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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.

Example 1.Let A = ( 1, 1) and B = . Show that f (x) = and g (x) = , x A

are functions from A into B.

Solution and Since the square of a real number and the

product of 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.

Example 2.Let A = ( 1, 1) and B = . Show that g (x) = is an onto function while

f (x) = is not an onto function from A into B.

Solution i) Let y . We need to check for the existence of an x in ( 1, 1) such that g (x) = y.

58Produced by Tekleyohannes Negussie, July 2009

Unit III Functions

Now y = x = . But y 1 1. Hence x ( 1, 1), so range

of g is B.

Therefore, g is onto.

ii) f (x) = x ( 1, 1) f (x) [0, 1). Hence any y \ [0, 1) doesn’t belong

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.

Example 3 Let A = ( 1, 1) and B = . Prove that g (x) = , x A is a one to one function

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) = = .

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

and these functions satisfy the condition that

f (x) = y if and only if (y) = x x A and y B.

Furthermore; the two functions have the following relations.


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i) (x) = f (x) x dom. f.

ii) = x x dom. f.

iii) = 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, : J I exists. Then i) If f is continuous on I, then is continuous on J. ii) If f is differentiable on I, then is differentiable on J and

(y) =

for x such that f (x) = y, provided that f (x) 0.

Example 5 Let f (x) = + 3x +2. Then find (2) and (6).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 = + 3x +2 + 3x + 2 = 2 x = 0. Hence (0) = 3.

y = 6 and y = + 3x +2 + 3x + 2 = 6 + 3x 4 = 0 x = 1. Hence (1) = 6.

Therefore, (2) = and (6) = .

Example 6 Find a formula for (y) if f (x) = .

Solution Now domain of f = range of f = .

If f (x) = , then f (x) = and y = if and only if x = y3. Moreover; f (x) =

Therefore, (y) = .

Exercises 3.1

In exercises 1-8, find an interval on which each of the following functions has an inverse.


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1. f (x) = 2. f (x) = 3. f (x) = 4. f (x) =

5. f (x) = 6. 7. 8.

9. Find (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

, x .

Restrict its domain to , so that the new function becomes strictly increasing on the restricted

domain and the cosine function is non-negative.

Definition 3.5 The inverse sine function

:

whose value at any x usually denoted by

arcsin x or , is defined by:

y = if and only if x = sin y y .

Example 1 Find the exact value of each of the following expressions.

i) ii) iii)

Solutions i) Let x = .


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x = and x sin x = 0 and x

x = n , where n Z and x x = 0.

Therefore, = 0.

ii) Let x = .

x = and x sin x = and x

x = + (4n + 1) , where n Z and x x = .

Therefore, = .

iii) Let x = .

x = and x sin x = and x

x = + (4n 1) , where n Z and x x = .

Therefore, = .

From the definition of the arcsine function we get:

y = if and only if x = sin y x and y .

Therefore, In general we can conclude that:

sin ( ) = x , x

and = x, x .

Now we need to find the derivative of the arcsine function and its applications on evaluating some integrals.

y = 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) =

But y , cos y > 0 and hence cos y = = .

Therefore


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= x .

From the nature of the derivative of the arcsine function we can observe that integrals of the form

, where a > 0 can be evaluated by substituting x = a sin t.

Example 2 Integrate , where a > 0.

Solution Let x = a sin t.

Then dx = a cos t dt and = a cos t. Hence = = t + c, where c .

Therefore, = + c, where c .

Example 3 Evaluate .

Solution Let x = 3 sin t. Then dx = 3 cos t dt and = 3 cos t.

Hence = = = .

Therefore, = .

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

, x .

Restrict its domain to , so that the new function becomes strictly increasing on the restricted

domain and the secant function is non-negative.

Definition 3.6 The inverse tangent function

:


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whose value at any x usually denoted by

arctan x or , is defined by:

y = if and only if x = tan y y .


i) ii) iii)

Solutions i) Let x = .

x = and x tan x = 0 and x

x = n , where n Z and x x = 0.

