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MTH 102
ELEMENTARY MATHEMATICS II (CALCULUS)
Course Code: MTH 102
Course Title: Elementary Mathematics II (Calculus)
Published by: Book Bank Team
Material: Book Bank Material

CHAPTER ONE

Functions of a Real Variable

1.1 Introduction to Functions

The concept of a function is fundamental to calculus and to mathematics generally. Calculus is concerned largely with quantities that vary and with the relationships between such quantities. A function provides a precise mathematical way of expressing one quantity in terms of another.

For example, the area of a square depends on the length of its side. If the side length is represented by x, then the area A is given by

A=x2.

Thus, every permitted value of x determines exactly one value of A. In this situation, the area is said to be a function of the side length.

Similarly, if an object moves at a constant speed of 20 metres per second, the distance travelled after t seconds is

d=20t.

The distance d depends on time t. Consequently, d is a function of t.

Functions occur naturally in science, engineering, economics, statistics, computing and everyday quantitative situations. In calculus, they are particularly important because limits, continuity, differentiation and integration are all developed in relation to functions.

1.2 Meaning and Definition of a Function

Definition of a Function

A function is a rule or correspondence that assigns to each element of a specified set, called the domain, exactly one element of another specified set, called the codomain.

If a function f maps elements of a set A into a set B, this may be written as

f:A→B.

If x∈A and the function assigns the element y∈B to x, then

y=f(x).

The element x is called the input or independent variable, while f(x) is the corresponding output or value of the function.

Essential Property of a Function

Each permissible input must have exactly one output. Different inputs are allowed to have the same output, but a single input cannot be assigned two different outputs by the same function.

Example

Consider the rule

f(x)=2x+3.

If x=1, then f(1)=5. If x=4, then f(4)=11. Every chosen value of x produces exactly one corresponding value.

1.3 Function Notation and Evaluation

The notation f(x) is read as “f of x.” It does not mean f multiplied by x. Rather, it denotes the value produced by the function f when the input is x.

Suppose

f(x)=3x2-2x+5.

To determine f(2), every occurrence of x is replaced by 2:

f(2)=3(2)2-2(2)+5=12-4+5=13.

Therefore,

f(2)=13.
Worked Example 1

Question: Given f(x)=2x2-5x+4, find f(3).

Given:

f(x)=2x2-5x+4.

Required: f(3).

Solution: Substitute x=3:

f(3)=2(3)2-5(3)+4.
=18-15+4=7.

Answer: f(3)=7.

Evaluation at Algebraic Inputs

The input of a function need not be a numerical value. It may itself be an algebraic expression.

Worked Example 2

Question: If f(x)=x2+2x-1, determine f(x+1).

Solution:

Replace every occurrence of x in the function by x+1:

f(x+1)=(x+1)2+2(x+1)-1.

Expand:

=x2+2x+1+2x+2-1.

Collect like terms:

f(x+1)=x2+4x+2.

Answer: x2+4x+2.

Difference Quotient

An expression that will become particularly important when differentiation is introduced is

[f(x+h)-f(x)]/h,   h≠0.

It measures an average rate of change over an interval of length h.

Worked Example 3

Question: If f(x)=x2, simplify

[f(x+h)-f(x)]/h.

Solution:

f(x+h)=(x+h)2 =x2+2xh+h2.

Hence,

[f(x+h)-f(x)]/h = [x2+2xh+h2-x2]/h.
=(2xh+h2)/h.

Since h≠0, divide by h:

[f(x+h)-f(x)]/h=2x+h.

Answer: 2x+h.

1.4 Domain, Codomain and Range

A function is not completely described merely by writing a formula. The values that may be used as inputs and the values that may occur as outputs are also important.

Domain

The domain of a function is the set of all permissible input values for which the function is defined.

Codomain

The codomain is the specified set into which the outputs of the function are required to fall.

Range

The range is the set of actual output values produced by the function as the input varies throughout its domain.

The range is therefore contained in the codomain, but the two need not be equal.

Finding the Domain of an Algebraic Function

When no domain is explicitly stated, the natural real domain is usually taken to be all real numbers for which the formula has a real, defined value.

For polynomial functions there is no restriction on real x. Rational functions exclude values that make a denominator zero. Functions involving an even root require the quantity inside the root to be non-negative.

Worked Example 4

Question: Find the domain of

f(x)=1/(x-4).

Solution:

The denominator must not equal zero:

x-4≠0.

Therefore,

x≠4.

Answer:

Domain = ℝ \ {4}.
Worked Example 5

Question: Determine the domain of

g(x)=√(2x-6).

Solution:

For the square root to be real,

2x-6≥0.
2x≥6.
x≥3.

Answer:

Domain=[3,∞).
Worked Example 6

Question: Find the domain of

h(x)=√(x+2)/(x-5).

Solution:

Two restrictions must be satisfied.

For the square root:

x+2≥0

so that

x≥-2.

For the denominator:

x-5≠0,

so that x≠5.

Answer:

Domain=[-2,5)∪(5,∞).

Determining a Range

Worked Example 7

Question: Find the range of f(x)=x2+1, where x∈ℝ.

Solution:

For every real number x,

x2≥0.

Therefore,

x2+1≥1.

The minimum value of the function is 1, attained when x=0. Every value greater than 1 can also be produced.

Answer:

Range=[1,∞).

1.5 Functions of a Real Variable

Function of a Real Variable

A real-valued function of a real variable is a function whose domain is a subset of the real numbers and whose values are real numbers.

Such a function may be represented in the form

f:D→ℝ,

where D⊆ℝ.

For example,

f(x)=x3-2x+7

defines a real-valued function for every real number x. Its natural domain is ℝ.

In contrast,

g(x)=1/√(x-2)

is real-valued only when x-2>0. The inequality is strict because the expression inside the square root must be positive and the resulting denominator must not be zero. Hence,

Domain=(2,∞).

Functions of a real variable form the main objects studied in elementary calculus.

1.6 Types of Real Functions

Real functions can be classified according to the form of the rule that defines them. Familiarity with these classes is important because later techniques of limits, differentiation and integration depend partly on the type of function involved.

Constant Functions

A constant function has the form

f(x)=c,

where c is a fixed real number. Its value does not change when x changes.

For example,

f(x)=5.

Identity Function

The identity function is defined by

f(x)=x.

Every input is returned unchanged.

Linear Functions

A linear function in elementary coordinate form is commonly written as

f(x)=mx+c,

where m and c are constants. Its graph is a straight line, and m represents the slope.

Example
f(x)=3x-2.

Polynomial Functions

A polynomial function has the general form

P(x)=anxn+an-1xn-1+...+a1x+a0,

where n is a non-negative integer and the coefficients are constants.

Examples include

f(x)=x2-5x+6

and

g(x)=2x4-3x2+7.

Rational Functions

A rational function is a quotient of two polynomial functions:

f(x)=P(x)/Q(x),

where Q(x)≠0.

Example
f(x)=(x+1)/(x2-4).

Since

x2-4=(x-2)(x+2),

the domain excludes x=2 and x=-2.

Radical Functions

A radical function contains a variable within a root. For example,

f(x)=√(x+4).

For a real-valued square-root function,

x+4≥0,

so the domain is [-4,∞).

Absolute-Value Functions

The absolute-value function is defined by

|x| = x, if x≥0;    -x, if x<0.

Thus |x| represents the non-negative distance of x from zero on the real number line.

Exponential Functions

An exponential function has the form

f(x)=ax,

where a>0 and a≠1.

Examples include 2x, 10x and ex.

Logarithmic Functions

The logarithmic function

f(x)=logax

is the inverse of the exponential function ax, where a>0 and a≠1. Its real domain is

x>0.

Trigonometric Functions

The basic trigonometric functions include

sin x,   cos x,   tan x.

They are especially important in calculus because they describe periodic behaviour and arise naturally in geometry, waves, oscillations and rotational motion.

Function Type Typical Form Basic Domain Information
Constant f(x)=c Usually all real numbers
Polynomial P(x) All real numbers
Rational P(x)/Q(x) Exclude zeros of Q(x)
Square-root √g(x) Require g(x)≥0
Exponential ax All real numbers
Logarithmic logax x>0

1.7 Equality of Functions

Equality of Functions

Two functions are equal if they have the same domain and produce the same output for every input in that domain.

For example, consider

f(x)=(x+1)2

and

g(x)=x2+2x+1.

Since expansion gives

(x+1)2=x2+2x+1,

and both functions have domain ℝ, the two functions are equal.

However, formulas that simplify to the same algebraic expression need not define equal functions if their domains differ.

Example

Consider

f(x)=(x2-1)/(x-1)

and

g(x)=x+1.

Since

x2-1=(x-1)(x+1),

f(x)=x+1 whenever x≠1. Nevertheless, f is undefined at x=1, whereas g is defined there. Therefore, with their natural domains, the functions are not equal.

1.8 Operations on Functions

Functions can be combined using the ordinary algebraic operations of addition, subtraction, multiplication and division.

If f and g are functions, then

(f+g)(x)=f(x)+g(x),
(f-g)(x)=f(x)-g(x),
(fg)(x)=f(x)g(x),

and

(f/g)(x)=f(x)/g(x),

provided g(x)≠0.

Worked Example 8

Question: Let

f(x)=x2+1,   g(x)=2x-3.

Find (f+g)(x) and (fg)(x).

Solution:

For the sum:

(f+g)(x) = (x2+1)+(2x-3).
(f+g)(x)=x2+2x-2.

For the product:

(fg)(x)=(x2+1)(2x-3).

Expanding:

(fg)(x)=2x3-3x2+2x-3.

Answer:

(f+g)(x)=x2+2x-2

and

(fg)(x)=2x3-3x2+2x-3.

1.9 Composite Functions

A function can be applied to the output of another function. This process is called composition of functions.

Composite Function

If f and g are functions for which the relevant values are defined, the composite function f∘g is defined by

(f∘g)(x)=f(g(x)).

The function g acts first, and its output becomes the input of f.

Worked Example 9

Question: Given

f(x)=2x+1

and

g(x)=x2-3,

find (f∘g)(x) and (g∘f)(x).

Solution:

First,

(f∘g)(x)=f(g(x)).

Substitute g(x)=x2-3 into f:

f(g(x))=2(x2-3)+1.
(f∘g)(x)=2x2-5.

Next,

(g∘f)(x)=g(f(x)).
g(f(x))=(2x+1)2-3.
=4x2+4x+1-3.
(g∘f)(x)=4x2+4x-2.

Answer:

(f∘g)(x)=2x2-5,
(g∘f)(x)=4x2+4x-2.

This example shows that function composition is generally not commutative:

f∘g ≠ g∘f.

Domain of a Composite Function

For f(g(x)) to be defined, x must first belong to the domain of g, and the resulting value g(x) must belong to the domain of f.

Worked Example 10

Question: Let

f(x)=√x,   g(x)=x-5.

Determine (f∘g)(x) and its domain.

Solution:

(f∘g)(x)=√(x-5).

For this expression to be real,

x-5≥0.

Therefore,

x≥5.

Answer:

(f∘g)(x)=√(x-5),   Domain=[5,∞).

1.10 One-to-One and Many-to-One Functions

One-to-One Functions

One-to-One Function

A function f is one-to-one, or injective, if different inputs always produce different outputs.

Equivalently,

f(x1)=f(x2) ⇒ x1=x2.
Worked Example 11

Question: Show that f(x)=3x-7 is one-to-one on ℝ.

Solution:

Suppose

f(x1)=f(x2).

Then

3x1-7=3x2-7.

Add 7 to both sides:

3x1=3x2.

Divide by 3:

x1=x2.

Therefore, f is one-to-one.

Many-to-One Functions

A function is many-to-one when two or more distinct inputs can produce the same output.

For example,

f(x)=x2

gives

f(2)=4

and

f(-2)=4.

Thus, f(x)=x2 is not one-to-one on the whole real line.

1.11 Inverse Functions

An inverse function reverses the action of the original function. If f takes an input x to an output y, then its inverse takes y back to x.

Inverse Function

If f is a one-to-one function with a suitable domain and range, its inverse is denoted by f-1 and satisfies

f-1(f(x))=x

and

f(f-1(x))=x

for values belonging to the appropriate domains.

Notation

The notation f-1(x) means the inverse function. It does not mean 1/f(x).

Finding an Inverse Function

A convenient algebraic procedure is:

  • write y=f(x);
  • interchange x and y;
  • solve the resulting equation for y;
  • write the result as f-1(x).
Worked Example 12

Question: Find the inverse of

f(x)=3x+5.

Solution:

Write

y=3x+5.

Interchange x and y:

x=3y+5.

Solve for y:

x-5=3y,
y=(x-5)/3.

Therefore,

f-1(x)=(x-5)/3.

Verification:

f(f-1(x)) = 3[(x-5)/3]+5=x.

Answer:

f-1(x)=(x-5)/3.

Restricting a Domain to Obtain an Inverse

A function that is not one-to-one on its original domain may sometimes become one-to-one when its domain is restricted.

For example, f(x)=x2 is not one-to-one on ℝ. If its domain is restricted to x≥0, however, it becomes one-to-one and its inverse is

f-1(x)=√x.
Worked Example 13

Question: Let f(x)=x2+1 for x≥0. Find f-1(x).

Solution:

y=x2+1.

Interchange x and y:

x=y2+1.

Hence,

y2=x-1.

Because the original function is restricted to non-negative inputs, the inverse must return non-negative values:

y=√(x-1).

Therefore,

f-1(x)=√(x-1),   x≥1.

Answer: f-1(x)=√(x-1) with domain [1,∞).

1.12 Applications of Functions

Functions provide mathematical models for relationships between changing quantities. A mathematical model identifies the relevant variables and expresses one variable in terms of another.

Cost Function

Worked Example 14

Question: A small production firm has a fixed daily cost of ₦50,000 and an additional production cost of ₦1,200 per item. Express the total daily cost as a function of the number of items produced, and find the cost of producing 80 items.

Given:

Fixed cost = ₦50,000.
Variable cost per item = ₦1,200.
Let x represent the number of items.

Required: Cost function and cost of 80 items.

Solution:

Total cost equals fixed cost plus variable cost:

C(x)=50,000+1,200x.

For 80 items,

C(80)=50,000+1,200(80).
=50,000+96,000.
C(80)=146,000.

Answer: The cost function is C(x)=50,000+1,200x, and the cost of producing 80 items is ₦146,000.

Area as a Function

Worked Example 15

Question: A rectangle has perimeter 40 m. Express its area as a function of its length x.

Given:

2(length+width)=40.

Required: Area as a function of x.

Solution:

Let the length be x and width be w. Then

2(x+w)=40.

Therefore,

x+w=20,

so

w=20-x.

The area is

A=xw.

Hence,

A(x)=x(20-x).

Therefore,

A(x)=20x-x2.

For a non-degenerate rectangle, both dimensions must be positive, so the physically meaningful domain is

0<x<20.

Answer:

A(x)=20x-x2,   0<x<20.

Temperature Conversion as a Function

The Celsius temperature C corresponding to a Fahrenheit temperature F is given by

C(F)=5(F-32)/9.
Worked Example 16

Question: Find the Celsius temperature corresponding to 95°F.

Formula:

C(F)=5(F-32)/9.

Solution:

C(95)=5(95-32)/9.
=5(63)/9=35.

Answer: 95°F=35°C.

Functions and the Development of Calculus

The ideas introduced in this chapter provide the language needed for the remainder of calculus. When a function is represented graphically, its behaviour becomes visible. Limits will allow the behaviour of a function near a point to be studied precisely. Continuity will describe whether the function behaves without a break at a point. Differentiation will measure how rapidly the function changes, while integration will provide methods for accumulation and for determining quantities such as areas and volumes.

For this reason, a sound understanding of functions, domains, ranges, compositions and inverses is not merely preliminary algebra. These ideas form part of the conceptual foundation on which elementary calculus is built.

Chapter Summary

  • A function assigns exactly one output to each permissible input.
  • The notation f(x) denotes the value of the function f at the input x.
  • The domain is the set of permissible inputs, the codomain is the specified target set, and the range is the set of outputs actually produced.
  • A real-valued function of a real variable has a domain contained in ℝ and produces real-number outputs.
  • Important classes of functions include constant, identity, linear, polynomial, rational, radical, absolute-value, exponential, logarithmic and trigonometric functions.
  • Polynomial functions are defined for every real input, while rational and radical functions may have restrictions on their natural real domains.
  • Two functions are equal only when their domains agree and they give the same output at every point of that domain.
  • Functions may be combined by addition, subtraction, multiplication and division, subject to the appropriate domain restrictions.
  • The composition (f∘g)(x)=f(g(x)) applies g first and then f.
  • Function composition is generally not commutative.
  • A one-to-one function assigns different outputs to different inputs and can possess an inverse on the appropriate domain and range.
  • The inverse function reverses the action of the original function and is denoted by f-1.
  • Functions provide mathematical models for relationships among changing quantities and form the foundation for graphs, limits, continuity, differentiation and integration.

CHAPTER TWO

Graphs of Functions

2.1 Introduction to Graphs of Functions

A graph provides a visual representation of the relationship between the input and output values of a function. While an algebraic formula tells us how a function is defined, its graph often reveals features that are less obvious from the formula alone, such as where the function increases or decreases, where it crosses the coordinate axes, whether it is symmetric, and how rapidly it changes.

If a function is given by

y=f(x),

then its graph is the set of all points

(x,f(x))

in the Cartesian plane for values of x belonging to the domain of the function.

Graph of a Function

The graph of a function f is the set of all ordered pairs (x,f(x)) corresponding to permissible values of x.

