FlipWorkbook DUM30172

DUM30172
MATHEMATICS
FOR
MECHANICAL
ENGINEERING

Workbook & Tutorial

UNIT 1
COORDINATE GEOMETRY

UNIT 2
APPLICATIONS OF DIFFERENTIATION

UNIT 3
STATISTICS AND PROBABILITY

UNIT 4
ORDINARY DIFFERENTIAL EQUATION

UNIT 1 DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

INTRODUCTION

We are all familiar with the concepts of coordinate geometry if we have
used a road map to locate a place. In coordinate geometry, locations are made in
terms of two
numbers or letters.

In this unit, we will learn how to obtain properties of straight lines such as
the distance, midpoint and gradient and also to obtain the equation of straight
lines. This will be followed properties of parallel and perpendicular lines and lastly
to obtain the point of intersection of two lines.

CONTENT

1.1 DISTANCE AND MIDPOINT
1.2 GRADIENT AND EQUATION OF STRAIGHT LINE
1.3 PARALLEL AND PERPENDICULAR
1.4 INTERSECTION LINES

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

UNIT LEARNIN After completing the unit, students should be able to:
G 1.1.1 Determine the distance and midpoint between two points using
formula
OUTCOMES

1.1 DEFINITION

1.1 DISTANCE AND MIDPOINT
CARTESIAN COORDINATES
y

• P(x, y)

x
O

The Cartesian coordinates system is a system that enables a position or
location of a point to be determined on a plane formed from two number lines
which intersect at a right angle.

The horizontal line is called the x-axis and the vertical line is called the y-axis.
The point where they intersect is called the origin and is marked as O. Distances
to the right and above the origin are taken as positive while distances to the left
and below the origin are taken as negative.

The notation (x, y) denotes a point whose perpendicular distances from the y-
axis and x-axis are x and y respectively. The Cartesian coordinates system is
also known as rectangular coordinates system.

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

Example 1.1 : Find the coordinates of P, Q, R, S, T and U below.

y

5•- T

-

4-

- •P

3- ||| x
| •| S | | | | | |
345
-5 -4 -3 -2 -1 0 1 2
•Q
• R -1 --

-2 --

-3 --

-4 --

-5 -- • U

Figure 1.1

Example 1.2 : Plot the following points on the graph below.

1. A(5,0 ) 2. B( 4,−2 ) 3. C( 2,−3 ) 4. D( −1,5 ) 5. E( −4,−5 ) 6. F( −5,2 )

y

5- x
-

4-
-

3-

|||||||||||

-5 -4 -3 -2 -1 0 1 2 3 4 5

-1 --
-2 --
-3 --
-4 -- Figure 1.2
-5 --

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

DISTANCE BETWEEN TWO POINTS • Q(x2, y2)
y y2, – y1

d

P(x1, y1) • x2, – x1

x

Figure1.3

As shown in Figure 1.2, the distance between two points (x1, y1) and (x2, y2) is the
length of the straight line connecting them. Using the Pythagoras Theorem, the
distance is given by the formula

DISTANCE

d=

Example 1.3 : To find the distance between two points

Find the distance between points P(-1, -3) and Q(2, 3).

Solution :

PQ = (2 − (−1))2 + (3 − (−3))2

= 9 + 36
= 45
= 6.71

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

Example 1.4 : To find one of the point if the distance is given
Given the distance between two points A(α, 11) and B(4, 3) is 17. Find the
values of α.
Solution :
AB2 = (α – 4)2 + (11 – 3)2
172 = α2 – 8α + 16 + 64
α2 – 8α – 209 = 0
(α – 19)(α + 11) = 0
α = 19 or α = -11

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

MIDPOINT y

• Q(x2, y2)

• M(xm, ym)

P(x1, y1) •

x

Figure 1.4

As shown in Figure 1.4, the midpoint of the line joining two points P(x1, y1) and Q(x2, y2)
is given by the formula

MIDPOINT

M(xm, ym) =

Example 1.4 : To find the midpoint between two points

Find the midpoint of the line joining P(1, -6) and Q(-8, -4).

Solution :

Midpoint PQ =  1+ (−8) , − 6 + (−4) 
2 2

=  − 7 ,− 5
2 

Example 1.5 : To find one of the point if the midpoint is given

Given P(9, p), Q(q, -1) and R(8, -3). If R is the midpoint of PQ, find the value of p
and q.

