13 Variations Case Study 13.1 Direct Variations 13.2 Inverse Variations 13.3 Joint Variations 13.4 Partial Variations Chapter Summary.
Download ReportTranscript 13 Variations Case Study 13.1 Direct Variations 13.2 Inverse Variations 13.3 Joint Variations 13.4 Partial Variations Chapter Summary.
Slide 1
13 Variations
Case Study
13.1 Direct Variations
13.2 Inverse Variations
13.3 Joint Variations
13.4 Partial Variations
Chapter Summary
Slide 2
Case Study
The cost of making 200 badges is
$1000. Thus, the average cost of
making a badge is $5.
Will the average cost per badge be
lowered if we make 500 badges?
Andy and Betty are responsible for making the school badges.
After comparing the cost of different companies, they find that the cost
of making school badges consists of two parts, one is fixed for design
while the other part depends on the number of badges made.
So, the cost per badge can be lowered if more school badges are made.
In this case, the cost of making school badges and the quantity to be
made demonstrate a partial variation.
P. 2
Slide 3
13.1 Direct Variations
Variation describes a relation between two variables.
There are different kinds of variations. In this section, we will
discuss direct variations first.
Suppose Mrs. Chan goes to a store and buys some flour for baking cakes.
The following table shows the total payment required for buying different
amounts of flour.
Total payment ($P)
8
24
40
56
72
Amount of flour (x kg)
1
3
5
7
9
From the above table, we observe that the ratio of the total payment to
the amount of flour bought is a constant, that is,
8
1
24
3
40
5
56
7
72
8
9
P. 3
Slide 4
13.1 Direct Variations
The relation can be expressed as
P 8x,
where $P is the total payment and x kg is the amount of flour.
The relation between the total payment and the
amount of flour can also be represented by a
graph.
From the graph, we note that the total payment
$P increases as the amount of flour x kg
increases.
P. 4
Slide 5
13.1 Direct Variations
This kind of relation is an example of direct variation.
Direct Variation
Both statements ‘y varies directly as x’ and ‘y is directly proportional
to x’ mean that y kx for a non-zero constant k.
Symbolically, we write y x.
Notes:
1. k is called the variation constant. Since y kx, if y1 is the
corresponding value of x1, then
y1
x1
k.
2. ‘’ means ‘varies directly as’.
In general, if y is directly proportional to x, that is, y kx, then the
graph of y against x is a straight line that passes through the origin
with slope k.
P. 5
Slide 6
13.1 Direct Variations
Example 13.1T
Suppose y is directly proportional to x 3 and y 32 when x 5.
(a) Find an equation connecting x and y.
(b) Find the values of
(i) y when x 4;
(ii) x when y 48.
Solution:
(a) Since y x 3, we have
(b) (i) When x 4,
y k(x 3), where k 0.
y 4(4 3)
Substituting y 32 and x 5 into
28
the equation, we have
(ii) When y 48,
32 k(5 3)
48 4(x 3)
32 8k
12 x 3
k 4
x9
y 4(x 3)
P. 6
Slide 7
13.1 Direct Variations
Example 13.2T
Suppose 2y 1 x and y 14 when x 4. Find the value of x
when y 5.
Solution:
Since 2y 1 x, for any points of (x1, y1) and (x2, y2), we have:
2 y1 1
2 y2 1
x1
x2
Substituting x1 4, y1 14 and y2 5 into the
equation, we have
2 (14 ) 1
2 (5) 1
Since
When y 5 , x
4
k
for any
x
pairs of the corresponding
values of x any y, we can
find x without finding the
variation constant k.
4
x2
27 x 2 36
4
x2
3
2y 1
.
3
P. 7
Slide 8
13.1 Direct Variations
Example 13.3T
The height h cm of an average person varies directly with their
foot length f cm. If the height of a person is 168 cm, then his/her foot length
is 24 cm. What is the height of David if his foot length is 26.5 cm?
Solution:
Since h f , we have h kf, where k 0.
Substituting h 168 and f 24 into the equation, we have
168 24k
k 7
h 7f
When f 26.5, h 7(26.5)
185.5
The height of David is 185.5 cm.
P. 8
Slide 9
13.1 Direct Variations
Example 13.4T
A ball is dropped onto the ground. Let v m/s be the speed of the
ball after it falls a distance of s m. Suppose v s . If s increases
by 44%, find the percentage change in v.
Solution:
Since v s , we have v k s , where k 0.
Let v1 and s1 be the original speed and the original distance respectively.
Then v1 k s1 .
