Download Key to PS1

Assignment: Problem Set 1
Name: Juan Wang
Program: E&E
Instructor: Dr. Ming Xu
Due Date: Sept. 13, 2004
P14: 3,4,5,6
3. Solve the following equations
(a) 2 x  4  6
Solution: 2x  4  6
2x  6  4
2x  10 , or 2 x  2
So, x  5 , or x  1
(b) x  3  2
Solution: x  3  2
x  2  3
So, x  5 , or x  1
(c) 2 x  3  5
Solution: 2x  3  5
2x  5  3
2 x  2 , or 2 x  8
So, x  1, or x  4
(d) 7  3x  2
No solutions for this problem because 7  3x  0 .
4. Solve the following equations
(a) 2 x  4  5x  2
Solution: 2x  4  5x  2 , or 2 x  4  (5 x  2)
3x  6 , or 7 x  2
2
So, x  2 , or x  
7
(b) 5  3u  3  2u
Solution: 5  3u  3  2u , or 5  3u  (3  2u )
5u  2 , or u  8
2
So, u  , or u  8
5
1
t
3
 t2
2
2
t 3
t
3
Solution: 4   t  2 , or 4   ( t  2)
2 2
2
2
t  6 ,or 2t  2
So, t  6 ,or t  1
(c) 4 
(d) 2s  3  7  s
Solution: 2s  3  7  s , or 2s  3  (7  s)
3s  10 ,or s  4
10
So, s  , or s  4
3
5. Solve the following inequalities
(a) 5x  2  4
Solution:  4  5x  2  4
 2  5x  6
2
6
So,   x 
5
5
(b) 1  3x  8
Solution: 1  3x  8 , or 1  3x  8
 3x  7 , or  3x  9
7
So, x   , or x  3
3
(c) 7 x  4  3
Solution: 7 x  4  3 , or 7 x  4  3
7 x  1, or 7 x  7
1
So, x   , or x  1
7
(d) 6  5x  7
Solution:  7  6  5x  7
 13  5x  1
1
13
So,   x 
5
5
2
6. Solve the following inequalities
(a) 2 x  3  6
Solution:  6  2x  3  6
 9  2x  3
9
3
So,   x 
2
2
(b) 3  4 x  2
Solution: 3  4x  2 , or 3  4x  2
 4x  1, or  4x  5
1
5
So, x  , or x 
4
4
(c) x  5  1
Solution:  1  x  5  1
So,  6  x  4
(d) 7  2 x  0
No solutions for this problem.
P15: 46-53
46. Assume Seeds= a (biomass)
Substitute biomass= 217g, seeds=17 into the equation
So, 17 = 217a
a = 0.0783
The equation is Seeds = 0.0783(biomass).
47. 1 ft = 0.305 m
1
ft
1m =
0.305
1 2 2
12 m 2  (
) ft
0.305
1m 2  10.750 ft 2
48. 1ha  10,000m 2 ,1acre  4046.86m 2
So,
1ha
10,000m 2

1acre 4046.86m 2
1ha
 2.47
1acre
1ha  2.47acre
49. 1liter  33.81ounces
3
1liter
 33.81
1ounce
(a) ( y )(liters )  ( y  33.81)(ounces)  ( x)(ounces)
x
y
33.81
(b) Substitute x  12ounces into the above equation,
12
 0.35liters
We can get y 
33.81
50. 1mile  1.609kilometers
1kilometer
1

1mile
1.609
(a) ( y )( kilomaters)  ( y )(
1
miles )  ( x)( miles )
1.609
y  1.609 x
(b) Substitute x  261miles into the equation,
We can get y  419.949kilometers
51. (a) We can use the equation we get from problem 50 and substitute x  55miles into
the equation. Then, y  1.609 * 55  88.495kilometers . So the speed limit on many U.S.
highways is 88.49 kilometers per hour.
(b) We can substitute y  130kilometers into the equation, then we get
y
130
x