Therefore, = 0.

ii) Let x = .

x = and x tan x = and x

x = + n, where n Z and x x = .

Therefore, = .

iii) Let x = .

x = and x tan x = and x

x = + 2n, where n Z and x x = .

Therefore, = .

From the definition of the arctangent function we get:

y = if and only if x = tan y , y .

Therefore, In general we can conclude that:

tan ( ) = x , x

and = x, x .


Unit III Functions

We need to find the derivative of the arctangent function and its applications on evaluating some integrals.

Now y = if and only if x = tan y. Hence taking the derivative of both sides of x = tan y, we get:

1 = ( ) =

But y , sec y > 0 and hence = = .

Therefore, = , x .

From the nature of the derivative of the arctan function we can observe that integrals of the form

, where a > 0 can be evaluated by substituting x = a tan t.

Example 5 Integrate , where a > 0.

Solution Let x = a tan t. Then dx = a dt and = .

Hence = = t + c, where c .



Solution Let x = 3 tan t. Then dx = 3 dt and = 9 .

Hence = = = .

Therefore, = .

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

Now we need to choose an interval, around the origin, on which the sine function is non-negative.

If we restrict its domain to , in which sine function is non-negative, then this new


Unit III Functions

function becomes one-to-one in the restricted domain.

Definition 3.7 The inverse secant function

:

whose value at any x usually denoted by arcsec x or , is defined by:

y = if and only if x = sec y y .


i) ii) Solutions i) Let x = .

x = and x

sec x = 2 and x

x = 2n , where n Z and x x = .

Therefore, = .

ii) Let x = .

x = and x sec x = and x

x = 2n , where n Z and x x = .

Therefore, = .

From the definition of the arcsecant function we get:

y = if and only if x = sec y x .

Therefore, sec ( ) = x , x

and = x, x .

Furthermore, taking the derivative of both sides of x = sec y with respect to x, we get:

1 = (tan y) (sec y) =

But y , tan y = = and sec y tan y > 0.


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Therefore,

= x

.

From the nature of the derivative of the arcsecant function we can observe that integrals of the form

, where a > 0 can be evaluated by substituting x = a sec t.

Example 8 Integrate .

Solution Let x = 3 sec t. Then dx = 3 sec t tan t dt and = 3 tan t.

Hence = = t + c where c .



Solution Let x = 3 sec t. Then dx = 3 sec t tan t dt and = 3 tan t.

Hence = = = .

Therefore, = .

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

inverse functions

Domain Range Derivative

Cosiney = if and only if x = cos y

[ 1 , 1] [0, ] = x ( 1 ,

1)

Cotangenty = if and only if x = cot y

(0, ) = x

Cosecanty = if and only if x = csc y

\ 0 = x


Unit III Functions

Basic relations between the inverse trigonometric functions

1. = 2. = 3.

=

4. 5. 6.

Negative argument formulas

1. 2. 3.

4. 5. 6.

Exercises 3.2

In exercises 1-5, simplify the given expressions.

1. 2. 3. cos (2 arcsin x) 4. sin (2 arcsin x) 5.

In exercises 6-10, find the derivative of each of the following functions.

6. 7. 8.

9. 10. In exercises 11-16, integrate the given indefinite integrals.

11. 12. 13.

14. 15. , where a, b > 0 16.

In exercises 17-20, evaluate the given definite integrals.


Unit III Functions

17. 18. 19. 20.

21. Let f (x) = . Show that = 0 and conclude that = .

22. Let f (x) = . Solve = 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 = and sinh t =

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) = 1, t .The other four hyperbolic functions, namely the hyperbolic tangent, the hyperbolic secant and the

hyperbolic cosecant are defined by:

tanh t = , sech t = , coth t = and csch t = .