Solution : (8,−3) =  9 + q , p + (−1) 
→ 2 2 
9+q =
8 p − 1 = -3
2 = 2
9+q =
16 p – 1 = -6
q 7 p = -5

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

UNIT EXERCISE 1.1

1. Find the distance between the following pair of points.
(a) M(0, -2) and N(5, 10)
(b) P(-2, 6) and R(4, -1)
(c) V(-5, -2) and W(-6, -1)
(d) T(-4, -1) and U(-2, 5)

2. Given the distance between A(3, p) with B(-2, 3) are 13 unit, find the
value of p.

3. Given that the distance of point V(8, t) from the origin is 17 units.
Find the value t.

4. Find the midpoint for the following pair of points.
(a) T(-8, 3) and U(5, -4)
(b) X(8, 4) and Y(-4, 9)
(c) R(-4, 9) and S(3, -6)
(d) K(11, -1) and L(3, -9)

5. If (m, 5) is the midpoint for a line which joint points (4, -2) and
(-5, n), find the value of m and n.

6. Given C(2, 5) is the midpoint 0f the point B(h, 3) and the point
D(-4, k), find the value of h and k.

7.

P(1, 2) RS
Q(7, 4)

In the diagram, PGRS is a straight line. Q is the midpoint of PS and R
is the midpoint of QS. Find the coordinates of R.

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

After completing the unit, students should be able to:

1.2.1 Determine the gradient of the straight lines joining two points using

UNIT LEARNING formula.
OUTCOMES
1.2.2 Form the equation of the straight line in the form
• y = mx + c

• y − y1 = m( x + x1 )
y − y1 y2 − y1
• x − x1 = x2 − x1

1.2 GRADIENT AND EQUATION OF STRAIGHT LINES

Gradient is a quantity which measures how steep a straight line is sloping with
reference to the horizontal. It is defined as the ratio of the vertical rise to the
horizontal run of the straight line.

y

• Q(x2, y2)

m y2, – y1

P(x1, y1) • x2, – x1

x

Figure 1.5

As shown in Figure 1.4, the gradient of the straight line that passes through
points P(x1, y1) and Q(x2, y2) is denoted by m and given by the formula

GRADIENT

m= =

8

Possibilities of Gradient DUM 30172 ENGINEERING MATHEMATICS 3
y UNIT 1: COORDINATES GEOMETRY
m>0
y
m=0

x x
Line rises from left to right
Line is horizontal
y y
m<0
m undefined
x
Line falls from left to right x

Line is vertical

Example 1.6 : To find the gradient if two points are given

Find the gradient of the line passing through each pair of points.
(a) (-3, -1) and (-2, 4)
(b) (-3, 4 ) and (2, -2)

Solution :

(a) m = 4 − (−1) = 5
− 2 − (−3)

(b) m = − 2 − 4 = −6
2 − (−3) 5

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

Example 1.7 : To find the point if the gradient is given

Given the gradient of a line through two points P(1, p) and Q(4p, 9) is -2, find
the value of p.

Solution :

Gradient PQ, m = -2
9−p =
4p − 1 -2

9–p = -8p + 2
7p = -7
p= -1

Generally, the equation of a straight line can be determined when the gradient
and the coordinates of a point or the coordinates of two points are known. We can use
the gradient of a line to obtain various forms of the line’s equation.

1.2.1 EQUATION OF A STRAIGHT LINE GIVEN THE GRADIENT AND A
POINT
Equation of a straight line passing through the point (x1, y1) and having
gradient m is given by the formula

EQUATION OF STRAIGHT LINE

y – y1 = m( x – x1)

Example 1.8 : To write an equation if a point and a gradient is given

Write the equation of the line passing through (1, 5) with gradient 2.

Solution :

Use the equation y – y1 = m( x – x1)
with m = 2
x1 =1 y1 = 5

y – 5 = 2( x – 1)
y – 5 = 2x – 2
y = 2x – 2 + 5
y = 2x + 3

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

1.2.2 EQUATION OF A STRAIGHT LINE PASSING THROUGH TWO POINTS
Equation of a straight line passing through the points (x1, y1) and (x2, y2) is
given by the formula

EQUATION OF STRAIGHT LINE

=

Example 1.9 : To write an equation if the two points are given

Write the equation of the line passing through the points (-1, 2) and
(3, 10).