 80.796miles
1.609 1.609
So, the recommended speed limit on German highways is 80.796 miles per hour.
52. (a) Assume the linear relation between temperature measured in Celsius and
temperature measured in Fahrenheit is y  ax  b
Where
x is the temperature measured in Fahrenheit;
y is the temperature measured in Celsius.
Then we can substitute y  0  C , x  32  F , and y  100  C , x  212  F into the equation.
We get
0  32a  b and 100  212a  b
100 5
160
 , b
Solving this set of equations, we know that a 
180 9
9
5
160
So the linear equation is y  x 
9
9

(b) Substitute x  97.6 F and x  99.6  F into the equation,
We get y  36.4  C and y  37.56  C
So the normal body temperature in humans ranges from 36.4  C to 37.56  C .
4
53.(a) Assume the linear relation between temperature measured in Celsius and
temperature measured in Kelvin is y  ax  b
Where
x is the temperature measured in Kelvin;
y is the temperature measured in Celsius.
We know that 1K denotes the temperature difference as 1  C , so we can get
y  1  a( x  1)  b
y  ax  a  b  1
Combined with the equation y  ax  b , we get a  1
So the equation can be simplified as y  x  b
Then we substitute y  273.15 C , x  0K into the simpler equation,
We get
b  273.15
So, the equation relates the Kelvin and Celsius scales is
y  x  273.15
(b) Substitute x nitrogen 77.4 K , and xoxygen  90.2 K into the equation, respectively.
We get the boiling point temperature of nitrogen is -195.75  C , and the boiling point
temperature of oxygen is  182.95 C .
(c) Substitute ynitrogen  195.75 C , and yoxygen  182.95 C into the equation
5
160
x
9
9
x is the temperature measured in Fahrenheit;
y is the temperature measured in Celsius.
y
Where
We get xnitrogen  320.35 F , and xoxygen  297.31 F
(d) Nitrogen gets distilled first because its boiling point is lower than that of oxygen.
P16: 63-76
63.(a) 75   75  *
(b)

180
180 
17
17
  *
12
12

64.(a)  15   15  *


5

12
 255 

180



12

(b)
3
3
180
  *
 135
4
4

5
3
3

2
65.(a) sin(   )  sin(  )  sin(    )  sin( ) 
4
4
4
4
2
5


3
(b) cos(  )   cos(  )   cos( )  
6
6
6
2
5

3

3  2  3
(c) tan( ) 

1
3
cos( )
3
2
sin(
)
3

2
66.(a) sin(  )  sin( ) 
4
4
2
13
13


3
(b) cos(   )  cos(  )  cos( 2  )  cos( ) 
6
6
6
6
2
4


(c) tan(  )  tan(   )  tan( )  3
3
3
3
1
3,
2
4
5
We find    , or     [0,2 )
3
3
4 5
The solution set is therefore {  ,  }
3 3
(b) Solving tan   3

4
We find   , or     [0,2 )
3
3
 4
The solution set is therefore { ,  }
3 3
67.(a) Solving sin   
1
2,
2
3
5
We find    , or     [0,2 )
4
4
3 5
The solution set is therefore {  ,  }
4 4
(b) Solving sec   2 ,
1
2
We find sec  
cos 
1

5
So cos   ,   , or   
2
3
3
 5
The solution set is therefore { ,  }
3 3
68.(a) Solving cos   
6
69.  sin 2   cos 2   1
sin 2   cos 2 
1


2
cos 
cos 2 
1  tan 2   sec 2 
70.  sin 2   cos 2   1
sin 2   cos 2 
1


2
sin 
sin 2 
2
2
1  cot   csc 
71. Solve 2 cos sin   sin  on [0,2 ) .
Solution:
sin  (2 cos   1)  0
That is, sin   0 or 2 cos  1  0
Solving sin   0 , we find
  0 or 0  
Solving 2 cos  1  0 , we get
1
cos  
2