The Derivatives of the Hyperbolic Functions

The derivatives of the hyperbolic functions are presented in the following table.

Function Cosh t Sinh t Tanh t Coth t Sech t Csch t

Derative Sinh t Cosh 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 = 2x =


Unit III Functions

2x = = 0 .

Since 0 for any t , . Hence t = .

Therefore, the inverse of the hyperbolic sine function f (x) = sinh t is given by the formula

Example 1 Show that x .

Solution using the above formula and the rule for taking derivatives of quotient of function we get:

= = = .

Therefore, 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; 1, so the inverse function will be defined for 1.

Now let x = tanh t.

x = tanh t x = x = x + x = 1 + x =

t = .

Therefore, the inverse of the function tanh t is given by the formula

= for 1.

Example 2 Show that for 1.

Since sech t is not one-to-one we need to restrict the domain to [0, ). Moreover 0 sech t 1 and so the

inverse function will be defined only for 0 x 1. Now let x = sech t.

x = sech t x = x + x = = 0

.

Since 1 for any t [0, ), t = .

Therefore, the inverse of the function sech t is given by the formula

= for 0 x 1.

Example 3 Show that for 0 x 1.


Unit III Functions

Solution using the above formula and taking the derivative of a composition of functions we get:

=

= = for 0 x 1.

Therefore, for 0 x 1.

The table below presents the inverses, derivatives and domains of the remaining three hyperbolic functions.

No Function Formula Domain Derivative

1 x 1

2 1.

3 x 0

Hyperbolic Identities

I Addition

a) c)

b) d)

II Sum, difference and product

a) d)

b) e)

c) f)


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III Half and Double angle Formula

a) d) sinh 2x = 2 sinh x cosh x

b) e) ) cosh 2x = =

=

c) f)

Exercises 3.3

In exercises 1-6, prove that:

1. = coth x 2. = for x ≥ 1.

3. = for x ≥ 0. 4. If y = , then .

5. for t > 0,

6. If a = c cosh x and b = c sinh x, then show that .

In exercises 7 and 8 find the derivative of f if

7. 8.

In exercises 9 and 10 evaluate each of the following integrals.

9. 10.

3.4 The L’Hopital’s Rule

In this section we describe a technique for evaluating many indeterminate limits such as:

; ; ; etc.

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:

Proof Define the function h on [a, b] by:


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:

, since h (a) = h (b), = 0.

Now = { f (b) f (a) } g (x) { g (b) g (a) } f (x) and since g (c) 0 and g (a) g (b) we get:

If we let g (x) = x for a ≤ x ≤ b, then

But this is the mean value theorem.

The Intermediate Form .

If = 0 = , then we say that has the indeterminate form .

The same notion can be applied if is replaced by , , and .

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.

If = 0 = and = L, then

= L = .

An analogous result holds if is replaced by or where c (a, b).

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.

If = 0 = and = L, then

= L = .

An analogous result holds if is replaced by .

Example 1 Find .

Solution Since = 0 = , applying L’Hopital’s rule we get:


Unit III Functions

= = = .

Therefore, = .

Example 2 Find .


= = = .

Therefore, = .

Example 3 Find .


= = = 1.

Therefore, = 1.

Example 4 Find .


= . But = 0 = .

Applying L’Hopital’s rule once more we get:

= = = .

Therefore, = .

The Intermediate Form


Unit III Functions

If = or and = = or , then we say that has the

indeterminate form . The same notion can be applied if is replaced by , ,

and .

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.

If = or , = or and = L,

then

= L = .

An analogous result holds if is replaced by or where c (a, b).

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.

If = or , = or and = L, then

= L = .

An analogous result holds if is replaced by .

Example 5 Find .

Solution. = = . Hence apply L’Hopital’s rule.

= = = 0.

Therefore, = 0.


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Example 6 Find .

Example 7 = and = = .

Hence by apply L’Hopital’s rule we get:

= = 0.

Therefore, = 0.

Example 8 Find .