Solution :

Use the equation y − y1 = y2 − y1 y2 = 10
with x1 = -1 x − x1 x2 − x1
y1 = 2
x2 = 3

y−2 = 10 − 2
x − (−1) 3 − (−1)

=2

(y – 2) = 2(x + 1)
y – 2 = 2x + 2
y = 2x + 4

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

1.2.3 TYPES OF STRAIGHT LINE EQUATION
There are three types of equation,

TYPES OF EQUATIONS

1. Gradient Form : y = mx + c
2. General Form : ax + by + c = 0
x + y =1
3. Intercept Form : ab

Example 1.10 : To change the given to gradient form

Rewrite the following equations in gradient form. Find the value of
gradient, m and y-intercept, c
(a) 3x + y – 5 = 0
(b) 4y + 28 = 0

Solution : 3x + y – 5 = 0
(a) Rearrange :

(b) y = -3x + 5

Compare with y = mx + c :
m = -3 and c = 5

4y + 28 = 0
4y = -28
y = -7

Compare with y = mx + c :
m = 0 and c = -7

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

Example 1.11 : To change the equation given to general form

Rewrite the given equation in general form. Find the value of a, b and c
(a) y = -2x + 1
(b) x + y = 1

23

Solution :

(a) y = -2x + 1
2x + y – 1 = 0
Compare with ax + by + c = 0 :
a = 2, b = 1, c = -1

(b) x + y = 1
23
Multiply with 6 :
3x + 2y = 6
3x + 2y – 6 = 0

Compare with ax + by + c = 0 :
a = 3, b = 2, c = -6

Example 1.12 : To write an equation if the two points are given and change
the equation to another form

Write the equation of the line passing through the points (2, 4) and
(-1, 10).

Solution :

m = 10 − 4 = −2 Gradient form
−1− 2

y – y1 = m( x – x1)
y – 4 = -2( x – 2)

= -2x + 4
y = -2x + 8 :

2x + y – 8 = 0 : General form

2x + y = 8
8 88

x + y =1 : Intercept form
4 8

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

UNIT EXERCISE 1.2

1. Find the gradient of a straight line joining each of the following pairs
of points.

(a) (1, 3) and (4, 9)
(b) (-1, 2) and (1, -8)

2. Find the value of h if the straight line joining the points (2h, -3) and
(-2, -h +2) has a gradient of 2.

3. Given P(3a-1, -a), Q(-5, 3) and R(1, 6) are three points on a straight
line. Find the value of a.

4. Given that the gradient of a line that passes through the points
(-5, k) and (k – 3, 3) is 1 , find the value of k.
4

5. Express the following equations in general form.

(a) y = 2 x + 7

(b) y = − 1 x + 3
3
x y
(c) 3 + 5 =1

(d) 2y − 3 = 1 x
3

6. Find the equation of a straight line that has a gradient of -2 and
passes through (2, -1)

7. Find the equation of a straight line that has a gradient of 1 and
2

passes through (-1, 6)

8. Find the equation of the straight line which passes through each of
the following pairs of point.

(a) (1, 6) and (5, 14) (c) (5, 7) and (2, 8)
(b) (-4, 2) and (4, -6) (d) (-8, 0) and (0, -5)

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

8. Given A(6, -4), B(4, -2) and C(-1, -1). Find :
(a) the distance between AB
(b) midpoint BC
(c) gradient AC
(d) the equation that passes through the points A and B
(in general form).

9. Write the equation for the following lines.
(a) gradient, m = 2 and passing through (3, 5).
(b) gradient, m = - 3 and passing through (10, -4).
5
(c) passing through (-3, -1) and (2, 4).
(d) passing through (2, 3) and (4, 8).
(e) passing through (1, -3) with x-intercept = -1.

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

1.3 PARALLEL AND PERPENDICULAR LINES

1.3.1 PARALLEL LINES

Two lines that do not intersect or meet and lie in the same plane are said
to be parallel.

m1
m2

L1
L2
Figure 1.5

PARALLEL LINE

In Figure 1.5, if L1 is parallel to L2 then they
have the same gradient, that is m1 = m2

•
Example 1.13 : To determine whether two lines are parallel or not

Determine whether the two lines are parallel or not.
(a) y = 2x + 3 and y = 2x - 8

(b) 3y = x + 3 and y – 2 x = 5
3

Solution :

(a) y = 2x + 3 → m1 = 2

y = 2x – 8 → m2 = 2
Since m1 = m2 = 2, so they are parallel.

(b) 3y = x + 3 → m1 = 1
y– 2x=5 → 3
3
m2 = 2
3

Since m1 ≠ m2 , so they are not parallel.

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

Example 1.14 : To write an equation that passes through a point and
parallel to another line

Write the equation of the line passing through (-3, 2) and parallel to the
line whose equation is y = 2x + 1.