5

Which yields
or   
3
3

5
The solution set is therefore {0, ,  ,  }.
3
3
72. Solve sec 2 x  3 tan x  1 on [0,  ) .
Solution:
1
3 sin x cos x


0
2
cos x
cos x
cos x
1  3 sin x cos x  cos 2 x
0
cos 2 x
sin 2 x  cos 2 x  3 sin x cos x  cos 2 x
0
cos 2 x
sin 2 x  3 sin x cos x
0
cos 2 x
sin x(sin x  3 cos x)
0
cos 2 x
That is, sin x  0 or sin x  3 cos x  0 and cos x  0
Solving sin x  0 , we find x  0
Solving sin x  3 cos x  0 , we get tan x  3 and cos x  0
7
Which yields x 

3

The solution set is therefore {0, }.
3
73.(a) 43 4 2 / 3  4 (32 / 3)  4 7 / 3
3 2 31 / 2
(b) 1 / 2  3( 21 / 21 / 2)  33
3
5 k 5 2 k 1
(c)
 5 ( k  2 k 11 k )  5 ( 4 k 2)
51k
74.(a) (2 4 2 3 / 2 ) 2  (2 42 / 3 ) 2  (210 / 3 ) 2  2 20 / 3
65 / 2 6 2 / 3 3
)  (6 5 / 2 2 / 31 / 3 ) 3  (617 / 6 ) 3  617 / 2
61 / 3
32 k 3
(c) ( 4 k ) 3  (32 k 34k ) 3  (33k 1 ) 3  39 k 3
3
(b) (
75.(a) x  4  2 
1
16
1
(b) x  ( ) 3  27
3
1
(c) x  10  2 
100
1
76. (a) x  ( )  4  16
2
1 2 1
(b) x  ( ) 
4
16
3
(c) x  5  125
77. (a) x  5
(b) x  6
(c) x  3
78. (a) x  3
(b) x  4
(c) x  4
1
79. (a)  ln( )  (ln 1  ln 3)  ln 3  ln 1
3
2
(b) log 4 ( x  4)  log 4 ( x  2)( x  2)  log 4 ( x  2)  log 4 ( x  2)
(c) log 2 4 3 x 1  (3x  1) log 2 4  2(3x  1)  6 x  2
8
1
80. (a)  ln( )  (ln 1  ln 5)  ln 5  ln 1
5
2
x  y2
( x  y)( x  y)
1
(b) ln
 ln
 ln( x  y)  ln( x  y)  ln x
2
x
x
2 x 1
 2x  1
(c) log 3 3
81. (a) Solve e 3 x 1  2
Solution:
ln e 3 x 1  ln 2
3x  1  ln 2
1  ln 2
x
3
2 x
(b) Solve e  10
Solution:
ln e 2 x  ln 10
 2x  ln 10
ln 10
x
2
x 2 1
(c) Solve e
 10
Solution:
ln e x 1  ln 10
x 2  1  ln 10
x   1 ln 10
2
82. (a) Solve 3 x  81
Solution:
3 x  34
x4
2 x 1
 27
(b) Solve 9
Solution:
3 4 x  2  33
4x  2  3
1
x
4
5x
(c) Solve 10  1000
Solution:
10 5 x  10 3
5x  3
3
x
5
9
83.(a) Solve ln( x  3)  5
Solution:
x  3  e5
x  e5  3
(b) Solve ln( x  2)  ln( x  2)  1
Solution:
ln( x  2)( x  2)  1
x2  4  e
x   e4
Since x  2  0 and x  2  0 , the solution should be x  e  4
(c) Solve log 3 x 2  log 3 2 x  2
Solution:
x2
log 3
 log 3 9
2x
x2
9
2x
x 2  18 x  0
x  0 or x  18
Since both of x 2 and x should be more 0, the solution should be x  18 .
84. (a) Solve ln( 2 x  3)  0
Solution:
2x  3  1
x2
(b) Solve log 2 (1  x)  3
Solution:
1 x  8
x  7
(c) Solve ln x 3  2 ln x  1
Solution:
x3
Ln 2  ln e
x
xe
85. (3  2i)  (2  5i)  (3  2)  (2  5)i  5  7i
86. (7  i )  4  (7  4)  i  3  i
87. (4  2i)  (9  4i )  (4  9)  (4  2)i  13  2i
88. (6  4i )  (2  5i )  (6  2)  (5  4)i  8  i
89. 3(5  3i )  15  9i
90. (2  3i)(5  2i)  10  4i  15i  6  16  11i
91. (6  i)(6  i)  36  i 2  36  1  37
10
92. (4  3i)( 4  2i)  16  8i  12i  6  10  20i
93. z  3  2i  3  2i
94. z  u  (3  2i )  (4  3i)  1  i
95. z  v  (3  2i )  (3  5i )  6  3i  6  3i
96. v  w  (3  5i )  (1  i )  2  6i  2  6i
97. vw  (3  5i )(1  i )  3  5  3i  5i  8  2i
98. uz  (4  3i )(3  2i )   12  6  8i  9i  6  17i
99. z  z  a  bi  a  bi  a  bi  (a  bi )  2a
z  z  (a  bi )  a  bi  (a  bi )  (a  bi )  2bi
100. z  a  bi  a  bi
( z )  a  bi  a  bi
So, (z ) is the same as the value of z .
101. Solving 2 x 2  3x  2  0
 (3)  (3) 2  4  2  2 3
7
 
