Download Unit III, Functions

Unit III
Functions
Unit III
Functions
In this unit, we discuss on some basic concepts on functions and types of functions, the conditions on which
a function has an inverse, the method of finding the derivative of the inverse of a function without finding
the formula for the inverse function, how to define the inverse trigonometric functions, derivatives of the
inverse trigonometric functions, Hyperbolic Functions, inverse hyperbolic functions and their derivatives,
the L’Hopital’s rules and some of the applications of these inverse functions in integrating some special
types of integrals.
3.1 Basic Ideas
Before we discuss on the conditions for the existence of an inverse functions, we state the definition of a
function and explore different types of functions.
Definition 3.1 f is said to be a function from set A into set B, denoted by
f: A  B if and only if
i) Domain of f = A
ii) No two elements in B have the same pre-images in A.

  
x ,  x A
,  . Show that f (x) = x 2 and g (x) =
2
 2 2
Example 1.Let A = ( 1, 1) and B =  
are functions from A into B.


Solution  x(1, 1), x2  [0, 1)   
 
,  and Since the square of a real number and the product of
2 2
two real numbers are unique, every member
of A has a unique image under both functions in B.
Therefore, both f and g are functions from A into B.
Definition 3.2 (Onto function)
A function f is said to be onto from set A into set B, denoted by
f: A  B if and only if range of f = B.

  
x is an onto function while
,  . Show that g (x) =
2
 2 2
Example 2.Let A = ( 1, 1) and B =  
f (x) = x 2 is not an onto function from A into B.
  
,  . We need to check for the existence of an x in ( 1, 1) such that g (x) = y.
 2 2
Solution i) Let y   
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Now y =

2
2
  
x x=
y . But y    ,    1 
y  1. Hence x  ( 1, 1), so range
2


 2 2
of g is B.
Therefore, g is onto.
  
ii) f (x) = x 2  x  ( 1, 1)  f (x)  [0, 1). Hence any y    ,  \ [0, 1) doesn’t belong

2 2
to range of f .
Therefore, f is not an onto function.
Definition 3.3 (One to one function)
A function f is said to be a one to one function from set A into set B if and
only if no element in B has more than one pre-image in A.

  
x ,  x A is a one to one function
,  . Prove that g (x) =
2
 2 2
Example 3 Let A = ( 1, 1) and B =  
from A into B.
Solution Proof by contradiction
Suppose there exists an element in B that has two distinct pre-images in A.
Let f (x1) = y and f (x2) = y for some y  B. Then f (x1) = f (x2).
f (x1) = f (x2) 

2
x1 =

x2  x1 = x2 .
2
Therefore, f is a one to one function from A into B.
Definition 3.4 (1  1 function)
A function f is said to be a 1  1 function from set A into set B if and
only if f is both onto and one to one from A into B.
Example 4 Show that the function given in example 3 is a 1  1 function from A into B.
Solution Follows from the results of example 2 and example 3.
The inverse of a function f from A into B is a function from B into A if and only if f is a 1  1
function from A into B. Hence the inverse of a real valued function f on any subset of the set of real
numbers is a function if and only if f is either strictly increasing or strictly decreasing on that interval.
Properties of Inverses
Let f be a function from A into B. If the inverse of f from B into A is a function it is usually denoted by
f
1
and these functions satisfy the condition that
f (x) = y if and only if f  1 (y) = x x  A and y  B.
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Furthermore; the two functions have the following relations.
i)
f 
1 1
1
ii) f
f
iii) f
(x) = f (x) x  dom. f.
 f ( x) = x x  dom. f.
1

( y) = y  y  range of f.
Even if we know that a 1  1 function has an inverse it may be difficult to find a formula for the
inverse function. Now let us state some simple method of finding the relations on the continuity and
differentiability of these functions with out finding the formula for the inverse function.
Continuity and Differentiability of Inverse Functions
Theorem 3.1 Let f: I  J be invertible, that is, f
i) If f is continuous on I, then f
1
ii) If f is differentiable on I, then f
 f  ' (y) =
1
1
: J  I exists. Then
is continuous on J.
1
is differentiable on J and
1
f ' ( x)
for x such that f (x) = y, provided that f  (x)  0.
 
 
1
1
' (2) and f
' (6).
Example 5 Let f (x) = x 3 + 3x +2. Then find f
Solution f being a polynomial, it is differentiable with f  (x) = 3x2 + 3 > 0  x  .
Thus f is strictly increasing and hence it has an inverse. Now let y = f (x). Then
y = 2 and y = x 3 + 3x +2  x 3 + 3x + 2 = 2  x = 0. Hence f ' (0) = 3.
y = 6 and y = x 3 + 3x +2  x 3 + 3x + 2 = 6  x 3 + 3x  4 = 0  x = 1. Hence f ' (1) = 6.
  ' (2) = 13 and  f  ' (6) = 16 .
 ' (y) if f (x) =
Example 6 Find a formula for  f
Therefore, f
1
1
1
3
x.
Solution Now domain of f = range of f = .
2
1 3
1 2
If f (x) = x , then f  (x) = x
and y = 3 x if and only if x = y3. Moreover; f  (x) = x
3
3
1 '
2
Therefore, f
(y) = 3 x .
3
 
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Exercises 3.1
In exercises 1-8, find an interval on which each of the following functions has an inverse.
1. f (x) = x
3
 5 x 1
5. f (x) =  4x 3  1
9. Find ( f
2. f (x) = sin 2 x
6. f ( x) 3 x  x
3. f (x) =
1
1 x
x
7. f ( x) 
1 x
4. f (x) =
2
x
1 x2
8. f ( x)  x 5  x 3
1
) ' (2e2), where f (x) = x ℓn x.
3.2 Inverse Trigonometric Functions
This section mainly focuses on how to develop a method of defining the inverses of the trigonometric
functions and then on how to find their derivatives and their applications in integrating some special types
of integrals.
We fully discuss on the sine, tangent and secant functions and the other trigonometric functions are left to
the reader.
I. Let f (x) = sin x.
The sine function f (x) = sin x is neither strictly increasing nor strictly decreasing in its entire domain, and
hence the sine function has no inverse in its entire domain. Now we need to choose an interval on which the
sine function has an inverse on the restricted domain. To do so, choose an interval around the origin on
which the sine function is strictly increasing and the cosine function is non-negative, since the sine and the
cosine functions are related by the identity
sin 2 x  cos 2 x 1 ,  x  .
  
Restrict its domain to  ,  , so that the new function becomes strictly increasing on the restricted
 2 2
domain and the cosine function is non-negative.
Definition 3.5 The inverse sine function
  
, 
 2 2
whose value at any x   1, 1 usually denoted by
f
1
:  1, 1 
arcsin x or sin 1 x , is defined by:
  
y = sin 1 x if and only if x = sin y  y   ,  .

2
2
Example 1 Find the exact value of each of the following expressions.
i) sin 1 0
ii) sin 1
3
2

2 
iii) sin 1  
2 

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Solutions i) Let x = sin 1 0 .
  
  
x = sin 1 0 and x   ,   sin x = 0 and x   , 
 2 2
 2 2
  
,   x = 0.
 2 2
 x = n , where n  Z and x  
Therefore, sin 1 0 = 0.
ii) Let x = sin 1
3
.
2
3
 
and x   ,
2
 2


  
x=
+ (4n + 1) , where n  Z and x   ,   x =
6
2
 2 2
 3

 =
Therefore, sin 1 
.

3
 2 
x = sin 1
3
  
and x   ,   sin x =
2
 2 2

2 

.
3

2 
iii) Let x = sin 1  
.
2 


2
2 
  
  
x = sin 1  
and x   ,   sin x = 
and x   , 

2 
2
 2 2
 2 2




  
+ (4n  1) , where n  Z and x   ,   x =  .
4
2
4
 2 2


2 
Therefore, sin 1  
=

.
2 
4

x=
From the definition of the arcsine function we get:
  
y = sin 1 x if and only if x = sin y  x   1, 1 and  y   ,  .