Percentage change in v
New distance s2 (1 44%)s1
1.44s1
New speed v2 k s 2
k 1 . 44 s1
1 .2 k
1 . 2 v1
s1
v 2 v1
100 %
v1
1 . 2 v1 v1
100 %
v1
20 %
v increases by 20%.
P. 9
Slide 10
13.2 Inverse Variations
A florist is going to construct a rectangular greenhouse with
area 84 m2.
He can build the greenhouse with materials of different lengths and widths.
The following table shows several pairs of length and width for the
greenhouse:
Width (w)
2
4
6
8
10
Length (l)
42
21
14
10.5
8.4
From the above table, we observe that w increases when l decreases and
the product of l and w is a constant, that is, lw 84 or l
84
w
P. 10
.
Slide 11
13.2 Inverse Variations
The relation between the length and the width can also be
represented graphically.
Width (w)
2
4
6
8
10
Length (l)
42
21
14
10.5
8.4
The relation between the length and the width
is called an inverse variation.
Inverse Variation
Both statements ‘y varies inversely as x’ and ‘y is inversely
proportional to x’ mean that xy k or y
1
k
x
for a non-zero
constant k. Symbolically, we write y .
x
P. 11
Slide 12
13.2 Inverse Variations
Suppose y
x
1
y
x
2
.
4
0.5
x
24
6
8
0.25 0.17 0.13
12
6
4
3
10
0.1
2.4
Remark:
1
We observe that the graph of y against is
x
a straight line. In fact, from the relation of
y
k
x
1
, we have y k . This means y varies
directly as
x
1
.
x
P. 12
Slide 13
13.2 Inverse Variations
Example 13.5T
Suppose y varies inversely as the square of x 1 and y 8
when x 3. Find the values of
(a) y when x 10;
(b) x when y 2.
Solution:
Since y
1
( x 1)
2
, we have y
k
( x 1)
2
, where k 0.
Substituting y 8 and x 3 into the equation, we have
8
k
( 3 1)
(a) When x 10,
2
y
k 32
y
32
( x 1)
2
32
(10 1)
2
32
81
P. 13
Slide 14
13.2 Inverse Variations
Example 13.5T
Suppose y varies inversely as the square of x 1 and y 8
when x 3. Find the values of
(a) y when x 10;
(b) x when y 2.
Solution:
(b) When y 2, 2
Alternative Solution:
32
( x 1)
32
( x 1)
2
( x 1)
2
2
16
2
x 1 4
x 4 1 or 4 1
x 3 or 5
1
y
y ( x 1) is a constant.
( x 1)
2
2
Let x0 be the desired value of
the variable when y 2.
2(x0 1)2 8(3 1)2
(x0 1)2 16
x0 3 or 5
When y 2, x 3 or 5.
P. 14
Slide 15
13.2 Inverse Variations
Example 13.6T
Suppose y
3
x
. Find the percentage change in y when x increases
by 15%. (Give the answer correct to 3 significant figures.)
Solution:
Since y
3
, we have y
x
3k
, where k 0 .
x
Let x and y be the new values of x and y respectively.
Then x (1 15%)x
1.15x
y
3k
x
3k
1 . 15 x
y
1 . 15
Percentage change
y y
y
y
100 %
y
1 . 15
y
100 %
13 . 0 %
(cor. to 3 sig. fig.)
y decreases by 13.0% when x increases
by 15%.
P. 15
Slide 16
13.2 Inverse Variations
Example 13.7T
At a fixed temperature, the pressure P of any gas with a fixed mass
varies inversely as its volume V. A gas is compressed to 80% of its
original volume without a temperature change. Find the percentage
change in the pressure of the gas.
Solution:
Let P
k
, where k 0.
Percentage change in the
pressure of the gas
V
Let V and P be the new volume and
the new pressure respectively.
Then V 0.8V.
P
k
V
k
0 . 8V
P
0 .8
P P
100 %
P
P
P
0 .8
100 %
P
25 %
The pressure of the gas
increases by 25%.
P. 16
Slide 17
13.3 Joint Variations
Consider the volume V of a cylinder with base radius r and
height h.
The volume of the cylinder can be calculated by the formula
V pr2h.
In this formula, p is a constant.
1. If h is kept constant, then V varies directly as r2;
2. If r is kept constant, then V varies directly as h;
3. If neither r nor h is kept constant, then V varies directly as r2h.
We say that the volume V varies jointly as the height h and the square
of the base radius r. Such a relation is called a joint variation.
The variation can be represented by V r2h or V kr2h, where k is the
variation constant.
Joint Variation
If one variable z varies jointly as two (say x and y) or more other
variables (either directly or inversely), it is called a joint variation.