i
2 2
4 4
102. Solving 3x 2  2 x  1  0
Solution: x 
 (2)  (2) 2  4  3  1 2  2 2i 1
2

 
i
23
6
3 3
103. Solving  x 2  x  2  0
Solution: x 
 1  12  4  (1)  2  1  3

2  (1)
2
So, x  2 or x  1
104. Solving  2 x 2  x  3  0
Solution: (2 x  3)( x  1)  0
3
So, x  or x  1
2
2
105. Solving 4 x  3x  1  0
Solution: x 
3  (3) 2  4  4  1 3
7
Solution: x 
 
i
2 4
8 8
106 Solving  2 x 2  4 x  3  0
Solution : x 
 4  4 2  4(2)( 3)  4  2 2i
2

 1
i
2(2)
4
2
11
107 Solving 3 x 2  4 x  7  0
Solution :   b 2  4ac  (4) 2  4 * 3 * (7)  100  0
So, the solutions are real numbers, that is ,
x  7/3
or
x  1
108 Solving 3 x 2  4 x  7  0
Solution :   b 2  4ac  (4) 2  4 * 3 * 7  68  0
So, the solutions are complex numbers, that is,
x
4  68i 2
17
 
i
2*3
3
3
109 Solving  x 2  2 x  1  0
Solution :   b 2  4ac  2 2  4 * (1) * (1)  0
So, the solutions are two equal real numbers, that is
x1  x 2 
2
1
2 * (1)
110 4 x 2  x  1  0
  b 2  4ac  (1) 2  4 * 4 *1  15  0
So, the solutions are real complex numbers, that is
x
 (1)  15i
1
15
 
i
2*4
8
8
111 Solving 3 x 2  5 x  6  0
Solution :   b 2  4ac  (5) 2  4 * 3 * 6  47  0
So, the solutions are real complex numbers, that is
x
 (5)  47 i
5
47
 
i
2*3
6
6
112 Solving  x 2  7 x  2  0
Solution :   b 2  4ac  7 2  4 * (1) * (2)  41  0
So, the solutions are real real numbers, that is
x
 7  41
7  41

2 * (1)
2
12
113 Suppose z  a  bi
( z )  (a  bi )  a  bi  a  bi  z
o
114 Suppose z  a  bi , w  c  di
z  w  (a  bi )  (c  di )  (a  c )  (b  d )i  (a  c )  (b  d )i
z  w  a  bi  c  di  (a  bi )  (c  di )  (a  c )  (b  d )i
So, z  w  z  w
115 Suppose z  a  bi , w  c  di
zw  (a  bi )(c  di )  (ac  bd )  (ad  bc )i  (ac  bd )  (ad  bc )i
z w  a  bi * c  di  (a  bi )(c  di )  (ac  bd )  (ad  bc )i
So, zw  z w
P43: Problems of odd number:
1. f ( x )  x 2 , x  R
The range of f is [0,).
Graph:
13
3.
f ( x )  x 2 ,1  x  0
The range of f is [0,1)
Graph:
x 2 1
 x 1
x 1
x 2  1 ( x  1)( x  1)