Solution = = = 1.

Therefore, = 1.

Example 9 Find .

Solution 10 = = .

Thus = = = 0.

Therefore, = 0.

Other Indeterminate Forms

Indeterminate forms, such as: , , , and can usually be converted into the

indeterminate forms or .

Example 11 Find .

Solution Since = 0 and = , = .


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But = and = . Thus = = 0.

Therefore, = 0.

Exercises 3.4

In exercises 1-10, apply L ‘Hospital’s Rule to evaluate each of the following.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

DEFENCE ENGINEERING COLLEGE

DEPARTMENT OF BASIC AND APPLIED COURSES

Applied Mathematics I (Math. 201)

Work sheet III (On Functions)

June 2007


Unit III Functions

1. Find an interval on which f has an inverse

i) f (x) = ii) f (x) = sin 2 x

iii) f (x) = iv) f (x) =

2. Integrate each of the following integrals

i) ii) iii) iv)

3. Evaluate the definite integrals

i) ii)

4. Integrate , where a and b are positive real numbers.

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) ii)

7. Prove that:

i) = for x ≥ 1.

ii) = for x 1and = for x ( 1, 1).

(Hint: use the fact that ).

iii) = for x ≥ 0.

8. Evaluate each of the following

i) ii) iii)

iv) v)

9. Evaluate each of the following

i) ii) iii)

Unit I

Vectors, Lines and Planes

1.1 Vectors


Unit III Functions

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 , 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 , etc.

Example 3

Notations or - vector A or vector a. - a vector with initial point P and terminal point Q.

- magnitude (or norm) of vector A.

1.1.2 Equality of two Vectors

Definition 1.1 Two non-zero vectors and are said to be equal, denoted = , 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 , and , if = and = , then = .

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.

79

a

initial point

Produced by Tekleyohannes Negussie, July 2009

terminal point

Unit III Functions

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 .

Note that: For any non-zero vector A , is a unit vector in the direction of that of vector 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

Definition 1.7 Let and be two vectors in a plane. Then the sum + is the

vector represented by .

Triangle Law of Vector Addition

Let and be any two vectors. To find + , join and head to tail. + is the vector whose

initial point is that of and terminal point that of . This law of vector addition is called triangle law of

vector addition.

For any vector , + = .

Theorem. 1.1

a) For any two vectors and

+ = + .

b) For any three vectors , and

+ ( + ) = ( + )+ .


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For any vector there exists a vector – such that + (– ) = . – , called the opposite of vector ,

has the same magnitude and opposite in direction to that of .

Subtraction of Vectors

For any two vectors and , – is the vector defined by adding and – pictorially

illustrated as follows:

Scalar Multiplication

Definition 1.8 Let be any vector and k be any scalar. k is a vector whose magnitude is

times and its direction is that of if k > 0, opposite to that of if

k < 0 and indeterminate if k = 0.

Definition 1.9 For any two vectors and and any two scalars m and n

i) m ( + ) = m + m ii) (m + n) = m + n

iii) (m n) = m (n ) iv) 1 = and 0 =

1.1.4 Components and Coordinate Representation of Vectors

Definition 1.10 Two vectors and are said to be parallel if = t for some real

number t.

Let and be any two non-zero vectors which are not parallel. Then any vector in the plane of

and can be uniquely expressed as

= s + t

where s and t are scalars.

In this case, we say that is expressed as a linear combination of and . The vectors s and t are

called the component vectors of relative to and respectively and is called a base.

81

b

a

b

a)( ba


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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 and (0, 1) by . The vectors and , being

perpendicular , is called an orthogonal base.

Note that: For any non-zero free vector there is a unique position vector such that = .

Now let (x, y) be the terminal point of a position vector . Then can be expressed as:

= x + y .

Similarly, if is a free vector with initial point P(x1, y1) and terminal point Q(x2, y2) can be expressed as:

= (x2 – x1) + (y2 – y1) .