P. 17
Slide 18
13.3 Joint Variations
Example 13.8T
Suppose P varies jointly as u2 and v . When u 3 and v 16,
P 54.
(a) Express P in terms of u and v.
(b) Find the value of P when u 6 and v 64.
Solution:
(a ) Since P u
2
v , we have P ku
2
v , where k 0.
Substituting u 3, v 16 and P 54
into the equation, we have
2
54 k ( 3 ) ( 16 )
54 36 k
k
P
3
(b) When u 6 and v 64,
P
3
2
( 6 ) ( 64 )
2
3
36 8
3
2
2
432
u
2
v
2
P. 18
Slide 19
13.3 Joint Variations
Example 13.9T
Suppose z varies inversely as x2 and directly as y .
1
When x 4 and y 25, z .
4
(a) Express z in terms of x and y.
(b) Find the value of z when x 6 and y 64.
Solution:
y
(a) Since z
x
2
, we have z
k
y
x
2
, where k 0.
1
Substituting x 4, y 25 and z
4
into the above equation,
1
k
4
25
4
z
2
k
y
5x
2
z
4
5
4
(b) When x 6 and y 64,
4 64
5(6)
32
2
180
8
45
P. 19
Slide 20
13.3 Joint Variations
Example 13.10T
The mass M kg that a wooden bar of a certain thickness can
support varies directly as its width W cm and inversely as its length L m.
If a bar of width 3 cm and length 15 m can support a mass of 300 kg,
what mass can be supported by a bar of width 4 cm and length 30 m?
Solution:
Let M
kW
, where k 0.
L
Substituting W 3, L 15 and M 300 into the equation,
300
3k
When W 4 and L 30,
15
k 1500
M
1500 W
L
M
1500 ( 4 )
30
200
The bar can support a mass
of 200 kg.
P. 20
Slide 21
13.3 Joint Variations
Example 13.11T
The current I flowing through a resistor varies directly as the
voltage V across it and inversely as its resistance R.
(a) Write down an equation connecting I, V and R.
(b) Find the percentage change in I if V increases by 10% and R decreases
by 10%. (Give the answer correct to 1 decimal place.)
Solution:
(a) The equation is I
kV
, where k 0.
R
(b) Let I, V and R be the new current, voltage and resistance respectively.
Percentage change in I
V (1 10 %) V 1 . 1V
R (1 10 %) R 0 . 9 R
k (1 . 1V )
I
0 .9 R
1 .1I
0 .9
I I
I
1 .1I
100 %
I
0 .9
100 %
I
22 % (cor. to 1 d. p.)
P. 21
Slide 22
13.4 Partial Variations
In many practical situations, a variable is the sum of two or
more parts; each part may be either fixed (a constant) or may
vary as other variables.
For example, an association organizes a seminar. The organizer has to
book a hall for running this function which involves a fixed expense.
Moreover, the organizer needs to prepare refreshments for the
participants.
If the rent of the hall is $5000 and the cost of refreshments for one
participant is $30, then the total expense $E can be expressed as
E 5000 30N, where N is the number of participants.
We see that E is partly constant, and it partly varies directly as N.
Such a relation is called partial variation.
P. 22
Slide 23
13.4 Partial Variations
Consider another example of partial variation.
The solid in the figure consists of two parts, one part is a
sphere with radius r while the other part is a cube with sides s.
The total volume V of the solid is given by:
V
4
3
πr s
3
3
In this case, V is the sum of two parts such that
one part varies directly as the cube of r and the
other part varies directly as the cube of s.
P. 23
Slide 24
13.4 Partial Variations
Example 13.12T
Suppose x is partly constant and partly varies directly as y .
When y 16, x 116; when y 64, x 132.
(a) Find an equation connecting x and y.
(b) Find the value of x when y 25.
Solution:
(a) Since x is partly constant and partly varies directly as y , we have
x c k y , where c and k are non-zero constants.
Substituting y 16 and x 116
(2) (1): 16 4k
into the equation,
k4
Substituting k 4 into (1),
116 c k 16
116 c 4 k .......... .....( 1)
Substituting y 64 and x 132
into the equation,
132 c k 64
132 c 8 k .......... ....( 2 )
116 c 4 ( 4 )
c 100
x 100 4 y
(b) When y 25, x 100 4 25
120
P. 24
Slide 25
13.4 Partial Variations
Example 13.13T
Suppose Q partly varies directly as the square of x and partly
varies inversely as x. Q 26 when x 1 or 3.
(a) Express Q in terms of x.
(b) Find the value of Q when x 6.