2
2
Therefore, In general we can conclude that:
sin ( sin 1 x ) = x ,  x   1, 1
  
and sin 1 (sin x) = x,  x   ,  .

2
2
Now we need to find the derivative of the arcsine function and its applications on evaluating some integrals.
y = sin 1 x if and only if x = sin y.
Hence taking the derivative of both sides of x = sin y with respect to x, we get:
1 = (cos y)
dy
dy
1

=
dx
dx
cos y
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  
,  , cos y > 0 and hence cos y = 1  sin 2 y = 1  x 2 .
 2 2
But  y   
Therefore
sin x' =
1
1
1 x2
 x   1, 1 .
From the nature of the derivative of the arcsine function we can observe that integrals of the form
dx

a
2
 x
, where a > 0 can be evaluated by substituting x = a sin t.
2
Example 2 Integrate
dx

a
2
, where a > 0.
2
 x
Solution Let x = a sin t.
Then dx = a cos t dt and a 2  x 2 = a cos t. Hence
dx

a
Therefore,
dx

a
2
= sin
 x
9 x2
0
 dt = t + c, where c  .
x
  + c, where c  .
a
dx

Example 3 Evaluate
 x
=
2
1 
2
3 3
2
2
.
Solution Let x = 3 sin t. Then dx = 3 cos t dt and 9  x 2 = 3 cos t.
3 3
2
Hence
dx

9 x2
0
3 3
2
Therefore,

0

3
=
dx
9 x2



dt = t 3 = .
0
0
=
3

.
3
II Let f (x) = tan x
The tangent function f (x) = tan x is neither strictly increasing nor strictly decreasing, and hence it has no
inverse in its entire domain. Now we need to choose an interval, around the origin, on which the tangent
function is strictly increasing and the secant function is non-negative, since the tangent and the secant
functions are related by the identity
  
1  tan 2 x  sec 2 x ,  x    ,  .
 2 2
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

Restrict its domain to  
 
,  , so that the new function becomes strictly increasing on the restricted
2 2
domain and the secant function is non-negative.
Definition 3.6 The inverse tangent function
  
f 1 :  ,     , 
 2 2
whose value at any x   ,  usually denoted by
arctan x or tan 1 x , is defined by:
  
, .
 2 2
y = tan 1 x if and only if x = tan y  y   
Example 4 Find the exact value of each of the following expressions.
 1 

iii) tan 1  
3

ii) tan 1 3
i) tan 1 0
Solutions i) Let x = tan 1 0 .
  
  
,   tan x = 0 and x    , 
 2 2
 2 2
x = tan 1 0 and x   
  
,   x = 0.
 2 2
 x = n , where n  Z and x   
Therefore, tan 1 0 = 0.
ii) Let x = tan 1 3 .
  
x = tan 1 3 and x    ,   tan x =

2
2
  
, 
 2 2
3 and x   


  
+ n, where n  Z and x    ,   x = .
3
3
 2 2

Therefore, tan 1 3 = .
3
x=
 1 
 .
iii) Let x = tan 1  
3

1
 1 
  
  
 and x    ,   tan x = 
x = tan 1  
and x    , 
3
3
 2 2
 2 2



  
+ 2n, where n  Z and x    ,   x =  .
6
6
 2 2

 1 
 =  .
Therefore, tan 1  
6
3

x=
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From the definition of the arctangent function we get:
  
, .
 2 2
y = tan 1 x if and only if x = tan y , y   
Therefore, In general we can conclude that:
tan ( tan 1 x ) = x ,  x   , 
  
and tan 1 (tan x) = x,  x    ,  .

2
2
We need to find the derivative of the arctangent function and its applications on evaluating some integrals.
Now y = tan 1 x if and only if x = tan y. Hence taking the derivative of both sides of x = tan y, we get:
1 = ( sec 2 x )
dy
dy
1

=
dx
dx
sec 2 y
  
,  , sec y > 0 and hence sec 2 y = 1  tan 2 y = 1  x 2 .
 2 2
But  y   

Therefore, tan
1

x '=
1
1 x2
,  x   ,  .
From the nature of the derivative of the arctan function we can observe that integrals of the form
dx
 a 2  x 2 , where a > 0 can be evaluated by substituting x = a tan t.
Example 5 Integrate
dx
 a 2  x 2 , where a > 0.
Solution Let x = a tan t. Then dx = a sec 2 t dt and a 2  x 2 = a 2 sec 2 t .
Hence
Therefore,
dx
 a2  x 2
dx
 a2  x 2 =
3
Example 6 Evaluate
=
1
a
 dt =
1
t + c, where c  .
a
1
1  x 
tan   + c, where c  .
a
a
dx
 9 x 2 .
0
Solution Let x = 3 tan t. Then dx = 3 sec 2 t dt and 9  x 2 = 9 sec 2 t .

1 4
3
dx
Hence 
=
2
3
9

x
0
3
Therefore,
dx
0

1

.
dt = t 4 =
3 0 12

 9  x 2 = 12 .
0
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III. Let f (x) = sec x.
The secant function f (x) = sec x is neither strictly increasing nor strictly decreasing, and hence it has no
inverse in its entire domain. The secant function being the reciprocal of the cosine function and the cosine
function is related to the sine function by
sin 2 x  cos 2 x  1
Now we need to choose an interval, around the origin, on which the sine function is non-negative.


If we restrict its domain to  0,



   ,   , in which sine function is non-negative, then this new
2 2 
function becomes one-to-one in the restricted domain.
Definition 3.7 The inverse secant function
f
1


:  1, 1 '   0,



  ,  
2 2 
whose value at any x usually denoted by arcsec x or sec 1 x , is defined by:


y = sec 1 x if and only if x = sec y  y   0,



  ,   .
2 2 
Example 7 Find the exact value of each of the following expressions.
ii) sec 1
i) sec 1 2
2
Solutions i) Let x = sec 1 2 .





  ,  
2 2 
    
 sec x = 2 and x   0,    ,  
 2 2 


    
 x = 2n   , where n  Z and x   0,    ,    x = .
3
3
 2 2 

1
x = sec 1 2 and x   0,
Therefore, sec
2=
ii) Let x = sec 1
2.
3
.
    


   ,    sec x = 2 and x   0,    ,  
2 2 
 2 2 


    
 x = 2n   , where n  Z and x   0,    ,    x = .
4
4
 2 2 

1
x = sec 1
Therefore, sec


2 and x   0,
2 =
4

.
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From the definition of the arcsecant function we get:
y = sec 1 x if and only if x = sec y  x  1, 1 ' .
Therefore, sec ( sec 1 x ) = x ,  x   1, 1 '


and sec 1 (sec x) = x,  x   0,



  ,   .
2 2 
Furthermore, taking the derivative of both sides of x = sec y with respect to x, we get:
1 = (tan y) (sec y)


But  y   0,
dy
dy
1

=
sec
y
tan y
dx
dx



   ,   , tan y =  sec 2 y  1 =  x 2  1 and sec y tan y > 0.
2 2 
Therefore,
sec 1 x ' =
1
x21
x
 x   1, 1' .
From the nature of the derivative of the arcsecant function we can observe that integrals of the form
dx
, where a > 0 can be evaluated by substituting x = a sec t.
2
2
x x a

Example 8 Integrate
dx

.
x x2  9
Solution Let x = 3 sec t. Then dx = 3 sec t tan t dt and x 2  9 = 3 tan t.
Hence
Therefore,

dx
=
1
3
=
1
 x
sec 1   + c, where c  .
3
3
x x2  9

dx
x x2  9
2 3

Example 9 Evaluate
1
 dt = 3 t + c where c  .
dx
.
2
3 2x x 9
Solution Let x = 3 sec t. Then dx = 3 sec t tan t dt and x 2  9 = 3 tan t.
2 3
Hence


dx
=
2
3 2x x 9
2 3
Therefore,

dx
2
3 2x x 9
6

6
4
4
1
1

dt = t
= 
.
3
3 
36

= 

36
.
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Unit III
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The other inverse trigonometric functions and their derivatives can be defined in a similar way. The table
below presents the domain, range and derivatives of these inverse functions.
Function
Definition of the
Domain
Range
Derivative
[ 1 , 1]
[0, ]

(0, )
1, 1 '
  
  2 , 2  \ 0


inverse functions
cos x' = 
1
Cosine
y = cos x if and
only if x = cos y
Cotangent
y = cot x if and
only if x = cot y
Cosecant
y = csc x if and
only if x = csc y
1
1
csc x' = 
1
 x  ( 1 , 1)
1 x2
cot x' =  1 1x
1
1
1
2
 x 
1
x 1 x
2
 x  1, 1'
Basic relations between the inverse trigonometric functions
1. arcsin x  arccos x =

2. arctan x  arccot x =
2

3. arc sec x  arc csc x =
2
1
 
5. arcsec x  arccos 
 x
1
4. arc csc x  arcsin  
 x

2
1
6. arc cot x  arctan 
 x
Negative argument formulas
1. arcsin ( x)   arcsin x
2. arccos( x)    arccos x
3. arctan( x)   arctan x
4. arcsec( x)    arcsec x
5. arc csc( x)   arc csc x
6. arc cot ( x)    arc cot x
Exercises 3.2
In exercises 1-5, simplify the given expressions.


