Transcript Ch5 Formulas - Wah Yan College, Kowloon

5
Formulas
Case Study
5.1 Sequences
5.2 Introduction to Functions
5.3 Simple Algebraic Fractions
5.4 Formulas and Substitution
5.5 Change of Subject
Chapter Summary
Case Study
Body Mass Index (BMI) is frequently used to check whether a person’s
body weight and body height are in an appropriate proportion. It can be
calculated by using the following formula:
Body Mass Index (BMI) 
Body weight (in kg)
Body height 2 (in m2 )
For example, if a person’s weight and height are 50 kg and 1.6 m
respectively, then the BMI  50  1.62  19.5.
The following table shows the range of the BMI
and the corresponding body condition.
BMI
Less than 18.5
18.5 – 22.9
23 – 24.9
25 – 29.9
30 or over
Condition
Underweight
Ideal
Overweight
Obese
Severely obese
P. 2
Is my weight normal?
Let us check your body
mass index and see whether
you are normal or not.
5.1 Sequences
A. Introduction to Sequences
John recorded the weights of 10 classmates. He listed their weights
(in kg) in the order of their class numbers.
We call such a list of numbers a sequence.
Each number in a sequence is called a term.
We usually denote the first term as T1, the second term as T2, and
so on.
In the above sequence, T1  46, T2  42, T3  50, ... , etc.
P. 3
5.1 Sequences
B. Some Common Sequences
Some sequences have certain patterns, but some do not.
1.
Consider the sequence 1, 5, 9, 13, ... . +4 +4
We can guess the subsequent terms: 17, 21, 25, ...
When the difference between any 2 consecutive terms is a
constant, such a sequence is called an arithmetic sequence,
and the difference is called a common difference.
2.
Consider the sequence 8, 16, 32, 64, ... . 2 2
We can guess the subsequent terms: 128, 256, 512, ...
When the ratio of each term (except the first term) to the
preceding term is a constant, such a sequence is called a
geometric sequence, and the ratio is called a common ratio.
P. 4
5.1 Sequences
B. Some Common Sequences
3.
We can arrange some dots to form some squares.
The number of dots used in each square
is called a square number.
4.
We can arrange some dots to form some triangles.
The number of dots used in each triangle
is called a triangular number.
5.
Consider the sequence 1, 1, 2, 3, 5, 8, ... .
In this sequence, starting from the third term, each term is
the sum of the 2 preceding terms. This sequence is called
the Fibonacci sequence.
P. 5
5.1 Sequences
C. General Terms
For a sequence that shows a certain pattern, we can use Tn to
represent the nth term.
Tn is called the general term of the sequence.
It is a common practice
to write the general term
of a sequence as an
algebraic expression in
terms of n.
For example, since the sequence 2, 4, 6, 8, 10, ... has consecutive
even numbers, we can deduce that the general term of this sequence
is Tn  2n.
Once the general term of a sequence is obtained, we can use it to
describe any term in a sequence.
P. 6
5.1 Sequences
C. General Terms
Example 5.1T
Find the general terms of the following sequences.
(a) 7, 14, 21, 28, ...
(b) 1, 2, 4, ...
1 1 1 1
(c) , , , , ...
(d) 15, 14, 13, 12, ...
2 4 8 16
Solution:
(a) The sequence 7, 14, 21, 28, ... can be written as
7(1), 7(2), 7(3), 7(4), ...
∴ The general term of the sequence is 7n.
(b) The sequence 1, 2, 4, ... can be written as
21  1, 22  1, 23  1 ...
∴ The general term of the sequence is 2n  1.
P. 7
5.1 Sequences
C. General Terms
Example 5.1T
Find the general terms of the following sequences.
(a) 7, 14, 21, 28, ...
(b) 1, 2, 4, ...
1 1 1 1
(c) , , , , ...
(d) 15, 14, 13, 12, ...
2 4 8 16
Solution:
1 1 1 1
, , , , ... can be written as
2 4 8 16
1
2
3
4
 1  ,  1  ,  1  ,  1  , ...
n
 2  2  2  2
1
 