Solution:
(a) Since Q partly varies directly as x2 and partly varies inversely as x,
we have Q k 1 x 2
k2
x
, where k1 and k2 are non-zero constants.
Substituting x 1 and Q 26
into the equation,
k1 k 2 26 .......... ....( 1)
Substituting x 3 and Q 26
into the equation,
2
k1 ( 3 )
k2
26
3
27 k1 k 2 78 ......( 2 )
(2) (1): 26k1 52
k1 2
Substituting k1 2 into (1),
2 k 2 26
k 2 24
2
Q 2x
24
x
P. 25
Slide 26
13.4 Partial Variations
Example 13.13T
Suppose Q partly varies directly as the square of x and partly
varies inversely as x. Q 26 when x 1 or 3.
(a) Express Q in terms of x.
(b) Find the value of Q when x 6.
(c) Find the possible values of x when Q 20.
(Give the answers in surd form if necessary.)
Solution:
(b) When x 6,
2
Q 2(6)
24
6
76
(c) When Q 20,
2
2x
24
20
x
3
2 x 24 20 x
3
x 10 x 12 0
Let P(x) x3 10x 12.
P(2) 23 10(2) 12
x2
0
x 2 is a factor of P(x).
By long division,
2
x 2x 6
2
x 2 or 1
2
3
2
x 2x
( x 2 )( x 2 x 6 ) 0
3
x 0 x 10 x 1 2
2
2 x 10 x 1 2
2
2x 4x
6 x 12
6 x 12
7
P. 26
Slide 27
13.4 Partial Variations
Example 13.14T
The total cost $C of printing a magazine is partly constant and
partly varies directly as the number of copies N printed. If 500 copies are
printed, the cost of printing per copy is $15.5. If 1200 copies are printed,
the cost of printing per copy is $8.5.
(a) Express C in terms of N.
(b) What is the cost of printing per copy if 1000 copies are printed?
Solution:
(a) Since C is partly constant and partly varies directly as N, we have
C k1 k2N, where k1 and k2 are non-zero constants.
Substituting N 500 and C 15.5 500 into the equation,
15 . 5 500 k1 500 k 2
7750 k1 500 k 2 ........( 1)
Substituting N 1200 and C 8.5 1200 into the equation,
8 . 5 1200 k1 1200 k 2
10 200 k1 1200 k 2 ......( 2 )
P. 27
Slide 28
13.4 Partial Variations
Example 13.14T
The total cost $C of printing a magazine is partly constant and
partly varies directly as the number of copies N printed. If 500 copies are
printed, the cost of printing per copy is $15.5. If 1200 copies are printed,
the cost of printing per copy is $8.5.
(a) Express C in terms of N.
(b) What is the cost of printing per copy if 1000 copies are printed?
Solution:
7750 k 1 500 k 2 .......... ...( 1)
10 200 k1 1200 k 2 ........( 2 )
(2) (1): 2450 700k2
k2 3.5
Substituting k2 3.5 into (1),
7750 k1 500(3.5)
k1 6000
C 6000 3.5N
(b) When N 1000,
C 6000 3 . 5 (1000 )
9500
The total cost is $9500.
Cost of printing per copy
9500
$
1000
$ 9 .5
P. 28
Slide 29
13.4 Partial Variations
Example 13.15T
The cost $C of making a wooden cube is partly constant and partly varies
directly as the surface area of the cube. When the side s cm of the cube is
4 cm, the cost is $18.4; when the side of the cube is 7 cm, the cost is $31.6.
(a) Express C in terms of s.
Solution:
(a) Since C is partly constant and partly varies directly as the surface area,
we have C k1 k2s2, where k1 and k2 are non-zero constants.
Substituting s 4 and C 18.4
into the equation,
(2) (1): 33k2 13.2
2
k2 0.4
k 1 k 2 ( 4 ) 18 . 4
Substituting k2 0.4 into (1),
k1 16 k 2 18 . 4 ......( 1)
Substituting s 7 and C 31.6
18.4 k1 16(0.4)
into the equation,
k1 12
2
k 1 k 2 ( 7 ) 31 . 6
C 12 0 . 4 s
2
k1 49 k 2 31 . 6 ......( 2 )
P. 29
Slide 30
13.4 Partial Variations
Example 13.15T
The cost $C of making a wooden cube is partly constant and partly varies
directly as the surface area of the cube. When the side s cm of the cube is
4 cm, the cost is $18.4; when the side of the cube is 7 cm, the cost is $31.6.
(a) Express C in terms of s.
(b) In order to make a profit percentage of 25%, what is the selling price
of a cube with sides of 6 cm?