Solution :

y = 2x + 1 → m1 = 2
Hence m2 = 2

Therefore, equation of the line passing through (-3, 2) and parallel to the
line whose equation is y = 2x + 1 is

y – 2 = 2(x + 3)
= 2x + 6

y = 2x + 8

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

Exercise 1.3.1

1. Determine whether the two lines are parallel or not.

a. 4y −1 = 2x and y − 1 x = 45 b. 2 x + y = 8 and y = 2 x + 5
2

c. 2 x − y = 6 and 6 x − 3y + 3 = 0 d. y = 3 x and 2 x + y − 3 = 0

2. The straight lines 3 x + ky = 2 and 2y + x −7 = 0 are parallel. Find the value of k.

2. Write down the equation of each of the following lines

a. Parallel to y - 6x = 3 and passing b. parallel to x + 2y = 4 and passing
through (10, 5) through the point (0, 5)

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

c. Parallel to y = −2x + 10 and d. Parallel to x − 3y − 6 = 0 and

passing through the origin. passing through (−9, −2).

3. Find the equation of a straight line that passes through point (-2, -5) and is
parallel to the line containing the points (−4, 5) and (2, −1).

4. . Find the equation of a straight line that passes through point (-4, 1) and is
parallel to the straight line joining points (1, 8) and (-3, -2).

5. . Find the equation of the straight line which passes through the point
P(-3, 6) and is parallel to the straight line 4 x − 2y +1 = 0 .

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

1.3.2 PERPENDICULAR LINES

Two lines that intersect or meet at right angle are said to be
perpendicular.

L1
m2

L2 m1
Figure 1.6

PERPENDICULAR LINE

In Figure 1.6, if L1 and L2 are perpendicular then

m1 x m2 = -1 so that m1 = .

Example 1.15 : To determine whether two lines are perpendicular or not

Determine whether the following pair of lines are perpendicular or not.

(a) y = -5x + 3 and y = 1 x – 8
5

(b) 2y = 3x + 5 and y – 2 x = 8
3

Solution :

(a) y = -5x + 3 → m1 = -5

y= 1x–8 → m2 = 1
5 5

Since m1 x m2 = -5 x 1 = -1, so they are perpendicular.
5

(b) 2y = 3x + 5 → m1 = 3
y– 2x = 5 2
3
→ m2 = 2
3

Since m1 x m2 = 3x 2 =1 , so they are not perpendicular.
2 3

20

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

Example 1.16 : To find the gradient of any line that is perpendicular to the
given line

Find the gradient of any line that is perpendicular to the line whose
equation is x + 4y – 8 = 0

Solution :

Given line : x + 4y – 8 = 0
4y = -x + 8
- 1x + 2
y= → m1 = -1
4 4
-1
Perpendicular line : m1 x m2 =
-1
- 1 x m2 =
4 4

m2 =

Example 1.17 : To write an equation that passes through a point and
perpendicular to given line

Write the equation of the line passing through (2, -3) and perpendicular to
the line whose equation is 5y = 3x + 11.

Solution :

Given line : 5y = 3x + 11

y= 3 x + 11 → m1 = 3
55 5

Perpendicular line : m1 x m2 = -1

3 x m2 = -1 , m2 = -5
5 3

Hence, equation of the line passing through (2, -3) and perpendicular to

the line 5y = 3x + 11 is

y + 3 = - 5 (x − 2)
3

y + 3 = − 5 x + 10
33

y = -5x+ 1
33

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

Exercise 1.3.2

1. Determine whether the two lines are perpendicular or not.

a. 3y − x − 2 = 0 and y + 3 x + 4 = 0 b. 2 x − y + 4 = 0 and y − 3 x + 5

c. 2y = x +1 and y + 2 x = 4 d. 4 x − y − 5 = 0 and x + 4y +1 = 0

2. Find the value of h if straight line y − hx + 2 = 0 is perpendicular to straight line
5y + x +3 = 0 .

3. Write down the equation of each of the following lines

a. Perpendicular to y = 2x − 1 and b. Perpendicular to x + 4y − 6 = 0 and
passing through (4, 0). passing through (1, 8).

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

c. perpendicular to y = 1 x − 3 which d. perpendicular to 5 y + 2 x − 5 = 0
2 and passing through (6, 7).

passes through the point (–5, 3).

4. Find the equation of a straight line that passes through point (-2, -5) and is
parallel to the line containing the points (−4, 5) and (2, −1).

5. Find the equation of a straight line that passes through point (1, 3) and is
perpendicular to the straight line joining points P(-2, 5) and Q(2, -1).

6. Find the equation of the straight line which passes through the point
P(2, -5) and is perpendicular to the straight line y = −3 x +1 .

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

UNIT EXERCISE 1.3

1. Find the equation of the line that passes through point A and
parallel to the equation given.

(a) point A(-1, 2) ; equation : x + y – 2 = 0

(b) point A(5, -1) ; equation : 2x - 3y – 2 = 0

(c) point A(2, -3) ; equation : 3y + x = 4

2. Find the equation of the line that passes through B and
perpendicular to the equation given.

(a) point B(-2, 3) ; equation : y = 5 – 4x

(b) point B(1, 8) ; equation : 2x + 3y = 4

(c) point B(-8, 2) ; equation : x + y = 1
34

3. Find the equation of a straight line that passes through (-2, -3) and

parallel to the line which connect points (2, 1) and (-2, 4).