Graphs are especially important in calculus because the ideas of limits, continuity, derivatives, maxima, minima and area under a curve all have clear graphical interpretations.

2.2 The Cartesian Coordinate Plane

The Cartesian coordinate plane is formed by two perpendicular number lines:

  • the horizontal axis, called the x-axis;
  • the vertical axis, called the y-axis.

The point at which the axes intersect is called the origin and is denoted by

O=(0,0).

Every point in the plane is represented by an ordered pair

(x,y).

The first coordinate specifies horizontal position and the second coordinate specifies vertical position.

The Four Quadrants

Quadrant Sign of x Sign of y
I Positive Positive
II Negative Positive
III Negative Negative
IV Positive Negative
Examples of Points

(3,4) lies in Quadrant I.

(-2,5) lies in Quadrant II.

(-4,-1) lies in Quadrant III.

(6,-3) lies in Quadrant IV.

2.3 Plotting Points and Reading Coordinates

To plot a point (x,y), begin at the origin. Move horizontally according to the value of x, then move vertically according to the value of y.

For example, to plot

P=(4,3),

move four units to the right and then three units upward.

To plot

Q=(-2,-5),

move two units to the left and then five units downward.

To graph a function numerically, one may first construct a table of selected input values and their corresponding outputs.

Worked Example 1

Question: Construct a table of values for y=x+2 when x=-2,-1,0,1,2.

Solution:

x y=x+2 Point
-20(-2,0)
-11(-1,1)
02(0,2)
13(1,3)
24(2,4)

Plotting these points and joining them produces a straight line.

2.4 Graphs of Linear Functions

A linear function in coordinate form is commonly written as

y=mx+c,

where m is the slope or gradient and c is the y-intercept.

Slope of a Straight Line

If two distinct points on a line are (x1,y1) and (x2,y2), then the slope is

m=(y2-y1)/(x2-x1),   x2≠x1.

A positive slope means the line rises from left to right, while a negative slope means it falls.

Worked Example 2

Question: Find the slope of the line passing through (1,3) and (5,11).

Given:

(x1,y1)=(1,3),   (x2,y2)=(5,11).

Formula:

m=(y2-y1)/(x2-x1).

Solution:

m=(11-3)/(5-1)=8/4=2.

Answer: The slope is 2.

Equation of a Line Through a Point

If a line has slope m and passes through (x1,y1), then

y-y1=m(x-x1).
Worked Example 3

Question: Find the equation of the line with slope 3 passing through (2,5).

Formula:

y-y1=m(x-x1).

Solution:

y-5=3(x-2).
y-5=3x-6.
y=3x-1.

Answer: y=3x-1.

Parallel and Perpendicular Lines

Two non-vertical lines are parallel if they have equal slopes:

m1=m2.

Two non-vertical lines are perpendicular if

m1m2=-1.
Worked Example 4

Question: Find the slope of a line perpendicular to y=4x+7.

Solution:

The given line has slope 4. If the perpendicular slope is m, then

4m=-1.

Hence,

m=-1/4.

Answer: -1/4.

2.5 Graphs of Quadratic Functions

A quadratic function has the form

y=ax2+bx+c, &prev; a≠0.

Its graph is called a parabola.

If a>0, the parabola opens upward. If a<0, it opens downward.

The Vertex and Axis of Symmetry

For

y=ax2+bx+c,

the x-coordinate of the vertex is

x=-b/(2a).

The corresponding y-coordinate is obtained by substituting this value into the function.

The vertical line

x=-b/(2a)

is the axis of symmetry.

Worked Example 5

Question: Find the vertex of

y=x2-6x+5.

Given:

a=1,   b=-6.

Formula:

x=-b/(2a).

Solution:

x=-(-6)/(2×1)=3.

Substitute x=3:

y=32-6(3)+5.
y=9-18+5=-4.

Answer: The vertex is (3,-4).

Intercepts of a Quadratic Graph

The y-intercept is obtained by setting x=0. The x-intercepts are obtained by solving

ax2+bx+c=0.
Worked Example 6

Question: Find the intercepts of

y=x2-5x+6.

For the y-intercept:

x=0 ⇒ y=6.

Hence the y-intercept is

(0,6).

For the x-intercepts:

x2-5x+6=0.
(x-2)(x-3)=0.

Therefore,

x=2 or x=3.

Answer: The x-intercepts are (2,0) and (3,0), while the y-intercept is (0,6).

Completing the Square and Graph Form

A quadratic function can often be written in the form

y=a(x-h)2+k.

The vertex is then immediately visible as

(h,k).
Worked Example 7

Question: Rewrite y=x2-4x+7 in vertex form.

Solution:

y=x2-4x+7.

Complete the square:

x2-4x=(x-2)2-4.

Therefore,

y=(x-2)2-4+7.
y=(x-2)2+3.

Answer: The vertex form is y=(x-2)2+3, so the vertex is (2,3).

2.6 Graphs of Polynomial Functions

A polynomial function may have degree 0, 1, 2, 3 or higher. The degree strongly influences the general shape of its graph.

A polynomial of degree n can have at most n real zeros and at most n-1 turning points.

For example, the cubic function

y=x3

passes through the origin, is negative for negative x, and positive for positive x.

By contrast,

y=x4

is non-negative for every real value of x and is symmetric about the y-axis.

Zeros of a Polynomial and Graph Intersections

A zero of a polynomial function is a value of x for which

f(x)=0.

Graphically, a real zero corresponds to an x-intercept.

Worked Example 8

Question: Find the x-intercepts of

f(x)=x3-4x.

Solution:

Set f(x)=0:

x3-4x=0.

Factorise:

x(x2-4)=0.
x(x-2)(x+2)=0.

Hence,

x=-2,0,2.

Answer: The graph crosses or meets the x-axis at (-2,0), (0,0) and (2,0).

2.7 Graphs of Rational Functions

A rational function is of the form

f(x)=P(x)/Q(x),

where P and Q are polynomials and Q(x)≠0.

Rational functions may have values of x at which the function is undefined. These excluded values are often associated with vertical asymptotes or removable holes.

Vertical Asymptotes

For the function

f(x)=1/(x-2),

the denominator is zero at x=2. The function becomes arbitrarily large in magnitude as x approaches 2 from either side. The line

x=2

is a vertical asymptote.

Horizontal Asymptotes

For

f(x)=1/x,

the function approaches zero as |x| becomes large. Thus,

y=0

is a horizontal asymptote.

Worked Example 9

Question: Identify the basic asymptotes of

f(x)=3/(x+1)+2.

Solution:

The denominator is zero when

x+1=0,

so

x=-1

is the vertical asymptote.

As |x| becomes very large, 3/(x+1) approaches zero, so

y=2

is the horizontal asymptote.

Answer: Vertical asymptote: x=-1; horizontal asymptote: y=2.

2.8 Graphs of Absolute-Value Functions

The basic absolute-value function is

y=|x|.

It may be written piecewise as

|x|= x, if x≥0;    -x, if x<0.

Its graph is V-shaped, with vertex at the origin.

A more general function

y=|x-h|+k

has vertex

(h,k).
Worked Example 10

Question: State the vertex of y=|x-3|-2.

Solution:

Comparing with

y=|x-h|+k,

we have h=3 and k=-2.

Answer: The vertex is (3,-2).

2.9 Graphs of Exponential and Logarithmic Functions

Exponential Functions

An exponential function has the form

y=ax,

where a>0 and a≠1.

If a>1, the function increases as x increases. If 0<a<1, the function decreases as x increases.

Every exponential graph of this form passes through

(0,1),

because

a0=1.

Its range is (0,∞), and y=0 is a horizontal asymptote.

Worked Example 11

Question: Evaluate y=2x for x=-2,-1,0,1,2.

Solution:

x 2x
-21/4
-11/2
01
12
24

These values show the rapid growth characteristic of an exponential function with base greater than 1.

Logarithmic Functions

The logarithmic function

y=logax

is the inverse of

y=ax.

Its domain is (0,∞) and its range is ℝ.

The graph passes through

(1,0),

because

loga1=0.

The line x=0 is a vertical asymptote.

Inverse Relationship

The graphs of y=ax and y=logax are reflections of each other in the line y=x.

2.10 Graphs of Trigonometric Functions

The basic trigonometric functions are periodic. Their graphs therefore repeat after fixed intervals.

The Sine Function

The function

y=sin x

has range

[-1,1]

and period

2π.

Its zeros occur at

x=nπ,

where n is an integer.

x 0 π/2 π 3π/2 2π
sin x 0 1 0 -1 0

The Cosine Function

The function

y=cos x

also has range [-1,1] and period 2π.

x 0 π/2 π 3π/2 2π
cos x 1 0 -1 0 1

The Tangent Function

The function

y=tan x

has period

π.

It is undefined when

cos x=0,

that is, at

x=π/2+nπ,

where n is an integer.

These values correspond to vertical asymptotes of the tangent graph.

2.11 Intercepts and Symmetry

x-Intercepts

An x-intercept occurs where the graph meets the x-axis. At such a point,

y=0.

Therefore, the x-intercepts of y=f(x) are obtained by solving

f(x)=0.

y-Intercepts

A y-intercept occurs where the graph meets the y-axis. At such a point,

x=0.

Therefore, if 0 belongs to the domain, the y-intercept is

(0,f(0)).
Worked Example 12

Question: Find the intercepts of

y=2x-6.

For the x-intercept:

0=2x-6.
x=3.

Hence the x-intercept is (3,0).

For the y-intercept:

x=0 ⇒ y=-6.

Answer: x-intercept = (3,0); y-intercept = (0,-6).

Symmetry About the y-Axis

A function is even if

f(-x)=f(x).

The graph of an even function is symmetric about the y-axis.

Example

For f(x)=x2,

f(-x)=(-x)2=x2=f(x).

Therefore, f(x)=x2 is even.

Symmetry About the Origin

A function is odd if

f(-x)=-f(x).

The graph of an odd function is symmetric about the origin.

Worked Example 13

Question: Determine whether f(x)=x3-2x is even, odd or neither.

Solution:

f(-x)=(-x)3-2(-x).
f(-x)=-x3+2x.

But

-f(x)=-(x3-2x) =-x3+2x.

Therefore,

f(-x)=-f(x).

Answer: The function is odd.

2.12 Transformations of Graphs

Many graphs can be obtained from a familiar basic graph by shifting, reflecting or stretching it. Understanding transformations reduces the need to construct every graph from a large table of values.

Vertical Translation

If

y=f(x)

is shifted upward by k>0, the new graph is

y=f(x)+k.

A downward shift by k is represented by

y=f(x)-k.
Example

The graph of

y=x2+4

is the graph of y=x2 shifted upward by 4 units.

Horizontal Translation

The graph of

y=f(x-h)

is obtained by shifting y=f(x) to the right by h units.

The graph of

y=f(x+h)

is shifted to the left by h units.

Worked Example 14

Question: Describe the transformation from y=x2 to y=(x-3)2+2.

Solution:

The replacement x→x-3 shifts the graph 3 units to the right.

The addition of 2 shifts the graph 2 units upward.

Answer: Shift 3 units right and 2 units upward.

Reflection in the x-Axis

The graph of

y=-f(x)

is the reflection of y=f(x) in the x-axis.

Reflection in the y-Axis

The graph of

y=f(-x)

is the reflection of y=f(x) in the y-axis.

Vertical Stretch and Compression

For

y=af(x),

all y-values are multiplied by a.

If |a|>1, the graph is stretched vertically. If 0<|a|<1, it is compressed vertically.

Horizontal Stretch and Compression

For

y=f(bx),

the horizontal scale is altered by the factor 1/|b|.

If |b|>1, the graph is compressed horizontally. If 0<|b|<1, it is stretched horizontally.

Transformation Effect on y=f(x)
y=f(x)+k Shift upward by k
y=f(x)-k Shift downward by k
y=f(x-h) Shift right by h
y=f(x+h) Shift left by h
y=-f(x) Reflect in x-axis
y=f(-x) Reflect in y-axis
y=af(x) Vertical scaling
y=f(bx) Horizontal scaling by factor 1/|b|

2.13 Reading and Interpreting Graphs

A graph should be interpreted as mathematical information rather than merely as a picture. Important features include the domain, range, intercepts, intervals of increase or decrease, turning points, maximum and minimum values, symmetry and asymptotic behaviour.

Increasing and Decreasing Behaviour

A function is increasing on an interval if its output values rise as the input values increase. It is decreasing on an interval if its output values fall as the input values increase.

For example,

f(x)=x2

decreases as x moves from negative values toward zero, reaches a minimum at x=0, and then increases for positive x.

Maximum and Minimum Values

A graph may possess a highest or lowest value over part or all of its domain.

Worked Example 15

Question: Determine the minimum value of

f(x)=x2-4x+7.

Solution:

Complete the square:

f(x)=x2-4x+4+3.
f(x)=(x-2)2+3.

Since

(x-2)2≥0,

the smallest possible value of f(x) is obtained when x=2.

f(2)=3.

Answer: The minimum value is 3, occurring at x=2.

Intersections of Two Graphs

If two functions

y=f(x)

and

y=g(x)

intersect, their y-values are equal at the point of intersection. Hence, the x-coordinates of intersections satisfy

f(x)=g(x).
Worked Example 16

Question: Find the points of intersection of

y=x2

and

y=2x+3.

Solution:

At an intersection,

x2=2x+3.

Rearrange:

x2-2x-3=0.

Factorise:

(x-3)(x+1)=0.

Therefore,

x=3 or x=-1.

When x=3:

y=32=9.

When x=-1:

y=(-1)2=1.

Answer: The graphs intersect at (3,9) and (-1,1).

Graphs as Preparation for Calculus

The graph of a function provides an intuitive introduction to several central ideas in calculus. A graph can suggest whether a function approaches a particular value near a point, which leads to the concept of a limit. A graph without a break at a point suggests continuity. The steepness of a graph introduces the idea of a derivative, while the region between a curve and an axis leads naturally to integration.

For this reason, graphical understanding complements algebraic manipulation. A reliable study of calculus requires both: algebra provides exact symbolic methods, while graphs reveal the behaviour and structure of functions.

Chapter Summary

  • The graph of a function f is the set of points (x,f(x)).
  • The Cartesian plane consists of the x-axis, y-axis, origin and four quadrants.
  • A linear function y=mx+c has a straight-line graph with slope m and y-intercept c.
  • The slope between two points is (y2-y1)/(x2-x1).
  • A quadratic function has a parabolic graph, with axis of symmetry x=-b/(2a).
  • Real zeros of a function correspond to x-intercepts of its graph.
  • Rational functions may have vertical and horizontal asymptotes.
  • The graph of an absolute-value function is characteristically V-shaped.
  • Exponential and logarithmic functions are inverse functions and their graphs are reflections in the line y=x.
  • Sine and cosine have period 2π, while tangent has period π.
  • x-intercepts are found by solving f(x)=0, while the y-intercept is found from f(0) when defined.
  • Even functions are symmetric about the y-axis, while odd functions are symmetric about the origin.
  • Translations, reflections and scalings produce new graphs from familiar parent functions.
  • Graphs reveal important behaviour such as increasing and decreasing intervals, turning points, extrema, intercepts, symmetry and asymptotes.
  • The graphical study of functions provides essential preparation for limits, continuity, differentiation and integration.

CHAPTER THREE

Limits of Functions

3.1 Introduction to Limits

The concept of a limit is one of the central ideas of calculus. It provides a precise way of describing the value that a function approaches when the independent variable approaches a particular point. Limits make it possible to study the behaviour of functions even at points where direct substitution is impossible or where the function itself is not defined.

The ideas of continuity, differentiation and integration all depend fundamentally on limits. In differentiation, a derivative is defined as a limit of an average rate of change. In integration, definite integrals can be constructed from limits of sums.

Consider the function

f(x)=x+2.

As x approaches 3, the value of f(x) approaches 5. Symbolically,

limx→3(x+2)=5.

This statement means that by choosing x sufficiently close to 3, the value of x+2 can be made as close to 5 as desired.

3.2 Intuitive Meaning of a Limit

Limit of a Function

If the values of f(x) approach a number L as x approaches a number a, then L is called the limit of f(x) as x approaches a.

The notation is

limx→a f(x)=L.

A limit describes what happens to the function near a point. It does not necessarily depend on the actual value of the function at that point.

Example

Consider

f(x)=(x2-4)/(x-2).

The function is undefined at x=2 because the denominator is zero. However,

x2-4=(x-2)(x+2).

Therefore, for x≠2,

f(x)=x+2.

As x approaches 2, x+2 approaches 4. Hence,

limx→2 (x2-4)/(x-2)=4.
Important Idea

The existence of a limit at x=a does not require the function to be defined at a. A limit concerns the behaviour of the function as x approaches a.

3.3 Limit Notation

The expression

limx→af(x)=L

is read as “the limit of f(x) as x approaches a is L.”

The symbol x→a means that x is allowed to take values increasingly close to a, but it need not actually equal a.

Examples
limx→2(3x+1)=7.
limx→-1x2=1.
limx→0sin x=0.

3.4 Limits from Tables and Graphs

A limit can often be estimated numerically by constructing a table of function values near the point of interest.

Worked Example 1

Question: Estimate

limx→2 (x2-4)/(x-2).

Solution:

x (x²-4)/(x-2)
1.93.9
1.993.99
1.9993.999
2.0014.001
2.014.01
2.14.1

From both sides of x=2, the function values approach 4.

Answer:

limx→2 (x2-4)/(x-2)=4.