Solution:
(b) For a cube with sides of 6 cm,
cost $( 12 0 . 4 6 2 )
$ 26 . 4
Selling price $[26.4 (1 25%)]
$33
P. 30
Slide 31
Chapter Summary
13.1 Direct Variations
Variation describes a relation between two changing quantities
and we can use an equation to express the relation.
Both statements ‘y varies directly as x’ and ‘y is directly proportional
to x’ mean that y kx, where k is a non-zero constant. Symbolically,
we write y x.
P. 31
Slide 32
Chapter Summary
13.2 Inverse Variations
Both statements ‘y varies inversely as x’ and ‘y is inversely
proportional to x’ mean that xy k or y
constant. Symbolically, we write y
1
x
k
x
, where k is a non-zero
.
P. 32
Slide 33
Chapter Summary
13.3 Joint Variations
If one variable z varies jointly as two (say x and y) or more
other variables (either directly or inversely), it is a joint variation.
Symbolically, if z varies jointly as x and y, we write z xy.
P. 33
Slide 34
Chapter Summary
13.4 Partial Variations
1. If z is partly constant and partly varies directly as x, then
z c kx,
where c is a constant and k is the variation constant.
2. If z partly varies directly as x and partly varies directly as y, then
z k1x k2y,
where k1 and k2 are variation constants.
P. 34
13 Variations
Case Study
13.1 Direct Variations
13.2 Inverse Variations
13.3 Joint Variations
13.4 Partial Variations
Chapter Summary
Slide 2
Case Study
The cost of making 200 badges is
$1000. Thus, the average cost of
making a badge is $5.
Will the average cost per badge be
lowered if we make 500 badges?
Andy and Betty are responsible for making the school badges.
After comparing the cost of different companies, they find that the cost
of making school badges consists of two parts, one is fixed for design
while the other part depends on the number of badges made.
So, the cost per badge can be lowered if more school badges are made.
In this case, the cost of making school badges and the quantity to be
made demonstrate a partial variation.
P. 2
Slide 3
13.1 Direct Variations
Variation describes a relation between two variables.
There are different kinds of variations. In this section, we will
discuss direct variations first.
Suppose Mrs. Chan goes to a store and buys some flour for baking cakes.
The following table shows the total payment required for buying different
amounts of flour.
Total payment ($P)
8
24
40
56
72
Amount of flour (x kg)
1
3
5
7
9
From the above table, we observe that the ratio of the total payment to
the amount of flour bought is a constant, that is,
8
1
24
3
40
5
56
7
72
8
9
P. 3
Slide 4
13.1 Direct Variations
The relation can be expressed as
P 8x,
where $P is the total payment and x kg is the amount of flour.
The relation between the total payment and the
amount of flour can also be represented by a
graph.
From the graph, we note that the total payment
$P increases as the amount of flour x kg
increases.
P. 4
Slide 5
13.1 Direct Variations
This kind of relation is an example of direct variation.
Direct Variation
Both statements ‘y varies directly as x’ and ‘y is directly proportional
to x’ mean that y kx for a non-zero constant k.
Symbolically, we write y x.
Notes:
1. k is called the variation constant. Since y kx, if y1 is the
corresponding value of x1, then
y1
x1
k.
2. ‘’ means ‘varies directly as’.
In general, if y is directly proportional to x, that is, y kx, then the
graph of y against x is a straight line that passes through the origin
with slope k.
P. 5
Slide 6
13.1 Direct Variations
Example 13.1T
Suppose y is directly proportional to x 3 and y 32 when x 5.
(a) Find an equation connecting x and y.
(b) Find the values of
(i) y when x 4;
(ii) x when y 48.
Solution:
(a) Since y x 3, we have
(b) (i) When x 4,
y k(x 3), where k 0.
y 4(4 3)
Substituting y 32 and x 5 into
28
the equation, we have
(ii) When y 48,
32 k(5 3)
48 4(x 3)
32 8k
12 x 3
k 4
x9
y 4(x 3)
P. 6
Slide 7
13.1 Direct Variations
Example 13.2T
Suppose 2y 1 x and y 14 when x 4. Find the value of x
when y 5.
Solution:
Since 2y 1 x, for any points of (x1, y1) and (x2, y2), we have:
2 y1 1
2 y2 1
x1
x2
Substituting x1 4, y1 14 and y2 5 into the
equation, we have
2 (14 ) 1
2 (5) 1
Since
When y 5 , x
4
k
for any
x
pairs of the corresponding
values of x any y, we can
find x without finding the
variation constant k.
4
x2
27 x 2 36
4
x2
3
2y 1
.