 x  1, x  1
x 1
x 1
x2 1
(b) f ( x ) 
, x  1 and g ( x )  x  1, x  R are not equal since they have difference
x 1
domains and ranges.
5(a) Show that for x  1,
7. f ( x )  2 x
Graph :
From the graph, we can see that f ( x )  2 x is an odd function.
Check : f ( x )  2 * ( x )  2 x   f ( x )
14
9. f ( x)  3 x
Graph:
From the graph, we can see that f ( x )  3 x is an even function.
Check : f ( x )  3( x )  3 x  f ( x )
11. f ( x)   x
Graph:
From the graph, we can see that f ( x )   x is an even function.
Check : f ( x )    x   x  f ( x )
p
15
15. Suppose that
f ( x )  1  x 2 , x  R and g ( x )  2 x , x  0
(a)( f  g )( x )  f [ g ( x )]  1  [ g ( x )] 2  1  (2 x ) 2  1 - 4 x 2 , x  0
(b) ( g  f )( x )  g[ f ( x )]  2 f ( x ), f ( x )  0
f ( x)  1  x 2
}
}
 ( g  f )( x )  2(1  x 2 )  2  2 x 2 , 1  x 2  0
 ( g  f )( x )  2(1  x 2 )  2  2 x 2 ,  1  x  1
P44: Problems of even number:
16. Suppose that
1
, x  1 and g ( x )  2 x 2 , x  R
x 1
1
(a) ( f  g )( x )  f [ g ( x )] 
, g( x)  1
g( x)  1
f ( x) 
}
g( x)  2 x 2
 ( f  g )( x ) 
1
, 2x2  1
2x 1
1
2
 ( f  g )( x ) 
, x
2
2
2x 1
2
(b) ( g  f )( x )  g[ f ( x )]  2[ f ( x )] 2
}
1
,x 1
x 1
1 2
2
 ( g  f )( x )  2(
)  2
,x 1
x 1
x  2x 1
f ( x) 
18. Suppose that
f ( x )  x 4 , x  3 and g ( x ) 
x  1, x  3
Solution :
( f  g )( x )  f [ g ( x )]  [ g ( x )] 4 , g ( x )  3
g( x) 
x  1, x  3
}
 ( f  g )( x)  ( x  1) 4  ( x  1) 2  x 2  2 x  1, x  1  3
 ( f  g )( x)  x 2  2 x  1, x  8
16
20. Suppose that f ( x )  x 4 , x  0,
Solution : ( f  g )( x )  f [ g ( x )]  [ g ( x )] 4 , g ( x )  0
( g  f )( x )  g[ f ( x )]  g ( x 4 )
( f  g )( x )  ( g  f )( x )
}
 [ g( x)] 4  g( x 4 )
Let’s assume that g( x)  (a  bx) n  0
So, [(a  bx ) n ] 4  (a  bx 4 ) n
 (a  bx ) 4  a  bx 4
 C 40 a 4  C 41 a 3 (bx )  C 42 a 2 (bx ) 2  C 43 a (bx ) 3  C 44 (bx ) 4  a  bx 4
 a 4  a , b 4  b, a 3 b  0, ab 3  0, a 2 b 2  0
 a  0, b  1 or a  1, b  0
If a  0, b  1, we can get f  g  [ g ( x )] 4  x 4 n , x n  0  x  [0,) if n is odd,
and x  (, ) if n is even
If a  1, b  0, we can get f  g  [ g( x)]  1
4
Thus, g ( x ) 
{
x n , x  [0,) if n is odd
1,
x  (,) if n is even
xR
22.Graph:
f ( x)  x 3 , x  0
g( x)  x 5 , x  0
When x  (0,1), f ( x )  g ( x );
when x  (1,), f ( x )  g ( x )
17
24.(a) Gragh :
f ( x )  x, x  0
g( x)  x 2 , x  0
(b) When x  [0,1], f ( x )  g ( x );
when x  [1,), f ( x )  g ( x )
26.(a) 0  x  1  ln x  ln 1( 0)
mn
}
 m ln x  n ln x
 ln x m  ln x n
 xm  xn
(b) x  1  ln x  0
mn
}
 m ln x  n ln x
 ln x m  ln x n
 xm  xn
30.(a) Let' s assume n  2m , m is integer.
f ( x)  y  x n  x 2m
f ( x )  ( x ) n  ( x ) 2 m  x 2 m  f ( x )
So, y  x n , x  R is an even function w hen n is an even integer.
18
(b) Let' s assume n  2m  1, m is integer.
f ( x )  y  x n  x 2 m 1
f ( x )  ( x ) n  ( x ) 2 m 1  x 2 m ( x )   x 2 m 1   f ( x )
So, y  x n , x  R is an odd function w hen n is an odd integer.
30.(a) R( x )  kx (a  x )  akx  kx 2   kx 2  akx
So, it is a polynomial and its degree is 2
(b) Substitute k  2, a  6 into R( x ), we get
R( x )  2 x (6  x )  2 x 2  12 x ,0  x  6
Graph: When x  3, the reaction rate is maximal.
32. The function is A   (10t ) 2  100t 2
Where A is the affected area
t is the time measured in days.
Substitute t  2, 4, 8 into the function, respective ly,
we get
A2  400 ( ft ) after 2 days
A4  1,600 ( ft ) after 4 days
A8  6,400 ( ft ) after 8 days
19
34
2x
( x  2)( x  3)
Solution : Domain : { x | x  2 and x  3}
f ( x) 
Range : (-,)
36
1
x 1
Solution : Domain : {x | x  R}
f ( x) 
2
Range : (0,1]
P45: problems of odd number
37 Graph:
f ( x )  1 / x, x  0
f ( x)  1 / x 2 , x  1
From the curves, we find that when x  1, the two curves intersected; the function
f ( x)  1 / x 2 is greater for small values of x; for large values of x, function f ( x)  1 / x is greater.
20
39
f ( x) 
1
, x  1
x 1
(a) Graph:
(b) Based on the graph in (a), the range of f(x) is (0,+  ).
(c) If f ( x)  2, x  0.5
(d) Based on the graph in (a), only one solution f ( x)  a has, where
a is in the range of f ( x).
41
f ( x) 
3x
, x0
1 x
(a) Graph:
21
(b) The range of f ( x) is [0,3)
(c) iIf f ( x)  2, x  2
(d) Based on the graph in (a), explain in words (?).
f ( x)  a

