Engineering Mathematics 4 Episode 3 pptx

Transposition of formulae 10.1 Introduction to transposition of formulae When a symbol other than the subject is required to be calculated it is usual to rearrange the formula to make a

9.4 More difficult worked problems on

In this type of equation the solution is easier if a

substitution is initially made Let 1

x Daand

1

y DbThus equation (1) becomes: 2a C 3b D 7 3

and equation (2) becomes: a 4b D 2 4

Multiplying equation (4) by 2 gives:



C10

3

2

2C

3

5y D43

bDy then b D

1

y D

15

Hence the solution is a = 1

4

To eliminate fractions, both sides of equation (1) are

multiplied by 27x C y giving:

27x C y

1

x C y



D27x C y

427

27 D 45 C 4yfrom which 4y D 27  20 D 7

4 D1

34

Hence x =5, y = 1 3

4 is the required solution,

which may be checked in the original equations

Now try the following exercise

Exercise 34 Further more difficult

prob-lems on simultaneous tions

equa-In problems 1 to 5, solve the simultaneous

equations and verify the results



p D 1

4, q D

15

Problem 11 The law connecting friction Fand load L for an experiment is of the form

F D aL C b, where a and b are constants.When F D 5.6, L D 8.0 and when F D 4.4,

L D2.0 Find the values of a and b and thevalue of F when L D 6.5

Substituting F D 5.6, L D 8.0 into F D aL C bgives:

Substituting F D 4.4, L D 2.0 into F D aL C bgives:

Substituting a D 1

5 into equation (1) gives:

5.6 D 8.0

15



Cb5.6 D 1.6 C b

5.6  1.6 D b

Checking, substituting a D 1

5 and b D 4 in tion (2), gives:

equa-RHS D 2.0

15

Problem 12 The equation of a straight line,

of gradient m and intercept on the y-axis c,

is y D mx C c If a straight line passes

through the point where x D 1 and y D 2,

and also through the point where x D 312 and

y D1012, find the values of the gradient and

the y-axis intercept

Substituting x D 1 and y D 2 into y D mx C c

212

Removing the brackets from equation (2) gives:

26 D 2I28I1C8I2Rearranging gives:

Substituting I1 D2 into equation (3) gives:

Hence the solution is I1 =2 and I2 =−1

(which may be checked in the original equations)

Problem 14 The distance s metres from a

fixed point of a vehicle travelling in a

straight line with constant acceleration,

am/s2, is given by s D ut C12at2, where u is

the initial velocity in m/s and t the time in

seconds Determine the initial velocity and

the acceleration given that s D 42 m when

t D2 s and s D 144 m when t D 4 s Find

also the distance travelled after 3 s

42 D 2u C 215

42  30 D 2u

u D12

2 D6Substituting a D 15, u D 6 in equation (2) gives:

RHS D 46 C 815 D 24 C 120 D 144 D LHS

Hence the initial velocity, u= 6 m/s and the

acceleration, a= 15 m/s 2.Distance travelled after 3 s is given by s D utC1

2at

2

where t D 3, u D 6 and a D 15Hence s D 63 C1

2153

2D18 C 67.5

i.e distance travelled after 3 s = 85.5 m

Problem 15 The resistance R  of a length

of wire at t°C is given by R D R01 C ˛t,where R0 is the resistance at 0°C and ˛ isthe temperature coefficient of resistance in/°C Find the values of ˛ and R0 if R D 30 

at 50°C and R D 35  at 100°C

Substituting R D 30, t D 50 into R D R01 C ˛tgives:

Substituting R D 35, t D 100 into R D R01 C ˛tgives:

Although these equations may be solved by theconventional substitution method, an easier way is toeliminate R0by division Thus, dividing equation (1)

1250 D

1

250 or 0.004Substituting ˛ D 1

250 into equation (1) gives:

30 D R0



1 C 50

1250



30 D R01.2

R0D 301.2 D25

Checking, substituting ˛ D 1

250 and R0 D 25 inequation (2) gives:

RHS D 25



1 C 100

1250



D251.4 D 35 D LHS

Thus the solution isa =0.004 =°C and R0 = 25 Z

Problem 16 The molar heat capacity of a

solid compound is given by the equation

c D a C bT, where a and b are constants

When c D 52, T D 100 and when c D 172,

T D400 Determine the values of a and b

52 D a C 1000.4

a D 52  40 D 12

Hence a= 12 and b= 0.4

Now try the following exercise

Exercise 35 Further practical problems

involving simultaneous tions

equa-1 In a system of pulleys, the effort P

required to raise a load W is given by

P D aW C b, where a and b are constants

If W D 40 when P D 12 and W D 90when P D 22, find the values of a and b

[I1D6.47, I2 D4.62]

3 Velocity v is given by the formula v D

u C at IfvD20 when t D 2 andvD40when t D 7 find the values of u and a.Hence find the velocity when t D 3.5

[u D 12, a D4, vD26]

4 y D mx C c is the equation of a straightline of slope m and y-axis intercept c Ifthe line passes through the point where

x D 2 and y D 2, and also throughthe point where x D 5 and y D 12,find the slope and y-axis intercept of thestraight line

a C bT When c D 52, T D 100 and when

c D 172, T D 400 Find the values of a

Transposition of formulae

10.1 Introduction to transposition of

formulae

When a symbol other than the subject is required

to be calculated it is usual to rearrange the formula

to make a new subject This rearranging process is

called transposing the formula or transposition.

The rules used for transposition of formulae are

the same as those used for the solution of

sim-ple equations (see Chapter 8) — basically, that the

equality of an equation must be maintained.

10.2 Worked problems on

transposition of formulae

Problem 1 Transpose p D q C r C s to

make r the subject

The aim is to obtain r on its own on the left-hand

side (LHS) of the equation Changing the equation

around so that r is on the LHS gives:

It is shown with simple equations, that a quantity

can be moved from one side of an equation to

the other with an appropriate change of sign Thus

equation (2) follows immediately from equation (1)

The result of multiplying each side of the equation

by 1 is to change all the signs in the equation

It is conventional to express answers with positivequantities first Hence rather than x D a  b C

w C y, x =wYy − a − b, since the order of terms

connected by C and  signs is immaterial

Problem 3 Transpose vDthe subject

R DI

Multiplying both sides by R gives:

R

VR

Dividing both sides by R gives:

aR

lRi.e a= r l

Now try the following exercise

Exercise 36 Further problems on

transpo-sition of formulae

Make the symbol indicated the subject of each

of the formulae shown and express each in itssimplest form





x D y  cm





T D IPR

or 1 aS

m

ftm



Dmvu i.e ft D mvuDividing both sides by t gives:

Problem 9 The final length, l2 of a piece

of wire heated through #°C is given by the

formula l2Dl11 C ˛# Make the

coefficient of expansion, ˛, the subject

Problem 10 A formula for the distance

moved by a body is given by: s D 1

2vCut.

Rearrange the formula to make u the subject

2vCut D sMultiplying both sides by 2 gives: vCut D2s

Dividing both sides by t gives:

vCut

2st

tHence u= 2s

squared term, that term is isolated on the LHS, and

then the square root of both sides of the equation is

taken

Multiplying both sides by 2 gives: mv2 D2k

Dividing both sides by m gives: mv

2

2km

Whenever the prospective new subject is within asquare root sign, it is best to isolate that term on theLHS and then to square both sides of the equation.Rearranging gives: 2 l

g Dt

Dividing both sides by 2 gives: l

gD

t2

Squaring both sides gives: l

g D

t2

2

D t242

Cross-multiplying, i.e multiplying each term by42g, gives:

Problem 14 The impedance of an a.c

circuit is given by Z DpR2CX2 Make thereactance, X, the subject

Rearranging gives:

R2CX2DZSquaring both sides gives: R2CX2DZ2

Dividing both sides by 2 gives:

Taking the cube root of both sides gives:

Now try the following exercise

Exercise 37 Further problems on

transpo-sition of formulae

Make the symbol indicated the subject of each

of the formulae shown and express each in its

Rearranging gives: a

2x C a2y

Multiplying both sides by r gives: a2x C a2y D rp

Factorising the LHS gives: a2x C y D rp

Dividing both sides by x C y gives:

bd C be D

x  ya

Rearranging gives: b

1 C b DaMultiplying both sides by 1 C b gives:

b D a1 C b

Removing the bracket gives: b D a C abRearranging to obtain terms in b on the LHS gives:

b  ab D aFactorising the LHS gives: b1  a D aDividing both sides by (1  a) gives:

Er D VR C r

Removing the bracket gives: Er D VR C VrRearranging to obtain terms in r on the LHS gives:

Er  Vr D VRFactorising gives: rE  V D VRDividing both sides by E  V gives:

Rearranging gives: f C p

f  p D

DdSquaring both sides gives:

Factorising gives: pd2CD2 D fD2d2

Dividing both sides by (d2CD2) gives:

p= f D2− d2/

d2 YD2/

Now try the following exercise

Exercise 38 Further problems on

transpo-sition of formulae

Make the symbol indicated the subject of each

of the formulae shown in Problems 1 to 7, and

express each in its simplest form

L D150, d D 0.30 and g D 9.81



2dgh0.03L, 0.965



11 The sag S at the centre of a wire is given

by the formula: S D

3dl  d

Quadratic equations

11.1 Introduction to quadratic

equations

As stated in Chapter 8, an equation is a statement

that two quantities are equal and to ‘solve an

equa-tion’ means ‘to find the value of the unknown’.

The value of the unknown is called the root of the

equation

A quadratic equation is one in which the highest

power of the unknown quantity is 2 For example,

x2
3x C 1 D 0 is a quadratic equation

There are four methods of solving quadratic

equations.

These are: (i) by factorisation (where possible)

(ii) by ‘completing the square’

(iii) by using the ‘quadratic formula’

or (iv) graphically (see Chapter 30)

11.2 Solution of quadratic equations

by factorisation

Multiplying out 2xC1x
3 gives 2x2
6xCx
3,

i.e 2x2
5x
3 The reverse process of moving

from 2x2
5x
3 to 2x C 1x
3 is called

factorising.

If the quadratic expression can be factorised this

provides the simplest method of solving a quadratic

x2+ 2x − 8 = (x + 4)(x − 2)

(Note that the product of the two inner termsadded to the product of the two outer termsmust equal the middle term, C2x in thiscase.)

The quadratic equation x2C2x
8 D 0 thusbecomes x C 4x
2 D 0

Since the only way that this can be true is foreither the first or the second, or both factors to

be zero, theneither x C4 D 0 i.e x D
4

or x
2 D 0 i.e x D2

Hence the roots of x2 Y2x −8 = 0 are

x =−4 and 2

(b) 3x2
11x
4 D 0The factors of 3x2 are 3x and x These areplaced in brackets thus: (3x )(x )

The factors of
4 are
4 and C1, or C4 and

1, or
2 and 2

Remembering that the product of the two innerterms added to the product of the two outerterms must equal
11x, the only combination

to give this is C1 and
4, i.e

3x2
11x
4 D 3x C 1x
4

The quadratic equation 3x2
11x
4 D 0 thus

(a) x2
6x C 9 D 0 Hence x
3x
3 D 0, i.e

x
32 D0 (the left-hand side is known as a

perfect square) Hence x = 3 is the only root

of the equation x2
6x C 9 D 0

(b) 4x2
25 D 0 (the left-hand side is the

dif-ference of two squares, 2x2 and 52) Thus

(a) 4x2C8x C 3 D 0 The factors of 4x2 are 4x

and x or 2x and 2x The factors of 3 are 3

and 1, or
3 and
1 Remembering that the

product of the inner terms added to the product

of the two outer terms must equal C8x, the only

combination that is true (by trial and error) is:

which may be checked in the original equation

(b) 15x2C2x
8 D 0 The factors of 15x2 are 15xand x or 5x and 3x The factors of
8 are
4and C2, or 4 and
2, or
8 and C1, or 8 and