3
P. 7
Slide 8
13.1 Direct Variations
Example 13.3T
The height h cm of an average person varies directly with their
foot length f cm. If the height of a person is 168 cm, then his/her foot length
is 24 cm. What is the height of David if his foot length is 26.5 cm?
Solution:
Since h f , we have h kf, where k 0.
Substituting h 168 and f 24 into the equation, we have
168 24k
k 7
h 7f
When f 26.5, h 7(26.5)
185.5
The height of David is 185.5 cm.
P. 8
Slide 9
13.1 Direct Variations
Example 13.4T
A ball is dropped onto the ground. Let v m/s be the speed of the
ball after it falls a distance of s m. Suppose v s . If s increases
by 44%, find the percentage change in v.
Solution:
Since v s , we have v k s , where k 0.
Let v1 and s1 be the original speed and the original distance respectively.
Then v1 k s1 .
Percentage change in v
New distance s2 (1 44%)s1
1.44s1
New speed v2 k s 2
k 1 . 44 s1
1 .2 k
1 . 2 v1
s1
v 2 v1
100 %
v1
1 . 2 v1 v1
100 %
v1
20 %
v increases by 20%.
P. 9
Slide 10
13.2 Inverse Variations
A florist is going to construct a rectangular greenhouse with
area 84 m2.
He can build the greenhouse with materials of different lengths and widths.
The following table shows several pairs of length and width for the
greenhouse:
Width (w)
2
4
6
8
10
Length (l)
42
21
14
10.5
8.4
From the above table, we observe that w increases when l decreases and
the product of l and w is a constant, that is, lw 84 or l
84
w
P. 10
.
Slide 11
13.2 Inverse Variations
The relation between the length and the width can also be
represented graphically.
Width (w)
2
4
6
8
10
Length (l)
42
21
14
10.5
8.4
The relation between the length and the width
is called an inverse variation.
Inverse Variation
Both statements ‘y varies inversely as x’ and ‘y is inversely
proportional to x’ mean that xy k or y
1
k
x
for a non-zero
constant k. Symbolically, we write y .
x
P. 11
Slide 12
13.2 Inverse Variations
Suppose y
x
1
y
x
2
.
4
0.5
x
24
6
8
0.25 0.17 0.13
12
6
4
3
10
0.1
2.4
Remark:
1
We observe that the graph of y against is
x
a straight line. In fact, from the relation of
y
k
x
1
, we have y k . This means y varies
directly as
x
1
.
x
P. 12
Slide 13
13.2 Inverse Variations
Example 13.5T
Suppose y varies inversely as the square of x 1 and y 8
when x 3. Find the values of
(a) y when x 10;
(b) x when y 2.
Solution:
Since y
1
( x 1)
2
, we have y
k
( x 1)
2
, where k 0.
Substituting y 8 and x 3 into the equation, we have
8
k
( 3 1)
(a) When x 10,
2
y
k 32
y
32
( x 1)
2
32
(10 1)
2
32
81
P. 13
Slide 14
13.2 Inverse Variations
Example 13.5T
Suppose y varies inversely as the square of x 1 and y 8
when x 3. Find the values of
(a) y when x 10;
(b) x when y 2.
Solution:
(b) When y 2, 2
Alternative Solution:
32
( x 1)
32
( x 1)
2
( x 1)
2
2
16
2
x 1 4
x 4 1 or 4 1
x 3 or 5
1
y
y ( x 1) is a constant.
( x 1)
2
2
Let x0 be the desired value of
the variable when y 2.
2(x0 1)2 8(3 1)2
(x0 1)2 16
x0 3 or 5
When y 2, x 3 or 5.
P. 14
Slide 15
13.2 Inverse Variations
Example 13.6T
Suppose y
3
x
. Find the percentage change in y when x increases
by 15%. (Give the answer correct to 3 significant figures.)
Solution:
Since y
3
, we have y
x
3k
, where k 0 .
x
Let x and y be the new values of x and y respectively.
Then x (1 15%)x
1.15x
y
3k
x
3k
1 . 15 x
y
1 . 15
Percentage change
y y
y
y
100 %
y
1 . 15
y
100 %
13 . 0 %
(cor. to 3 sig. fig.)
y decreases by 13.0% when x increases
by 15%.
P. 15
Slide 16
13.2 Inverse Variations
Example 13.7T
At a fixed temperature, the pressure P of any gas with a fixed mass
varies inversely as its volume V. A gas is compressed to 80% of its
original volume without a temperature change. Find the percentage
change in the pressure of the gas.
Solution:
Let P
k
, where k 0.
Percentage change in the
pressure of the gas
V
Let V and P be the new volume and
the new pressure respectively.
Then V 0.8V.