2. arccos(sin  )
3. cos (2 arcsin x)
4. sin (2 arcsin x)
6
4
In exercises 6-10, find the derivative of each of the following functions.
1. arcsin (sin
)
 x 1
7. f ( x)  tan 1

 x 1 
6. f ( x)  cos1  x 2  1  x 


8. f ( x) 
5. cos (arctan x)
1  arctan x
2  3 arctan x
9. f ( x)  x arcsin x  1  x 2
10. f ( x)  x arctan x  2
In exercises 11-16, integrate the given indefinite integrals.
11.
14.
dx
 x 2  4x  7
x
dx
4 x 4  25
12.
15.
e
x
dx
 1  e  2x

13.
dx
, where a, b > 0
a 2sin 2 x  b 2 cos 2 x
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16.

arctan 2 x dx
1  4x 2
x3
 2 x 2  4 x  3 dx
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Unit III
Functions
In exercises 17-20, evaluate the given definite integrals.
4
17.

4
3
2 32
dx
3
x
x
2

2
18.
4
1
dx
x
2
19.
 4x  8
0
1
x2
4 x
2
dx
20.
dx
0 x 2  4x  7

21. Let f (x) = arcsin x  arccos x . Show that f ' ( x) = 0 and conclude that arcsin x  arccos x = .
2
22. Let f (x) = 2 x 10 arc cot x . Solve f ' ( x) = 0 for x.
3.3 Hyperbolic Functions
Other types of functions called the hyperbolic functions defined below plays fundamental roles in various
disciplines, in particular in integrating some special types of integrals.
Definition 3.8 Let t be a real number. The hyperbolic cosine of t, denoted cosh t
and the hyperbolic sine of t, denoted sinh t are defined by:
cosh t =



1 t
1 t
t
t
and sinh t =
e e
e e
2
2

Note that: i) Dom. Cosh = dom. Sinh =  and range of sinh =  while range of cosh = [1, ).
ii) Hyperbolic cosine is an even function while hyperbolic sine is an odd function.
iii) cosh 2 t  sinh 2 t = 1,  t  .
The other four hyperbolic functions, namely the hyperbolic tangent, the hyperbolic secant and the
hyperbolic cosecant are defined by:
tanh t =
sinh t
1
cosh t
1
, sech t =
, coth t =
and csch t =
.
sinh t
cosh t
cosh t
sinh t
The Derivatives of the Hyperbolic Functions
The derivatives of the hyperbolic functions are presented in the following table.
Function
Derative
Cosh t
Sinh t
Sinh t
Tanh t
Coth t
Sech t
Csch t
Cosh t
sec h 2 t
 csc h 2 t
 Sech t Tanh t
 Csch t Coth t
Inverse Hyperbolic Functions
Since the hyperbolic sine function f (x) = sinh t is one-to-one in its entire domain, there is no need to restrict
the domain of the hyperbolic sine function. Now let x = sinh t.
x = sinh t  x =


1 t
t
t
 2x = e t  e
e e
2
 2x e t = e 2t  1  e 2t  2 x e t  1 = 0  e t  x 
x2 1.
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Unit III
Functions
x 2  1 . Hence t = n  x 

Since e t  0 for any t  , e t  x 
x 2  1  .

Therefore, the inverse of the hyperbolic sine function f (x) = sinh t is given by the formula
x  n  x 

1 '
Example 1 Show that sinh x 
x 2  1 

1
 x  .
x 2 1
Solution using the above formula and the rule for taking derivatives of quotient of function we get:
1
sinh


x
1
sinh x ' =
 x  x 2 1

= 

2
2
x  1 
x  1 
2
x 1
1

Therefore, sinh
1
x
x
' 
1
x 2 1


1

 =


2
 x  x 1
1
2
x 1
.
 x  .
Since the hyperbolic tangent function f (x) = tanh t is one-to-one in its domain, hence there is no need to
restrict its domain. However; tanh t  1, so the inverse function will be defined for x  1.
Now let x = tanh t.
x = tanh t  x =
et  e
t
e e
t
t
 x=
e 2t  1
e 2t  1
 e 2t 
 x e 2t + x = e 2t  1  1 + x = e 2t 1 x 
 1 x
1
1 x
.
 t=
n 
1 x
2  1  x 
Therefore, the inverse of the function tanh t is given by the formula
tanh
1
x =

 1
1
n 
2  1
1

x ' 
x
 for x  1.
x 
1
for x  1.
1 x2
Since sech t is not one-to-one we need to restrict the domain to [0, ). Moreover 0  sech t  1 and so the
Example 2 Show that tanh
inverse function will be defined only for 0  x  1. Now let x = sech t.
x = sech t  x =
2
et  e
t 
1
x e 2 t + x = 2 e t  x e 2t  2 e t  x = 0  e t 
1 x 2
.
x
 1  1 x 2 
.
Since e t  1 for any t  [0, ), t = n 

x



Therefore, the inverse of the function sech t is given by the formula
 1 1 x2
x


sec h  1 x = n 

 for 0  x  1.


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Unit III
Functions

1
 ' 
1
for 0  x  1.
x 1 x2
Solution using the above formula and taking the derivative of a composition of functions we get:


x2

 1  1  x 2  



1 x2 
x
1



sec h x = 
 1  1  x 2  

x2







2 
2
2 
 x   1 x 1 x  



2




1 x
x
1

 = 
= 
for 0  x  1.
2

2
 1 1 x2 
x

x
1

x







1
1
Therefore, sec h x '  
for 0  x  1.
x 1 x2
The table below presents the inverses, derivatives and domains of the remaining three hyperbolic functions.
Example 3 Show that sec h
x


No
1
Function
cosh
2
coth
3
csc h
1
1
1
x
Formula
n  x 

Domain
x 2  1 

x
 x  1
1

n 
2  x  1 
x
1
n  
x

1
Derivative
1
x1
x 21
1
x  1.
1 
x 2 
x0
1  x2
1

x
x 2 1
Hyperbolic Identities
I Addition
a) sinh ( x  y)  sinh x cosh y  cosh x sinh y
b) cosh ( x  y)  cosh x cosh y  sinh x sinh y
tanh x  tanh y
1  tanh x tanh y
coth x coth y  1
d) coth ( x  y) 
coth y  coth x
c) tanh ( x  y) 
II Sum, difference and product
 x y 
 x y 
a) sinh x  sinh y  2 sinh 
 cosh 

 2 
 2 
 x y 
 x y 
b) sinh x  sinh y  2 cosh 
 sinh 

 2 
 2 
 x y 
 x y 
c) cosh x  cosh y  2 cosh 
 cosh 

 2 
 2 
 x y 
 x y 
d) cosh x  cosh y  2 sinh 
 sinh 

 2 
 2 
1
e) sinh x sinh y  cosh ( x  y)  cosh ( x  y)
2
1
f) cosh x cosh y  cosh ( x  y)  cosh ( x  y)
2
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Unit III
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III Half and Double angle Formula
cosh x 1  if x  0