4. Given the points A(-2, 7), B(6, 1) and C(9, 4), find the equation of a
line that passes through B and perpendicular to AC.

5. Find the value of p if each of the following equation are parallel

(a) 2x + py + 5 = 0 and 3x – y + 1 = 0

(b) 3y = 5 – 2x and x − y =1
3 p

(c) 2y = px − 1 and 3x + 2y = -6
2

6. Find the value of k if each of the following equation are
perpendicular to each other.
(a) kx – y = 2 and x – 2y + 5 = 0

(b) x − y = 1 and 2x + ky = 4
23

(c) 3x + 2y + 18 = 0 and x + y = 2
k4

24

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

1.4 INTERSECTION OF LINES

When two lines intersect, the point of intersection lies on both lines. This means
that the coordinates of the point satisfy both the equations of the lines. Therefore,
we need to solve the equation simultaneously in order to determine the point of
intersection.

Example 1.18 : To find the point of intersection of two lines

Find the intersection point in Figure 1.7.

y

x + 2y = 2 x

●P
x – 2y = 6

Figure 1.7

Solution :

L1 : x + 2y = 2 . . . (1)
L2 : x – 2y = 6 . . . (2)

(1) – (2) : 4y = -4
y = -1

Substituting y = -1 into (1) :
x + 2(-1) = 2
x–2 = 2
x=4

Therefore P (4, -1)

25

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

Exercise 1.4.1

a. Find the point of intersection of the straight lines y = −2 x +1 and y = 1 x + 6 .
2

b. The straight line AB intersecting with the

C straight line CD at point E. The equation of the
straight line CD is y = −x − 4 . Points B and D lie
E B(0, 6) on the y-axis. Find

O a. the y-intercept of the straight line CD.
b. the equation of line AB

A(-9, -1) D c. the coordinates of point E

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

c. y Straight lines AB and CD intersecting at point R.
B(-1, 4)
C(4, 4) Find
R
a. the equations of AB and CD.
A(5, 1)
x b. the coordinates of R

c. the equation of the straight line that passes
3
D(1, -2) through point R and has a gradient of − 4 .

d. Two straight lines y − x =1 and ky = −x +12 intercept the y-axis at the same point.
6 2

Find

a. the value of k
b. the gradient of the straight line ky = −x +12 .

27

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

UNIT EXERCISE 1.4

1. Find the point of intersection for each of the following pairs of
straight lines.

a. 2 x + 3y = 8 and x + 2y = 5 .
b. 2 x + y = 4 and x − y = −1 .
c. x + 3y + 5 = 0 and x + 5 y + 7 = 0 .

2. The straight line which has a gradient of 3 and passes through the
point(1, 4) intersects with the straight line y = −x + 5 at the point P.
Find the coordinates of point P.

The line y = x intersects with 2x2 + y2 = 27 at the points R and S. Find
(a) point R and S
(b) the length of RS

2. The line y = 2x – 2 intersects y = x2 – x – 6 at two points A and B.
Find
(a) the coordinates of A and B
(b) the midpoint of AB

3. y
C(6, 5)

D x
A
B
Figure 1

Figure 1, shows a triangle ABC with A on the y-axis, B on x-axis
and C (6, 5). Given the equation of straight line ADC is y – x + 1 = 0
and BD is 2y + x – 4 = 0. Find
(a) the coordinate of A, B and D
(b) the distance between A and C
(c) the equation of straight line which is passes through C and

perpendicular to ADC

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DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 1: COORDINATES GEOMETRY

4. R(8, 6)
Q(1, 7)

x–y+6=0

P(-4, 2)

S

Figure 2

Figure 2 shows a rhombus, PQRS with P(-4, 2), Q(1, 7) and R(8,
6). Diagonal PR and QS are perpendicular and intercept at point
M(h, k). Given equation PQ is x – y + 6 = 0. Find :
(a) coordinate M and S
(b) the distance between P and S
(c) the equation of straight line QR

29

UNIT 2 DUM 30172 ENGINEERING MATHEMATICS 3

UNIT 2: APPLICATION OF DIFFERENTIATION
APPLICATIONS OF

DIFFERENTIATION

INTRODUCTION

Differentiation is a branch of mathematics that is concerned with changing variable. By
differentiating a variable, we can find the rate at which it is changing. It is widely used by
mathematicians, scientists and engineers in many fields to solve problems.