P
k
V
k
0 . 8V
P
0 .8
P P
100 %
P
P
P
0 .8
100 %
P
25 %
The pressure of the gas
increases by 25%.
P. 16
Slide 17
13.3 Joint Variations
Consider the volume V of a cylinder with base radius r and
height h.
The volume of the cylinder can be calculated by the formula
V pr2h.
In this formula, p is a constant.
1. If h is kept constant, then V varies directly as r2;
2. If r is kept constant, then V varies directly as h;
3. If neither r nor h is kept constant, then V varies directly as r2h.
We say that the volume V varies jointly as the height h and the square
of the base radius r. Such a relation is called a joint variation.
The variation can be represented by V r2h or V kr2h, where k is the
variation constant.
Joint Variation
If one variable z varies jointly as two (say x and y) or more other
variables (either directly or inversely), it is called a joint variation.
P. 17
Slide 18
13.3 Joint Variations
Example 13.8T
Suppose P varies jointly as u2 and v . When u 3 and v 16,
P 54.
(a) Express P in terms of u and v.
(b) Find the value of P when u 6 and v 64.
Solution:
(a ) Since P u
2
v , we have P ku
2
v , where k 0.
Substituting u 3, v 16 and P 54
into the equation, we have
2
54 k ( 3 ) ( 16 )
54 36 k
k
P
3
(b) When u 6 and v 64,
P
3
2
( 6 ) ( 64 )
2
3
36 8
3
2
2
432
u
2
v
2
P. 18
Slide 19
13.3 Joint Variations
Example 13.9T
Suppose z varies inversely as x2 and directly as y .
1
When x 4 and y 25, z .
4
(a) Express z in terms of x and y.
(b) Find the value of z when x 6 and y 64.
Solution:
y
(a) Since z
x
2
, we have z
k
y
x
2
, where k 0.
1
Substituting x 4, y 25 and z
4
into the above equation,
1
k
4
25
4
z
2
k
y
5x
2
z
4
5
4
(b) When x 6 and y 64,
4 64
5(6)
32
2
180
8
45
P. 19
Slide 20
13.3 Joint Variations
Example 13.10T
The mass M kg that a wooden bar of a certain thickness can
support varies directly as its width W cm and inversely as its length L m.
If a bar of width 3 cm and length 15 m can support a mass of 300 kg,
what mass can be supported by a bar of width 4 cm and length 30 m?
Solution:
Let M
kW
, where k 0.
L
Substituting W 3, L 15 and M 300 into the equation,
300
3k
When W 4 and L 30,
15
k 1500
M
1500 W
L
M
1500 ( 4 )
30
200
The bar can support a mass
of 200 kg.
P. 20
Slide 21
13.3 Joint Variations
Example 13.11T
The current I flowing through a resistor varies directly as the
voltage V across it and inversely as its resistance R.
(a) Write down an equation connecting I, V and R.
(b) Find the percentage change in I if V increases by 10% and R decreases
by 10%. (Give the answer correct to 1 decimal place.)
Solution:
(a) The equation is I
kV
, where k 0.
R
(b) Let I, V and R be the new current, voltage and resistance respectively.
Percentage change in I
V (1 10 %) V 1 . 1V
R (1 10 %) R 0 . 9 R
k (1 . 1V )
I
0 .9 R
1 .1I
0 .9
I I
I
1 .1I
100 %
I
0 .9
100 %
I
22 % (cor. to 1 d. p.)
P. 21
Slide 22
13.4 Partial Variations
In many practical situations, a variable is the sum of two or
more parts; each part may be either fixed (a constant) or may
vary as other variables.
For example, an association organizes a seminar. The organizer has to
book a hall for running this function which involves a fixed expense.
Moreover, the organizer needs to prepare refreshments for the
participants.
If the rent of the hall is $5000 and the cost of refreshments for one
participant is $30, then the total expense $E can be expressed as
E 5000 30N, where N is the number of participants.
We see that E is partly constant, and it partly varies directly as N.
Such a relation is called partial variation.
P. 22
Slide 23
13.4 Partial Variations
Consider another example of partial variation.
The solid in the figure consists of two parts, one part is a
sphere with radius r while the other part is a cube with sides s.
The total volume V of the solid is given by:
V
4
3
πr s
3
3
In this case, V is the sum of two parts such that
one part varies directly as the cube of r and the
other part varies directly as the cube of s.
P. 23
Slide 24
13.4 Partial Variations
Example 13.12T
Suppose x is partly constant and partly varies directly as y .
When y 16, x 116; when y 64, x 132.
(a) Find an equation connecting x and y.