Notation: = (x, y) represents the position vector with terminal point (x, y).

Note that: = (1, 0) and = (0, 1) are unit position vectors determined by the coordinates of their

terminal points.

Similarly, = (1, 0, 0), = (0, 1, 0) ) and = (0, 0, 1) are mutually perpendicular

unit position vectors in 3.

Now let (x, y, z) be the terminal point of a position vector in 3. Then can be expressed as:

= x + y + z .

The length (norm) of a vector = (x, y, z) is denoted and defined by:

Similarly, if is a free vector with initial point P(x1, y1, z1) and terminal point Q(x2, y2, z2) can be expressed

as:

= (x2 – x1) + (y2 – y1) + (z2 – z1)

Example 4 Let = (2, 0, 5). Find the coordinates of the terminal point of the vector that is

equal to if P (2, 3, 1) is its initial point.

Solution Let Q (x, y, z) be the terminal point of the required vector.

Then = (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 = (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 = t

Solution = t (2, 0, 5) = t ( 4, 0, 10) 4 t = 2 and 10 t = 5 t = 0.5.


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Therefore, t = 0.5 is the required solution.

1.1.5 The Scalar (Dot) Product

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 and be any two non-zero free vectors, and let (x1, y1) and (x2, y2) be

position vectors associated to and respectively. The angle between and

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.

Definition 1.12 Let and be any two non-zero free vectors. The scalar (dot or inner) product

of and , denoted is defined by:

= Where is the angle between and .

Note that: , and are numbers and hence scalar (dot) product of any two

non-zero vectors is a scalar quantity.

Note that: i) For any vector , and hence , since = 0

and cos 0 = 1.

ii) The scalar product of any two non-zero perpendicular vectors and is zero.

i.e. = 0, since = and cos = 0.

iii) If is the angle between two non-zero vectors and , then

cos =

Theorem 1.2 For any vectors , and , and any scalar k


Unit III Functions

i) = ii) k ( ) = =

iii) = +

Theorem 1.3 If = and = , then

=

Corollary 1.3.1 If = , = are non-zero vectors and is

the angle between and , then

Cos =

1.1.6 The Two Important Inequalities

Let A and B be two vectors.

1. (Cauchy-Schwarz Inequality)

2. + (Triangle Inequality)

Example 6 Compute the scalar product of

i) = 3 + 4 and = 4 3

i) = 4 + 3 and = 8 6

Solutions Using the above definition we get:

i) = (3, 4) (4, 3) = 12 12 = 0.

and ii) = ( 4, 3) (8, 6) = 32 18 = 50.

Example 7 Given: The angle between two unit vectors and is 60. Then find

i) ii) the angle between and +

Solutions i) = ( + ) ( + )

= + 2 cos 60 +

= 2 (1 + cos 60)

= 3.

Therefore, = .


Unit III Functions

ii) Let be the angle between and + .

Then = + cos 60

= 1 + cos 60

= 1.5. i)

On the other hand = cos

= cos ii)

From i) and ii) we get:

cos = 1.5 cos =

= 30.

Therefore, the angle between and + 30.

Example 8 Given: = and = + . Find the value of k such that

i) + k is orthogonal to

ii) + k is orthogonal to

Solutions = 1 and = .

i) + k is orthogonal to if and only if ( + k ) = 0.

Now ( + k ) = 0 + k = 0 1 + k = 0 k = 1.

Therefore, k = 1.

ii) + k is orthogonal to if and only if ( + k ) = 0.

Now ( + k ) = 0 k + = 0 1 + 2k = 0 k = 0.5.

Therefore k = 0.5.

Example 9 Find the angle between ( , 1, 1) and the positive x axis.

Solution Let A = ( , 1, 1) and B = (1, 0, 0) be two position vectors.

Then = cos

Hence, cos = cos = = .

Therefore, the angle between ( , 1, 1) and the positive x axis is .