Graphical Interpretation

On a graph, the statement

limx→af(x)=L

means that the points on the graph approach a height of L as their x-coordinates approach a from both sides.

The actual point (a,f(a)) may lie at (a,L), may lie somewhere else, or may not exist at all. The limit depends on the nearby behaviour.

3.5 Left-Hand and Right-Hand Limits

A two-sided limit examines the behaviour of a function as x approaches a point from both directions.

Left-Hand Limit

Left-Hand Limit

The left-hand limit of f(x) as x approaches a is the value approached by f(x) when x approaches a through values less than a.

limx→a-f(x).

Right-Hand Limit

Right-Hand Limit

The right-hand limit of f(x) as x approaches a is the value approached by f(x) when x approaches a through values greater than a.

limx→a+f(x).

A finite two-sided limit exists precisely when both one-sided limits exist and are equal:

limx→af(x)=L

if and only if

limx→a-f(x) = limx→a+f(x) = L.
Worked Example 2

Question: Consider

f(x)= 2x+1, if x<2;    x+4, if x≥2.

Determine the left-hand limit, right-hand limit and two-sided limit at x=2.

Solution:

From the left:

limx→2-f(x) = 2(2)+1 = 5.

From the right:

limx→2+f(x) = 2+4 = 6.

Since

5≠6,

the two-sided limit does not exist.

Answer:

limx→2-f(x)=5,   limx→2+f(x)=6,

therefore

limx→2f(x)

does not exist.

3.6 Laws of Limits

Limit laws allow complicated limits to be evaluated from simpler ones. Suppose

limx→af(x)=L

and

limx→ag(x)=M.

Then, whenever the resulting expressions are defined, the following laws apply.

Law Formula
Constant Law limx→a c = c
Sum Law lim(f+g)=L+M
Difference Law lim(f-g)=L-M
Constant Multiple Law lim(cf)=cL
Product Law lim(fg)=LM
Quotient Law lim(f/g)=L/M, provided M≠0
Power Law lim[f(x)]n=Ln
Root Law lim √[n]{f(x)}=√[n]{L}, where defined
Worked Example 3

Question: Evaluate

limx→2(3x2-4x+7).

Solution:

Polynomial functions satisfy the limit laws directly, so substitute x=2:

3(2)2-4(2)+7 = 12-8+7 = 11.

Answer:

limx→2(3x2-4x+7)=11.

3.7 Direct Substitution

For many elementary functions, the first method to try when evaluating a finite limit is direct substitution.

If substituting x=a produces an ordinary real number, then that number is often the required limit.

Worked Example 4

Question: Evaluate

limx→3(2x3-x+4).

Solution:

=2(3)3-3+4.
=54-3+4.
=55.

Answer: 55.

Worked Example 5

Question: Evaluate

limx→1 (x2+3)/(2x+5).

Solution:

The denominator does not vanish at x=1, so direct substitution is valid:

=(12+3)/(2(1)+5) = 4/7.

Answer: 4/7.

3.8 Indeterminate Forms and Algebraic Simplification

Sometimes direct substitution produces the expression

0/0.

This is called an indeterminate form. It does not mean that the limit is zero, nor does it mean that the limit is automatically undefined. Instead, it indicates that further simplification is required before the limit can be determined.

Interpretation of 0/0

The form 0/0 is not a numerical answer. It signals that the numerator and denominator are both tending to zero and that their relative behaviour must be examined.

Common techniques for resolving elementary 0/0 forms include factorisation, cancellation and rationalisation.

3.9 Limits Involving Factorisation

Worked Example 6

Question: Evaluate

limx→3 (x2-9)/(x-3).

Solution:

Direct substitution gives

0/0.

Factorise the numerator:

x2-9=(x-3)(x+3).

Therefore, for x≠3,

(x2-9)/(x-3)=x+3.

Hence,

limx→3(x+3)=6.

Answer: 6.

Worked Example 7

Question: Evaluate

limx→2 (x3-8)/(x-2).

Solution:

Use the difference of cubes:

x3-8=(x-2)(x2+2x+4).

Therefore,

(x3-8)/(x-2) = x2+2x+4

for x≠2.

Now substitute x=2:

22+2(2)+4 = 4+4+4 = 12.

Answer: 12.

Worked Example 8

Question: Evaluate

limx→1 (x2+x-2)/(x2-1).

Solution:

Factorise both numerator and denominator:

x2+x-2=(x-1)(x+2),
x2-1=(x-1)(x+1).

Hence, for x≠1,

(x2+x-2)/(x2-1) = (x+2)/(x+1).

Therefore,

limx→1(x+2)/(x+1) = 3/2.

Answer: 3/2.

3.10 Limits Involving Rationalisation

When square roots occur and direct substitution gives 0/0, multiplication by a conjugate often removes the indeterminate form.

Worked Example 9

Question: Evaluate

limx→4 (√x-2)/(x-4).

Solution:

Direct substitution produces 0/0. Multiply numerator and denominator by the conjugate √x+2:

[(√x-2)/(x-4)] × [(√x+2)/(√x+2)].

The numerator becomes

(√x-2)(√x+2)=x-4.

Hence,

(√x-2)/(x-4) = 1/(√x+2)

for x≠4.

Therefore,

limx→41/(√x+2) = 1/(2+2) = 1/4.

Answer: 1/4.

Worked Example 10

Question: Evaluate

limx→0 [√(x+9)-3]/x.

Solution:

Multiply by the conjugate:

[√(x+9)-3]/x × [√(x+9)+3]/[√(x+9)+3].

The numerator becomes

(x+9)-9=x.

Therefore, for x≠0,

[√(x+9)-3]/x = 1/[√(x+9)+3].

Thus,

limx→01/[√(x+9)+3] = 1/(3+3) = 1/6.

Answer: 1/6.

3.11 Limits Involving Trigonometric Functions

Two important limits used frequently in elementary calculus are

limx→0 (sin x)/x=1

and

limx→0 (1-cos x)/x=0,

where the angle x is measured in radians.

Radian Requirement

The standard trigonometric limit limx→0(sin x)/x=1 holds in this form when x is measured in radians.

Worked Example 11

Question: Evaluate

limx→0 (sin 5x)/x.

Solution:

Write

(sin 5x)/x = 5(sin 5x)/(5x).

As x→0, 5x→0. Therefore,

limx→05(sin 5x)/(5x) = 5(1) = 5.

Answer: 5.

Worked Example 12

Question: Evaluate

limx→0 (sin 3x)/(sin 2x).

Solution:

Rewrite as

(sin 3x)/(sin 2x) = [(sin 3x)/(3x)] × (3/2) × [(2x)/(sin 2x)].

As x→0,

(sin 3x)/(3x)→1

and

(2x)/(sin 2x)→1.

Hence,

limx→0(sin 3x)/(sin 2x) = 1×(3/2)×1 = 3/2.

Answer: 3/2.

Worked Example 13

Question: Evaluate

limx→0 (1-cos x)/x2.

Solution:

Multiply numerator and denominator by 1+cos x:

(1-cos x)/x2 × (1+cos x)/(1+cos x).

Using

(1-cos x)(1+cos x)=1-cos2x=sin2x,

we obtain

sin2x/[x2(1+cos x)].

This can be written as

[(sin x)/x]2 × 1/(1+cos x).

Therefore,

limx→0(1-cos x)/x2 = 12 × 1/2 = 1/2.

Answer: 1/2.

3.12 Infinite Limits and Vertical Asymptotes

A function may increase or decrease without bound as x approaches a particular value. Such behaviour is described using an infinite limit.

For example,

f(x)=1/x2

becomes arbitrarily large and positive as x approaches zero from either side. Hence,

limx→01/x2=+∞.

This notation indicates unbounded growth; it does not mean that infinity is an ordinary real number.

Vertical Asymptote

If f(x) becomes unbounded as x approaches a from one or both sides, the line x=a is called a vertical asymptote of the graph.

Worked Example 14

Question: Examine the one-sided limits of

f(x)=1/(x-3)

as x→3.

Solution:

As x→3+, the denominator is a small positive number, so

limx→3+1/(x-3)=+∞.

As x→3-, the denominator is a small negative number, so

limx→3-1/(x-3)=-∞.

Therefore, x=3 is a vertical asymptote.

3.13 Limits at Infinity

Limits at infinity describe the behaviour of a function when x becomes arbitrarily large in the positive or negative direction.

For example,

limx→∞1/x=0.

As the denominator grows without bound, the fraction approaches zero.

Rational Functions at Infinity

For a rational function, the relative degrees of the numerator and denominator often determine the limit as x→∞ or x→-∞.

Degrees Typical Behaviour
Degree of numerator < degree of denominator Limit is 0
Degrees are equal Limit is the ratio of leading coefficients
Degree of numerator > degree of denominator The function is generally unbounded or requires further analysis
Worked Example 15

Question: Evaluate

limx→∞ (3x+2)/(x2+1).

Solution:

Divide numerator and denominator by x2:

(3/x+2/x2)/(1+1/x2).

As x→∞,

3/x→0,   2/x2→0,   1/x2→0.

Therefore,

limx→∞(3x+2)/(x2+1)=0.

Answer: 0.

Worked Example 16

Question: Evaluate

limx→∞ (5x2-3x+1)/(2x2+7).

Solution:

Divide numerator and denominator by x2:

(5-3/x+1/x2)/(2+7/x2).

As x→∞, the terms involving 1/x and 1/x2 approach zero. Hence,

=5/2.

Answer: 5/2.

Worked Example 17

Question: Evaluate

limx→∞ (2x3-x)/(x2+1).

Solution:

The numerator has degree 3 while the denominator has degree 2. For large positive x, the dominant behaviour is

(2x3)/(x2)=2x.

Since 2x→+∞ as x→∞,

limx→∞(2x3-x)/(x2+1)=+∞.

Answer: The function increases without bound.

Horizontal Asymptotes

If

limx→∞f(x)=L

or

limx→-∞f(x)=L,

then the line

y=L

is a horizontal asymptote in the corresponding direction.

Example

For

f(x)=(3x+1)/(x+2),

the numerator and denominator have the same degree. The ratio of leading coefficients is 3, so

limx→∞f(x)=3.

Thus, y=3 is a horizontal asymptote.

3.14 Applications and Significance of Limits

The importance of limits extends beyond the direct evaluation of expressions. Limits provide the mathematical foundation for describing instantaneous change, continuity and accumulation.

Average Rate of Change

For a function f(x), the average rate of change between x=a and x=a+h is

[f(a+h)-f(a)]/h, &prev; h≠0.

When h is allowed to approach zero, the limiting value, when it exists, gives the instantaneous rate of change. This will lead directly to the derivative.

Worked Example 18

Question: For f(x)=x2, determine the limit of the average rate of change at x=3 as h→0.

Given:

f(x)=x2.

Required:

limh→0[f(3+h)-f(3)]/h.

Solution:

f(3+h)=(3+h)2 = 9+6h+h2.
f(3)=9.

Therefore,

[f(3+h)-f(3)]/h = [9+6h+h2-9]/h.
=(6h+h2)/h.
=6+h.

Now take the limit as h→0:

limh→0(6+h)=6.

Answer: The limiting rate of change is 6.

Limits and Continuity

A function behaves continuously at a point when its limiting value agrees with its actual function value there. In symbols, continuity at x=a requires

limx→af(x)=f(a).

This relationship will be developed fully in the next chapter.

Limits and Differentiation

The derivative of f at a point x is defined through the limit

f'(x)=limh→0[f(x+h)-f(x)]/h,

provided the limit exists. Thus, the theory of differentiation rests directly on the concept of a limit.

Limits and Integration

A definite integral can also be interpreted through a limiting process in which a region is divided into many narrow subregions and the corresponding sums approach a fixed value. In this way, limits also provide the theoretical basis for the calculation of areas and volumes.

Chapter Summary

  • A limit describes the value approached by a function as the independent variable approaches a specified point.
  • The notation limx→af(x)=L means that f(x) approaches L as x approaches a.
  • A function need not be defined at a point for its limit at that point to exist.
  • Limits may be estimated numerically from tables or interpreted graphically.
  • The left-hand limit describes behaviour as x approaches from smaller values, while the right-hand limit describes behaviour from larger values.
  • A finite two-sided limit exists when the left-hand and right-hand limits exist and are equal.
  • The sum, difference, product, quotient, power and root laws allow limits to be combined algebraically under appropriate conditions.
  • Direct substitution is the simplest method when substitution produces a defined real number.
  • The form 0/0 is indeterminate and usually requires algebraic simplification rather than being treated as an answer.
  • Factorisation and rationalisation are important methods for resolving elementary indeterminate forms.
  • The fundamental trigonometric limit limx→0(sin x)/x=1 is valid when angles are measured in radians.
  • Infinite limits describe unbounded behaviour and are associated with vertical asymptotes.
  • Limits at infinity describe the long-term behaviour of a function and may identify horizontal asymptotes.
  • For rational functions with numerator degree less than denominator degree, the limit at infinity is zero.
  • If numerator and denominator have the same degree, the limit at infinity is the ratio of their leading coefficients.
  • Limits provide the mathematical foundation for continuity, differentiation and definite integration.

CHAPTER FOUR

Continuity of Functions

4.1 Introduction to Continuity

Continuity is one of the fundamental ideas used to describe the behaviour of functions. Informally, a function is continuous over a region when its values change without sudden breaks, jumps or interruptions. Although this intuitive description is useful, calculus requires a more precise definition based on limits.

Consider the polynomial function

f(x)=x2+2x-3.

At x=2,

f(2)=22+2(2)-3=5.

Also,

limx→2f(x)=5.

Thus, the value approached by the function as x approaches 2 is exactly the value of the function at 2. This agreement between the limiting value and the actual function value is the essential idea of continuity at a point.

Continuity is important because many of the standard procedures of calculus rely on functions behaving predictably. It also provides the bridge between the theory of limits developed in the preceding chapter and the study of derivatives that follows.

4.2 Continuity at a Point

Continuity at a Point

A function f is continuous at x=a if f(a) is defined and

limx→af(x)=f(a).

This definition states that the value approached by the function near a must coincide with the actual value assigned to the function at a.

Worked Example 1

Question: Determine whether f(x)=3x2-2x+1 is continuous at x=2.

Given:

f(x)=3x2-2x+1.

Required: Determine whether f is continuous at x=2.

Solution:

First calculate the function value:

f(2)=3(2)2-2(2)+1.
=12-4+1=9.

Next calculate the limit:

limx→2(3x2-2x+1) = 3(2)2-2(2)+1 = 9.

Therefore,

limx→2f(x)=f(2)=9.

Answer: The function is continuous at x=2.

4.3 The Three Conditions for Continuity

The definition of continuity at x=a can be expressed through three conditions. All three conditions must be satisfied.

Condition Mathematical Statement Meaning
1 f(a) exists The function is defined at the point.
2 limx→af(x) exists The function approaches one finite value from both sides.
3 limx→af(x)=f(a) The limiting value equals the actual function value.

If any one of these conditions fails, the function is not continuous at the point.

Worked Example 2

Question: Consider

f(x)=(x2-9)/(x-3).

Is f continuous at x=3?

Solution:

At x=3,

f(3)=0/0,

which is undefined. Therefore, the first condition for continuity fails.

However, for x≠3,

f(x) = [(x-3)(x+3)]/(x-3) = x+3.

Hence,

limx→3f(x)=6.

Although the limit exists, f(3) does not exist.

Answer: The function is not continuous at x=3.

Repairing a Removable Discontinuity

In the preceding example, continuity can be restored by defining

f(3)=6.

Then the function value and limiting value agree:

limx→3f(x)=f(3)=6.

This illustrates an important feature of a removable discontinuity: the discontinuity can be eliminated by assigning the function the appropriate value at the missing point.

4.4 Continuity on an Interval

Continuity may be considered not only at a single point but also over an interval.

Continuity on an Open Interval

A function is continuous on an open interval (a,b) if it is continuous at every point of that interval.

For a closed interval [a,b], the endpoints require one-sided conditions.

Continuity on a Closed Interval

A function f is continuous on [a,b] if it is continuous at every point of (a,b) and satisfies

limx→a+f(x)=f(a)

and

limx→b-f(x)=f(b).
Example

The function

f(x)=√x

is continuous throughout its natural domain [0,∞). At the endpoint x=0, continuity is understood from the right:

limx→0+√x=0=f(0).

4.5 One-Sided Continuity

At endpoints of intervals, and especially for piecewise-defined functions, one-sided continuity is useful.

A function is right-continuous at x=a if

limx→a+f(x)=f(a).

Similarly, it is left-continuous at x=a if

limx→a-f(x)=f(a).
Worked Example 3

Question: Let

f(x)= x+1, if x<2;    5, if x=2;    2x+1, if x>2.

Examine continuity at x=2.

Solution:

The function value is

f(2)=5.

The left-hand limit is

limx→2-f(x) = 2+1 = 3.

The right-hand limit is

limx→2+f(x) = 2(2)+1 = 5.

Thus,

limx→2+f(x)=f(2),

so the function is right-continuous at 2. However,

limx→2-f(x)≠f(2),

and the two one-sided limits are unequal.

Answer: The function is right-continuous at x=2, but it is not continuous there.

4.6 Continuity of Elementary Functions

Many functions commonly encountered in elementary calculus are continuous throughout their natural domains. Recognising this fact allows many continuity and limit problems to be handled efficiently.

Polynomial Functions

Every polynomial function is continuous for every real number.

For example,

f(x)=4x5-3x2+7x-9

is continuous on

(-∞,∞).

