Solution manual for introduction to MATLAB for engineers 3rd edition by palm

Thus the values of x, y, and z are correct... Thus the values of a, b, c, and d are correct... The area is given by The perimeter is given by... a If we neglect drag, then conservation o

Solutions Manual c

to accompany

Introduction to MATLAB for Engineers, Third Edition

by

William J Palm III

University of Rhode Island

Solutions to Problems in Chapter One

c reserved No part of this manual may be displayed, reproduced, or distributed

in any form or by any means without the written permission of the publisher

or used beyond the limited distribution to teachers or educators permitted by McGraw-Hill for their individual course preparation Any other reproduction

or translation of this work is unlawful.

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Test Your Understanding Problems T1.1-1 a) 6*10/13 + 18/(5*7) + 5*9^2 Answer is 410.1297.

T1.3-1 The session is:

x = cos(0):0.02:log10(100);

length(x)ans =

51

x(25)ans =1.4800

T1.3-2 The session is:

roots([1, 6, -11, 290]

ans =-10.0002.0000 + 5.0000i2.0000 - 5.0000i

The roots are −10 and 2 ± 5i.

T1.3-3 The session is:

T1.3-4 The session is:

T1.3-5 The session is:

A = [6, -4, 8; -5, -3, 7; 14, 9, -5];

b = [112; 75; -67];

x = A\b

x =2.0000-5.000010.0000

To check the answer, compute the right-hand sides:

A*xans =

T1.4-1 The session is:

r = input(0Enter the sphere radius: 0);

v = x*y

v =30

w = x/y

w =3.3333

z = sin(x)

z =-0.5440

r = 8*sin(y)

r =1.1290

s = 5*sin(2*y)

s =-1.3971

2 The session is:

y*x^3/(x-y)ans =

-13.3333

3*x/(2*y)ans =

0.6000

3*x*y/2ans =15

x^5/(x^5-1)ans =

3 The session is:

x = 3; y = 4;

1/(1-1/x^5)ans =

1.0041

3*pi*x^2ans =84.8230

3*y/(4*x-8)ans =

3

4*(y-5)/(3*x-6)ans =

4 a) x = 3;y = 6*x^3 + 4/x The answer is y = 163.3333.

b) x = 7;y = (x/4)*3 The answer is y = 5.2500.

c) x = 9;y = (4*x)^2/25 The answer is y = 51.8400.

d) x = 4;y = 2*sin(x)/5 The answer is y = −0.3027.

e) x = 30;y = 7*x^(1/3) + 4*x^0.58 The answer is y = 50.5107.

5 The session is:

a = 1.12; b = 2.34; c = 0.72;d = 0.81;f = 19.83;

x = 1 + a/b + c/f^2

x =1.4805

s = (b-a)/(d-c)

s =13.5556

r = 1/(1/a + 1/b + 1/c + 1/d)

r =0.2536

y = a*b/c*f^2/2

y =715.6766

6 The session is:

7 The session is:

The required radius is 4.958

8 The session is

x = -7-5i;y = 4+3i;

x+yans =-3.0000 - 2.0000i

x*yans =-13.0000 -41.0000i

x/yans =-1.7200 + 0.0400i

9 The session is:

(3+6i)*(-7-9i)ans =

33.0000 -69.0000i

(5+4i)/(5-4i)ans =

0.2195 + 0.9756i

3i/2ans =

0 + 1.5000i

3/2ians =

10 The session is:

x = 5+8i;y = -6+7i;

u = x + y;v = x*y;

w = x/y;z = exp(x)

r = sqrt(y);s = x*y^2;

The answers are u = −1 + 15i, v = −86 − 13i, w = 0.3059 − 0.9765i, z = −21.594 + 146.83i,

r = 1.2688 + 2.7586i, and s = 607 − 524i.

11 The session is:

n = 1;R = 0.08206;T = 273.2;V=22.41;

a = 6.49;b = 0.0562;

Pideal = n*R*T/VPideal =

1.0004

P1 = n*R*T/(V - nb)P1 =

1.0029

P2 = (a*n^2)/V^2P2 =

0.0129

Pwaals = P1 + P2Pwaals =

1.0158

The ideal gas law predicts a pressure of 1.0004 atmospheres, while the van der Waals modelpredicts 1.0158 atmospheres Most of the difference is due to the P2 term, which modelsthe molecular attractions

12 The ideal gas law gives

13 The session is:

x=1:0.2:5;

y = 7*sin(4*x);

length(y)ans =

21

y(3)ans =-4.4189

There are 21 elements The third element is −4.4189.