∴ The general term of the sequence is   .
2
(d) The sequence 15, 14, 13, 12, ... can be written as
16  1, 16  2, 16  3, 16  4, ...
∴ The general term of the sequence is 16  n.
(c) The sequence
P. 8
5.1 Sequences
C. General Terms
Example 5.2T
Find the general term and the 9th term of each of the following sequences.
(a) 1, 2, 5, 8, ...
(b) 1, 4, 9, 16, ...
+3 +3 +3
Solution:
(a) T1  1  3  4  3(1)  4
T2  2  6  4  3(2)  4
T3  5  9  4  3(3)  4
T4  8  12  4  3(4)  4
∴ Tn  3n  4
T9  3(9)  4
 23
P. 9
(b) T1  1  12
T2  4  2 2
T3  9  3 2
T4  16  42
∴ Tn  n 2
T9  9 2
 81
Rewrite the terms of
the sequence as
expressions in which
the order of the terms
can be observed.
5.2 Introduction to Functions
In the previous section, we learnt how to find the values of the terms
in a sequence from the general term by substituting different values
of n in the general term.
Consider a sequence with the general term Tn  5n  2.
The above figure shows an ‘input-process-output’ relationship,
which is called a function.
For each input value of
In this example, we call Tn a function of n.
n, there is exactly one
output value of Tn.
P. 10
5.2 Introduction to Functions
The idea of function is common in our daily lives.
Suppose that each can of cola costs $5.
Let x be the number of cans of cola, and $y be the corresponding
total cost.
Since the total cost of x cans of cola is $5x, the equation y  5x
represents the relationship between x and y.
The following table shows some values of x and the corresponding
values of y.
x
y
1
5
2
10
3
15
4
20
5
25
For every value of x, there is only one corresponding value of y.
We say that y is a function of x.
P. 11
5.2 Introduction to Functions
Example 5.3T
If p is a function of q such that p  4q  5, find the values of p for the
following values of q.
2
(a) 6
(b) 5
(c)
5
Solution:
(a) When q  6, p  4(6)  5
 19
(b) When q  5, p  4(5)  5
 25
2
 2
(c) When q  , p  4   5
5
5
8
 5
5
17

5
P. 12
Substitute q  6 into
the expression 4q  5.
5.3 Simple Algebraic Fractions
A. Simplification
a
We learnt at primary level that numbers in the form , where
b
a and b are integers and b  0, are called fractions.
When both the numerator and the denominator of a fractional
expression are polynomials, where the denominator is not a
constant, such as:
3
5b
6x2
,
or
,
4b 2b  1 ( x  1)( x  2)
we call such an expression an algebraic fraction.
P. 13
3x  4 y
is
5
not an algebraic
fraction because the
denominator is a
constant.
Note that
5.3 Simple Algebraic Fractions
A. Simplification
We can simplify a numerical fraction, whose numerator and
denominator both have common factors, by cancelling the
common factors.
24 12  2

For example:
36 12  3
2

3
For algebraic fractions, we can simplify them in a similar way,
when the common factors are numbers, variables or polynomials.
For example:
4a
2 a ( 2)

6a 3 2a (3a 2 )
2
 2
3a
P. 14
5.3 Simple Algebraic Fractions
A. Simplification
Example 5.4T
Simplify the following algebraic fractions.
5ua  10uc
14 y 2  7 y
(a)
(b)
2ua  4uc
21y
Solution:
5ua  10uc 5u (a  2c)

2ua  4uc 2u (a  2c)
5

2
14 y 2  7 y 7 y (2 y  1)

(b)
7 y (3)
21y
2 y 1  1  2 y 

 or

3 
3 
(a)
P. 15
First factorize the
numerator and the
denominator. Then
cancel out the common
factors.
We cannot cancel
common factors from
the terms 7y and 21y
only, i.e., the fraction
cannot be simplified as
14 y 2  7 y 14 y 2

.
21y 3
3
5.3 Simple Algebraic Fractions
A. Simplification
Example 5.5T
Simplify the following algebraic fractions.
mh  2nh  2km  4kn
3 x
(a)
(b)
2k  h
(6  2 x ) 2
Solution:
(a)
mh  2nh  2km  4kn
2k  h
h( m  2n)  2k ( m  2n)

2k  h
(h  2k )(m  2n)

2k  h
 m  2n
P. 16
(b)
3 x
(6  2 x ) 2
3 x

[2(3  x)]2
3 x

4(3  x) 2
1

4(3  x)
First factorize the
numerator. You may
check if 2k + h (the
denominator) is a
factor of the numerator.
5.3 Simple Algebraic Fractions
B. Multiplication and Division
When multiplying or dividing a fraction, we usually try to cancel
out all common factors before multiplying the numerator and the
denominator separately to get the final result.
3 25
3 5(5)
 

10 9 2(5) 3(3)
5

6
We can perform the multiplication or division of algebraic fractions
in a similar way.
3a 25b3
3a 5b 2 (5b)
 2 

For example:
10b 9a
2(5b) 3a(3a)
5b 2

6a
For example:
P. 17
5.3 Simple Algebraic Fractions
B. Multiplication and Division
Example 5.6T
Simplify the following algebraic fractions.
6q
24q 2
2k 2
8t
(a)
(b)

 3
2
n

4
5n  20
64t  8t
k
Solution:
2k 2
8t
2k 2
8t


(a)
 3
2
8t (8  t ) k 3
64t  8t
k
1

4tk
6q
24q 2
6q 5(n  4)
(b)



n  4 5n  20 n  4 24q 2
5

4q
P. 18
5.3 Simple Algebraic Fractions
C. Addition and Subtraction
The method used for the addition and subtraction of algebraic
fractions is similar to that for numerical fractions.
2 5 25


For example:
9a 9a
9a
7

9a
When the denominators of algebraic fractions are not equal, first
we have to find the lowest common multiple (L.C.M.) of the
denominators.
For example:
1
2 1(3)  2(2)


6a 9a
18a
7

18a
P. 19
5.3 Simple Algebraic Fractions
C. Addition and Subtraction
Example 5.7T
Simplify
7
3
 .
2 h 8h
Solution:
7
3 7 ( 4) 3