Rational Functions

A rational function

f(x)=P(x)/Q(x)

is continuous at every point where

Q(x)≠0.
Worked Example 4

Question: Determine the intervals on which

f(x)=(x+2)/(x2-9)

is continuous.

Solution:

The denominator must not equal zero:

x2-9≠0.

Factorising,

(x-3)(x+3)≠0.

Therefore,

x≠3,   x≠-3.

Answer: The function is continuous on

(-∞,-3)∪(-3,3)∪(3,∞).

Radical Functions

A square-root function is continuous wherever its radicand is non-negative.

Example

For

f(x)=√(x-4),

the natural domain is

[4,∞).

The function is continuous throughout this domain.

Exponential and Logarithmic Functions

Exponential functions of the form

f(x)=ax, &prev; a>0, &prev; a≠1,

are continuous for every real number.

Logarithmic functions of the form

f(x)=logax

are continuous for

x>0.

Trigonometric Functions

The sine and cosine functions are continuous for all real values of their arguments:

sin x,   cos x.

The tangent function

tan x=sin x/cos x

is continuous wherever cos x≠0. Thus, it is discontinuous at

x=π/2+nπ, &prev; n∈ℤ.
Function Type Continuity
Polynomial Continuous for all real numbers
Rational Continuous wherever the denominator is non-zero
Square-root Continuous throughout its real domain
Exponential Continuous for all real numbers
Logarithmic Continuous for positive arguments
Sine and Cosine Continuous for all real numbers
Tangent Continuous wherever cos x≠0

4.7 Algebra of Continuous Functions

Continuity is preserved under the standard algebraic operations. Suppose f and g are continuous at x=a. Then the following functions are also continuous at a whenever they are defined:

f+g,
f-g,
cf,
fg,

and

f/g,

provided g(a)≠0.

Worked Example 5

Question: Explain why

f(x)=(x2+1)/(x+4)

is continuous at x=2.

Solution:

Both x2+1 and x+4 are polynomial functions and are therefore continuous at x=2.

Also,

2+4=6≠0.

Therefore, their quotient is continuous at x=2.

Indeed,

f(2)=(22+1)/(2+4)=5/6.

Answer: The function is continuous at x=2.

4.8 Continuity of Composite Functions

Continuity is also preserved under suitable composition.

Continuity of a Composite Function

If g is continuous at x=a and f is continuous at g(a), then the composite function f∘g is continuous at a.

In limit form,

limx→af(g(x)) = f(limx→ag(x)) = f(g(a)).
Worked Example 6

Question: Determine whether

h(x)=√(x2+5)

is continuous on ℝ.

Solution:

The inner function

g(x)=x2+5

is a polynomial and is continuous for all real x.

Furthermore,

x2+5≥5>0.

The square-root function is continuous for non-negative inputs. Hence the composition is defined and continuous for every real number.

Answer: h(x)=√(x2+5) is continuous on ℝ.

4.9 Discontinuity and Its Types

Discontinuity

A function is discontinuous at x=a if it fails to satisfy one or more of the conditions required for continuity at that point.

Several forms of discontinuity occur frequently in elementary calculus.

Removable Discontinuity

A removable discontinuity occurs when the limit exists but the function is either undefined at the point or has a value different from the limiting value.

For example,

f(x)=(x2-1)/(x-1)

is undefined at x=1. For x≠1,

f(x)=x+1.

Therefore,

limx→1f(x)=2.

The discontinuity can be removed by defining f(1)=2.

Jump Discontinuity

A jump discontinuity occurs when the left-hand and right-hand limits both exist as finite numbers but are unequal.

Example

Suppose

f(x)= 1, if x<0;    3, if x≥0.

Then

limx→0-f(x)=1

while

limx→0+f(x)=3.

Since these limits differ, the function has a jump discontinuity at x=0.

Infinite Discontinuity

An infinite discontinuity occurs when the function becomes unbounded as the variable approaches a point.

For example,

f(x)=1/(x-2)

has an infinite discontinuity at x=2, where the graph has the vertical asymptote

x=2.

Oscillatory Discontinuity

A function may also fail to approach a single value because it oscillates indefinitely near a point. A standard example is

f(x)=sin(1/x), &prev; x≠0.

As x→0, the quantity 1/x grows in magnitude and the sine function oscillates between -1 and 1 increasingly rapidly. Consequently,

limx→0sin(1/x)

does not exist.

Type Characteristic Behaviour
Removable The finite limit exists, but the function value is missing or incorrect.
Jump Finite left-hand and right-hand limits exist but are unequal.
Infinite The function becomes unbounded near the point.
Oscillatory The function oscillates without approaching a single limiting value.

4.10 Continuity of Piecewise-Defined Functions

A piecewise-defined function uses different formulas on different parts of its domain. The main points requiring examination are usually the boundary points where the defining formula changes.

Worked Example 7

Question: Determine whether

f(x)= x2+1, if x≤2;    3x-1, if x>2

is continuous at x=2.

Solution:

Since x≤2 in the first part,

f(2)=22+1=5.

The left-hand limit is

limx→2-f(x) = 22+1 = 5.

The right-hand limit is

limx→2+f(x) = 3(2)-1 = 5.

Thus,

limx→2-f(x) = limx→2+f(x) = f(2) = 5.

Answer: The function is continuous at x=2.

Worked Example 8

Question: Determine the continuity at x=1 of

f(x)= 2x+3, if x<1;    x2+1, if x≥1.

Solution:

The left-hand limit is

limx→1-f(x) = 2(1)+3 = 5.

The right-hand limit is

limx→1+f(x) = 12+1 = 2.

Since

5≠2,

the two-sided limit does not exist.

Answer: The function is discontinuous at x=1. The discontinuity is a jump discontinuity.

4.11 Determining Unknown Constants for Continuity

A common application of the definition of continuity is to determine an unknown constant in a piecewise-defined function. The value of the constant is chosen so that the left-hand limit, right-hand limit and function value agree at the joining point.

Worked Example 9

Question: Find the value of k that makes the function

f(x)= 2x+k, if x<3;    x2-1, if x≥3

continuous at x=3.

Required: The value of k.

Solution:

For continuity at x=3,

limx→3-f(x) = limx→3+f(x) = f(3).

From the left,

limx→3-f(x)=2(3)+k=6+k.

From the right,

limx→3+f(x)=32-1=8.

Also, because the second rule includes x=3,

f(3)=8.

For continuity,

6+k=8.

Therefore,

k=2.

Answer: k=2.

Worked Example 10

Question: Find a so that

f(x)= (x2-4)/(x-2), if x≠2;    a, if x=2

is continuous at x=2.

Solution:

For x≠2,

(x2-4)/(x-2) = [(x-2)(x+2)]/(x-2) = x+2.

Therefore,

limx→2f(x)=4.

Continuity requires

f(2)=limx→2f(x).

Since f(2)=a,

a=4.

Answer: a=4.

Worked Example 11

Question: Determine the values of a and b for which

f(x)= ax+b, if x<1;    3, if x=1;    2x+a, if x>1

is continuous at x=1.

Solution:

Continuity requires both one-sided limits to equal f(1)=3.

From the right,

2(1)+a=3.

Hence,

a=1.

From the left,

a(1)+b=3.

Using a=1,

1+b=3.

Therefore,

b=2.

Answer:

a=1,   b=2.

4.12 The Intermediate Value Property

One of the important consequences of continuity is that a continuous function cannot move from one value to another without passing through all intermediate values.

Intermediate Value Theorem

If f is continuous on a closed interval [a,b] and N is any number between f(a) and f(b), then there exists at least one number c∈[a,b] such that

f(c)=N.

A particularly useful case occurs when f(a) and f(b) have opposite signs. In that situation, zero lies between the two function values, so the continuous function must have at least one zero between a and b.

Worked Example 12

Question: Show that the equation

x3-x-1=0

has at least one real root between 1 and 2.

Solution:

Let

f(x)=x3-x-1.

Since f is a polynomial, it is continuous on [1,2].

Evaluate the endpoints:

f(1)=1-1-1=-1.
f(2)=8-2-1=5.

Since

f(1)<0<f(2),

the Intermediate Value Theorem guarantees that there exists at least one c∈(1,2) such that

f(c)=0.

Answer: The equation has at least one real root in the interval (1,2).

Meaning of the Result

The Intermediate Value Theorem guarantees the existence of a root in the interval, but it does not by itself give the exact value of that root.

4.13 Continuity and Differentiability

Continuity and differentiability are closely related, but they are not identical concepts. A derivative measures instantaneous rate of change, while continuity describes whether the function value agrees with its limiting behaviour.

An important theorem states:

Differentiability Implies Continuity

If a function f is differentiable at x=a, then f is continuous at x=a.

Thus,

Differentiable at a ⇒ Continuous at a.

The converse is not generally true. A function can be continuous at a point without being differentiable there.

A Continuous Function That Is Not Differentiable

Consider

f(x)=|x|.

The function is continuous at x=0 because

limx→0|x|=0=f(0).

However, the graph has a sharp corner at the origin. The slope from the left is -1, while the slope from the right is 1. Therefore, there is no single derivative at x=0.

Hence,

Continuous does not necessarily imply differentiable.

This distinction becomes particularly important when derivatives are studied in the following chapters.

4.14 Applications and Significance of Continuity

Continuity provides a mathematical description of smooth and predictable variation. Many physical quantities are modelled by continuous functions over suitable intervals. Examples include the position of a moving object as a function of time, temperature variation over time, the volume of water in a steadily filled container, and many idealised relationships used in science and engineering.

Using Continuity to Evaluate Limits

If f is known to be continuous at x=a, then

limx→af(x)=f(a).

This means that the limit can be evaluated simply by substitution.

Worked Example 13

Question: Evaluate

limx→4√(x2+9).

Solution:

The polynomial x2+9 is continuous for all real x and is always positive. The square-root function is continuous for positive inputs. Therefore, the composite function is continuous at x=4.

Hence, direct substitution gives

limx→4√(x2+9) = √(42+9).
=√25=5.

Answer: 5.

Continuity in Mathematical Modelling

Worked Example 14

Question: The height of water in a tank during a controlled filling process is modelled by

h(t)=0.5t+2,

where h is measured in metres and t in hours. Determine the height approached as t→6.

Given:

h(t)=0.5t+2.

Required:

limt→6h(t).

Solution:

The function is linear and therefore continuous for all real values of t. Hence,

limt→6h(t)=h(6).
h(6)=0.5(6)+2=3+2=5.

Answer: The water height approaches 5 metres.

Continuity as Preparation for Differentiation

The derivative is defined through a limiting process. Before a function can be differentiable at an ordinary point, it must first be continuous there. A break, jump or infinite discontinuity prevents the derivative from existing at that point.

For this reason, continuity is an essential checkpoint when analysing the behaviour of a function. In later chapters, continuity will support the study of derivatives, tangents, rates of change, maxima and minima, and curve sketching.

Chapter Summary

  • A function f is continuous at x=a when limx→af(x)=f(a).
  • Continuity at a point requires that f(a) exists, the limit exists, and the limit equals f(a).
  • A function is continuous on an open interval if it is continuous at every point of that interval.
  • Continuity on a closed interval also requires appropriate one-sided continuity at its endpoints.
  • Right-continuity uses a right-hand limit, while left-continuity uses a left-hand limit.
  • Polynomial functions are continuous for all real numbers.
  • Rational functions are continuous wherever their denominators are non-zero.
  • Radical, logarithmic and trigonometric functions are continuous throughout the appropriate parts of their natural domains.
  • Sums, differences, products and constant multiples of continuous functions remain continuous; quotients are continuous where the denominator is non-zero.
  • Under suitable conditions, the composition of continuous functions is continuous.
  • Important types of discontinuity include removable, jump, infinite and oscillatory discontinuities.
  • A removable discontinuity can be repaired when the missing or incorrect function value is replaced by the existing limiting value.
  • For a piecewise-defined function to be continuous at a joining point, its left-hand limit, right-hand limit and function value must agree.
  • Unknown constants in piecewise functions can be determined by imposing the conditions for continuity.
  • The Intermediate Value Theorem guarantees that a continuous function assumes every intermediate value between its values at the endpoints of an interval.
  • If a continuous function changes sign over an interval, it must have at least one zero between the two endpoints.
  • Differentiability at a point implies continuity at that point, but continuity does not necessarily imply differentiability.
  • Continuity provides an essential foundation for differentiation, curve analysis and later applications of calculus.

CHAPTER FIVE

The Derivative and Principles of Differentiation

5.1 Introduction to the Derivative

The derivative is one of the central concepts of differential calculus. It measures how rapidly one quantity changes with respect to another. When a quantity y depends on another quantity x, the derivative expresses the instantaneous rate at which y changes as x changes.

If

y=f(x),

then the derivative of f with respect to x may be written as

f'(x)

or

dy/dx.

The derivative has both an algebraic and a geometrical interpretation. Algebraically, it is defined by a limit. Geometrically, it gives the slope of the tangent to the graph of a function at a point.

5.2 Average Rate of Change

Before studying instantaneous change, it is useful to consider average change over an interval.

Suppose a function y=f(x) changes from f(x1) to f(x2) as x changes from x1 to x2. The average rate of change is

[f(x2)-f(x1)]/(x2-x1).
Average Rate of Change

The average rate of change of f from x=a to x=b is

[f(b)-f(a)]/(b-a),   a≠b.
Worked Example 1

Question: Find the average rate of change of f(x)=x2+1 from x=2 to x=5.

Formula:

Average rate of change=[f(5)-f(2)]/(5-2).

Solution:

f(5)=52+1=26,
f(2)=22+1=5.

Therefore,

[26-5]/[5-2] = 21/3 = 7.

Answer: The average rate of change is 7.

Average Rate of Change as Secant Slope

If the points

P=(a,f(a))

and

Q=(b,f(b))

lie on the graph of f, then the average rate of change is also the slope of the secant line joining P and Q.

5.3 Instantaneous Rate of Change

An average rate of change describes change over an interval. In many applications, however, the rate of change at a particular instant or at a particular point is required.

Suppose the interval begins at x and has width h. The average rate of change over the interval is

[f(x+h)-f(x)]/h.

If h is made smaller and smaller, the second point approaches the first. When the limit exists,

limh→0[f(x+h)-f(x)]/h

gives the instantaneous rate of change of the function at x.

Derivative

The derivative of f at x is defined by

f'(x)=limh→0[f(x+h)-f(x)]/h,

provided the limit exists.

5.4 Derivative from First Principles

Differentiation from first principles means finding a derivative directly from its limit definition rather than using an already established differentiation rule.

The basic procedure is:

  • find f(x+h);
  • form f(x+h)-f(x);
  • divide by h;
  • simplify;
  • take the limit as h→0.
Worked Example 2

Question: Differentiate f(x)=x2 from first principles.

Formula:

f'(x)=limh→0[f(x+h)-f(x)]/h.

Solution:

f(x+h)=(x+h)2 = x2+2xh+h2.

Therefore,

f(x+h)-f(x) = x2+2xh+h2-x2.
=2xh+h2.

Divide by h:

[f(x+h)-f(x)]/h = (2xh+h2)/h.
=2x+h.

Now take the limit as h→0:

f'(x) = limh→0(2x+h) = 2x.

Answer:

f'(x)=2x.
Worked Example 3

Question: Differentiate f(x)=3x+4 from first principles.

Solution:

f(x+h)=3(x+h)+4=3x+3h+4.

Then,

f(x+h)-f(x) = 3x+3h+4-(3x+4) = 3h.

Hence,

[f(x+h)-f(x)]/h = 3h/h = 3.

Therefore,

f'(x)=limh→03=3.

Answer: f'(x)=3.

Worked Example 4

Question: Find the derivative of f(x)=x3 from first principles.

Solution:

f(x+h)=(x+h)3.

Expanding,

(x+h)3 = x3+3x2h+3xh2+h3.

Thus,

f(x+h)-f(x) = 3x2h+3xh2+h3.

Divide by h:

[f(x+h)-f(x)]/h = 3x2+3xh+h2.

Taking the limit,

f'(x) = limh→0 (3x2+3xh+h2) = 3x2.

Answer:

f'(x)=3x2.

5.5 Geometrical Meaning of the Derivative

The derivative has a direct geometrical interpretation. Consider a curve y=f(x) and a point

P=(a,f(a)).

A second point nearby may be written as

Q=(a+h,f(a+h)).

The slope of the secant line through P and Q is

[f(a+h)-f(a)]/h.

As h→0, the point Q approaches P. If the secant slopes approach a definite value, that value is the slope of the tangent to the curve at P.

Slope of tangent=f'(a).
Worked Example 5

Question: Find the slope of the tangent to y=x2 at x=3.

Given:

f(x)=x2.

From first principles,

f'(x)=2x.

At x=3,

f'(3)=2(3)=6.

Answer: The slope of the tangent is 6.

Equation of a Tangent

If the tangent to y=f(x) at x=a has slope f'(a), then its equation is

y-f(a)=f'(a)(x-a).
Worked Example 6

Question: Find the equation of the tangent to y=x2 at x=2.

Solution:

The point on the curve is

(2,22)=(2,4).

The derivative is

dy/dx=2x.

Hence the slope at x=2 is

m=2(2)=4.

Using the point-slope form,

y-4=4(x-2).

Therefore,

y=4x-4.

Answer: y=4x-4.

Equation of a Normal

The normal is perpendicular to the tangent. If the tangent slope is m, where m≠0, then the normal slope is

mn=-1/m.
Worked Example 7

Question: Find the equation of the normal to y=x2 at x=2.

Solution:

The tangent slope at x=2 is 4. Therefore,

mn=-1/4.