14 The session is:

x = sin(-pi/2):0.05:cos(0);

length(x)ans =

41

x(10)ans =-0.5500

There are 41 elements The tenth element is 0.55

15 The session is:

16 The session is:

a = 6*pi*atan(12.5)+4;

b = 5*tan(3*asin(13/5));

c = 5*log(7);

d = 5*log10(7);

The results are a = 32.1041, b = 0.0000 − 5.0006i, c = 9.7296, and d = 4.2255 In part (b)

note that complex results are obtained from asin(x) if |x| > 1 and from tan(x) if x iscomplex

17 The session is:

18 The session is:

p = ([13,182,-184,2503];

r = roots(p)

r =-15.68500.8425 + 3.4008i0.8425 - 3.4008i

The roots are x = −15.685 and x = 0.8425 ± 3.4008i.

19 The session is:

p = [70, 24, -10, 20];

roots(p)ans =-0.87710.2671 + 0.5044i0.2671 - 0.5044i

The roots are −0.8771 and 0.2671 ± 0.5044i.

21 The session is:

t = 1:0.005:3;

T = 6*log(t) - 7*exp(-0.2*t);

plot(t,T),title(0Temperature Versus Time0),

xlabel(0Time t (min)0),ylabel(0Temperature T (◦ C)0)

22 The session is:

x = 0:0.01:2;

u = 2*log10(60*x+1);

v = 3*cos(6*x);

plot(x,u,x,v,0 0),ylabel(0Speed (mi/hr)0),

xlabel(0Distance x (mi)0)

25 The session is:

A = [7,14,-6;12,-5,9;-5,7,15];

b = [95;-50;145];

x = A\b

The answers are −3, 10, and 4, which correspond to x = −3, y = 10, and z = 4 Typing

A*x gives the result 95, -50, 145, which are the right-hand sides of the equations Thus the

values of x, y, and z are correct.

26 Substituting the four (x, y) values into the function gives the following equations.

The answers are 7, 5, −6, and 4, which correspond to a = 7, b = 5, c = −6, and d = 4.

Typing A*x gives the result 8, 4, 10, 68, which are the right-hand sides of the equations

Thus the values of a, b, c, and d are correct.

27 The script file is:

a = 95;b = -50;c = 145;

A = [7, 14, -6;12, -5, 9;-5, 7, 15];

bvector = [a;b;c];

x = A\bvector

The answers for x are −3, 10, 4, which correspond to x = −3, y = 10, and z = 4 Typing

A*x gives the result 95, −50, 145, which are the right-hand sides of the equations Thus the

values of x, y, and z are correct.

28 Note that

W2= D2+ D2= 2D2Thus D = W/√2 The area is given by

The perimeter is given by

29 Applying the law of cosines to the two triangles gives

a2 = b21+ c21− 2b1c1 cos A1

a2 = b22+ c22− 2b2c2 cos A2With the given values we can solve the first equation for a, then solve the second equation for c2 The second equation is a quadratic in c2, and can be written as

c22− (2b2 cos A2)c2+ b22− a2= 0

The script file is:

b1 = 180;b2 = 165;c1 = 115;A1 = 120*pi/180;A2 = 100*pi/180;

30 (a) If we neglect drag, then conservation of mechanical energy states that the kinetic

energy at the time the ball is thrown must equal the potential energy when the ball reachesthe maximum height Thus

For h = 20 feet, we get v =p

2(32.2)(20) = 35.9 feet/second Because speed measured

in miles per hour is more familiar to most of us, we can convert the answer to miles per

hour as a “reality check” on the answer The result is v = 35.9(3600)/5280 = 24.5 miles

per hour, which seems reasonable

(b) The issues here are the manner in which the rod is thrown and the effect of drag onthe rod If the drag is negligible and if we give the mass center a speed of 35.9 feet/second,then the mass center of the rod will reach a height of 20 feet However, if we give the rodthe same kinetic energy, but throw it upward by grasping one end of the rod, then it willspin and not reach 20 feet The kinetic energy of the rod is given by

of its energy is contained in the spinning motion The inertia I increases with the length

and radius of the rod In addition, a longer rod will have increased drag, and will thus reach

a height smaller than that predicted using conservation of mechanical energy

31 (a) When A = 0◦, d = L1 + L2 When A = 180◦, d = L1− L2 The stroke is the

difference between these two values Thus the stroke is L1 + L2 − (L1 − L2) = 2L2 and

0 20 40 60 80 100 120 140 160 180 0.8

1 1.2 1.4 1.6 1.8 2

A (degrees)

Figure : for Problem 31

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