1 By trial and error the only combinationthat works is:

15x2C2x
8 D 5x C 43x
2Hence 5x C 43x
2 D 0 from whicheither 5x C 4 D 0

Hence x=−4

5 or x=

2 3

which may be checked in the original equation

Problem 4 The roots of a quadraticequation are 1

3 and
2 Determine theequation

If the roots of a quadratic equation are ˛ and ˇ then



x

2 D 0

x
13

(b) If 1.2 and
0.4 are the roots of a quadratic

equation then:

x
1.2x C 0.4 D 0i.e x2
1.2x C 0.4x
0.48 D 0

i.e x2−0 8x −0.48 = 0

Now try the following exercise

Exercise 39 Further problems on solving

quadratic equations by torisation

fac-In Problems 1 to 10, solve the given equations



5 6x2
5x C 1 D 0

1

2,

13



6 10x2C3x
4 D 0

1

2,

45



9 6x2
5x
4 D 0

4

3,

12



10 8x2C2x
15 D 0

5

4,

32



In Problems 11 to 16, determine the quadratic

equations in x whose roots are:

11.3 Solution of quadratic equations

by ‘completing the square’

An expression such as x2 or x C 22 or x
32 iscalled a perfect square

a quadratic equation into a perfect square before

solving is called ‘completing the square’.

x C a2Dx2C2ax C a2Thus in order to make the quadratic expression

x2C2ax into a perfect square it is necessary to add(half the coefficient of x2 i.e

2a2

2

, i.e

x2C3x C

32

2

D

x C32

2

The method is demonstrated in the following workedproblems

Problem 6 Solve 2x2C5x D 3 by

‘completing the square’

The procedure is as follows:

1 Rearrange the equation so that all terms are

on the same side of the equals sign (and thecoefficient of the x2 term is positive)

Hence 2x2C5x
3 D 0

2 Make the coefficient of the x2 term unity Inthis case this is achieved by dividing throughout

2x

3

2 D0

3 Rearrange the equations so that the x2 and x

terms are on one side of the equals sign and the

constant is on the other side Hence

x2C5

2x D

32

4 Add to both sides of the equation (half the

coefficient of x2 In this case the coefficient of

2

D 3

2C

54

6 Taking the square root of both sides of the

equation (remembering that the square root of a

number gives a š answer) Thus

4 D š

74

7 Solve the simple equation Thus

x D
5

4š

74i.e x D
5

Problem 7 Solve 2x2C9x C 8 D 0, correct

to 3 significant figures, by ‘completing thesquare’

Making the coefficient of x2 unity gives:

x2C9

2x C4 D 0and rearranging gives: x2C9

2x D
4Adding to both sides (half the coefficient of x2

gives:

x2C9

2x C

94

2

D

94

2

D 81

16
4 D

1716Taking the square root of both sides gives:

x C9

4 D

17

1.754.6 D0

and rearranging gives: y2C3.5

4.6y D

1.754.6

Adding to both sides (half the coefficient of y2

gives:

y2C 3.5

4.6y C

3.59.2

2

D 1.754.6 C

3.59.2

y C3.5

9.2D

p0.5251654 D š0.7246830

9.2š0.7246830i.e y= 0.344 or −1.105

Now try the following exercise

Exercise 40 Further problems on solving

quadratic equations by pleting the square’

‘com-Solve the following equations by completing

the square, each correct to 3 decimal places

a perfect square gives:

x2Cb

ax C

b2a

2

D

b2a

2

caRearranging gives:

x Cba

2

D b24a2
c

aD

b2
4ac4a2

Taking the square root of both sides gives:

x C b2a D

2a š

p

b2
4ac2ai.e the quadratic formula is: x D
b š

p

b2
4ac2a(This method of solution is ‘completing thesquare’ — as shown in Section 10.3.) Summarising:

if ax2Cbx C c D0

then x= − b ±pb2− 4ac

2a

This is known as the quadratic formula.