The point is (2,4). Hence,

y-4=-(1/4)(x-2).

Answer:

y-4=-(1/4)(x-2).

5.6 Derivative Notation

Several notations are commonly used for derivatives. If y=f(x), then the first derivative may be written as

f'(x),
y',
dy/dx,

or

d[f(x)]/dx.

The notation dy/dx is particularly useful when applying differentiation rules because it clearly indicates the variable with respect to which differentiation is being performed.

Notation Meaning
f'(x) Derivative of the function f
y' Derivative of y
dy/dx Derivative of y with respect to x
d/dx[f(x)] Differentiation operator applied to f(x)

5.7 Differentiability

Differentiability

A function f is differentiable at x=a if the finite limit

limh→0[f(a+h)-f(a)]/h

exists.

Differentiability is a stronger property than continuity. If a function is differentiable at a point, then it must be continuous at that point.

Differentiable ⇒ Continuous.

However, continuity does not always imply differentiability.

Failure of Differentiability at a Corner

Consider

f(x)=|x|.

At x=0, the function is continuous. But the slope to the left is -1 while the slope to the right is 1. Therefore, the derivative does not exist at the origin.

A derivative may fail to exist at:

  • a discontinuity;
  • a sharp corner;
  • a cusp;
  • a point with a vertical tangent in the ordinary finite-derivative sense.

5.8 Derivatives of Constant and Power Functions

Derivative of a Constant

If

f(x)=c,

where c is a constant, then

f'(x)=0.

This is expected geometrically because the graph of a constant function is horizontal.

The Power Rule

Power Rule

If

y=xn,

then

dy/dx=nxn-1,

for values of n and x for which the expression is defined.

Worked Example 8

Question: Differentiate y=x7.

Solution:

dy/dx=7x6.

Answer: 7x6.

Worked Example 9

Question: Differentiate y=1/x3.

Solution:

Rewrite as

y=x-3.

Applying the power rule,

dy/dx=-3x-4.

Therefore,

dy/dx=-3/x4.

Answer: -3/x4.

Worked Example 10

Question: Differentiate y=√x.

Solution:

Write

y=x1/2.

Then,

dy/dx=(1/2)x-1/2.

Therefore,

dy/dx=1/(2√x).

Answer: 1/(2√x).

5.9 Basic Rules of Differentiation

Constant Multiple Rule

If c is a constant, then

d[cf(x)]/dx=c f'(x).

Sum Rule

d[f(x)+g(x)]/dx=f'(x)+g'(x).

Difference Rule

d[f(x)-g(x)]/dx=f'(x)-g'(x).
Worked Example 11

Question: Differentiate

y=4x5-3x3+6x-9.

Solution:

dy/dx = 20x4-9x2+6.

Answer:

dy/dx=20x4-9x2+6.
Worked Example 12

Question: Differentiate

y=5√x+4/x2.

Solution:

Rewrite using indices:

y=5x1/2+4x-2.

Differentiate term by term:

dy/dx = (5/2)x-1/2-8x-3.

Thus,

dy/dx = 5/(2√x)-8/x3.

Answer:

dy/dx=5/(2√x)-8/x3.

5.10 Higher-Order Derivatives

The derivative of a function may itself be differentiated. This produces a second derivative. Repeating the process gives higher-order derivatives.

If

y=f(x),

then the second derivative may be written as

f''(x)

or

d2y/dx2.

The third derivative may be written as

f'''(x)

or

d3y/dx3.
Worked Example 13

Question: Find the first, second and third derivatives of

y=2x4-3x3+5x.

Solution:

First derivative:

y'=8x3-9x2+5.

Second derivative:

y''=24x2-18x.

Third derivative:

y'''=48x-18.

Answer:

y'=8x3-9x2+5,
y''=24x2-18x,
y'''=48x-18.

5.11 Derivative as Velocity and Acceleration

A major application of differentiation occurs in motion. If the position of a particle at time t is represented by s(t), then velocity is the derivative of position:

v(t)=ds/dt.

Acceleration is the derivative of velocity:

a(t)=dv/dt.

Equivalently,

a(t)=d2s/dt2.
Worked Example 14

Question: A particle moves along a straight line with position

s(t)=t3-6t2+9t

metres, where t is measured in seconds. Find its velocity and acceleration at t=2.

Given:

s(t)=t3-6t2+9t.

Velocity:

v(t)=ds/dt = 3t2-12t+9.

At t=2,

v(2)=3(4)-12(2)+9 = 12-24+9 = -3.

Thus,

v(2)=-3 m/s.

Acceleration:

a(t)=dv/dt=6t-12.

At t=2,

a(2)=12-12=0.

Answer: Velocity = -3 m/s; acceleration = 0 m/s2.

5.12 Significance of the Derivative in Calculus

The derivative provides a precise mathematical description of local change. It determines the slope of a curve, the instantaneous velocity of a moving object, the rate at which one variable changes with respect to another, and the behaviour of a function near a point.

The derivative also provides the foundation for many later applications, including:

  • finding tangents and normals;
  • determining increasing and decreasing intervals;
  • locating maximum and minimum values;
  • solving optimization problems;
  • analysing motion;
  • sketching curves accurately;
  • studying rates in physical, economic and biological models.

Chapter Summary

  • The derivative measures instantaneous rate of change.
  • The average rate of change from a to b is [f(b)-f(a)]/(b-a).
  • The derivative is defined by f'(x)=limh→0[f(x+h)-f(x)]/h.
  • Differentiation from first principles uses the limit definition directly.
  • Geometrically, f'(a) is the slope of the tangent to the graph at x=a.
  • The equation of the tangent at x=a is y-f(a)=f'(a)(x-a).
  • If the tangent slope is non-zero, the normal slope is its negative reciprocal.
  • Common derivative notations include f'(x), y' and dy/dx.
  • Differentiability implies continuity, but continuity does not always imply differentiability.
  • The derivative of a constant is zero.
  • The power rule is d(xn)/dx=nxn-1.
  • Sum, difference and constant multiple rules allow polynomial-type expressions to be differentiated efficiently.
  • Higher-order derivatives are obtained by repeated differentiation.
  • If s(t) is position, then velocity is ds/dt and acceleration is d2s/dt2.

CHAPTER SIX

Techniques of Differentiation

6.1 Introduction to Differentiation Techniques

Differentiation from first principles provides the theoretical foundation of the derivative, but it is not the most efficient method for differentiating complicated functions. Once the fundamental derivative rules have been established, they can be applied systematically to polynomial, rational, trigonometric, exponential, logarithmic and composite functions.

The choice of rule depends on the structure of the function. A product requires the product rule, a quotient usually requires the quotient rule, and a function nested inside another function requires the chain rule.

Rule Selection

Before differentiating, first examine the structure of the function. Identify whether it is a sum, product, quotient, power, composition, exponential, logarithmic or trigonometric expression. This determines the most efficient technique.

6.2 Review of the Power, Sum and Difference Rules

For a power function,

d(xn)/dx=nxn-1.

If several differentiable terms are added or subtracted, they may be differentiated separately.

d[f(x)±g(x)]/dx=f'(x)±g'(x).
Worked Example 1

Question: Differentiate

y=7x6-5x4+3x2-8x+10.

Solution:

dy/dx = 42x5-20x3+6x-8.

Answer:

dy/dx=42x5-20x3+6x-8.
Worked Example 2

Question: Differentiate

y=4√x-3/x+2/x2.

Solution:

Write in index form:

y=4x1/2-3x-1+2x-2.

Differentiate term by term:

dy/dx = 2x-1/2 +3x-2 -4x-3.

Therefore,

dy/dx = 2/√x+3/x2-4/x3.

Answer:

dy/dx=2/√x+3/x2-4/x3.

6.3 The Product Rule

When two functions are multiplied together, differentiating each separately and multiplying the results is generally incorrect. The product rule must be used.

Product Rule

If

y=u(x)v(x),

then

dy/dx=u(dv/dx)+v(du/dx).

Equivalently,

(uv)'=u'v+uv'.
Worked Example 3

Question: Differentiate

y=(x2+3)(2x-5).

Given:

u=x2+3,   v=2x-5.

Then,

du/dx=2x,   dv/dx=2.

Formula:

dy/dx=u(dv/dx)+v(du/dx).

Solution:

dy/dx = (x2+3)(2) +(2x-5)(2x).
=2x2+6+4x2-10x.
=6x2-10x+6.

Answer:

dy/dx=6x2-10x+6.
Worked Example 4

Question: Differentiate

y=x3sin x.

Solution:

Let

u=x3,   v=sin x.

Then,

u'=3x2,   v'=cos x.

By the product rule,

dy/dx = 3x2sin x+x3cos x.

Answer:

dy/dx=3x2sin x+x3cos x.

6.4 The Quotient Rule

When one differentiable function is divided by another, the quotient rule provides the derivative.

Quotient Rule

If

y=u/v,

then

dy/dx= [v(du/dx)-u(dv/dx)]/v2,

provided v≠0.

Worked Example 5

Question: Differentiate

y=(x2+1)/(x-2).

Given:

u=x2+1,   v=x-2.

Then,

u'=2x,   v'=1.

Solution:

dy/dx = [(x-2)(2x)-(x2+1)(1)]/(x-2)2.

Simplify the numerator:

2x2-4x-x2-1 = x2-4x-1.

Therefore,

dy/dx= (x2-4x-1)/(x-2)2.

Answer:

dy/dx= (x2-4x-1)/(x-2)2.
Worked Example 6

Question: Differentiate

y=(sin x)/x.

Solution:

Let

u=sin x,   v=x.

Then,

u'=cos x,   v'=1.

Hence,

dy/dx = [x cos x-sin x]/x2.

Answer:

dy/dx= [x cos x-sin x]/x2.

6.5 The Chain Rule

The chain rule is used when one function is contained inside another. Such a function is called a composite function.

Suppose

y=f(u)

and

u=g(x).

Then y=f(g(x)).

Chain Rule
dy/dx=(dy/du)(du/dx).
Worked Example 7

Question: Differentiate

y=(3x+1)5.

Solution:

Let

u=3x+1.

Then,

y=u5.

Differentiate with respect to u:

dy/du=5u4.

Also,

du/dx=3.

Therefore,

dy/dx = 5u4(3) = 15u4.

Substitute u=3x+1:

dy/dx=15(3x+1)4.

Answer:

dy/dx=15(3x+1)4.
Worked Example 8

Question: Differentiate

y=√(x2+4).

Solution:

Rewrite:

y=(x2+4)1/2.

Using the chain rule,

dy/dx = (1/2)(x2+4)-1/2(2x).

Simplifying,

dy/dx= x/√(x2+4).

Answer:

dy/dx=x/√(x2+4).
Worked Example 9

Question: Differentiate

y=(2x2-3x+5)4.

Solution:

The outer function is a fourth power and the inner function is 2x2-3x+5.

dy/dx = 4(2x2-3x+5)3(4x-3).

Answer:

dy/dx= 4(4x-3)(2x2-3x+5)3.

6.6 Differentiation of Trigonometric Functions

When angles are measured in radians, the fundamental derivatives of the basic trigonometric functions are

Function Derivative
sin x cos x
cos x -sin x
tan x sec2x
cot x -csc2x
sec x sec x tan x
csc x -csc x cot x
Worked Example 10

Question: Differentiate

y=4sin x-3cos x+2tan x.

Solution:

dy/dx = 4cos x+3sin x+2sec2x.

Answer:

dy/dx= 4cos x+3sin x+2sec2x.

Trigonometric Functions with the Chain Rule

Worked Example 11

Question: Differentiate

y=sin(5x).

Solution:

The outer function is sine and the inner function is 5x.

dy/dx = cos(5x)×5.

Hence,

dy/dx=5cos(5x).

Answer: 5cos(5x).

Worked Example 12

Question: Differentiate

y=cos(x2+1).

Solution:

Using the chain rule,

dy/dx = -sin(x2+1)(2x).

Therefore,

dy/dx=-2x sin(x2+1).

Answer:

dy/dx=-2x sin(x2+1).
Worked Example 13

Question: Differentiate

y=tan(3x2).

Solution:

dy/dx = sec2(3x2)×6x.

Answer:

dy/dx=6x sec2(3x2).

6.7 Differentiation of Exponential Functions

The natural exponential function has the important property

d(ex)/dx=ex.

For a general exponential function,

d(ax)/dx=axln a,

where a>0 and a≠1.

Worked Example 14

Question: Differentiate

y=5ex.

Solution:

dy/dx=5ex.

Answer: 5ex.

Worked Example 15

Question: Differentiate

y=e3x.

Solution:

Apply the chain rule:

dy/dx=e3x(3).

Therefore,

dy/dx=3e3x.

Answer: 3e3x.

Worked Example 16

Question: Differentiate

y=2x.

Solution:

dy/dx=2xln2.

Answer: 2xln2.

6.8 Differentiation of Logarithmic Functions

The natural logarithm satisfies

d(ln x)/dx=1/x, &prev; x>0.

More generally,

d[ln u]/dx=(1/u)(du/dx).
Worked Example 17

Question: Differentiate

y=ln(x2+1).

Solution:

Let

u=x2+1.

Then,

du/dx=2x.

Therefore,

dy/dx= 1/(x2+1) × 2x.

Hence,

dy/dx= 2x/(x2+1).

Answer:

dy/dx=2x/(x2+1).

For logarithms to base a,

d(logax)/dx=1/(x ln a).
Worked Example 18

Question: Differentiate

y=log10x.

Solution:

dy/dx= 1/(x ln10).

Answer: 1/(x ln10).

6.9 Implicit Differentiation

Not every relationship between x and y is conveniently written in the explicit form y=f(x). Sometimes the variables appear together in one equation. In such cases, implicit differentiation is useful.

The key principle is to differentiate both sides with respect to x, remembering that y is itself a function of x.

Worked Example 19

Question: If

x2+y2=25,

find dy/dx.

Solution:

Differentiate both sides with respect to x:

2x+2y(dy/dx)=0.

Rearrange:

2y(dy/dx)=-2x.

Therefore,

dy/dx=-x/y.

Answer:

dy/dx=-x/y.
Worked Example 20

Question: If

x2+xy+y2=7,

find dy/dx.

Solution:

Differentiate term by term:

2x+d(xy)/dx+2y(dy/dx)=0.

For xy, use the product rule:

d(xy)/dx = x(dy/dx)+y.

Hence,

2x+x(dy/dx)+y+2y(dy/dx)=0.

Collect the derivative terms:

(x+2y)(dy/dx)=-(2x+y).

Therefore,

dy/dx= -(2x+y)/(x+2y).

Answer:

dy/dx=-(2x+y)/(x+2y).

6.10 Logarithmic Differentiation

Logarithmic differentiation is useful when a function involves complicated products, quotients or variables occurring in both the base and the exponent.

The method involves taking the natural logarithm of both sides and then differentiating implicitly.

Worked Example 21

Question: Differentiate

y=xx,   x>0.

Solution:

Take natural logarithms:

ln y=ln(xx).

Using the logarithm law,

ln y=x ln x.

Differentiate both sides:

(1/y)(dy/dx)=ln x+1.

Therefore,

dy/dx=y(ln x+1).

Since y=xx,

dy/dx=xx(ln x+1).

Answer:

dy/dx=xx(ln x+1).

6.11 Differentiation of Composite Algebraic Functions

More complicated functions may require more than one differentiation rule. The most efficient method is to work from the outermost structure inward.

Worked Example 22

Question: Differentiate

y=(x2+1)3(2x-5).

Solution:

The expression is a product, so use the product rule. Let

u=(x2+1)3,   v=2x-5.

Differentiate u using the chain rule:

u' = 3(x2+1)2(2x) = 6x(x2+1)2.

Also,

v'=2.

Therefore,

dy/dx = 6x(x2+1)2(2x-5) +2(x2+1)3.

Answer:

dy/dx = 6x(2x-5)(x2+1)2 +2(x2+1)3.
Worked Example 23

Question: Differentiate

y=[sin(2x)]3.

Solution:

This function contains several layers. The outer function is a cube, followed by sine, followed by 2x.

dy/dx = 3[sin(2x)]2 ×cos(2x) ×2.

Therefore,

dy/dx= 6sin2(2x)cos(2x).

Answer:

dy/dx=6sin2(2x)cos(2x).
Worked Example 24

Question: Differentiate

y=ln(1+x2)ex.

Solution:

This is a product. Let

u=ln(1+x2),   v=ex.

Then,

u'=2x/(1+x2)

and

v'=ex.

Using the product rule,

dy/dx = [2x/(1+x2)]ex +ln(1+x2)ex.

Factorising ex,

dy/dx = ex [2x/(1+x2)+ln(1+x2)].

Answer:

dy/dx = ex [2x/(1+x2)+ln(1+x2)].

6.12 Higher-Order Derivatives of Standard Functions

Higher-order derivatives are often required in the study of acceleration, curvature, maxima, minima and curve sketching.

Worked Example 25

Question: Find y'' if

y=e2x.

Solution:

First derivative:

y'=2e2x.

Second derivative:

y''=4e2x.

Answer: y''=4e2x.

Worked Example 26

Question: Find the first four derivatives of y=sin x.

Solution:

y'=cos x,
y''=-sin x,
y'''=-cos x,
y''''=sin x.

The derivatives then repeat in a cycle of four.

Answer:

cos x,\;-sin x,\;-cos x,\;sin x.

6.13 Choosing an Appropriate Differentiation Rule

Efficient differentiation depends on recognising the structure of a function before beginning the calculation.

