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1.1.Determine the quotient and remainder of the following algebraic fractions and reduce the
remaining proper fraction to a sum of partial fractions.
a)
2
5 1
2
x
x x

 
Since the degree of the numerator is less than the degree of the denominator, we see
that the above algebraic fraction is proper, i.e., the quotient is 0 and the remainder is
5 1x .

Now to compute its partial fraction expansion, we first need to factor the
denominator. We have

  2
5 1 5 1
2 1 2
1 2
x x
x x x x
A B
x x
 

   
 
 
for some constants A and B. Multiplying both sides by   1 2x x  , we obtain
   
   
5 1 2 1
2
2 .
x A x B x
Ax A Bx B
A B x A B
    
   
   
Equating corresponding coefficients, we obtain the following linear system:

5;
2 1.
A B
A B
 
 
Adding these equations, we obtain
3 6,A 
whence
2.A 
Plugging this value into the first equation, we obtain
2 5,B 
whence
2.B 

Thus we have the following partial fraction expansion:

2
5 1 2 3
.
2 1 2
x
x x x x

 
   
b)
2
2
2 8 7
.
5 6
x x
x x
 
 
Since the degrees of the numerator and denominator are equal, the quotient is equal to
the quotient of the coefficient of the numerator and the coefficient of the
denominator, which is 2 /1 2. Now we have
 2 2
2 2
2 8 7 2 5 6
2 8 7 2 10 12
2 5.
x x x x
x x x x
x
    
     
  
Thus, the quotient is 2 and the remainder is 2 5x  , whence
  
2
2 2
2 8 7 2 5
2
5 6 5 6
2 5
2 .
2 3
x x x
x x x x
x
x x
  
 
   

 
 
Next we write

  
2 5
.
2 3 2 3
x A B
x x x x

 
   
Multiplying both sides by   2 3x x  , we obtain
   
   
2 5 3 2
3 2
3 2 .
x A x B x
Ax A Bx B
A B x A B
    
   
   

Equating corresponding coefficients, we obtain

2;
3 2 5.
A B
A B
 
 
Multiplying the first equation by 3, we obtain
3 3 6.A B 

Subtracting the second equation from this one yields
1.B 

Substituting this equation into the first one, we obtain
1 2,A 
whence
1.A

Thus we have

  
2 5 1 1
,
2 3 2 3
x
x x x x

 
   
whence

2
2
2 8 7 1 1
2 .
5 6 2 3
x x
x x x x
 
  
   
1.2.We wish to solve the following engineering problems:
a) Consider an RL circuit with resistance R, inductance L, and an initial current of 0.i The
current decays according to the formula
  /0 .
Rt Li t i e
We wish to calculate the current after 1.7st  given 0 14A,i  4 ,R   and 8H.L 
We have
 
  
/
0
4 1.7 /8
14
5.98A.
Rt Li t i e
e





b) The curve assumed by a heavy chain or cable is

cosh .
x
y C
C

If 60,C  we wish to calculate the value of y when 120.x 
We have
cosh
120
60cosh
60
60cosh 2
225.73.
x
y C
C




c) We are given a time-varying current  i t given by

   35cos 0.7 , 0.i t t t  

We wish to find the time t in which the current is first zero.
We have
   35cos 0.7 0,i t t  

whence
 cos 0.7 0.t  

Thus we have

2
0.7 ,t  
whence

2
0.7
2.27s.
t  

1.3.We wish to solve the following scientific problems which involve arithmetic and geometric
series:
(I) A lathe tool has to accommodate work between 25 mm and 300 mm in diameter. Six
spindle speeds are required and the cutting speed is to be 25...