2
 if x  0
cosh x 1
 x
b) cosh    
2
2
 x
a) sinh    
2
d) sinh 2x = 2 sinh x cosh x
2
e) ) cosh 2x = cosh x  sinh 2 x = 1  2 sinh 2 x
= 2 cosh 2 x  1
cosh x 1  if x  0

cosh x 1  if x  0
 x
c) tanh    
2
f) tanh 2 x 
2 tanh x
1  tanh 2 x
Exercises 3.3
In exercises 1-6, prove that:
1.
1  sinh 2 x  cosh 2 x
=  coth x
1  sinh 2 x  cosh 2 x
2. sinh
3. cosh  1 x 2  1 = sinh 1 x for x ≥ 0.
4. If y =
1
cosh
x
cosh t
5. for t > 0,

x 2  1 dx  
1
x 2  1 = cosh
2
1
x
1
1
x for x ≥ 1.
, then (1  x 2)
dy
 xy  1 .
dx
t 1
 sinh 2t
2 4
6. If a = c cosh x and b = c sinh x, then show that (a  b) 2 e
2 x
 a2  b2  c2.
In exercises 7 and 8 find the derivative of f if


1 

8. f ( x)  cosh 1 tanh 
2x 

In exercises 9 and 10 evaluate each of the following integrals.
7. f ( x)  coth ln
x
1
4
0
9. t cosh t sinh t dt
10.
1
cosh x
x
dx
3.4 The L’Hopital’s Rule
In this section we describe a technique for evaluating many indeterminate limits such as:
x
x
im sin x ; im e  x  1 ; im e ; im x n x etc.
x2
x 0
x   x 2 x 0 
x  x
Theorem 3.2 (The Generalized mean Value Theorem)
Let f and g be continuous on [a, b] and differentiable on (a, b). If g  (x)  0
for a < x < b, then there is a number c in (a, b) such that:
f (b)  f (a ) f  (c)

g (b)  g (a )
g  (c )
Proof Define the function h on [a, b] by:
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Unit III
Functions
h (x) = { f (b)  f (a) } g (x)  { g (b)  g (a) } f (x).
Now h is continuous on [a, b] and differentiable on (a, b). Thus by the mean value theorem there is a
number c in (a, b) such that:
h (b)  h (a)
 h (c) , since h (a) = h (b), h (c) = 0.
ba
Now h (x) = { f (b)  f (a) } g  (x)  { g (b)  g (a) } f  (x) and since g  (c)  0 and g (a)  g (b) we get:
f (b)  f (a ) f  (c)

g (b)  g (a )
g  (c )
f (b)  f (a )
 f  (c )
If we let g (x) = x for a ≤ x ≤ b, then
ba
But this is the mean value theorem.
The Intermediate Form 0 .
0
If
f ( x)
im
g (x) , then we say that im
f (x) = 0 = im
has the indeterminate form 0 .
0
xa 
xa 
x  a  g ( x)
The same notion can be applied if
im
im
im , im
im
is replaced by
,
and
.
xa 
xa  xa x
x
Theorem 3.3 Let l be a real number or  or  .
a) Suppose f and g are differentiable on (a, b) and g  (x)  0 for a < x < b.
f  ( x)
im
g (x) and im
f (x) = 0 = im
= L, then
xa 
xa 
x  a  g  ( x)
f ( x)
f  ( x)
im

im
=L=
.
x  a  g ( x)
x  a  g  ( x)
If
An analogous result holds if
im
im
im where c  (a, b).
 is replaced by
 or
xa
x b
x c
In the latter case f and g need not be differentiable at c.
b) Suppose f and g are differentiable on (a, ) and g  (x)  0 for x > a.
f  ( x)

im

im

im
g
f
(x
)
(x
)
If
=0=
and
= L, then
x 
x
x   g  ( x)
f ( x)
f  ( x)
im
= L = im
.
x   g ( x)
x   g  ( x)
An analogous result holds if
im is replaced by im
.
x
x
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Unit III
Functions
5x  3x
.
x
x 0
Example 1 Find im
im 5 x  3 x = 0 = im x , applying L’Hopital’s rule we get:
x 0
x 0
Solution Since
x
x
im 5  3 = im 5 x n 5  3 x n 3 = im
 n 5   n 3 = n 5 .
x
3
x 0
x0
x 0
5x  3x
5
= n .
x
3
x 0
Therefore, im
Example 2 Find
im
cos x
  sin x  1 .
x
2
Solution Since
im
cos x = 0 = im  sin x  1, applying L’Hopital’s rule we get:
 
x
x
2
2
im
im
im
cos x
 sin x
tan x
= 
=  .
  sin x  1 =
 
 
cos
x
x
x
x
2
Therefore,
2
2
im
cos x
  sin x  1 =  .
x
2

2
Example 3 Find im 
x 1
 arcsin x
1 x2
.

im
Solution Since im  2  arcsin x = 0 =

x 1
x 1

2
im

x 1
 arcsin x
1 x2

Therefore, im 
x 1
Example 4 Find
Solution Since
2

im
=

x 1

1  x 2 , applying L’Hopital’s rule we get:
1
1 x2
x
=
1
im
= 1.

x 1
x
1 x2
 arcsin x
1 x
2
= 1.
x
im e  x  1 .
x2
x 0
im e x  x  1 = 0 = im x 2 , applying L’Hopital’s rule we get:
x0
x0
x
x
im e  x  1 = im e  1 . But im e x  1 = 0 = im 2 x .
x2
x 0
x 0
x0
x 0 2 x
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Applying L’Hopital’s rule once more we get:
x
x
x
im e  x  1 = im e  1 = im e = 1 .
x2
x 0
x0 2
2
x 0 2 x
e x  x 1 1

im
Therefore,
= .
x2
x 0
2
The Intermediate Form 

f ( x)

im

im

im
g
f
(x
)
(x
)
If
=  or   and =
=  or  , then we say that
has the
xa 
xa 
x  a  g ( x)
indeterminate form 

. The same notion can be applied if
im
im
im ,
is replaced by
,
xa 
xa  xa
im
im
and
.
x
x
Theorem 3.4 Let l be a real number or  or  .
a) Suppose f and g are differentiable on (a, b) and g  (x)  0 for a < x < b.
f  ( x)
im
g (x) =  or   and im
f (x) =  or  , im
= L, then
xa 
xa 
x  a  g  ( x)
f ( x)
f  ( x)
im
im
=
L
=
.
x  a  g ( x)
x  a  g  ( x)
If
im
im
im where c  (a, b).
is replaced by
or
xa 
x b 
x c
An analogous result holds if
In the latter case, neither f nor g will be differentiable at c.
b) Suppose f and g are differentiable on (a, ) and g  (x)  0 for x > a.
f  ( x)
im
f (x) =  or  , im g (x) =  or   and im
= L, then
x 
x
x   g  ( x)
f ( x)
f  ( x)
im

im
=L=
.
x   g ( x)
x   g  ( x)
If
An analogous result holds if
Example 5 Find
im
x 0 
1
x
im is replaced by im
.
x
x
1 .
x2
e
Solution.
1
im
im
e
 x ==
x 0
x 0 
1
x2
. Hence apply L’Hopital’s rule.
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1
x
im
x 0 
im
1 =
x 0 
x2
e
e
1
x
Therefore, im 
x 0

1
x2

1  2  =

x 2 
3

im
x 0 
x e
1
x2
2
= 0.
 x 
1 = 0.
x2
e
Example 6 Find
im
x n 1 .

x
x 0
n 1
x
1
im
im
x n 1 = im
n 1 =  = im
Example 7
and


 x.

x
x
1
x 0
x 0
x 0
x 0
x
Hence by apply L’Hopital’s rule we get:
im
x = 0.
x n 1 = im


x
x 0
x0
Therefore,
im
x n 1 = 0.

x
x 0
Example 8 Find
im
x 0 

1 x
 n x  .