Structure of Function Typical Form Recommended Rule
Simple power xn Power rule
Sum or difference f±g Differentiate term by term
Product fg Product rule
Quotient f/g Quotient rule or algebraic simplification
Composite function f(g(x)) Chain rule
Implicit relation F(x,y)=0 Implicit differentiation
Variable in base and exponent xx Logarithmic differentiation
Trigonometric expression sin x, cos x, tan x Standard trig derivatives, often with chain rule
Exponential expression eu, au Exponential derivative and chain rule
Logarithmic expression ln u Logarithmic derivative and chain rule
Worked Example 27

Question: Differentiate

y=(x2+1)/(x3+2).

Reasoning:

The entire expression is a quotient of two polynomial functions. Therefore, the quotient rule is appropriate.

Solution:

dy/dx = [(x3+2)(2x)-(x2+1)(3x2)] /(x3+2)2.

Expand the numerator:

2x4+4x -3x4-3x2.
=-x4-3x2+4x.

Hence,

dy/dx= (-x4-3x2+4x)/(x3+2)2.

Answer:

dy/dx= (-x4-3x2+4x)/(x3+2)2.
Worked Example 28

Question: Differentiate

y=ex2+3x.

Reasoning:

This is an exponential function whose exponent is itself a function of x. Therefore, use the exponential derivative together with the chain rule.

Solution:

dy/dx = ex2+3x(2x+3).

Answer:

dy/dx=(2x+3)ex2+3x.

Chapter Summary

  • The power, sum and difference rules form the basis for differentiating many algebraic expressions.
  • The product rule is (uv)'=u'v+uv'.
  • The quotient rule is (u/v)'=(vu'-uv')/v2.
  • The chain rule is used for composite functions and may be written as dy/dx=(dy/du)(du/dx).
  • The derivatives of sin x, cos x and tan x are cos x, -sin x and sec2x, respectively.
  • Trigonometric derivative formulas in elementary calculus assume that angles are measured in radians.
  • The derivative of ex is ex.
  • The derivative of ax is axln a.
  • The derivative of ln x is 1/x.
  • Implicit differentiation is used when x and y occur together in an equation that is not conveniently solved explicitly for y.
  • Logarithmic differentiation is especially useful when the variable occurs in both the base and exponent.
  • Complicated functions may require several differentiation rules in a single calculation.
  • Higher-order derivatives are obtained by differentiating repeatedly.
  • Correct rule selection begins by identifying the mathematical structure of the function.

CHAPTER NINE

Definite Integrals and the Fundamental Theorem of Calculus

9.1 Introduction to Definite Integration

Indefinite integration produces a family of antiderivatives. A definite integral, however, produces a numerical value associated with a function over a specified interval. It is one of the principal tools for measuring accumulated change.

If a quantity varies continuously with respect to another quantity, integration can be used to determine its total accumulated effect over an interval. Applications include area, displacement, total production, accumulated flow and volume.

A definite integral has two limits of integration and is written in the form

∫ab f(x) dx.

The numbers a and b are called the lower and upper limits of integration respectively.

9.2 Meaning and Notation of a Definite Integral

Definite Integral

If f is integrable on the interval [a,b], then

∫ab f(x)dx

represents the net accumulation of f(x) over the interval from x=a to x=b.

Unlike an indefinite integral, a definite integral does not require an arbitrary constant C. Once the limits are specified, the result is a definite number.

Example

The expression

∫13 2x dx

represents the net accumulation of the function 2x between x=1 and x=3.

9.3 Definite Integrals as Limits of Sums

The concept of a definite integral originates from the idea of dividing an interval into many small subintervals and adding the contributions from each part.

Suppose [a,b] is divided into n equal subintervals of width

Δx=(b-a)/n.

If xi* is a selected point in the ith subinterval, then the sum

Σ f(xi*)Δx

approximates the accumulated contribution over the interval.

When the number of subdivisions increases without bound and the widths approach zero, the limiting value is the definite integral:

∫ab f(x)dx = limn→∞ Σi=1n f(xi*)Δx.
Riemann Sum Idea

A finite sum gives an approximation to total accumulation. The definite integral is obtained by taking a limit in which the partition becomes arbitrarily fine.

9.4 Geometrical Interpretation of Definite Integrals

When f(x)≥0 on [a,b], the definite integral

∫abf(x)dx

is equal to the area between the graph y=f(x), the x-axis and the vertical lines x=a and x=b.

When the graph lies below the x-axis, the integral contributes negatively. Thus, a definite integral gives signed area or net area, not necessarily the total geometric area.

Signed Area

Regions above the x-axis contribute positively to a definite integral, while regions below the x-axis contribute negatively.

9.5 Properties of Definite Integrals

Definite integrals satisfy several useful algebraic properties.

Equal Limits

∫aa f(x)dx=0.

Reversal of Limits

∫ab f(x)dx = -∫ba f(x)dx.

Additivity over Intervals

If c lies between a and b, then

∫abf(x)dx = ∫acf(x)dx + ∫cbf(x)dx.

Constant Multiple Rule

∫abk f(x)dx = k∫abf(x)dx.

Sum and Difference Rules

∫ab[f(x)±g(x)]dx = ∫abf(x)dx ± ∫abg(x)dx.

Order Property

If

f(x)≤g(x)

for every x∈[a,b], then

∫abf(x)dx ≤ ∫abg(x)dx.

9.6 The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus establishes the central connection between differentiation and integration.

Fundamental Theorem of Calculus

If f is continuous on [a,b] and F is any antiderivative of f, so that

F'(x)=f(x),

then

∫ab f(x)dx = F(b)-F(a).

The notation

[F(x)]ab

means

F(b)-F(a).

This theorem is extremely important because it allows definite integrals to be evaluated using antiderivatives instead of calculating limits of sums directly.

Accumulation Form of the Theorem

If

A(x)=∫axf(t)dt,

where f is continuous, then

A'(x)=f(x).

Thus, differentiation and definite integration are inverse processes in a precise mathematical sense.

9.7 Evaluation of Definite Integrals

Worked Example 1

Question: Evaluate

∫132x dx.

Solution:

An antiderivative of 2x is

x2.

Therefore,

∫132x dx = [x2]13.
=32-12 = 9-1 = 8.

Answer: 8.

Worked Example 2

Question: Evaluate

∫02(3x2-4x+5)dx.

Solution:

An antiderivative is

F(x)=x3-2x2+5x.

Hence,

[x3-2x2+5x]02.

At x=2,

8-8+10=10.

At x=0,

0.

Therefore,

∫02(3x2-4x+5)dx=10.

Answer: 10.

Worked Example 3

Question: Evaluate

∫141/√x dx.

Solution:

Write

1/√x=x-1/2.

Then,

∫x-1/2dx = 2x1/2.

Therefore,

∫141/√x dx = [2√x]14.
=2√4-2√1.
=4-2=2.

Answer: 2.

9.8 Definite Integrals Involving Trigonometric Functions

Worked Example 4

Question: Evaluate

∫0π/2 cos x dx.

Solution:

An antiderivative of cos x is sin x.

∫0π/2 cos x dx = [sin x]0π/2.
=sin(π/2)-sin0.
=1-0=1.

Answer: 1.

Worked Example 5

Question: Evaluate

∫0π sin x dx.

Solution:

An antiderivative of sin x is -cos x.

∫0πsin x dx = [-cos x]0π.
=-cosπ-(-cos0).
=1+1=2.

Answer: 2.

Worked Example 6

Question: Evaluate

∫0π/4 sec2x dx.

Solution:

∫sec2x dx=tan x.

Hence,

[tan x]0π/4 = tan(π/4)-tan0.
=1-0=1.

Answer: 1.

9.9 Definite Integrals Involving Exponential Functions

Worked Example 7

Question: Evaluate

∫01exdx.

Solution:

∫exdx=ex.

Therefore,

∫01exdx = [ex]01.
=e-e0 = e-1.

Answer: e-1.

Worked Example 8

Question: Evaluate

∫023e2xdx.

Solution:

An antiderivative is

(3/2)e2x.

Thus,

[(3/2)e2x]02.
=(3/2)e4-(3/2).

Therefore,

=(3/2)(e4-1).

Answer:

(3/2)(e4-1).

9.10 Substitution in Definite Integrals

Substitution can be applied directly to definite integrals. When the variable is changed, the limits should also be changed to values corresponding to the new variable.

Worked Example 9

Question: Evaluate

∫012x(x2+1)3dx.

Solution:

Let

u=x2+1.

Then,

du=2x dx.

Change the limits:

When x=0,

u=02+1=1.

When x=1,

u=12+1=2.

Thus,

∫012x(x2+1)3dx = ∫12u3du.
= [u4/4]12.
=(16/4)-(1/4) = 15/4.

Answer: 15/4.

Worked Example 10

Question: Evaluate

∫0π/84cos(4x)dx.

Solution:

Let

u=4x,   du=4dx.

Change the limits:

When x=0, u=0.

When x=π/8,

u=π/2.

Hence,

∫0π/2cos u du.
=[sin u]0π/2.
=1.

Answer: 1.

9.11 Symmetry Properties of Definite Integrals

Symmetry can simplify many definite integrals.

Odd Functions

If f is odd, so that

f(-x)=-f(x),

then over a symmetric interval,

∫-aaf(x)dx=0.

Even Functions

If f is even, so that

f(-x)=f(x),

then

∫-aaf(x)dx = 2∫0af(x)dx.
Worked Example 11

Question: Evaluate

∫-33x3dx.

Solution:

Since x3 is an odd function and the interval is symmetric about zero,

∫-33x3dx=0.

Answer: 0.

Worked Example 12

Question: Evaluate

∫-22x2dx.

Solution:

Since x2 is even,

∫-22x2dx = 2∫02x2dx.
= 2[x3/3]02.
=2(8/3)=16/3.

Answer: 16/3.

9.12 Total Change from a Rate Function

If F'(x)=f(x), then the Fundamental Theorem of Calculus gives

F(b)-F(a) = ∫abf(x)dx.

This principle is often called the net change theorem. It states that the integral of a rate of change gives the net change in the original quantity.

Worked Example 13

Question: A particle has velocity

v(t)=3t2-4t+2

metres per second. Find its displacement from t=0 to t=3.

Formula:

Displacement = ∫03v(t)dt.

Solution:

∫03 (3t2-4t+2)dt.

An antiderivative is

t3-2t2+2t.

Therefore,

[t3-2t2+2t]03.
=27-18+6=15.

Answer: The displacement is 15 metres.

Displacement and Distance

Integrating velocity gives displacement, which is a signed quantity. Total distance travelled may require splitting the interval at points where the velocity changes sign and integrating the magnitude of the velocity.

9.13 Average Value of a Function

Average Value

If f is integrable on [a,b], its average value on the interval is

favg = 1/(b-a) ∫abf(x)dx.
Worked Example 14

Question: Find the average value of

f(x)=x2

on the interval [0,3].

Formula:

favg = 1/(3-0) ∫03x2dx.

Solution:

∫03x2dx = [x3/3]03 = 9.

Hence,

favg=9/3=3.

Answer: The average value is 3.

9.14 Significance of Definite Integration

Definite integration provides a rigorous method for calculating accumulated quantities over intervals. Its importance lies not only in area calculations but in its ability to convert rates into totals.

For example, integrating velocity over time gives displacement, integrating a flow rate gives total volume, and integrating a density over a region can give total mass under appropriate conditions.

The Fundamental Theorem of Calculus makes these calculations practical by connecting accumulated change to antiderivatives.

Chapter Summary

  • A definite integral gives the net accumulation of a function over a specified interval.
  • The notation ∫abf(x)dx contains lower and upper limits of integration.
  • Definite integrals can be defined as limits of sums.
  • When f(x)≥0, a definite integral represents the area beneath the graph over the interval.
  • Regions below the x-axis make negative contributions to a definite integral.
  • Reversing the limits changes the sign of the integral.
  • Definite integrals are additive over adjacent intervals.
  • The Fundamental Theorem of Calculus states that ∫abf(x)dx=F(b)-F(a) when F'(x)=f(x).
  • Substitution may be used in definite integrals, with corresponding changes to the limits.
  • The integral of an odd function over a symmetric interval is zero.
  • The integral of an even function over [-a,a] is twice its integral over [0,a].
  • Integrating a rate function over an interval gives the net change in the associated quantity.
  • The average value of f on [a,b] is 1/(b-a)∫abf(x)dx.

CHAPTER TEN

Applications of Definite Integrals to Areas and Volumes

10.1 Introduction to Applications of Definite Integrals

One of the most important applications of definite integration is the measurement of geometric quantities. Areas bounded by curves and volumes of solids generated by rotation can often be calculated exactly using definite integrals.

The central principle is accumulation. A region or solid is imagined as being divided into many extremely small parts. The contribution of each small part is expressed mathematically, and integration adds these contributions over the required interval.

In area problems, the essential quantity is usually a vertical or horizontal strip. In volume problems, the basic element may be a thin disk, washer or cylindrical shell.

10.2 Area Under a Curve

Suppose f(x)≥0 on [a,b]. The area under the curve y=f(x) and above the x-axis is

Area Under a Curve
A= ∫abf(x)dx.
Worked Example 1

Question: Find the area under the curve

y=x2

between x=0 and x=2.

Given:

f(x)=x2,   0≤x≤2.

Formula:

A=∫02x2dx.

Solution:

A= [x3/3]02.
=8/3.

Answer: The area is 8/3 square units.

Worked Example 2

Question: Find the area under

y=2x+1

from x=1 to x=4.

Solution:

A= ∫14(2x+1)dx.
= [x2+x]14.
=(16+4)-(1+1) = 20-2 = 18.

Answer: 18 square units.

10.3 Area Between a Curve and the x-Axis

If a function is entirely above the x-axis on the interval, its definite integral gives the geometric area directly.

If the function is entirely below the x-axis, however, the definite integral is negative. Geometric area must be positive, so the magnitude of the integral is taken.

Area= |∫abf(x)dx|.
Worked Example 3

Question: Find the area between the curve

y=x-3

and the x-axis from x=0 to x=2.

Solution:

On [0,2], the function x-3 is negative.

∫02(x-3)dx = [x2/2-3x]02.
=2-6=-4.

The negative sign indicates that the region lies below the axis. Therefore, the geometric area is

|-4|=4.

Answer: 4 square units.

10.4 Total Area When a Curve Crosses the x-Axis

If a curve crosses the x-axis within the interval, a single definite integral gives net signed area rather than total geometric area. The interval must therefore be divided at the zeros of the function.

Worked Example 4

Question: Find the total area between

y=x-1

and the x-axis from x=0 to x=3.

Solution:

First find where the curve crosses the axis:

x-1=0,
x=1.

The function is negative on [0,1] and positive on [1,3].

Area below the axis:

A1 = -∫01(x-1)dx.
= -[x2/2-x]01.
=-(-1/2)=1/2.

Area above the axis:

A2 = ∫13(x-1)dx.
= [x2/2-x]13.
=(9/2-3)-(1/2-1).
=3/2+1/2=2.

Total area:

A=A1+A2 = 1/2+2 = 5/2.

Answer: The total area is 5/2 square units.

10.5 Area Between Two Curves

Suppose two functions y=f(x) and y=g(x) bound a region between x=a and x=b, with

f(x)≥g(x).

Then the area between the curves is

Area Between Two Curves
A= ∫ab[f(x)-g(x)]dx.

In words: upper curve minus lower curve.

Worked Example 5

Question: Find the area enclosed by

y=2x

and

y=x2.

Solution:

First determine the points of intersection:

2x=x2.
x2-2x=0.
x(x-2)=0.

Hence,

x=0,   x=2.

Between 0 and 2, 2x lies above x2. Therefore,

A= ∫02(2x-x2)dx.
= [x2-x3/3]02.
=4-8/3 = 4/3.

Answer: The enclosed area is 4/3 square units.

Worked Example 6

Question: Find the area enclosed between

y=x+2

and

y=x2.

Solution:

Determine the intersections:

x2=x+2.
x2-x-2=0.
(x-2)(x+1)=0.

Thus,

x=-1,   x=2.

Over this interval, the line y=x+2 lies above the parabola.

A= ∫-12 [(x+2)-x2]dx.

An antiderivative is

x2/2+2x-x3/3.

At x=2,

2+4-8/3=10/3.

At x=-1,

1/2-2+1/3=-7/6.

Hence,

A= 10/3-(-7/6) = 20/6+7/6 = 27/6 = 9/2.

Answer: The area is 9/2 square units.

10.6 Determining the Limits of Integration

In many area problems, the limits of integration are not provided directly. They must be obtained by finding where the bounding curves intersect.

If the curves are

y=f(x)

and

y=g(x),

their intersection points satisfy

f(x)=g(x).
Worked Example 7

Question: Find the limits of integration for the region enclosed by

y=x2

and

y=4x.

Solution:

x2=4x.
x2-4x=0.
x(x-4)=0.

Thus,

x=0   and   x=4.

Answer: The limits of integration are 0 and 4.

10.7 Areas with Respect to y

Some regions are more conveniently described by horizontal strips rather than vertical strips. In that case, the curves are written in the form

x=f(y)

and

x=g(y).

If the right-hand curve is x=f(y) and the left-hand curve is x=g(y), then

A= ∫cd[f(y)-g(y)]dy.

In words: right curve minus left curve.

Worked Example 8

Question: Find the area bounded by

x=y2

and

x=2y.

Solution:

Determine the intersections:

y2=2y.
y(y-2)=0.

Hence,

y=0   and   y=2.

For 0≤y≤2, the right-hand curve is x=2y and the left-hand curve is x=y2.