1.1.7 Direction Angles and Direction Cosines


Unit III Functions

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 , and . From this definition we get:

cos = , cos = and cos =

Furthermore;

= cos , = cos = and = cos

Therefore A = ( cos + cos + 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 = = .

Hence cos = , cos = 0 and cos = .

= , = and = .

Therefore, , and are the direction cosines of A.

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

is defined as:

Note that: is a vector parallel to A.

Example 11 Let A = ( 2, 3, 1) and B = (0, 1, 1). Find and .

Solution = , = and = 2.

Therefore, = A and = B.


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Theorem 1.4 Let A be a non-zero vector. Then for any vector B,

Proof .

Therefore, .

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 = +

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

to A and B.

Solution = 0 and these three vectors lie on the yz plane.

= and = .

Now = 13, = 6, = = .

Hence = and = .

Therefore, C = + .

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 = 0 and these three vectors lie on the xz plane.

= and = .

Now = 18, = 1, = = .

Hence = and = .

Therefore, C = .

1.1.9 Cross Product

Definition 1.15 Let A = (a1, a2, a3) and B = (b1, b2, b3) be two vectors. The cross (Vector)


Unit III Functions

product of A and B, written A B is defined by:

A B = (a2 b3 b2a3) + (a3 b1 a1 b3) + (a1 b2 a2 b1)

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 =

ii) The second method.

Example 14 Let A = (5, 1, 0) and B = (0, 2, 2). Find A B and B A.

Solution A B =

= 2 + 10 + 10

Therefore, A B = 2 + 10 + 10 .

A B =

= 2 10 10

Therefore, A B = 2 10 10 .

Remark: = , = and = .

Properties of Cross Product

88

a1 a2 a3

b1 b2 b3

+ + +

a1 a2 a3 a1 a2

b1 b2 b3 b1 b2

5 1 0

0 2 2

0 2 2

5 1 0


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Let A, B, C be vectors and let m be a scalar. Then

i) A B = (B A)

ii) A A =

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.

Example 15 ( ) = while, ( ) = = .

Theorem 1.5 Let A and B be two non-zero vectors.

a) and

Consequently; if , then is orthogonal to both A and B.

b) If is the angle between A and B (0 ), then

= sin .

Proof i) = 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.

= 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) = (a2 b3 b2a3)2 + (a3 b1 a1 b3)2+ (a1 b2 a2 b1)2

= (a22 b3

2 2 a2a3b2b3 + b2

2 a32) + (a1

2 b32 2 a1a3b1b3 + b1

2 a32)

+ (a12 b2

2 2 a1a2b1b2 + a2

2b12)

= a12 (b2

2 + b3

2) + a22 (b1

2+ b32

) + a32 (b1

2+ b22 )

(2 a2a3b2b3 + 2 a1a3b1b3 + 2 a1a2b1b2)

= (a12 +a2

2 + a32) (b1

2 + b22 + b3

2 ) (a1b1 + a2b2 + a3b3)2

= ( cos ) 2

=


Unit III Functions

= (1 )

=

Therefore, = sin .

Corollary 1.5.1 Two non-zero vectors A and B are parallel if and only if = .

Proof = = 0

sin

sin = 0

= 0 or = .

A∥ B.

Therefore, Two non-zero vectors A and B are parallel if and only if = .

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) = (3 0) + (2 3) + ( 2 7) = 2, = and = .

Hence neither 0 nor .

Therefore, A and B are neither parallel nor orthogonal.

Remark: 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, 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

=

Solution = cos , where is the angle between A and .

= cos sin , where is the angle between B and

and = cos sin , where is the angle between C and

and is the angle between A and B.

Now cos = sin and sin = cos , because co-functions of complementary angles

are equal.

Therefore, = .

ii) Vector Triple Product


Unit III Functions

For any three vectors A, B and C, is called the Vector triple product of

A, B and C.