3x
a
1 x
 (3  a) x  a
x
43
a
3a
r(N )  5
N
, N 0
1 N
(1)
0 .1
 0.45
1  0. 1
0 .2
f ( N  0 .2 )  5
 0.83
1  0. 2
The percentage increase is (0.83  0.45) / 0.45 *100%  84%
f ( N  0.1)  5
(2)
10
 4.55
1  10
20
f ( N  20)  5
 4.76
1  20
The percentage increase is (4.76 - 4.55)/4.55 *100%  4.62%
Conclusion: the percentage increase when the nutrient concentration is doubled from
N=0.1 to N=0.2 is larger than that when the nutrient concentration is doubled from N=10
to N=20.
f ( N  10)  5
22
45
f ( x) 
x2
, x0
4  x2
(a) Graph:
(b) Based on the graph in (a), the range of f (x) is [0,1).
(c) As x gets larger, f ( x) gets larger, but the increase rate of f ( x) gets smaller.
47 y  x 3 / 2 , x  0
Graph:
23
49 y  x 1 / 4 , x  0
Graph:
51
(a) Graph:
f ( x)  x 1 / 2 , x  0
f ( x)  x 1 / 2 , x  0
24
(b) 0  x  1  ln x  0
1
1
  ln x  0, ln x  0
2
2
1
1
  ln x  ln x
2
2
1 / 2
 ln x
 ln x 1 / 2
 x 1 / 2  x 1 / 2
(c) x  1  ln x  0
1
1
  ln x  0, ln x  0
2
2
1
1
 ln x   ln x
2
2
1/ 2
 ln x  ln x 1 / 2
 x 1 / 2  x 1 / 2
P46: Problems of even numbers
54 Volume fraction  (leaf thickness) 0.49 , leaf thickness  0
Graph:
Based on the graph, as the leaf thickness increases, the volume fraction of spongy
mesophyll deceases.
25
56
V  L3 , M  0.35V
 M  0.35 L3
L(
M 1/ 3
) ,M  0
0.35
58 N (t )  40  2 t , t  0
(a) N (t  0)  40 * 2 0  40
(b)
ln 40et ln 2  ln 40  ln et ln 2  ln 40  ln eln 2  ln 40  ln 2t ln e  ln 40  ln 2t  ln 40 * 2t  ln N (t )
t
 N (t )  40e t ln 2 , t  0 ,
(c)
N (t )  40 * 2 t  1000
2 t  1000 / 40  25
t  log 2 25
60 W (t )  W 0 e  t , t  0
Solution :
(a ) It is the half time of C14 , which is 5730 years.
(b) W(t)  5, W0  20
5  20e t
e t  0.25
Since we know that
ln2
 0.00012095
5730years
ln 0.25
t
 11,462 years
 0.00012095