CONTENTS

2.1 Tangent and Normal
2.2 Maximum and Minimum
2.3 Rates of Change
2.4 Small Change
2.5 Motion of a Straight Line

1

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

After completing the unit, students should be able to:

• Determine the gradient of a curve, tangent and normal at a point

UNIT LEARNING on a curve and form their equations.

2.O1UTCOTMAENSGENTS AND NORMALS

2.1 TANGENT AND NORMAL

2.1.1 DEFINITION

y

y = f(x) Recall Back
mt × mn = −1
tangent

P(x1, y1 )

normal
x

Figure 2.1: Tangent and normal line of a curve

If y = f (x) is this function of a curve, then dy = f’(x) is the gradient of the

dx

curve. The gradient of a curve at any point is defined as the gradient of a

tangent (mt) to the curve at the point. If P(x1, y1) is point on the curve

y = f (x) then the gradient of the tangent at P is the value of dy when
dx

x = x1, i.e. f(x).
The normal to the curve y = f(x) at point P(x1, y1) is the straight line that

passes through the point P and it perpendicular to the tangent at the point,

thus gradient of the normal is mn = − 1 or mn = −1
mt
f' (x )

2

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

2.1.2 EQUATION OF TANGENT AND A NORMAL TO A CURVE

The equation of the tangent at point P(x1, y1) is y − y1 = mt (x − x1).

The equation of a normal at point P(x1, y1) is y − y1 = mn (x − x1) where

mn = − f 1 ) .

' (x

Example 2.1.1 :

Find the gradient of a curve y = 3x2 − 2 at the point (2,10)

Solution :
y = 3x2 – 2
dy = 6x
dx
= 6(2)
= 12

Example 2.1.2 :
Find the equation of a tangent and normal to a curve y = 4x3 + 5 at the

point (− 1,1) .

Solution :
y = 4x3 + 5
dy = 12x 2
dx
= 12(-1)2
= 12
Gradient of tangent = 12
The equation of the tangent to the curve
y – 1 = 12 (x + 1)
y – 1 = 12x + 12
y = 12x + 13
The gradient of the normal = − 1
12
The equation of the normal to the curve
y – 1 = − 1 (x + 1)
12
12y – 12 = -x – 1
x + 12y -11 = 0

3

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

EXERCISE 2.1

1. Find the gradient of the curve and gradient of normal at the given point.

a) y = 3x2 − 4x + 2 (2,6)

b) y = 1 x2 + 4x  − 1,− 7 
2  2

c) y = 4x + 3 (1,7)
x2

d) y = 2 sin3x x = π
3

4

e) y = 9x + 1 DUM 30172 ENGINEERING MATHEMATICS 3
x UNIT 2: APPLICATION OF DIFFERENTIATION

x =1

Answers: a) 8, − 1 , b) 3, − 1 , c) - 2, 1 , d) - 6, 1 , e) 8, − 1
8 32 6 8

2. Find the equation of the tangent and normal to the curve at the given point
in each case.

a) y = 2x3 − 3x + 1 at the point (2,3)

b) y = 1 x2 − 4x + 1 at the point (− 1,5)

22

5

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

c) y = 2x2 + 1 − 3 (1,0)
x

d) y = 2e2x + x at the point (0,2)

Answers: a) Tangent, y = 21x – 39, Normal, y = - x/21 + 65/21, b) Tangent, y = -5x, Normal,
y = x/5 + 26/5, c) Tangent, y = 3x - 3, Normal, y = -x/3 + 1/3, d) Tangent, y = 5x+2, Normal, y = -x/5 + 2

6

UNIT LEARNING DUM 30172 ENGINEERING MATHEMATICS 3
OUTCOMES UNIT 2: APPLICATION OF DIFFERENTIATION

After completing the unit, students should be able to:
• Determine maximum and minimum points of a curve.
• Solve maximum and minimum problems

2.2 MAXIMUM AND MINIMUM

2.2.1 SECOND ORDER DIFFERENTIATION

The second derivative of the function y = f (x) is the second order

differentiation and denoted by d2y or f" (x) .
dx 2

d2y or f" (x) means to differentiate dy or f' (x) another time.
dx 2 dx

Example 2.2.1 :

Find d2y for each of the following function
dx 2

a. y = x3 − 3 x2 − 6x + 1 b. y = x2 + 54
2 x

Solution :

a. y = x3 - 3 x2 – 6x + 1
2

dy = 3x2 – 3x – 6
dx

d2y = 6x - 3
dx 2

b. y = x2 + 54
x

dy = 2x – 54x-2
dx

d2y =2 + 108x-3
dx 2

= 2 + 108
x3

7

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

2.2.2 TURNING POINTS, MAXIMUM AND MINIMUM POINTS

MAX POINT dy y
=0
(SOD)
dx
d2y 〈 0 at x = a y = f(x)
dx2 dy
dy 〉 0 〉0
dx dy 〈 0
dx Q dx
Po x