A= ∫02(2y-y2)dy.
= [y2-y3/3]02.
=4-8/3=4/3.

Answer: 4/3 square units.

10.8 Introduction to Volumes of Solids of Revolution

A plane region can generate a three-dimensional solid when rotated about a line. Such a solid is called a solid of revolution.

For example, rotating a region under a curve about the x-axis may produce a solid whose cross-sections perpendicular to the axis are circular disks.

Definite integration allows the volumes of such solids to be obtained by summing the volumes of extremely thin elements.

10.9 Volume by the Disk Method

Suppose a region under the curve

y=f(x)

from x=a to x=b is rotated about the x-axis and f(x)≥0.

A typical cross-section is a circular disk of radius

R=f(x).

Its area is

A(x)=π[f(x)]2.

Therefore, the volume is

Disk Method
V= π∫ab[f(x)]2dx.
Worked Example 9

Question: Find the volume obtained when the region under

y=x

from x=0 to x=2 is rotated about the x-axis.

Formula:

V= π∫02y2dx.

Solution:

V= π∫02x2dx.
= π[x3/3]02.
=8π/3.

Answer: The volume is 8π/3 cubic units.

Worked Example 10

Question: Find the volume generated by rotating the region under

y=√x

from x=0 to x=4 about the x-axis.

Solution:

V= π∫04(√x)2dx.
= π∫04x dx.
= π[x2/2]04.
=8π.

Answer: 8π cubic units.

10.10 Volume by the Washer Method

If the rotating region does not extend all the way to the axis of rotation, each cross-section may have a hole. Such a cross-section is called a washer.

If R(x) is the outer radius and r(x) the inner radius, then the cross-sectional area is

A(x)=π[R(x)]2-π[r(x)]2.

Therefore,

Washer Method
V= π∫ab ([R(x)]2-[r(x)]2)dx.
Worked Example 11

Question: The region between

y=x+1

and

y=1

for 0≤x≤2 is rotated about the x-axis. Find the volume.

Solution:

Outer radius:

R=x+1.

Inner radius:

r=1.

Therefore,

V= π∫02 [(x+1)2-1]dx.

Expand:

(x+1)2-1 = x2+2x.

Hence,

V= π∫02 (x2+2x)dx.
= π[x3/3+x2]02.
= π(8/3+4) = 20π/3.

Answer: The volume is 20π/3 cubic units.

10.11 Rotation About the y-Axis

When a region is rotated about the y-axis, one possible method is to write x as a function of y and use horizontal disks or washers.

If

x=g(y)

is rotated about the y-axis, the disk formula becomes

V= π∫cd[g(y)]2dy.
Worked Example 12

Question: Find the volume obtained by rotating the region bounded by

x=y2,
x=0,

and 0≤y≤2 about the y-axis.

Solution:

The radius of a disk is

R=y2.

Hence,

V= π∫02 (y2)2dy.
= π∫02y4dy.
= π[y5/5]02.
=32π/5.

Answer: 32π/5 cubic units.

10.12 Volume by Cylindrical Shells

Another useful method for volumes of revolution is the cylindrical shell method. It is particularly convenient when vertical strips are rotated about the y-axis.

A vertical strip at position x generates a thin cylindrical shell with:

  • radius x;
  • height f(x);
  • circumference 2πx;
  • small thickness dx.

Therefore,

Shell Method About the y-Axis
V= 2π∫abx f(x)dx,

when the region lies between y=f(x) and the x-axis.

Worked Example 13

Question: Find the volume obtained when the region under

y=x

from x=0 to x=2 is rotated about the y-axis.

Formula:

V= 2π∫02x f(x)dx.

Solution:

Since f(x)=x,

V= 2π∫02x2dx.
= 2π[x3/3]02.
=16π/3.

Answer: 16π/3 cubic units.

Worked Example 14

Question: The region under

y=4-x2

for 0≤x≤2 is rotated about the y-axis. Find its volume using cylindrical shells.

Solution:

V= 2π∫02 x(4-x2)dx.
= 2π∫02 (4x-x3)dx.
= 2π[2x2-x4/4]02.
= 2π(8-4).
=8π.

Answer: 8π cubic units.

10.13 Applications to Physical and Geometrical Problems

The formulas for areas and volumes are not restricted to purely abstract curves. They provide models for real geometric and physical structures whenever the dimensions can be represented by suitable functions.

Cross-Sectional Interpretation

If a solid has cross-sectional area A(x) at position x, then its volume over the interval [a,b] is

V= ∫abA(x)dx.

The disk and washer methods are special cases of this general principle.

Worked Example 15

Question: A solid extends from x=0 to x=3. Its cross-sectional area at position x is

A(x)=x2+2

square centimetres. Find the volume of the solid.

Formula:

V= ∫03A(x)dx.

Solution:

V= ∫03 (x2+2)dx.
= [x3/3+2x]03.
=9+6=15.

Answer: The volume is 15 cm3.

Recovering a Physical Quantity from a Rate

Worked Example 16

Question: Water enters a tank at a rate

R(t)=2t+5

litres per minute. Determine the amount of water added during the first 4 minutes.

Formula:

Amount added = ∫04R(t)dt.

Solution:

∫04(2t+5)dt.
= [t2+5t]04.
=16+20=36.

Answer: 36 litres of water are added.

10.14 Choosing an Appropriate Area or Volume Method

Successful applications of integration depend on translating the geometry of the problem into an appropriate integral. The method should be chosen according to the orientation of the region and the axis of rotation.

Problem Structure Typical Integral Main Idea
Area under a non-negative curve ∫f(x)dx Curve minus axis
Area between two curves using vertical strips ∫[upper-lower]dx Upper function minus lower function
Area using horizontal strips ∫[right-left]dy Right function minus left function
Rotation around an axis with no hole π∫R2dx Disk method
Rotation producing a central hole π∫(R2-r2)dx Washer method
Vertical strips rotated about y-axis 2π∫(radius)(height)dx Cylindrical shell method
Known cross-sectional area ∫A(x)dx Add cross-sectional volumes
Setting Up the Integral

Before integrating, identify the boundaries of the region, determine the correct limits, decide which function is above or below the other, and identify the radius or height required for a volume calculation. A correctly evaluated integral cannot compensate for an incorrectly modelled region.

Worked Example 17

Question: Find the volume obtained when the region bounded by

y=x2,   y=0,   x=0,   x=1

is rotated about the x-axis.

Reasoning:

The region touches the axis of rotation, so each perpendicular cross-section is a disk rather than a washer.

Formula:

V= π∫01[f(x)]2dx.

Solution:

V= π∫01(x2)2dx.
= π∫01x4dx.
= π[x5/5]01.
=π/5.

Answer: The volume is π/5 cubic units.

Worked Example 18

Question: The region bounded by

y=4,
y=x2

between x=-2 and x=2 is rotated about the x-axis. Find the resulting volume.

Reasoning:

Rotation about the x-axis produces washers. The outer radius is 4 and the inner radius is x2.

Solution:

V= π∫-22 [42-(x2)2]dx.
= π∫-22 (16-x4)dx.

The integrand is even, so

V= 2π∫02 (16-x4)dx.
= 2π[16x-x5/5]02.
= 2π(32-32/5).
= 2π(128/5) = 256π/5.

Answer: The volume is 256π/5 cubic units.

Chapter Summary

  • Definite integrals can be used to calculate exact geometric areas and volumes.
  • If f(x)≥0, the area under y=f(x) from a to b is ∫abf(x)dx.
  • When a curve lies below the x-axis, its definite integral is negative but its geometric area is positive.
  • If a curve crosses the x-axis, total area should be found by dividing the interval at the zeros and adding positive regional areas.
  • The area between two curves using vertical strips is obtained from ∫[upper-lower]dx.
  • Using horizontal strips, the area between two curves is ∫[right-left]dy.
  • Intersection points often determine the required limits of integration.
  • A solid of revolution is formed when a plane region is rotated about an axis.
  • The disk method gives V=π∫R2dx.
  • The washer method gives V=π∫(R2-r2)dx.
  • Rotation about the y-axis can sometimes be handled conveniently using integration with respect to y.
  • The cylindrical shell method about the y-axis uses V=2π∫(radius)(height)dx.
  • If the cross-sectional area of a solid is known as A(x), its volume is ∫A(x)dx.
  • Correct modelling of the boundaries, limits, radii and heights is essential before evaluating an area or volume integral.

CHAPTER SEVEN

Systems of Particles, Centre of Mass and Relative Motion

7.1 Introduction to Systems of Particles

Many objects encountered in mechanics are not isolated particles. A car, a rigid body, a collection of interacting masses and even a planet may be regarded as a system made up of many particles. Instead of describing the motion of every individual particle separately, mechanics often simplifies the analysis by studying the system as a whole.

System of Particles

A system of particles is a collection of two or more particles regarded together for the purpose of mechanical analysis.

The particles within a system may exert forces on one another. They may also experience forces from objects outside the system. The distinction between these two kinds of force is fundamental.

7.2 Internal and External Forces

Internal Force

An internal force is a force exerted by one particle of a system on another particle within the same system.

External Force

An external force is a force exerted on the system by an object or agent that is not included in the chosen system.

Internal forces occur in equal and opposite pairs according to Newton's third law. Consequently, when the motion of the whole system is considered, internal forces cancel in pairs in the total force balance. The motion of the centre of mass is therefore controlled by the resultant external force.

Illustration

Consider two blocks connected by a spring. If both blocks and the spring are included in the system, the spring forces are internal. If only one block is selected as the system, the spring force on that block becomes an external force.

7.3 Total Mass and Total Momentum of a System

For a system of particles with masses m1, m2, ..., mn, the total mass is

M=Σmi.

The total linear momentum of the system is the vector sum of the momenta of all its particles:

P = Σpi = Σmivi.
Worked Example 1

Question: Two particles of masses 3 kg and 5 kg move east at 4 m/s and west at 2 m/s respectively. Determine the total momentum of the system.

Given: Take east as positive.

m1=3kg,   v1=4m/s
m2=5kg,   v2=-2m/s

Formula:

P=m1v1+m2v2

Solution:

P=3(4)+5(-2)
P=12-10=2kg m/s

Answer: The total momentum is 2 kg m/s east.

7.4 Centre of Mass of a System

The centre of mass provides a convenient single point with which to describe the overall translational motion of a system.

Centre of Mass

The centre of mass of a system is the point whose position is the mass-weighted average of the positions of all particles in the system.

For discrete particles,

rcm = Σmiri/Σmi.

The centre of mass need not coincide with the location of any actual particle. It may even lie outside the physical material of an object, as can occur for a ring.

7.5 Centre of Mass in One Dimension

For particles positioned along the x-axis,

xcm = Σmixi/Σmi.
Worked Example 2

Question: Three particles lie on the x-axis. A 2 kg particle is at x=0, a 3 kg particle is at x=4 m, and a 5 kg particle is at x=8 m. Find the centre of mass.

Formula:

xcm = (m1x1+m2x2+m3x3) /(m1+m2+m3)

Solution:

xcm = [2(0)+3(4)+5(8)]/(2+3+5)
xcm = (0+12+40)/10 = 5.2m

Answer: The centre of mass is at x=5.2 m.

Worked Example 3

Question: Two masses of 4 kg and 6 kg are separated by 5 m. Taking the 4 kg mass to be at x=0, find the centre of mass.

Solution:

xcm = [4(0)+6(5)]/(4+6)
xcm=30/10=3m

Answer: The centre of mass lies 3 m from the 4 kg mass, or 2 m from the 6 kg mass.

7.6 Centre of Mass in Two and Three Dimensions

For a system distributed in a plane, the coordinates of the centre of mass are

xcm=Σmixi/M

and

ycm=Σmiyi/M.

For three-dimensional systems,

zcm=Σmizi/M.
Worked Example 4

Question: A 2 kg particle is at (1,2) m and a 4 kg particle is at (7,5) m. Find the centre-of-mass coordinates.

Solution:

xcm = [2(1)+4(7)]/(2+4) = 30/6 = 5m
ycm = [2(2)+4(5)]/6 = 24/6 = 4m

Answer: The centre of mass is at (5 m, 4 m).

7.7 Centre of Mass of Continuous Bodies

For an extended body whose mass is continuously distributed, the summation used for particles becomes an integral:

rcm = (1/M)∫rdm.

In one dimension,

xcm = (1/M)∫x dm.

If the body is uniform and possesses symmetry, the centre of mass often lies at the geometric centre. For example, the centre of mass of a uniform straight rod lies at its midpoint, while that of a uniform circular disc lies at its geometric centre.

Worked Example 5

Question: A uniform rod has length L and mass M. Show that its centre of mass is at its midpoint.

Solution:

For uniform linear density,

λ=M/L

and

dm=λdx.

Therefore,

xcm = (1/M)∫0Lxλdx
xcm = λ/M [x2/2]0L
xcm = (M/L)/M × L2/2 = L/2

Answer: The centre of mass of a uniform rod is at L/2, its midpoint.

7.8 Motion of the Centre of Mass

The velocity of the centre of mass is obtained by differentiating its position:

vcm = drcm/dt.

For a system of particles,

vcm = Σmivi/M.

Therefore, the total momentum is

P=Mvcm.

Differentiating again gives

ΣFext = Macm.
Worked Example 6

Question: A system consists of a 2 kg particle moving east at 6 m/s and a 3 kg particle moving east at 2 m/s. Find the velocity of the centre of mass.

Formula:

vcm = (m1v1+m2v2)/(m1+m2)

Solution:

vcm = [2(6)+3(2)]/5
vcm = 18/5 = 3.6m/s

Answer: The centre of mass moves at 3.6 m/s east.

7.9 Centre of Mass and Momentum Conservation

If the net external force on a system is zero,

ΣFext=0.

Hence,

acm=0.

Therefore the centre of mass either remains at rest or moves with constant velocity. Internal interactions may be violent, as in an explosion, but they cannot alter the motion of the centre of mass of an isolated system.

Worked Example 7

Question: A stationary object explodes into several fragments. Describe the motion of the centre of mass if external forces are negligible during the explosion.

Solution:

Before the explosion, the total momentum is zero. Since external forces are negligible, total momentum remains zero.

P=Mvcm=0.

Therefore,

vcm=0.

Answer: The centre of mass remains at rest even though the fragments move apart.

7.10 Introduction to Relative Motion

Motion is always described with respect to a reference frame. An object may be stationary in one reference frame and moving in another. Relative motion provides a systematic way of relating observations made from different moving frames.

Relative Motion

Relative motion is the motion of one object as observed from a reference frame attached to another object or moving system.

For example, a passenger sitting in a moving bus has zero velocity relative to the bus but a non-zero velocity relative to the road.

7.11 Relative Position and Relative Velocity

Let rA and rB be the position vectors of objects A and B relative to a common reference frame. The position of A relative to B is

rAB = rA - rB.

Differentiating with respect to time,

vAB = vA - vB.
Worked Example 8

Question: Car A moves east at 30 m/s while Car B moves east at 18 m/s. Find the velocity of A relative to B.

Formula:

vAB=vA-vB

Solution:

vAB=30-18=12m/s

Answer: A moves at 12 m/s east relative to B.

Worked Example 9

Question: Car A travels east at 25 m/s while Car B travels west at 20 m/s. Find the velocity of A relative to B.

Solution:

Take east as positive:

vA=25m/s,   vB=-20m/s
vAB=25-(-20)=45m/s

Answer: A moves at 45 m/s east relative to B.

7.12 Relative Acceleration

Differentiating the relative velocity equation gives the acceleration of A relative to B:

aAB = aA - aB.

If both objects have the same acceleration, their relative acceleration is zero, and their relative velocity remains constant.

Worked Example 10

Question: Vehicle A accelerates east at 4 m/s2, while Vehicle B accelerates east at 1.5 m/s2. Find the acceleration of A relative to B.

Solution:

aAB=4-1.5=2.5m/s2

Answer: The relative acceleration is 2.5 m/s2 east.

7.13 Relative Motion in One Dimension

When two objects move along the same straight line, relative motion can often be analysed by simple addition or subtraction of signed velocities.

Same Direction

For objects moving in the same direction, their relative speed is the difference between their speeds.

Opposite Directions

For objects moving towards one another, their closing speed is the sum of their speed magnitudes.

Worked Example 11

Question: Two trains are 900 m apart and move directly toward each other at 20 m/s and 25 m/s. How long before they meet?

Relative Speed:

vrel=20+25=45m/s

Time:

t=distance/relative speed
t=900/45=20s

Answer: The trains meet after 20 s.

Worked Example 12

Question: A car travelling at 30 m/s is 200 m behind another car travelling in the same direction at 20 m/s. Assuming constant velocities, how long will the faster car take to catch the slower one?

Relative Speed:

vrel=30-20=10m/s

Time:

t=200/10=20s

Answer: The faster car catches the slower car after 20 s.

7.14 Relative Motion in Two Dimensions

When motion occurs in a plane, relative velocity must be treated as a vector. Components are usually resolved along perpendicular axes and combined.

Worked Example 13

Question: Object A moves east at 8 m/s, while object B moves north at 6 m/s. Find the velocity of A relative to B.

Vector Velocities:

vA = 8i
vB = 6j

Relative Velocity:

vAB = 8i-6j

Magnitude:

vAB = √(82+62) = 10m/s

Direction:

tanθ=6/8
θ≈36.9°

Answer: A moves relative to B at 10 m/s, directed approximately 36.9° south of east.

7.15 Boats, Rivers, Wind and Moving Media

Relative velocity is particularly important when an object moves through a medium that is itself moving. Examples include a boat crossing a flowing river and an aircraft flying through moving air.