Problem 9 Solve (a) x2C2x
8 D 0 and(b) 3x2
11x
4 D 0 by using the quadraticformula

(a) Comparing x2C2x
8 D 0 with ax2CbxCc D0gives a D 1, b D 2 and c D
8

Substituting these values into the quadraticformula

D
2 šp4 C 32

2 šp362

Hence x=4

2= 2 or

8

2 =−4 (as inProblem 1(a))

Problem 10 Solve 4x2C7x C 2 D 0 giving

the roots correct to 2 decimal places

D
7 C 4.123

7
4.1238

Hence, x =−0.36 or−1.39, correct to 2 decimal

places.

Now try the following exercise

Exercise 41 Further problems on solving

quadratic equations by mula

for-Solve the following equations by using the

quadratic formula, correct to 3 decimal places

There are many practical problems where a

quadratic equation has first to be obtained, fromgiven information, before it is solved

Problem 11 Calculate the diameter of asolid cylinder which has a height of 82.0 cmand a total surface area of 2.0 m2

Total surface area of a cylinder

Dcurved surface area

C2 circular ends (from Chapter 19)

D2 rh C 2 r2(where r D radius and h D height)Since the total surface area D 2.0 m2 and theheight h D 82 cm or 0.82 m, then

2.0 D 2 r0.82 C 2 r2i.e 2 r2C2 r0.82
2.0 D 0Dividing throughout by 2 gives:

r2C0.82r
1

D0Using the quadratic formula:

21

D
0.82 šp1.9456

0.82 š 1.39482

D0.2874 or
1.1074

Thus the radius r of the cylinder is 0.2874 m (the

negative solution being neglected)

Hence the diameter of the cylinder

D2 ð 0.2874

D0.5748 m or 57.5 cm

correct to 3 significant figures

Problem 12 The height s metres of a mass

projected vertically upwards at time t

seconds is s D ut
1

2gt

2 Determine howlong the mass will take after being projected

to reach a height of 16 m (a) on the ascent

and (b) on the descent, when u D 30 m/s and

D5.53 or 0.59

Hence the mass will reach a height of 16 m

after 0.59 s on the ascent and after 5.53 s on the

descent.

Problem 13 A shed is 4.0 m long and

2.0 m wide A concrete path of constant

width is laid all the way around the shed If

the area of the path is 9.50 m2 calculate its

width to the nearest centimetre

Figure 11.1 shows a plan view of the shed with its

surrounding path of width t metres

Figure 11.1

Hence t D
12.0 š



12.02
44
9.5024

D
12.0 šp296.0

8

D
12.0 š 17.20465

8Hence t D0.6506 m or
3.65058 mNeglecting the negative result which is meaningless,

the width of the path, t= 0.651 m or 65 cm, correct

to the nearest centimetre

Problem 14 If the total surface area of asolid cone is 486.2 cm2 and its slant height

is 15.3 cm, determine its base diameter

From Chapter 19, page 145, the total surface area A

of a solid cone is given by: A D ...

If A D 48 2.2 and l D 15 .3, then

48 2.2 D r15 .3 C r2i.e r2C15 .3 r
48 2.2 D

2

D
15 .3 šp 848 . 046 1

2

D
15 .3 š 29.121 23

2... 3. 5

4. 6y C

3. 59.2

2

D 1.7 54. 6 C

3. 59.2

y C3. 5

9.2D

p0.52516 54 D š0.7 246 830 ... =−4 and 2

(b) 3x2
11x
D 0The factors of 3x2 are 3x and x These areplaced in brackets thus: (3x )(x )

The factors of
4 are
4 and C1, or C4 and