Example 18 Show that for any three vectors A, B and C

Solution = sin , where is the angle between A and .

= sin sin , where is the angle between B and

and = sin sin , where is the angle between C and

and is the angle between A and B.

Now sin sin and sin sin .

Therefore, .

Remark: For any three vectors A, B and C

, and

are undefined operations.

Some Properties of Triple Products

For any three vectors A, B and C

i) = =

ii) = “ bac – cab” rule.

Remark: For any three non-zero vectors A, B and C; is the volume of a

parallelepiped with sides A, B and C.

1.2 Lines in

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 ∥ L

(x x0, y y0, z z0) = t L ; for some t and t 0.

(x, y, z) = (x0, y0, z0) +

= + t L; for some t and t 0;

where = (x, y, z) and = (x0, y0, z0).

Therefore, = + 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


Unit III Functions

Solution Now = and L = .

= ( ) + t ( )

Therefore, = is the required vector form of the

equation of the line.

Now let L = 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:

= + t L.

Hence (x, y, z) =

x = , y = and z = . (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

.Solution Now (x0, y0, z0) = ( 2, 1, 3) and L = .

Then x = , y = and z = .

Therefore, x = , y = and z = 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:

; where t .

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).

Solution Now take L = = and (x0, y0, z0) = (2, 3, 1).

Therefore ; where t is the required equation.

Example 22 Find the vector, parametric and symmetric equations of the line containing the point

P ( 3, 4, 5) which is parallel to .

Solution (x0, y0, z0) = ( 3, 4, 5).

i) Vector equation

Hence = ( 3, 4, 5) + t (4, 0, 3).


Unit III Functions

= ( 3 + 4 t, 4, 5 3 t)

= ( 3 + 4 t) + + (5 3 t)

Therefore, ; where t is the required equation.

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

; where t .

Therefore, ; where t is the required equation.

Example 23 Show that the line containing the points (0, 0, 5) and (1, 1, 4) is perpendicular to the

line with equation .

Solution Let P (0, 0, 5) and Q (1, 1, 4).

We need to show that and = (7, 4, 3) are perpendicular.

Now = 0.

Therefore, the two lines are perpendicular.

1.3 Distance from a Point to a Line

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 , where 0 ≤ ≤ π.

Now D = sin , but = sin .

Therefore, 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 = (4, 2, 0).

Now = 2 + 4 4 , = 6 and = .


Unit III Functions

Therefore, D = 3 units.

1.4 Planes in 3

Given a point P0 and a non-zero vector , there exists one and only one plane J containing P0 and

perpendicular to .

Let P0 (x0, y0, z0) be a given point, = (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 .

is perpendicular to the plane J. is perpendicular to .

(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 = (a, b, c).

is said to be normal to the plane J. Expanding and rearranging (*) , we get an equivalent equation of the form:

a x + b y + c z = d; where d = · .

Example 25 Find the equation of the plane that contains the point (5, 1, 2) and has normal to

.

Solution P0 (5, 1, 2) and = (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 = (2, 3, ).

Now (2, 2, 1) · (2, 3, ) = .

Therefore, 2 x + 3 y + z = is the required equation of the plane.

Note that: i) Three distinct points P0, P1 and P2 in 3 are collinear if and only if


Unit III Functions

= 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:

· ( ) = 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 = (3, 1, 0) and = (1, 0, 2).

Hence = (2, 6, 1).

Therefore, 2 x 6 y + z = 1 is the required equation of the plane.

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 non-zero

vector = (a, b, c). To do so choose a point P0 on J and let be the angle between

and , where 0 ≤ ≤ π.

D = cos .

Now = cos .

Hence D = .

Therefore, D = for any point P0 on J.

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 = ( 4, 2, 1) and = (1, 1, 1).

Hence = 3 and = .

Therefore, D = units.

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 = (5, 3, 3) and = = ( 3, 12, 6).


Unit III Functions

Hence = 39 and = .

Therefore, D = units.


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