.

(b)

r = 6:-1:

r = 6 5 4 3 2 1 0

F = [10 12 15 9 12 16]; x = [0.013 0.020 0.009 0.010 0.012 0.010]; k = F./x

k = 1.0e+003 * 0.7692 0.6000 1.6667 0.9000 1.0000 1.

U = .5k.x.^

U = 0.0650 0.1200 0.0675 0.0450 0.0720 0.

max(U)

ans =

TF = 32:3.6:93.2; TC = 5/9(TF-32); rho = 5.5289e-8TC.^3-8.5016e-6TC.^2+6.5622e-5TC+0.99987; plot(TC,rho)

A = [.035 .0001 10 2;

.02 .0002 8 1; .015 .001 20 1.5; .03 .0007 24 3; .022 .0003 15 2.5]

A = 0.0350 0.0001 10.0000 2. 0.0200 0.0002 8.0000 1. 0.0150 0.0010 20.0000 1. 0.0300 0.0007 24.0000 3. 0.0220 0.0003 15.0000 2.

U = sqrt(A(:,2))./A(:,1).(A(:,3).A(:,4)./(A(:,3)+2*A(:,4))).^(2/3)

U =

t = 10:10:60; c = [3.4 2.6 1.6 1.3 1.0 0.5]; tf = 0:70; cf = 4.84exp(-0.034tf); plot(t,c,'s',tf,cf,'--')

v = 10:10:80; F = [25 70 380 550 610 1220 830 1450]; vf = 0:100; Ff = 0.2741*vf.^1.9842; loglog(v,F,'d',vf,Ff,':')

x = linspace(0,3*pi/2); c = cos(x); cf = 1-x.^2/2+x.^4/factorial(4)-x.^6/factorial(6); plot(x,c,x,cf,'--')

3.3 The M-file can be written as

function annualpayment(P, i, n) nn = 1:n; A = Pi(1+i).^nn./((1+i).^nn-1); y = [nn;A]; fprintf('\n year annualpayment\n'); fprintf('%5d %14.2f\n',y);

This function can be used to evaluate the test case,

annualpayment(35000,.076,5)

year annualpayment 1 37660. 2 19519. 3 13483. 4 10473. 5 8673.

3.4 The M-file can be written as

function Tavg = avgtemp(Tmean, Tpeak, tstart, tend) omega = 2pi/365; t = tstart:tend; Te = Tmean + (Tpeak-Tmean)cos(omega*(t-205)); Tavg = mean(Te);

This function can be used to evaluate the test cases,

avgtemp(5.2,22.1,0,59)

ans = -10.

avgtemp(23.1,33.6,180,242)

ans =

3.5 The M-file can be written as

function vol = tankvol(R, d) if d < R vol = pi * d ^ 3 / 3; elseif d <= 3 * R v1 = pi * R ^ 3 / 3; v2 = pi * R ^ 2 * (d - R); vol = v1 + v2; else error('overtop') end

This function can be used to evaluate the test cases,

tankvol(1,0.5) ans =

tankvol(1,1.2) ans =

tankvol(1,3.0) ans =

tankvol(1,3.1) ??? Error using ==> tankvol overtop

3.6 The M-file can be written as

function [r, th] = polar(x, y) r = sqrt(x .^ 2 + y .^ 2); if x < 0 if y > 0 th = atan(y / x) + pi; elseif y < 0 th = atan(y / x) - pi; else th = pi; end else if y > 0 th = pi / 2; elseif y < 0 th = -pi / 2; else th = 0; end end th = th * 180 / pi;

This function can be used to evaluate the test cases. For example, for the first case,

[r,th]=polar(1,1)

r =

th = 90

The remaining cases are

3.8 The M-file can be written as

function grade = lettergrade(score) if score >= 90 grade = 'A'; elseif score >= 80 grade = 'B'; elseif score >= 70 grade = 'C'; elseif score >= 60 grade = 'D'; else grade = 'F'; end

This function can be tested with a few cases,

lettergrade(95) ans = A

lettergrade(45) ans = F

lettergrade(80) ans = B

3.9 The M-file can be written as

function Manning(A)

A(:,5) = sqrt(A(:,2))./A(:,1).(A(:,3).A(:,4)./(A(:,3)+2*A(:,4))).^(2/3);

fprintf('\n n S B H U\n');

fprintf('%8.3f %8.4f %10.2f %10.2f %10.4f\n',A');

This function can be run to create the table,

Manning(A)

n S B H U 0.035 0.0001 10.00 2.00 0. 0.020 0.0002 8.00 1.00 0. 0.015 0.0010 20.00 1.50 2. 0.030 0.0007 24.00 3.00 1. 0.022 0.0003 15.00 2.50 1.

3.10 The M-file can be written as

function beam(x) xx = linspace(0,x); n=length(xx); for i=1:n uy(i) = -5/6.(sing(xx(i),0,4)-sing(xx(i),5,4)); uy(i) = uy(i) + 15/6.sing(xx(i),8,3) + 75sing(xx(i),7,2); uy(i) = uy(i) + 57/6.xx(i)^3 - 238.25.*xx(i); end plot(xx,uy)

function s = sing(xxx,a,n) if xxx > a s = (xxx - a).^n; else s=0; end

This function can be run to create the plot,

beam(10)

3.11 The M-file can be written as

function cylinder(r, L) h = linspace(0,2r); V = (r^2acos((r-h)./r)-(r-h).sqrt(2rh-h.^2))L; plot(h, V)

This function can be run to the plot,

cylinder(2,5)