26
62 W (t )  W 0 e  t , t  0
Solution
W (5)  37% * W 0  W 0 e 5
ln 0.37
 0.1989
5
W (Th )  50% * W 0  W 0 e Th
 
W (5)
 e  (5Th )  0.74
W (Th )
 Th  5 
ln 0.74

 5
ln 0.74
 3(days)
0.1989
64 W (t )  W 0 e  t , t  0
Solution :
W (t )
 e  t
W (0)
1
t  ln
0.35
5730 years
1
t
ln
 8679 years
ln 2
0.35
So, we find that the wood was cut about 8,679 years ago.
0.35 
66 W (t )  W 0 e  t , t  0
Solution :
Let' s assume that the weight of the rock is W .
10
0.00047% * W
 e 5.335*10 *t
W0
W 0  0.00047% * W  0.000079% * W
 0.8561  e 5.335*10
10
*t
1
0.8561  291,224,156 years
t
5.335 * 10 10
So, we find that the rock is 291,224,156 years old.
ln
68 L( x)  L (1  e  kx ), x  0
(a) Graph for L  20, k  1 :
27
Graph for L  20, k  0.1
(b) For k=1, L( x)  90% L  L (1  e  x )
 0.9  1  e  x
 e  x  0.1
 x   ln 0.1
L( x)  99% L  L (1  e  x )
 0.99  1  e  x
 0.01  e  x
 x   ln 0.01  2 ln 0.1
The fish cannot attain length L .The meaning of L : The length of the fish can not
grow unlimited. L is the limit of the fish length.
28
(c) The growth curve for k=1 reaches 90% of L faster.
When k is varied, the growth rate of the curve becomes different. In particular, if k is
increased, the growth curve increases faster to approaches L (for fixed L ).
P47: problems for odd numbers
71 (a) Show that
f ( x)  x 2  1, x  0, is one to one and find its inverse together with its domain.
Solution:
Suppose x1  x 2 , and x1  0, x 2  0
 f ( x1 )  x1  1, f ( x 2 )  x 2  1, and f ( x1 )  f ( x 2 )
2
2
 f ( x ) is one to one function.
Inverse function is f ( x )  ( x  1)1 / 2 , x  1
(b) Graph:
f ( x )  x 2  1, x  0
f ( x )  ( x  1)1 / 2 , x  1
f ( x )  x, x  R
73 (a) Show that
f ( x)  1 / x 3 , x  0, is one to one and find its inverse together with its domain.
Solution :
Suppose x1  x 2 , and x1  0, x 2  0
 f ( x1 )  1 / x1 , f ( x 2 )  1 / x 2 , and f ( x1 )  f ( x 2 )
3
3
 f ( x ) is one to one function.
1
Inverse function is f 1 ( x)  ( )1 / 3 , x  0
x
29
(b) Graph:
f ( x)  1 / x 3 , x  0
f ( x)  x 1 / 3 , x  0
f ( x )  x, x  R
75 Function f ( x)  3 x is one to one function, so inverse function is
f 1 ( x)  log 3 x, x  0
Graph:
f ( x)  3 x , x  R
f ( x)  log 3 x, x  0
f ( x )  x, x  R
30
1