1  x = im
im

n
e

x 
x 0 
x 0  

1 x
im
Therefore,
  n x  = 1.
x 0 

Solution
Example 9 Find im
x 
Solution 10 im
x 
x 
x
=e
x n (n 1x )
im
x0 
= 1.
1
x 2 n (n x) .
n (n x) =  = im
x 
Thus im
x 
Therefore, im
x n (n 1x )
1
x 2 .
1
x 2 n (n x) = im
x 
1
x n x
= im
 2 x   1
x 


2
x n x = 0.
1
2 n (n x)
= 0.
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Other Indeterminate Forms
Indeterminate forms, such as: 0  , 0 0 , 1  ,  0 and    can usually be converted into the
.
indeterminate forms 0
Example 11 Find
Solution Since
But
Therefore,
0
or 

.
im
x n x .
x 0 
n x
im
x = 0 and im
n x =  , im
x n x = im
.
x 0 
x 0 
x 0 
x 0  1 x
1
im
im
n x =  and im
x n x = im
( x ) = 0.



 x = . Thus
x 0
x 0
x 0
x 0
im
x n x = 0.
x 0 
Exercises 3.4
In exercises 1-10, apply L ‘Hospital’s Rule to evaluate each of the following.
1.
im sin ( x n sin x)
x 0 
1 x  1 x
x 0
x
4. im
7.
10.
x
im 1  1 
x  
x
im x cot x
x  0
x3
3. im
x  0 1  cos x
n x

x  0 n (sin x)
x
x
im
x im  1  1 
5.
x 2 
x 0 
x   
1
8. im x tan
x 
x
2. im
11.
im
x 
sin x
x 
n (1 x)
x   n x
6. im
9.
x
im
x  1  x 2
im e x  e x  x 2  2
12.
x  0 sin 2 x  x 2
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DEFENCE ENGINEERING COLLEGE
DEPARTMENT OF BASIC AND APPLIED COURSES
Applied Mathematics I (Math. 201)
Work sheet III (On Functions)
June 2007
1. Find an interval on which f has an inverse
i) f (x) = x
iii) f (x) =
3
 5 x 1
ii) f (x) = sin 2 x
1
1 x
iv) f (x) =
2
x
1 x2
2. Integrate each of the following integrals
i)
dx
2
x  4x  7

e  x dx
ii) 
1  e  2x
iii)

arctan 2 x dx
1  4x 2
iv)

dx
x 4 x 4  25
3. Evaluate the definite integrals
4
i)

4
3
3
4. Integrate
2 3  2
dx
x 4
2
x
a
ii)
2

2
dx
x  4x  8
2
dx
, where a and b are positive real numbers.
sin x  b 2 cos 2 x
2
5. Simplify the expressions
i) cos (2 arcsin x)
ii) sin (2 arcsin x)
6. Show that for any real numbers x and y
i) sinh ( x  y)  sinh x cosh y  cosh x sinh y
ii) cosh ( x  y)  cosh x cosh y  sinh x sinh y
7. Prove that:
i) sinh
ii) cosh
1
1
x 2  1 = cosh  1 x for x ≥ 1.
x = n  x 



1
x 2  1  for x  1and cosh x ' =

(Hint: use the fact that sinh
iii) cosh  1 x 2  1 = sinh
1
1
x  n  x 

1
x 21
for x ( 1, 1).
x 2  1  ).

x for x ≥ 0.
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8. Evaluate each of the following
i)
n x

x  0 n (sin x)
ii) im
im sin ( x n sin x)
x 0 
1 x  1 x
x 0
x
iv) im
x3
x  0 1  cos x
iii) im
im e x  e x  x 2  2
x  0 sin 2 x  x 2
v)
9. Evaluate each of the following
n (1 x)
x   n x
i) im
ii)
im
x 
1

1  
x

x
iii) im
x 
x tan
1
x
Unit I
Vectors, Lines and Planes
1.1 Vectors
1.1.1 Scalar and Vector Quantities
A scalar is a quantity that is determine by its magnitude (its number of units measured in a suitable scale).
Examples 1 Mass, length, temperature, voltage are examples of scalar quantities.
Quantities that have both magnitude and direction are called vectors. A vector is usually represented by an
arrow, the length of the arrow represents the magnitude of the vector and the arrow head indicates the
direction of the vector.
Examples 2 Velocity, acceleration, displacement and force are examples of vector quantities.
When a vector is represented by an arrow, say AB , the point A is called the initial point (tail) and B is
called the terminal point (head) of the vector. Vectors can also be represented by a single letter (usually
small letter) with a bar over it such as a , b etc.
Example 3
terminal point
a
initial point
Notations
A or a - vector A or vector a.
PQ
A
- a vector with initial point P and terminal point Q.
- magnitude (or norm) of vector A.
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1.1.2 Equality of two Vectors
Definition 1.1 Two non-zero vectors a and b are said to be equal, denoted a = b , if
and only if they have the same direction and magnitude, regardless of the
position of their initial points.
Note that: Equality of vectors is transitive.
i.e. For three vectors a , b and c , if a = b and b = c , then a = c .
Definition 1.2 A Vector is called a free vector, provided that its magnitude and direction
are fixed, but its position is indeterminate. If the initial point of a free
vector is fixed, then it is called a localized vector.
Definition 1.3 Two ( free) Vector are equal if and only if they have the same magnitude
and direction.
Definition 1.4 A vector of magnitude ( modulus) unity ( one) is called a unit vector.
Definition 1.5 Any vector whose magnitude is zero and direction indeterminate
is called a null ( zero) vector. A null vector is denoted by 0 .
Note that: For any non-zero vector A ,
A
is a unit vector in the direction of that of vector A.
A
1.1.3 Vectors in 2 and 3
Definition 1.6 Position Vectors
A non-zero vector in 2 (or 3) is called a position vector if and only if its initial
point is at the origin and its terminal point is anywhere other than the origin.
From this definition, the initial point (tail) of a vector can be anywhere with out changing the direction and
the magnitude of the vector.
Vector Addition
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Definition 1.7 Let AB and BC be two vectors in a plane. Then the sum AB + BC is the
vector represented by AC .
Triangle Law of Vector Addition
Let a and b be any two vectors. To find a + b , join a and b head to tail. a + b is the vector whose
initial point is that of a and terminal point that of b . This law of vector addition is called triangle law of
vector addition.
For any vector a , a + 0 = a .
Theorem. 1.1
a) For any two vectors a and b
a +b=b +a.
b) For any three vectors a , b and c
a + ( b + c ) = ( a + b )+ c .
For any vector a there exists a vector – a such that a + (– a ) = 0 . – a , called the opposite of vector
a , has the same magnitude and opposite in direction to that of a .
Subtraction of Vectors
For any two vectors a and b , a – b is the vector defined by adding a and – b pictorially
illustrated as follows:
b
b
a
a  ( b )
a
Scalar Multiplication
Definition 1.8 Let a be any vector and k be any scalar. k a is a vector whose magnitude is
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k times a and its direction is that of a if k > 0, opposite to that of a if
k < 0 and indeterminate if k = 0.
Definition 1.9 For any two vectors a and b and any two scalars m and n
i) m ( a + b ) = m a + m b
ii) (m + n) a = m a + n a
iii) (m n) a = m (n a )
iv) 1 a = a and 0 a = 0
1.1.4 Components and Coordinate Representation of Vectors
Definition 1.10 Two vectors a and b are said to be parallel if a = t b for some real
number t.
Let a and b be any two non-zero vectors which are not parallel. Then any vector u in the plane of a
and b can be uniquely expressed as
u =s a +tb
where s and t are scalars.
In this case, we say that u is expressed as a linear combination of a and b . The vectors s a and
 
t b are called the component vectors of u relative to a and b respectively and a , b is called a base.
Any pair of non-collinear vectors may be chosen as a base, but the usual and the most convenient choice of
base is a pair of unit position vectors (vectors of unit length) along the positive x-axis and along the positive
y-axis.
Note that: Any position vector is uniquely determined by the coordinates of its terminal point.
Now the position vector (1, 0) is usually denoted by i and (0, 1) by j . The vectors i and j , being
 