MIN POINT (SOD)

TURNING POINT dy d2y 〉 0 at x =b
(x=a & x=b) =0 dx 2

dy = 0 dx
dx

Figure 2.2 : Turning point, maximum and minimum points

i. As x increase through the point P, the value of dy changes from positive
dx

to zero and then to negative. Point P is known as the maximum point or it

is said that y = f (x) have maximum values.

ii. As x increase through the point Q, the value of dy change from negative to
dx

zero and then to positive. Point Q is known as the minimum points or it is

said that y = f (x) have minimum value.

iii. The points on the curve y = f (x) where dy = f' (x) = 0 are known as turning
dx

point or stationary point.

iv. Turning points consist of maximum and minimum points. The tangents at

turning points are parallel to the x-axis. Therefore the points P and Q are

the turning points.

8

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

The following steps can be used to determine whether a turning point is a maximum

or minimum point.

STEP 1 : Find dy
dx

STEP 2 : Find the value (s) of x and its corresponding value(s) of y when

dy = 0
dx

(At turning point dy = 0 ).
dx

STEP 3: To determine whether the turning point is a minimum or

maximum we can use the second order differentiation method.

Example 2.2.2.1 :
Find the coordinate of the turning points of the curve y = 2x2 – 8x + 1.

Determine whether the point is maximum or minimum.

Solution : Find dy
STEP 1 : dx

y = 2x2 – 8x + 1

dy = 4x – 8
dx

STEP 2 : Find the value (s) of x and its corresponding value(s) of y when

dy = 0
dx

At turning point dy = 0
dx

Then 4x – 8 = 0

x=2
When x = 2, y = 2(2)2 – 8(2) + 1 = -7

Therefore the turning point is (2, -7)

STEP 3 : To determine whether the turning point is a minimum or maximum

we

can use the second order differentiation method.

d2y =4 > 0
dx 2

Therefore (2, -7) is the minimum point

9

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

Example 2.2.2.2 :

Find the coordinates of the turning points of the curve y = 4x3 + 9x2 – 12x +
13. Determine whether the point is maximum or minimum point.

Solution :

y = 4x3 + 9x2 – 12x + 13

dy = 12x2 + 18x - 12
dx

12x2 + 18x – 12 = 0 or 2x2 + 3x – 2 = 0
(2x - 1) (x + 2) = 0

x = 1 and x = -2
2

when x = 1 when x = -2
2
y = 4(-2)3 + 9(-2)2 – 12(-2) + 13
y = 4 1 3 + 9 1 2 − 12 1  + 13
2 2 2 = 41

The turning point is (-2, 41)

= 39 d2y = 24x + 18
4 dx 2

The turning point is ( 1, 39 ) = 24(-2) + 18
2 4
= -30 < 0

d2y = 24x + 18 = 30 > 0 Therefore (-2, 41) is the maximum point.
dx 2

Therefore ( 1 , 39 ) is the minimum point.
2 4

10

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

2.2.3 PROBLEMS INVOLVING MAXIMUM OR MINIMUM VALUES

Many real-life situations the knowledge of differentiation especially the
problems involving maximum or minimum values.

Steps in solving on maximum and minimum

1) Express the quantity that has to be maximized or minimized in terms

of

only one variable based on the given information. Let a (area) be the

quantity that has to be maximized or minimized. Thus express a in

terms of only one variable, let it be r.

2) Find dA and solve the equation dA =0 to determine the value of r.
dr dr

3) Find d2A substitute the value of r that is found into d2 A , where if
dr 2 dr
2

• d2A is negative, the value of is maximum
dr 2

• d2A is positive, the value of is minimum.
dr 2

4) Calculate the maximum or minimum value of a by substituting the

value of

r into A.

Example 2.2.3 :

A farmer used 200 meters of rope to marks the perimeter of a rectangle
enclosure to fence in this sheep. If the rectangular enclosure is to measure 2x
m by y m show that 2x + y = 100 and the area A m2, of the enclosure is

given by A = 200x − 4x2

a) Find the values x and y
b) Determine the maximum area of the enclosure

11

Solution : DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

2x

y

a. Perimeter: 4x + 2y = 200

2x + y = 100 is shown

y = 100 – 2x …………….(1)

Area : A = 2xy…………………………..(2)

Substitute (1) into (2)

A = 2x(100 – 2x)