77 Function f ( x )  ( ) x is one to one function, so inverse function is
4
f 1 ( x)  log 1 / 4 x, x  0
Graph:
1
f ( x)  ( ) x , x  R
4
f ( x)  log 1 / 4 x, x  0
f ( x )  x, x  R
79 Inverse function: f 1 ( x)  log 2 x, x  1
Graph:
f ( x)  2 x , x  0
f ( x)  log 2 x, x  1
f ( x )  x, x  R
81
(a) 2 5 log2 x  x 5
(b) 3 4 log3 x  x 4
(c) 55 log1 / 5 x  5 5 log5 x  x 5
31
(d) 4 2 log2 x  2 4 log2 x  x 4
(e) 2 3 log1 / 2 x  2 3 log2 x  x 3
( f ) 4 log1 / 2 x  2  2 log1 / 2 x  2 log2 x  x
83
(a ) ln x 2  ln x 3  ln( x 2 * x 3 )  ln x 5
x4
 ln x 6
x 2
( x  1)( x  1)
(c) ln( x 2  1)  ln( x  1)  ln
 ln( x  1)
x 1
(d ) ln x 1  ln x 3  ln( x 1 * x 3 )  ln x  4
(b) ln x 4  ln x  2  ln
85
(a ) 3 x  e x ln 3
(b) 4 x
2
1
 e 2 ( x 1)( x 1) ln 2
(c) 2  x 1  e ( x 1) ln 2
(d ) 3  4 x 1  e ( 4 x 1) ln 3
87 Solution:
y  (1 / 2) x  e x ln1 / 2  e x (ln1ln 2)  e  x ln 2  e  x
where   ln 2  0
89 Solution :
Using the substituti on scheme of Jukes and Cankor, K and P are related by
3
4
K   ln( 1  p)
4
3
The variable denotes the proportion of observed nucleotide s difference s,
which is 47/300  0.1567. We thus find
3
4 47
K   ln( 1  *
)  3.2008
4
3 300
90 H  ( p1 ln p1  p2 ln p2      p s ln p s )
(a) s  5, p1  p 2      p5  1 / 5
1 1 1 1 1 1 1 1 1 1
 H  ( ln  ln  ln  ln  ln )
5 5 5 5 5 5 5 5 5 5
1
  ln
5
 ln5
32
(b) S  10, p1  p 2      p10  1 / 10
1
1
1
1
1
1
1
1
1
1
ln  ln  ln  ln  ln
10 10 10 10 10 10 10 10 10 10
1
1
1
1
1
1
1
1
1
1
 ln  ln  ln  ln  ln )
10 10 10 10 10 10 10 10 10 10
1
  ln
10
 ln10
 H  (
(c) for S  5, H / ln S  ln 5 / ln 5  1
for S  10, H / ln S  ln 10 / ln 10  1
(d) If S  N , and all species are equally abundant
then
then
p1  p 2      p s  1 / N
H / ln S 
1
1
1
ln )
ln
N N   N  ln N  1
ln N
ln N ln N
 N(
92 y  sin x, and y  sin( 2 x)
Graph:
y  sin x
y  sin( 2 x )
94 y  cos x, and y  cos( 2 x)
Graph:
y  cos x
y  cos( 2 x )
33
96 y  tan x, and y  tan( 2 x)
Graph:
y  tan x
y  tan( 2 x )
x
98 f ( x)  2 sin( ), x  R
2
The amplitude is 2, and the period is
2
 4 .
(1 / 2)
3

100 f ( x)   sin( x), x  R
2
2
The amplitude is 3/2, and the period is

 4.
( )
2
102 f ( x)  7 cos( 2 x), x  R
The amplitude is 7, and the period is
2
2
 .
2
2
3
104 f ( x)   cos( x), x  R
3

The amplitude is 2/3, and the period is
2
2
  2.
3
3
( )

34