perpendicular , i , j is called an orthogonal base.
Note that: For any non-zero free vector a there is a unique position vector b such that a = b .
Now let (x, y) be the terminal point of a position vector u . Then u can be expressed as:
u =x i +y j .
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Similarly, if r is a free vector with initial point P(x1, y1) and terminal point Q(x2, y2) can be expressed as:
r = (x2 – x1) i + (y2 – y1) j .
Notation: u = (x, y) represents the position vector u with terminal point (x, y).
Note that: i = (1, 0) and j = (0, 1) are unit position vectors determined by the coordinates of their
terminal points.
Similarly, i = (1, 0, 0), j = (0, 1, 0) ) and k = (0, 0, 1) are mutually perpendicular
unit position vectors in 3.
Now let (x, y, z) be the terminal point of a position vector u in 3. Then u can be expressed as:
u =x i +y j +z k .
The length (norm) of a vector u = (x, y, z) is denoted and defined by:
u

x
2
 y2  z2
Similarly, if r is a free vector with initial point P(x1, y1, z1) and terminal point Q(x2, y2, z2) can be
expressed as:
r = (x2 – x1) i + (y2 – y1) j + (z2 – z1) k
Example 4 Let r = (2, 0,  5). Find the coordinates of the terminal point of the vector that is
equal to r if P (2, 3, 1) is its initial point.
Solution Let Q (x, y, z) be the terminal point of the required vector.
Then PQ = r  (x – 2, y – 3, z – 1) = (2, 0,  5)
 x – 2 = 2, y – 3 = 0 and z – 1 =  5
 x = 4, y = 3 and z = – 4.
Therefore, (4, 3, – 4) is the terminal point of the required vector.
Example 5 Let u = (2, 0,  5) and let P (0, 3, – 6) and Q (– 4, 3, 4) be the initial point and
terminal points of a vector. Find a real number t such that u = t PQ
Solution u = t PQ  (2, 0,  5) = t ( 4, 0, 10)   4 t = 2 and 10 t =  5  t =  0.5.
Therefore, t =  0.5 is the required solution.
1.1.5 The Scalar (Dot) Product
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Before we define the dot product of vectors, we need to define what we mean by the angle between two
vectors.
Definition 1.11 Let a and b be any two non-zero free vectors, and let (x1, y1) and (x2, y2) be
position vectors associated to a and b respectively. The angle between a and b
is defined to be the angle between the two position vectors (x1, y1) and (x2, y2).
Note that: The angle  between any two non-zero position vectors satisfies the condition
0
Further more; if  = 0 or  = , then the two position vectors are parallel and
if  =

, then the two position vectors are perpendicular.
2
Definition 1.12 Let a and b be any two non-zero free vectors. The scalar (dot or inner) product
of a and b , denoted a
.b
a
= a
.b
is defined by:
cos 
b
Where  is the angle between a and b .
Note that: a
, b
and cos  are numbers and hence scalar (dot) product of any two
non-zero vectors is a scalar quantity.
Note that: i) For any vector a , a
.a
 a
2
and hence
a 
a
.a
, since  = 0
and cos 0 = 1.
ii) The scalar product of any two non-zero perpendicular vectors a and b is zero.
i.e. a
.b
= 0, since  =


and cos
= 0.
2
2
iii) If  is the angle between two non-zero vectors a and b , then
cos  =
a .b
a
b
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Theorem 1.2 For any vectors a , b and c , and any scalar k
.b
i) a
= b
.( b
iii) a
.a
ii) k ( a
 c) = a
.
b
+ a
.b
.
) = (k a ) b = a
. (k
b )
.c
Theorem 1.3 If a = x1 i  y2 j and b = x2 i  y2 j , then
.b
a
= x1 x2  y1 y2
Corollary 1.3.1 If a = x1 i  y2 j , b = x2 i  y2 j are non-zero vectors and  is
the angle between a and b , then
x1 x2  y1 y2
Cos  =
x  y1
2
1
x2  y2
2
2
2
1.1.6 The Two Important Inequalities
Let A and B be two vectors.
.
1. A B
2. A
B

A

B
(Cauchy-Schwarz Inequality)
A + B
(Triangle Inequality)
Example 6 Compute the scalar product of
i) a = 3 i + 4 j and b = 4 i  3 j
i) a =  4 i + 3 j and b = 8 i  6 j
Solutions Using the above definition we get:
.
i) a
.b
= (3, 4)
and ii) a
.b
= ( 4, 3)
(4,  3) = 12  12 = 0.
. (8,  6) =  32  18 =  50.
Example 7 Given: The angle between two unit vectors p and q is 60. Then find
i)
Solutions i)
p  q
p  q
2
ii) the angle between p and p + q
.
= ( p +q) ( p +q)
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p
=
2
+2 p
q cos 60 +
q
2
= 2 (1 + cos 60)
= 3.
Therefore,
p  q
3.
=
ii) Let  be the angle between p and p + q .

Then p  p  q
=
p
2
p
+
q cos 60
= 1 + cos 60
= 1.5.

On the other hand p  p  q
i)
=
p
p  q
cos 
= 3 cos 
ii)
From i) and ii) we get:
3
2
3 cos  = 1.5  cos  =
  = 30.
Therefore, the angle between p and p + q 30.
Example 8 Given: a = i and b = i + j . Find the value of k such that
i) a + k b is orthogonal to a
ii) a + k b is orthogonal to b
Solutions a
= 1 and b
2.
=
i) a + k b is orthogonal to a if and only if a
Now a
.( a
+kb )=0
a
2
.( a

+k a  b
+ k b ) = 0.
=0 1+k=0
k =  1.
Therefore, k =  1.
ii) a + k b is orthogonal to b if and only if b
Now b
.( a
+kb )=0 k b
2

.( a
+ a  b
+ k b ) = 0.
 = 0  1 + 2k = 0 
k =  0.5.
Therefore k =  0.5.
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Example 9 Find the angle  between ( 6 , 1,  1) and the positive x axis.
Solution Let A = ( 6 , 1,  1) and B = (1, 0, 0) be two position vectors.

A
Then A B =
Hence, cos  =

B cos 
A B
 cos  =
A B

6
3
 =
.

6
2
8
Therefore, the angle between ( 6 , 1,  1) and the positive x axis is

.
6
1.1.7 Direction Angles and Direction Cosines
Definition 1.13 Let A = (a1, a2, a3) be a non-zero vector. The angles ,  and  (between 0 and
 inclusively) that A makes with the positive x, y and z axes respectively are called the
direction angles of A.
Now take the unit vectors i , j and k . From this definition we get:
cos  =
a1
A
, cos  =
a2
A
and cos  =
a3
A
Furthermore;
a1 = A cos , a2 = A cos  = and a3 = A cos 
Therefore A = A ( i cos  + j cos  + k cos )
cos , cos  and cos  are called the direction cosines of A.
Example 10 Let A = ( 2, 0, 3). Find the direction cosine of A.
Solution
A =
( 2)2  02  32 = 13 .
Hence cos  = 
  = cos
2
, cos  = 0 and cos  =
13
3
.
13
 1
 3 
2 
 .
 
 ,  = cos  1 0  and  = cos  1 
13
13




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 1
2  
 1 3 

 are the direction cosines of A.
 
