Solution Manual For Applied Numerical Methods with MATLAB for Engineers and Scientists 5e Steven Chapra

1



Solution Manual For
Applied Numerical Methods with MATLAB for Engineers and Scientists 5e Steven
Chapra
Chapters 1-24

CHAPTER 1

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

dv c
 g  d v2
dt m

Multiply both sides by m/cd

m dv m
 g  v2
cd dt cd


Define a  mg / cd

m dv
 a2  v2
cd dt

Integrate by separation of variables,

cd
a  m dt
dv

2
v 2



A table of integrals can be consulted to find that


a
dx 1 x
 tanh 1
2
x 2
a a

Therefore, the integration yields

1 v c
tanh 1  d t  C
a a m

If v = 0 at t = 0, then because tanh–1(0) = 0, the constant of integration C = 0 and the solution is

1 v c
tanh 1  d t
a a m

This result can then be rearranged to yield

gm  gcd 
v tanh 
 m 
t
cd  

, 2




1.2 (a) For the case where the initial velocity is positive (downward), Eq. (1.21) is

dv c
 g  d v2
dt m

Multiply both sides by m/cd

m dv m
 g  v2
cd dt cd


Define a = mg / cd

m dv
 a2  v2
cd dt

Integrate by separation of variables,

cd
a  m dt
dv

2
v 2


A table of integrals can be consulted to find that


a
dx 1 x
 tanh 1
2
x 2
a a

Therefore, the integration yields

1 v c
tanh 1  d t  C
a a m

If v = v0 at t = 0, then

1 v
C tanh 1 0
a a

Substitute back into the solution

1 v c 1 v
tanh 1  d t  tanh 1 0
a a m a a

Multiply both sides by a, taking the hyperbolic tangent of each side and substituting a gives,

mg  gcd cd 
v tanh  t  tanh 1 v0  (1)
cd  mg 
 m

, 3



(b) For the case where the initial velocity is negative (upward), Eq. (1.21) is

dv c
 g  d v2
dt m

Multiplying both sides of Eq. (1.8) by m/cd and defining a  mg / cd yields

m dv
 a2  v2
cd dt

Integrate by separation of variables,

cd
a  m dt
dv

2
v 2


A table of integrals can be consulted to find that


a
dx 1 x
 tan 1
2
x 2
a a

Therefore, the integration yields

1 v c
tan 1  d t  C
a a m

The initial condition, v(0) = v0 gives

1 v
C tan 1 0
a a

Substituting this result back into the solution yields

1 v c 1 v
tan 1  d t  tan 1 0
a a m a a

Multiplying both sides by a and taking the tangent gives

 c v 
v  a tan  a d t  tan 1 0 
 m a

or substituting the values for a and simplifying gives

mg  gcd cd 
v tan  t  tan 1 v0  (2)
cd  mg 
 m

(c) We use Eq. (2) until the velocity reaches zero. Inspection of Eq. (2) indicates that this occurs when the
argument of the tangent is zero. That is, when

, 4


gcd cd
t zero  tan 1 v0  0
m mg

The time of zero velocity can then be computed as

m cd
t zero   tan 1 v0
gcd mg

Thereafter, the velocities can then be computed with Eq. (1.9),
mg  gcd 
v tanh  (t  t zero )  (3)
cd  
 m 

Here are the results for the parameters from Example 1.2, with an initial velocity of –40 m/s.

68.1  0.25 
t zero   tan 1  (40)   3.470239 s
9.81(0.25)  
 68.1(9.81) 

Therefore, for t = 2, we can use Eq. (2) to compute

68.1(9.81)  9.81(0.25) 0.25  m
v tan  (2)  tan 1 ( 40)   14.8093
0.25  68.1 68.1(9.81)  s
 

For t = 4, the jumper is now heading downward and Eq. (3) applies

68.1(9.81)  9.81(0.25)  m
v tanh  (4  3.470239)   5.17952
0.25  68.1  s
 

The same equation is then used to compute the remaining values. The results for the entire calculation are
summarized in the following table and plot:

t (s) v (m/s)
0 -40
2 -14.8093
3.470239 0
4 5.17952
6 23.07118
8 35.98203
10 43.69242
12 47.78758