= 200x – 4x2 shown the area

dA = 200 – 8x
dx

Turning points occur at dA = 0
dx

Then 200 – 8x = 0

x = 25

When x = 25, then y = 100 – 2(25) = 50

b. Maximum area of the enclosure

d2A = -8 < 0 has a maximum values
dx 2

The maximum area = 200x – 4x2

= 200(25) – 4(25)2

= 2500 m2

12

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

EXERCISE 2.2

1. Find the turning points of the following curves and determines whether it is a
maximum or minimum point.
a) y = x 2 − 8 x +12

b) y = x3 − 3x2 − 9x + 7

13

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

c) y = 8x + 1
2x2

d) y = x(x − 6)2

e) y = 12 x − x 3

Answers: a) (4,-4) minimum, b) (3, -20) minimum, (-1, 12) maximum, c)  1 ,6  minimum,
 2 

d) (6, 0) minimum, (2, 32) maximum, e) (2, 16) maximum, (-2, -16) minimum

14

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

2. A piece of wire of length 120 cm is bent to form a trapezium as shown in the

figure. y cm

10x cm 10x cm

8x

12x + y cm

a) Express y in terms of x
b) Show that the area A cm2 enclosed by the wire is given by A = 480x − 80x 2
c) Find the values of x and y for which area is maximum.

Answers: a) y = 60 – 16x, b) A = 480x – 80x2, c) x = 3 , y = 12

15

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

3. A wire with length 2 meters is bent to form a rectangle with the maximum area. Find
the measurement of its sides.

Answers: 1 m
2

16

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

UNIT After completing the unit, students should be able to:
LEARNING • Determine the rate of change and related rate of change of a
OUTCOMES quality.

• Solve problems on rate of change and related rate of change.

2.3 RATES OF CHANGE

Rate of change problem usually involve time t as one of the variable. If y = f (t ) ,

where t is time, then dy represents the rate of change of y.
dt

Let we see these two situations:

• If A is the area in cm 2 and t is the time in seconds, then dA = 10 means the
dt

area is increasing at the rate of 10 cm2s −1

• If V is the volume in cm 3 and t is the time in seconds, then dv = −30 means
dt

the volume is decreasing at the rate of 30 cm3s −1 .

NOTE : Notice that a negative sign has to be inserted to represent the rate of
decrease.

Steps to solve problem on related rates of changes

Step 1 : Write the rate of change required by the question in the form of
mathematical symbol.

The rate of change of volume = dV
dt

The rate of change of area = dA
dt

The rate of change of radius = dr
dt

The rate of change of variable x = dx
dt

The rate of change of variable y = dy
dt

17

Step 2 : DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

The related rate of change can be determined by using the chain rule.

dy = dy × dx dV = dV × dr dA = dA × dr
dt dx dt dt dr dt dt dr dt
dy =rate of change of y dV =rate of change of V dA = rate of change of A
dt dt dt
dr =rate of change of r dr = rate of change of r
dx =rate of change of x dt dt

dt

Example 2.3.1 :
Given that y = 4x2 – 5x . If x increases at the rate of 0.4 unit per second. Find
the rate of change in y at the instant when x = 2 .

Solution : dx = 0.4 unit per second
y = 4x2 – 5x dt

dy = 8x – 5
dx

dy = dy × dx
dt dx dt

= [8(2) – 5] x (0.4)

= 4.4 unit s-1

Example 2.3.2 :
The radius r of a circle is decreasing at the rate of 0.6 cm per minute
when r = 9 cm . Find the corresponding rate of increase in the area, A cm2

of the circle.

Solution :
A = π r2

dA = 2 π r dr = -0.6 cm m-1
dr dt

dA = dA × dr
dt dr dt

= 2 π (9) x (- 0.6)

= -10.8 π cm2 m-1

18

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 2: APPLICATION OF DIFFERENTIATION

Example 2.3.3 :
Each side of a cubes is decreasing at the rate of 4 cm s-1. When the length of
each side is 10 cm. Find the rate of its
a. volume
b. Total surface area.

Solution : dx = -4 cm s-1
a. dt
The volume, V = x3
x

dV = 3x2 x
dx x
dV = dV × dx
dt dx dt

= 3(10)2 x (-4)

= -1200 cm3 s-1

Therefore the rate of decreasing of its volume is 1200 cm3 s-1

b. The total surface area
A = 6x2
dA = 12x
dx
dA = dA × dx
dt dx dt
= 12 (10) x (-4)
= -480
The rate of decreasing of its total surface area is 480 cm2 s-1

Example 2.3.4:

Air escape from a leaked spherical balloon if the volume of the balloon is
decreasing at the rate of 10π cm3s −1 . Find the rate of the radius of the
balloon when radius is 4 cm.

Solution :

The volume, V = 4 πr 3
3

dV = 4π r2
dr

19