 ,
and cos
 13 
 13  2
Therefore, cos
1.1.8 Projection and Resolution of Vectors
Definition 1.14 Let A be a non-zero vector. The projection of a vector B onto A, denoted by
Pr oj
B is defined as:
A
Pr oj
Note that: Pr oj
A
A
B 
A. B
A
A 2
B is a vector parallel to A.
Example 11 Let A = ( 2,  3,  1) and B = (0, 1,  1). Find Pr oj
Solution A =
A
B=
B and Pr oj A .
B
2 and A . B =  2.
14 , B =
Therefore, Pr oj
A
1
A and Pr oj A =  B.
B
7
Theorem 1.4 Let A be a non-zero vector. Then for any vector B,
Pr oj
Proof Pr oj
A
B 
Therefore, Pr oj
A
A
A. B
A
B  B
 B cos   B .
B  B .
Now let A and B be orthogonal vectors and let C be a vector in the same plane as A and B.
Then we can express C as a linear combination of vectors parallel to A and B as follows:
C = Pr oj C + Pr oj C
A
B
In this case, we say that vector C is resolved into vectors parallel to A and B.
Example 12 Let A = (0, 1,  2), B = (0, 2, 1) and C = (0, 5,  4). Resolve C into vectors parallel
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to A and B.
.
Solution A B = 0 and these three vectors lie on the yz plane.
A.C
B.C
Pr oj C =
Pr oj C =
and
A
B.
A
B
A 2
B 2
.
.
5.
Now A C = 13, B C = 6, A = B =
Hence Pr oj C =
A
Therefore, C =
13
6
A and Pr oj C =
B.
B
5
5
13
6
A+
B.
5
5
Example 13 Let A = (1, 0, 3), B = ( 3, 0, 1) and C = (2, 0, 5). Resolve C into vectors parallel to A and B.
.
Solution A B = 0 and these three vectors lie on the xz plane.
Pr oj C =
A
A.C
B.C
.
10 .
.
A and Pr oj C =
B
A 2
B.
B 2
Now A C = 18, B C =  1, A = B =
Hence Pr oj C =
A
9
1
A and Pr oj C = 
B.
B
10
5
9
1
A 
B.
10
5
1.1.9 Cross Product
Therefore, C =
Definition 1.15 Let A = (a1, a2, a3) and B = (b1, b2, b3) be two vectors. The cross (Vector)
product of A and B, written A  B is defined by:
A  B = (a2 b3  b2a3) i + (a3 b1  a1 b3) j + (a1 b2  a2 b1) k
A  B is read as “A cross B”.
Now let us see a simple method how to recall the formula for the cross product of A and B
i)
The first method.
AB=
i
j
k
a1
a2
a3
b1
b2
b3
ii) The second method.
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+
+

+


i
j
k
i
j
a1
a2
a3
a1
a2
b1
b2
b3
b1
b2
Example 14 Let A = (5,  1, 0) and B = (0, 2,  2). Find A  B and B  A.
Solution
AB =
i
j
k
5
1
0
0
2
2
= 2 i + 10 j + 10 k
Therefore, A  B = 2 i + 10 j + 10 k .
AB =
i
j
k
0
2
2
5
1
0
=  2 i  10 j  10 k
Therefore, A  B =  2 i  10 j  10 k .
Remark: i  j = k , j  k = i and k  i = j .
Properties of Cross Product
Let A, B, C be vectors and let m be a scalar. Then
i) A  B =  (B  A)
ii) A  A = 0
iii) A  (B + C) = (A  B) + (A  C)
and (A + B)  C) = (A  C) + (B  C)
iv) (m A)  B = m (A  B) = A  (m B).
Remark: Cross Product is not associative.
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Example 15 i  ( k  k ) = 0 while, ( i  k )  k =  j  k =  i .
Theorem 1.5 Let A and B be two non-zero vectors.
a) A . ( A  B)  0 and B . ( A  B)  0
Consequently; if A  B  0 , then A  B is orthogonal to both A and B.
b) If  is the angle between A and B (0    ), then
A  B = A B sin .
Proof i) A . ( A  B ) = a1 (a2 b3  b2a3) + a2 (a3 b1  a1 b3) + a3 (a1 b2  a2 b1)
= a1a2 b3  a1a3b2 + a2a3 b1  a1a2b3 + a1a3b2  a2 a3b1
= (a1a2 b3  a1a2b3) + (a1a3b2 a1a3b2) + (a2a3 b1 +  a2 a3b1)
= 0.
B . ( A  B) = b1 (a2 b3  b2a3) + b2 (a3 b1  a1 b3) + b3 (a1 b2  a2 b1)
= a2 b1b3  a3 b1b2 + a3b1b2  a1b2b3 + a1b2b3  a2b1b3
= (a2 b1b3  a2b1b3) + (a3b1b2  a3 b1b2) + (a1b2b3 a1b2b3)
= 0.
ii) A  B 2 = (a2 b3  b2a3)2 + (a3 b1  a1 b3)2+ (a1 b2  a2 b1)2
= (a22 b32  2 a2a3b2b3 + b22 a32) + (a12 b32  2 a1a3b1b3 + b12 a32)
+ (a12 b22  2 a1a2b1b2 + a22b12)
= a12 (b22 + b32) + a22 (b12+ b32 ) + a32 (b12+ b22 )
 (2 a2a3b2b3 + 2 a1a3b1b3 + 2 a1a2b1b2)
= (a12 +a22 + a32) (b12 + b22 + b32 )  (a1b1 + a2b2 + a3b3)2
=
A 2 B 2 ( A
2
B 2
= A
B cos ) 2
A 2 B 2 cos 2 
2
B 2 (1  cos 2  )
= A
2
B 2 sin 2 
= A
Therefore, A  B = A
B sin .
Corollary 1.5.1 Two non-zero vectors A and B are parallel if and only if A  B = 0 .
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Proof A  B = 0  A  B = 0
 A
B sin 
 sin  = 0
  = 0 or  = .
 A∥ B.
Therefore, Two non-zero vectors A and B are parallel if and only if A  B = 0 .
Example 16 Let A (3, 2,  2) and B (0, 3, 7).
i) Determine whether A and B are parallel or orthogonal or neither.
ii) Find a vector orthogonal to both A and B.
Solution i) A . B = (3  0) + (2  3) + ( 2  7) =  2, A =
Hence neither A . B  0 nor A . B  A
17 and B = 2 14 .
B .
Therefore, A and B are neither parallel nor orthogonal.
Remark: A  B is the area of a parallelogram with adjacent sides A and B.
1.1.10 Triple Product
There are two types of triple products.
i) Scalar triple product
For any three vectors A, B and C, A . ( B  C ) is called the triple (box or mixed triple)
product of A, B and C.
Example 17 Show that for any three vectors A, B and C
A . ( B  C ) = ( A  B) . C
Solution A . ( B  C ) = A
= A
B  C cos , where  is the angle between A and B  C .
B
and ( A  B ) . C = A B
C cos  sin , where  is the angle between B and C
C cos  sin , where  is the angle between C and
A  B and  is the angle between A and B.
Now cos  = sin  and sin  = cos , because co-functions of complementary angles
are equal.
Therefore, A . ( B  C ) = ( A  B ) . C .
ii) Vector Triple Product
For any three vectors A, B and C, A  ( B  C ) is called the Vector triple product of
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A, B and C.
Example 18 Show that for any three vectors A, B and C
A  ( B  C )  ( A  B)  C
Solution A  ( B  C ) = A
= A
B  C sin , where  is the angle between A and B  C .
B
C sin  sin , where  is the angle between B and C
and ( A  B )  C = A B
C sin  sin , where  is the angle between C and
A  B and  is the angle between A and B.
Now sin   sin  and sin   sin .
Therefore, A  ( B  C )  ( A  B )  C .
Remark: For any three vectors A, B and C
A  ( B . C ) , ( A . B )  C and ( A . B) . C
are undefined operations.
Some Properties of Triple Products
For any three vectors A, B and C
i) ( A  B )  C = B . (C  A) = A  ( B  C )
ii) A  ( B  C ) = B ( A . C )  C ( A . B)
“ bac – cab” rule.
Remark: For any three non-zero vectors A, B and C; A . ( B  C ) is the volume of a
parallelepiped with sides A, B and C.
1.2 Lines in  3
A line in space is determined by a point p0 (x0, y0, z0) on ℓ and a non-zero vector L parallel to it.
Now let ℓ be a line parallel to a non-zero vector L and let p0 (x0, y0, z0) be a fixed point on ℓ.
Let p (x, y, z) be an arbitrary point on ℓ. We need to express p in terms of p0 and L.
ℓ ∥ L  Po P ∥ L
 (x  x0, y  y0, z  z0) = t L ; for some t   and t  0.
 (x, y, z) = (x0, y0, z0) +
 r = r
0
+ t L; for some t   and t  0;
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where r = (x, y, z) and r = (x0, y0, z0).
0
Therefore, r = r
0
+ t L; for some t   and t  0 is the vector form of the equation of a line.
Example 19 Find a vector equation of the line that contains ( 1, 3, 5) and is parallel to
4 i 3 j  2 k
Solution Now r =  i  3 j  5 k and L = 4 i  3 j  2 k .
0
r = (  i 3 j 5 k ) + t ( 4 i 3 j  2 k )
Therefore, r = (1  4t ) i  (3  3t ) j  (5  2 t ) k is the required vector form of the
equation of the line.
Now let L = a i  b j  c k be a given non-zero vector and let (x0, y0, z0) be a point on ℓ. Then
for any point (x, y, z) on ℓ that is parallel to L, the vector equation form of ℓ is given by:
r = r0 + t L.
Hence (x, y, z) = ( x0  at ) i  ( y0  b t ) j  ( z0  c t ) k
 x = x0  at , y = y0  b t and z = z0  c t .
(i)
These equations are called the parametric equations of ℓ and t is called the parameter.
Example 20 Find the parametric equation of the line that contain ( 2, 1, 3) and is parallel to
4 i 2 k .
Solution Now (x0, y0, z0) = ( 2, 1, 3) and L = 4 i  2 k .
Then x =  2  4 t , y = 1 and z = 3  2 t .
Therefore, x =  2  4 t , y = 1 and z = 3  2 t is the required solution.
In the above parametric equations of a line ℓ if a, b, and c are non-zero real numbers then
We can express (i) as follows:
t
xx 0
a

