64 Thus, a root is located at about 0.0028. The fzero function can be used to refine this
estimate, vroot = fzerofvol,0.0028
vroot = 0.00280840865703
The mass of methane contained in the tank can be computed as
3
m 317
. 1068
0.0028084 3
mass =
= =
v V
6.15 The function to be evaluated is
L h
rh h
r r
h r
r V
h f
⎥ ⎦
⎤ ⎢
⎣ ⎡
− −
− ⎟
⎠ ⎞
⎜ ⎝
⎛ − −
=
− 2
1 2
2 cos
To use MATLAB to obtain a solution, the function can be written as an M-file function y = fhh,r,L,V
y = V - r2acosr-hr-r-hsqrt2rh-h2L; The fzero function can be used to determine the root as
format long r = 2;L = 5;V = 8;
h = fzerofh,0.5,[],r,L,V h =
0.74001521805594
65
6.16 a The function to be evaluated is
10 500
cosh 10
10
A A
A A
T T
T T
f +
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
− =
The solution can be obtained with the fzero function as format long
TA = fzeroinline10-x10cosh500x+x10,1000 TA =
1.266324360399887e+003 b
A plot of the cable can be generated as x = linspace-50,100;
w = 10;y0 = 5; y = TAwcoshwxTA + y0 - TAw;
plotx,y,grid
6.17 The function to be evaluated is
5 .
3 2
sin 9
− =
−
t e
t f
t
π
A plot can be generated with MATLAB, t = linspace0,2;
y = 9exp-t . sin2pit - 3.5; plott,y,grid
66 Thus, there appear to be two roots at approximately 0.1 and 0.4. The fzero function can be
used to obtain refined estimates, t = fzero9exp-xsin2pix-3.5,[0 0.2]
t = 0.06835432096851
t = fzero9exp-xsin2pix-3.5,[0.2 0.8] t =
0.40134369265980
6.18 The function to be evaluated is
2 2
1 1
1 ⎟
⎠ ⎞
⎜ ⎝
⎛ −
+ −
= L
C R
Z f
ω ω
ω
Substituting the parameter values yields
2 6
2 10
6 .
50625 1
01 .
⎟ ⎠
⎞ ⎜
⎝ ⎛
− ×
+ −
=
−
ω ω
ω
f The fzero function can be used to determine the root as
fzero0.01-sqrt150625+.6e-6x-2.x.2,[1 1000] ans =
220.0202
6.19 The fzero function can be used to determine the root as
67 format long
fzero240x525+0.540000x2-959.8x-959.80.43,1 ans =
0.16662477900186
6.20 If the height at which the throw leaves the right fielders arm is defined as y = 0, the y at 90
m will be –0.8. Therefore, the function to be evaluated is
180 cos
1 .
44 180
tan 90
8 .
2
πθ θ
π θ
− ⎟
⎠ ⎞
⎜ ⎝
⎛ +
= f
Note that the angle is expressed in degrees. First, MATLAB can be used to plot this function versus various angles.
Roots seem to occur at about 40
o
and 50
o
. These estimates can be refined with the fzero function,
theta = fzero0.8+90tanpix180-44.1.cospix180.2,0 theta =
37.8380 theta = fzero0.8+90tanpix180-44.1.cospix180.2,[40 60]
theta = 51.6527
Thus, the right fielder can throw at two different angles to attain the same result.
6.21
The equation to be solved is
68 V
h Rh
h f
− ⎟
⎠ ⎞
⎜ ⎝
⎛ −
=
3 2
3
π π
Because this equation is easy to differentiate, the Newton-Raphson is the best choice to achieve results efficiently. It can be formulated as
2 3
2 1
2 3
i i
i i
i i
x Rx
V x
Rx x
x π
π π
π −
− ⎟
⎠ ⎞
⎜ ⎝
⎛ −
− =
+
or substituting the parameter values,
2 3
2 1
10 2
1000 3
10
i i
i i
i i
x x
x x
x x
π π
π π
− −
⎟ ⎠
⎞ ⎜
⎝ ⎛
− −
=
+
The iterations can be summarized as
iteration x
i
fx
i
fx
i
| ε
a
| 0 10
1094.395 314.1593
1 6.516432 44.26917
276.0353 53.458
2 6.356057 0.2858 272.4442
2.523 3 6.355008
1.26E-05 272.4202
0.017
Thus, after only three iterations, the root is determined to be 6.355008 with an approximate relative error of 0.017.
6.22
r = [-2 6 1 -4 8]; a = polyr
a = 1 -9 -20 204 208 -384
polyvala,1 ans =
b = poly[-2 6] b =
1 -4 -12 [q,r] = deconva,b
q = 1 -5 -28 32
r = 0 0 0 0 0 0
69
x = rootsq x =
8.0000 -4.0000
1.0000 a = convq,b
a = 1 -9 -20 204 208 -384
x = rootsa x =
8.0000 6.0000
-4.0000 -2.0000
1.0000 a = polyx
a = 1.0000 -9.0000 -20.0000 204.0000 208.0000 -384.0000
6.23
a = [1 9 26 24]; r = rootsa
r = -4.0000
-3.0000 -2.0000
a = [1 15 77 153 90]; r = rootsa
r = -6.0000
-5.0000 -3.0000
-1.0000
Therefore, the transfer function is
1 3
5 6
2 3
4 +
+ +
+ +
+ +
= s
s s
s s
s s
s G
70
CHAPTER 7
7.1
Aug = [A eyesizeA]
Here’s an example session of how it can be employed.