For a boat,

vboat,ground = vboat,water + vwater,ground.

For an aircraft,

vplane,ground = vplane,air + vair,ground.
Worked Example 14

Question: A boat moves north through still water at 4 m/s. A river flows east at 3 m/s. Find the boat's velocity relative to the ground.

Components:

vx=3m/s,   vy=4m/s

Magnitude:

v=√(32+42)=5m/s

Direction:

tanθ=3/4
θ≈36.9°

Answer: The boat moves at 5 m/s, approximately 36.9° east of north relative to the ground.

Worked Example 15

Question: A river is 120 m wide and flows east at 3 m/s. A boat is directed due north across the river with speed 4 m/s relative to the water. Find the crossing time and the downstream drift.

Crossing Time:

The northward component determines the time to cross:

t=120/4=30s

Downstream Drift:

x=vrivert
x=3(30)=90m

Answer: The boat takes 30 s to cross and drifts 90 m downstream.

Worked Example 16

Question: An aircraft has an airspeed of 200 km/h due north. A wind blows due east at 50 km/h. Find the aircraft's ground speed and direction.

Magnitude:

vground = √(2002+502)
vground = √42500 ≈206.2km/h

Direction:

tanθ=50/200=0.25
θ≈14.0°

Answer: The ground speed is approximately 206.2 km/h, directed 14.0° east of north.

Connecting the Ideas

The centre-of-mass method simplifies the motion of multi-particle systems, while relative motion explains how observations depend on the chosen reference frame. Both concepts are essential for understanding more advanced translational and rotational mechanics.

Chapter Summary

  • A system of particles is a collection of particles analysed together.
  • Internal forces act between particles inside the system, while external forces originate outside the chosen system.
  • Total system momentum is the vector sum P=Σmivi.
  • The centre of mass is the mass-weighted average position of a system.
  • For discrete particles, rcm=Σmiri/M.
  • For continuous bodies, centre-of-mass sums are replaced by integrals.
  • The total momentum of a system is P=Mvcm.
  • The centre-of-mass acceleration is determined by the resultant external force through ΣFext=Macm.
  • If the net external force is zero, the centre of mass moves with constant velocity.
  • Relative position is the difference between position vectors measured in a common frame.
  • Relative velocity is vAB=vA-vB.
  • Relative acceleration is aAB=aA-aB.
  • Relative motion in two dimensions requires vector addition or subtraction.
  • Boat-river and aircraft-wind problems are direct applications of relative velocity.

CHAPTER EIGHT

Rotational Motion, Torque and Angular Momentum

8.1 Introduction to Rotational Motion

Translational motion describes the movement of an object from one position to another. Rotational motion describes motion in which points of a body move around an axis. A rotating wheel, a ceiling fan, the hands of a clock and a spinning disc are familiar examples.

Many ideas from linear mechanics have rotational counterparts. Position corresponds to angular position, velocity to angular velocity, acceleration to angular acceleration, force to torque, mass to rotational inertia and linear momentum to angular momentum.

Translational Quantity Rotational Analogue
Position x Angular position θ
Velocity v Angular velocity ω
Acceleration a Angular acceleration α
Force F Torque τ
Mass m Moment of inertia I
Linear momentum p Angular momentum L

8.2 Angular Position and Angular Displacement

The orientation of a rotating body can be described by an angular position θ, measured relative to a chosen reference line.

Angular Displacement

Angular displacement is the change in angular position of a rotating body.

Δθ=θf-θi.

The SI unit of angular displacement is the radian.

2π rad=360°.

Hence,

1rad=180°/π≈57.3°.
Worked Example 1

Question: Convert 120° to radians.

Formula:

θ(rad)=θ(deg)×π/180

Solution:

θ=120×π/180
θ=2π/3 rad

Answer: 120°=2π/3 rad.

8.3 Angular Velocity

Angular Velocity

Angular velocity is the rate of change of angular position with time.

Average angular velocity is

ωavg=Δθ/Δt.

Instantaneous angular velocity is

ω=dθ/dt.

Its SI unit is radian per second, rad/s.

Worked Example 2

Question: A wheel turns through 20 rad in 4 s at uniform rate. Find its angular velocity.

Formula:

ω=Δθ/Δt

Solution:

ω=20/4=5rad/s

Answer: The angular velocity is 5 rad/s.

8.4 Angular Acceleration

Angular Acceleration

Angular acceleration is the rate of change of angular velocity with time.

Average angular acceleration is

αavg=Δω/Δt.

Instantaneous angular acceleration is

α=dω/dt=d2θ/dt2.

Its SI unit is rad/s2.

Worked Example 3

Question: A disc increases its angular velocity uniformly from 4 rad/s to 16 rad/s in 3 s. Find its angular acceleration.

Formula:

α=(ω-ω0)/t

Solution:

α=(16-4)/3=4rad/s2

Answer: The angular acceleration is 4 rad/s2.

8.5 Rotational Kinematics

For constant angular acceleration, rotational motion obeys equations directly analogous to those of uniformly accelerated linear motion.

Linear Motion Rotational Motion
v=u+at ω=ω0+αt
s=ut+(1/2)at2 θ=ω0t+(1/2)αt2
v2=u2+2as ω2=ω02+2αθ
s=(u+v)t/2 θ=(ω0+ω)t/2
Worked Example 4

Question: A wheel starts from rest and accelerates uniformly at 3 rad/s2 for 5 s. Find its final angular velocity and angular displacement.

Given:

ω0=0,   α=3rad/s2,   t=5s

Final Angular Velocity:

ω=ω0+αt
ω=0+3(5)=15rad/s

Angular Displacement:

θ=ω0t+(1/2)αt2
θ=0+(1/2)(3)(25)=37.5rad

Answer: Final angular velocity = 15 rad/s; angular displacement = 37.5 rad.

8.6 Relationship Between Linear and Angular Quantities

For a point at distance r from a fixed axis of rotation, the arc length travelled is

s=rθ,

provided θ is measured in radians.

Differentiating with respect to time gives

v=rω.

For tangential acceleration,

at=rα.

A point moving in a circle also has radial or centripetal acceleration

ar=v2/r=ω2r.
Worked Example 5

Question: A point on the rim of a wheel of radius 0.40 m rotates with angular velocity 12 rad/s. Find its linear speed.

Formula:

v=rω

Solution:

v=0.40(12)=4.8m/s

Answer: The linear speed is 4.8 m/s.

Worked Example 6

Question: A rotating disc has radius 0.25 m and angular acceleration 8 rad/s2. Find the tangential acceleration of a point at the rim.

Formula:

at=rα

Solution:

at=0.25(8)=2m/s2

Answer: The tangential acceleration is 2 m/s2.

8.7 Torque and Moment of a Force

A force can change the translational motion of a body, but its ability to produce rotation depends on where and how the force is applied.

Torque

Torque, also called the moment of a force, measures the turning effect of a force about a point or axis.

The magnitude of torque is

τ=rFsinθ,

where r is the distance from the axis to the point of application, F is the force magnitude and θ is the angle between r and F.

Equivalently,

τ=F r⊥,

where r⊥ is the perpendicular moment arm.

Worked Example 7

Question: A force of 40 N acts perpendicular to a spanner at a distance of 0.30 m from the pivot. Find the torque.

Formula:

τ=rFsinθ

Solution:

τ=0.30(40)sin90°
τ=12N m

Answer: The torque magnitude is 12 N m.

Worked Example 8

Question: A 25 N force is applied to a 0.50 m lever at an angle of 30° to the lever. Find the torque magnitude.

Solution:

τ=rFsinθ
τ=0.50(25)sin30°
τ=12.5(0.5)=6.25N m

Answer: The torque magnitude is 6.25 N m.

8.8 Torque as a Vector Product

Torque is a vector quantity and is defined by the cross product

τ = r×F.

Its direction is perpendicular to the plane containing r and F, determined by the right-hand rule.

In two-dimensional mechanics, counterclockwise torque is often assigned a positive sign and clockwise torque a negative sign, although the sign convention may be reversed if used consistently.

Worked Example 9

Question: A force

F=4j N

acts at position

r=3i m.

Find the torque about the origin.

Formula:

τ = r×F

Solution:

τ = (3i)×(4j)
τ = 12(i×j) = 12kN m

Answer: τ=12k N m.

8.9 Rotational Form of Newton's Second Law

For a rigid body rotating about a fixed axis, the rotational analogue of Newton's second law is

Στ=Iα,

where I is the moment of inertia about the axis and α is the angular acceleration.

The moment of inertia measures resistance to angular acceleration. Its detailed calculation and standard forms will be developed further in the next chapter.

Worked Example 10

Question: A wheel has moment of inertia 2.5 kg m2. A net torque of 10 N m acts on it. Find its angular acceleration.

Formula:

τ=Iα

Solution:

α=τ/I
α=10/2.5=4rad/s2

Answer: The angular acceleration is 4 rad/s2.

8.10 Angular Momentum of a Particle

Angular Momentum

The angular momentum of a particle about a point is the vector product of its position vector and linear momentum.

L = r×p.

Since p=mv,

L = r×mv.

The magnitude is

L=rmv sinθ.

If the particle moves perpendicular to the radius vector,

L=rmv.

The SI unit of angular momentum is

kg m2/s.
Worked Example 11

Question: A 2 kg particle moves at 5 m/s perpendicular to a radius vector of magnitude 3 m. Find its angular momentum about the origin.

Formula:

L=rmv

Solution:

L=3(2)(5)=30kg m2/s

Answer: The angular momentum magnitude is 30 kg m2/s.

8.11 Angular Momentum of a Rigid Body

For a rigid body rotating about a fixed principal axis, the angular momentum may be written as

L=Iω.

This is analogous to the translational relation

p=mv.

The relation L=Iω is especially useful when the angular momentum is parallel to the rotation axis.

Worked Example 12

Question: A rotating body has moment of inertia 4 kg m2 and angular velocity 6 rad/s. Find its angular momentum.

Formula:

L=Iω

Solution:

L=4(6)=24kg m2/s

Answer: The angular momentum is 24 kg m2/s.

8.12 Conservation of Angular Momentum

Torque is related to the time rate of change of angular momentum:

Στext = dL/dt.

If the resultant external torque is zero,

dL/dt=0,

and therefore

L=constant.
Conservation of Angular Momentum

When the resultant external torque on a system is zero, the total angular momentum of the system remains constant.

For rotation about a fixed axis,

Iiωi = Ifωf.
Worked Example 13

Question: A rotating skater has moment of inertia 5 kg m2 and angular speed 2 rad/s. By drawing in the arms, the moment of inertia decreases to 2 kg m2. Neglect external torque. Find the new angular speed.

Formula:

Iiωi = Ifωf

Solution:

5(2)=2ωf
10=2ωf
ωf=5rad/s

Answer: The new angular speed is 5 rad/s.

Physical Interpretation

When external torque is negligible, reducing moment of inertia increases angular velocity so that angular momentum remains constant. This explains why a spinning skater rotates faster after drawing the arms inward.

8.13 Rotational Work, Energy and Power

A torque acting through angular displacement can perform rotational work. For a constant torque,

W=τθ,

provided torque and angular displacement correspond to the same rotational direction.

The rotational kinetic energy of a rigid body rotating about a fixed axis is

Krot = (1/2)Iω2.

The rotational form of the work-energy theorem is

Wnet = ΔKrot.

Rotational power is

P=τω.
Worked Example 14

Question: A wheel with moment of inertia 3 kg m2 rotates at 8 rad/s. Find its rotational kinetic energy.

Formula:

K=(1/2)Iω2

Solution:

K=(1/2)(3)(82)
K=1.5(64)=96J

Answer: The rotational kinetic energy is 96 J.

Worked Example 15

Question: A motor applies a torque of 15 N m to a shaft rotating at 20 rad/s. Find the mechanical power transmitted.

Formula:

P=τω

Solution:

P=15(20)=300W

Answer: The power transmitted is 300 W.

8.14 Rotation of Coordinate Axes

Sometimes it is useful to describe the same point using a new coordinate system whose axes have been rotated relative to the original axes. This does not change the physical point; it changes only the coordinates used to describe it.

Suppose the new axes x' and y' are obtained by rotating the original axes counterclockwise through an angle φ. The coordinates are related by

x'=x cosφ+y sinφ

and

y'=-x sinφ+y cosφ.

Conversely,

x=x' cosφ-y' sinφ

and

y=x' sinφ+y' cosφ.

The distance from the origin remains unchanged:

x2+y2 = x'2+y'2.
Worked Example 16

Question: A point has coordinates (x,y)=(3,4). The coordinate axes are rotated counterclockwise by 90°. Find (x',y').

Formula:

x'=x cosφ+y sinφ
y'=-x sinφ+y cosφ

Solution: For φ=90°,

cos90°=0,   sin90°=1
x'=3(0)+4(1)=4
y'=-3(1)+4(0)=-3

Answer: The coordinates in the rotated system are (4,-3).

8.15 Polar Coordinates

Cartesian coordinates locate a point using perpendicular distances x and y. For rotational and circular motion, polar coordinates are often more natural.

Polar Coordinates

Polar coordinates specify the position of a point in a plane by its radial distance r from the origin and its angular coordinate θ measured from a chosen reference direction.

The relationship between Cartesian and polar coordinates is

x=r cosθ

and

y=r sinθ.

Conversely,

r=√(x2+y2)

and, with the quadrant taken into account,

tanθ=y/x.
Worked Example 17

Question: Convert the Cartesian point (3,4) to polar coordinates.

Radial Coordinate:

r=√(32+42)
r=5

Angular Coordinate:

tanθ=4/3
θ=tan-1(4/3)≈53.1°

Answer: The polar coordinates are approximately (5, 53.1°).

Unit Vectors in Polar Coordinates

In planar polar coordinates, the unit vectors are er in the radial direction and eθ in the direction of increasing angle.

Unlike the fixed Cartesian unit vectors, these unit vectors change direction as the particle moves around the origin. Their time derivatives are

der/dt = θ̇eθ

and

deθ/dt = -θ̇er.

Velocity in Polar Coordinates

For position

r=rer,

the velocity is

v = ṙer + rθ̇eθ.

Thus, motion may contain both a radial component and a transverse component.

Acceleration in Polar Coordinates

The acceleration is

a = ( r̈-rθ̇2 )er + ( rθ̈+2ṙθ̇ )eθ.

For uniform circular motion, r is constant, ṙ=0, r̈=0 and θ̇=ω is constant. Therefore,

a = -rω2er,

which is the familiar centripetal acceleration directed toward the centre.

8.16 Applications of Rotational Dynamics

Rotational mechanics combines angular kinematics, torque, rotational inertia, angular momentum and energy methods. These concepts allow the analysis of wheels, pulleys, rotating shafts, gears, flywheels and many other mechanical systems.

Worked Example 18

Question: A rotating wheel has moment of inertia 5 kg m2. A constant net torque of 20 N m acts for 4 s. If the wheel starts from rest, find its angular acceleration, final angular velocity and angular displacement.

Angular Acceleration:

τ=Iα
α=20/5=4rad/s2

Final Angular Velocity:

ω=ω0+αt
ω=0+4(4)=16rad/s

Angular Displacement:

θ=ω0t+(1/2)αt2
θ=(1/2)(4)(16)=32rad

Answer: Angular acceleration = 4 rad/s2; final angular velocity = 16 rad/s; angular displacement = 32 rad.

Worked Example 19

Question: A wheel of radius 0.50 m rotates at 10 rad/s. Find the linear speed and centripetal acceleration of a point on its rim.

Linear Speed:

v=rω
v=0.50(10)=5m/s

Centripetal Acceleration:

ar=ω2r
ar=102(0.50)=50m/s2

Answer: Linear speed = 5 m/s; centripetal acceleration = 50 m/s2 toward the centre.

Worked Example 20

Question: A rigid body has rotational kinetic energy 200 J and moment of inertia 4 kg m2. Find its angular speed.

Formula:

K=(1/2)Iω2

Solution:

200=(1/2)(4)ω2
200=2ω2
ω2=100
ω=10rad/s

Answer: The angular speed is 10 rad/s.

Connecting the Ideas

Rotational motion has a mathematical structure closely parallel to translational motion. Torque produces angular acceleration, rotational inertia measures resistance to angular acceleration, and angular momentum is conserved when external torque is zero. These ideas lead naturally to the detailed study of circular motion and moments of inertia.

Chapter Summary

  • Rotational motion describes motion about an axis.
  • Angular displacement is measured in radians, with 2π rad=360°.
  • Angular velocity is ω=dθ/dt, while angular acceleration is α=dω/dt.
  • For constant angular acceleration, rotational kinematics has equations analogous to linear kinematics.
  • Linear and angular quantities are related by s=rθ, v=rω and at=rα.
  • Radial acceleration in circular motion is ar=ω2r=v2/r.
  • Torque measures the turning effect of a force and has magnitude τ=rFsinθ.
  • Vector torque is τ=r×F.
  • For rotation about a fixed axis, Στ=Iα.
  • The angular momentum of a particle is L=r×p.
  • For a rigid body rotating about a fixed principal axis, L=Iω.
  • Angular momentum is conserved when the resultant external torque is zero.
  • Rotational kinetic energy is K=(1/2)Iω2.
  • Rotational power is P=τω.
  • Rotating coordinate axes provide an alternative description of the same physical point.
  • Polar coordinates use radial distance and angle to describe positions in a plane.
  • Polar-coordinate velocity contains radial and transverse components, while polar-coordinate acceleration contains corresponding radial and transverse terms.