y  y0 z  z0

; where t  .
b
c
This form of the equations of a line is called the Symmetric form of the equation of a line.
Example 21 Find the symmetric equations of the line containing the points P1 (2, 3,  1) and
P2 (5, 0, 4).
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Solution Now take L = PP0 = 3 i  3 j  5 k and (x0, y0, z0) = (2, 3,  1).
Therefore t 
x2
y  3 z 1


; where t   is the required equation.
3
3
5
Example 22 Find the vector, parametric and symmetric equations of the line containing the point
P ( 3, 4, 5) which is parallel to 4 i  3 k .
Solution (x0, y0, z0) = ( 3, 4, 5).
i) Vector equation
Hence r = ( 3, 4, 5) + t (4, 0,  3).
= ( 3 + 4 t, 4, 5  3 t)
= ( 3 + 4 t) i + 4 j + (5  3 t) k
Therefore, t 
x2
y  3 z 1


; where t   is the required equation.
3
3
5
ii) Parametric equations
(x, y, z) = ( 3 + 4 t, 4, 5  3 t)
 x =  3 + 4 t, y = 4 and z = 5  3t ; where t   is the required equation.
iii) Symmetric equations
x =  3 + 4 t, y = 4 and z = 5  3t ; where t  
 t
x 3
z 5

and y  4 ; where t  .
4
3
x 3
z 5

and y  4 ; where t   is the required equation.
4
3
Example 23 Show that the line containing the points (0, 0, 5) and (1,  1, 4) is perpendicular to the
Therefore, t 
line with equation
x
y 5 z 9


.
7
4
3
Solution Let P (0, 0, 5) and Q (1,  1, 4).
We need to show that PQ and n = (7, 4, 3) are perpendicular.
Now (1, 1, 1) . (7, 4, 3) = 0.
Therefore, the two lines are perpendicular.
1.3 Distance from a Point to a Line
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Given a line ℓ and a non-zero vector L parallel to ℓ, we wish to determine the distance D between
ℓ and a point P1 (not on ℓ).
To do so choose a point P0 on ℓand let  be the angle between ℓ and P0 P1 , where 0 ≤  ≤ π.
Now D = P0 P1
Therefore, D 
sin , but L  P0 P1
L  P0 P1
L
= L
P0 P1
sin .
is the distance of P1 from the line ℓ.
Example 24 Find the distance D from the point (2, 1, 0) to the line with equation
x =  2, y + 1 = z = t.
Solution Take any point P0 on the line. Say P0 ( 2,  1, 0).
P1 (2, 1, 0) and L = (0, 1, 1). Hence P0 P1
= (4, 2, 0).
Now L  P0 P1 =  2 i + 4 j  4 k , L  P0 P1
= 6 and
L =
2.
Therefore, D = 3 2 units.
1.4 Planes in 3
Given a point P0 and a non-zero vector n , there exists one and only one plane J containing P0
and perpendicular to n .
Let P0 (x0, y0, z0) be a given point, n = (a, b, c) be a non-zero vector and let P (x, y, z) be an
arbitrary point on the plane J containing P0 and which is perpendicular to n .
n is perpendicular to the plane J.

n is perpendicular to P0 P1 .
 (a, b, c) · (x  x0, y  y0, z  z0) = 0.
 a (x  x0) + b (y  y0) + c ( z  z0) = 0.
(*)
Therefore, a (x  x0) + b (y  y0) + c (z  z0) = 0 is the equation of the plane containing P0 and
perpendicular to n = (a, b, c).
n is said to be normal to the plane J.
Expanding and rearranging (*) , we get an equivalent equation of the form:
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a x + b y + c z = d; where d = n
· P0
.
Example 25 Find the equation of the plane that contains the point (5,  1, 2) and has normal to
2 i 3 k .
Solution P0 (5,  1, 2) and n = (2, 0,  3).
Now (5,  1, 2)
· (2, 0,  3) = 4.
Therefore, 2 x 3 z = 4 is the required equation of the plane.
Example 26 Find the equation of the plane that contains the point (2,  2, 1) and which is
perpendicular to the line with equation has normal to
Solution P0 (2,  2, 1) and n = (2, 3,
Now (2,  2, 1) · (2, 3,
Therefore, 2 x + 3 y +
1 x
y 2

 3z .
2
3
1
).
3
1
29
)= 
.
3
3
1
29
z= 
is the required equation of the plane.
3
3
Note that: i) Three distinct points P0, P1 and P2 in 3 are collinear if and only if
P0 P1  P0 P2 = 0.
ii) Three distinct non-collinear points P0, P1 and P2 in 3 determine a unique plane.
Let P0, P1 and P2 be three distinct non-collinear points in 3. To determine the equation of the
plane J that contains these points, we need to solve:
P0 P1
· ( P0 P1
 P0 P2 ) = 0.
Example 27 Find the equation of the plane that contains (1, 0, 1), (2, 1, 1) and (2, 0, 3).
Solution Let P0 (1, 0, 1), P1 (2, 1, 1) and P2 (2, 0, 3).
Then P0 P1 = (3, 1, 0) and P0 P2 = (1, 0, 2).
Hence P0 P1  P0 P2 = (2,  6, 1).
Therefore, 2 x  6 y + z =  1 is the required equation of the plane.
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1.5 Distance from a Point to a Plane
We need to determine the distance D between a point P1 to a plane J whose normal is the nonzero vector n = (a, b, c). To do so choose a point P0 on J and let  be the angle between n
and P0 P1 , where 0 ≤  ≤ π.
D = P0 P1
Now n . P0 P1
Hence D =
= n
n . P0 P1
P0 P1
cos .
cos .
.
n
n . P0 P1
Therefore, D =
for any point P0 on J.
n
Example 28 Calculate the distance D between the point P1 (2, 3,  1) and the plane
 4 x + 2y + z = 0.
Solution Let P0 (1, 2, 0) be a point on the plane. Now n = ( 4, 2, 1) and P0 P1
Hence n . P0 P1
Therefore, D =
= 3 and n
=
= (1, 1,  1).
21 .
3
units.
7
Example 29 Calculate the distance D between the point P1 (2, 3,  1) and the plane that passes
through A( 3, 0, 2), B(1, 1, 2) and C( 1,  1,  1).
Solution Let P0 ( 3, 0, 2). Now P0 P1 = (5, 3,  3) and n = AB . AC = ( 3, 12,  6).
Hence n . P0 P1
Therefore, D =
= 39 and n
=
189 .
13
units.
21
Produced by Tekleyohannes Negussie, July 2009
98
Unit III
Functions
Produced by Tekleyohannes Negussie, July 2009
99