 
2 h 8h 8 h 8 h
28  3

8h
25

8h
P. 20
5.3 Simple Algebraic Fractions
C. Addition and Subtraction
Example 5.8T
Simplify
5
3

.
2u  6v 3v  u
Solution:
5
3
5
3



2u  6v 3v  u 2(u  3v)  (u  3v)
5  3(2)

2(u  3v)
1

2(u  3v)
1

2(3v  u )
P. 21
For any number x  0,
1
1 1
 
x
x x
5.3 Simple Algebraic Fractions
C. Addition and Subtraction
Example 5.9T
Simplify
4
3
 .
3p  9 p
Solution:
4
3
4
3
 

3 p  9 p 3( p  3) p
4 p  3  3( p  3)

3 p( p  3)
4 p  9 p  27

3 p( p  3)
5 p  27

3 p ( p  3)
P. 22
Since 3(p  3) and p
have no common
factors other than 1, the
L.C.M. of 3(p  3) and
p is 3p(p  3).
5.4 Formulas and Substitution
Consider the volume (V) of a cuboid:
V  lwh
where l is the length, w is the width
and h is the height.
If the values of the variables l, w and h are already known, we can
find V by the method of substitution.
Similarly, if the values of V, l and w are known, we can find h.
Actually, in any given formula, we can find any one of the variables
by the method of substitution if all the others are known.
P. 23
5.4 Formulas and Substitution
Example 5.10T
Consider the formula T  r 2 h 
and r  2.5.
3h
. Find the value of h if T  121
4
Solution:
3h
4
3

T  h r 2  
4

3

121  h 2.52  
4

T  r 2h 
 5  2 3 
121  h    
 2  4 
P. 24
Factorize the expression.
 25 3 
121  h  
 4 4
11h
121 
2
121 2
h
11
h  22
5.5 Change of Subject
Given the area (A) of a triangle:
bh
A
where b is the base length and h is the height.
2
In the formula, A is the only variable on the left-hand side.
We call A the subject of the formula.
This formula can be used if we want to find A, when b and h are known.
In another case, if we need to find h when A and b are known, it is more
2A
h

convenient to use another formula
, with h being the subject.
b
2A
bh
The process of obtaining the formula h 
from A 
is called the
b
2
change of subject.
We can use the method of solving equations to change the subject of a
formula.
P. 25
5.5 Change of Subject
Example 5.11T
Make u the subject of the formula 3k  4  5u.
Solution:
3k  4  5u
3k  4  5u
5u  3k  4
3k  4
u
5
 4 is transposed to the L.H.S. to become  4.
Rewrite the formula such that only the
variable u is on the L.H.S.
P. 26
First move the other
terms to one side such
that only the variable u
remains on the other
side.
5.5 Change of Subject
Example 5.12T
Make m the subject of the formula
1 2 3
  .
p m r
Solution:
1 2 3
 
p m r
2 3 1
 
m r p
2 3p  r

m
pr
2 pr  m(3 p  r )
m(3 p  r )  2 pr
2 pr
m
3p  r
P. 27
Simplify the fractions.
The L.C.M. of r and p is pr.
Multiply both sides by mpr.
5.5 Change of Subject
Example 5.13T
Make m the subject of the formula k 
h  2m
.
3  2m
Solution:
h  2m
3  2m
k (3  2m)  h  2m
3k  2mk  h  2m
3k  h  2m  2mk
3k  h  2m(1  k )
2m(1  k )  3k  h
3k  h
m
2(1  k )
k
P. 28
Remove the brackets
Take out the common factor m.
First move all the terms
that involve m to one
side.
5.5 Change of Subject
Example 5.14T
A bag of food is put into a refrigerator. The temperature T (in C) of
9t
T

24

the food after time t (in hours) is given by the formula
.
2
(a) Make t the subject of the formula.
(b) How long will it take for the temperature of the food to become
–3C?
Solution:
(a)
T  24 
9t
2
9t
 24  T
2
2
t  (24  T )
9
P. 29
(b)
2
t  [24  (3)]
9
2
 (27)
9
6
∴
It takes 6 hours for the temperature
of the food to become 3C.
Chapter Summary
5.1 Sequences
A list of numbers arranged in an order is called a sequence.
Each number in a sequence is called a term.
For a sequence with a certain pattern, we can represent the sequence
by its general term.
P. 30
Chapter Summary
5.2 Introduction to Functions
A function describes an ‘input-process-output’ relationship between
2 variables. Each input gives only one output.
P. 31
Chapter Summary
5.3 Simple Algebraic Fractions
The manipulations of algebraic fractions are similar to those of
numerical fractions.
P. 32
Chapter Summary
5.4 Formulas and Substitution
A formula is any equation that describes the relationship of 2 or
more variables.
By the method of substitution, we can find the value of a variable
in a formula when the other variables are known.
P. 33
Chapter Summary
5.5 Change of Subject
When there is only one variable on one side of a formula, this
variable is called the subject of the formula.
P. 34