(b) Find the value of x when y 25.
Solution:
(a) Since x is partly constant and partly varies directly as y , we have
x c k y , where c and k are non-zero constants.
Substituting y 16 and x 116
(2) (1): 16 4k
into the equation,
k4
Substituting k 4 into (1),
116 c k 16
116 c 4 k .......... .....( 1)
Substituting y 64 and x 132
into the equation,
132 c k 64
132 c 8 k .......... ....( 2 )
116 c 4 ( 4 )
c 100
x 100 4 y
(b) When y 25, x 100 4 25
120
P. 24
Slide 25
13.4 Partial Variations
Example 13.13T
Suppose Q partly varies directly as the square of x and partly
varies inversely as x. Q 26 when x 1 or 3.
(a) Express Q in terms of x.
(b) Find the value of Q when x 6.
Solution:
(a) Since Q partly varies directly as x2 and partly varies inversely as x,
we have Q k 1 x 2
k2
x
, where k1 and k2 are non-zero constants.
Substituting x 1 and Q 26
into the equation,
k1 k 2 26 .......... ....( 1)
Substituting x 3 and Q 26
into the equation,
2
k1 ( 3 )
k2
26
3
27 k1 k 2 78 ......( 2 )
(2) (1): 26k1 52
k1 2
Substituting k1 2 into (1),
2 k 2 26
k 2 24
2
Q 2x
24
x
P. 25
Slide 26
13.4 Partial Variations
Example 13.13T
Suppose Q partly varies directly as the square of x and partly
varies inversely as x. Q 26 when x 1 or 3.
(a) Express Q in terms of x.
(b) Find the value of Q when x 6.
(c) Find the possible values of x when Q 20.
(Give the answers in surd form if necessary.)
Solution:
(b) When x 6,
2
Q 2(6)
24
6
76
(c) When Q 20,
2
2x
24
20
x
3
2 x 24 20 x
3
x 10 x 12 0
Let P(x) x3 10x 12.
P(2) 23 10(2) 12
x2
0
x 2 is a factor of P(x).
By long division,
2
x 2x 6
2
x 2 or 1
2
3
2
x 2x
( x 2 )( x 2 x 6 ) 0
3
x 0 x 10 x 1 2
2
2 x 10 x 1 2
2
2x 4x
6 x 12
6 x 12
7
P. 26
Slide 27
13.4 Partial Variations
Example 13.14T
The total cost $C of printing a magazine is partly constant and
partly varies directly as the number of copies N printed. If 500 copies are
printed, the cost of printing per copy is $15.5. If 1200 copies are printed,
the cost of printing per copy is $8.5.
(a) Express C in terms of N.
(b) What is the cost of printing per copy if 1000 copies are printed?
Solution:
(a) Since C is partly constant and partly varies directly as N, we have
C k1 k2N, where k1 and k2 are non-zero constants.
Substituting N 500 and C 15.5 500 into the equation,
15 . 5 500 k1 500 k 2
7750 k1 500 k 2 ........( 1)
Substituting N 1200 and C 8.5 1200 into the equation,
8 . 5 1200 k1 1200 k 2
10 200 k1 1200 k 2 ......( 2 )
P. 27
Slide 28
13.4 Partial Variations
Example 13.14T
The total cost $C of printing a magazine is partly constant and
partly varies directly as the number of copies N printed. If 500 copies are
printed, the cost of printing per copy is $15.5. If 1200 copies are printed,
the cost of printing per copy is $8.5.
(a) Express C in terms of N.
(b) What is the cost of printing per copy if 1000 copies are printed?
Solution:
7750 k 1 500 k 2 .......... ...( 1)
10 200 k1 1200 k 2 ........( 2 )
(2) (1): 2450 700k2
k2 3.5
Substituting k2 3.5 into (1),
7750 k1 500(3.5)
k1 6000
C 6000 3.5N
(b) When N 1000,
C 6000 3 . 5 (1000 )
9500
The total cost is $9500.
Cost of printing per copy
9500
$
1000
$ 9 .5
P. 28
Slide 29
13.4 Partial Variations
Example 13.15T
The cost $C of making a wooden cube is partly constant and partly varies
directly as the surface area of the cube. When the side s cm of the cube is
4 cm, the cost is $18.4; when the side of the cube is 7 cm, the cost is $31.6.
(a) Express C in terms of s.
Solution:
(a) Since C is partly constant and partly varies directly as the surface area,
we have C k1 k2s2, where k1 and k2 are non-zero constants.
Substituting s 4 and C 18.4
into the equation,
(2) (1): 33k2 13.2
2
k2 0.4
k 1 k 2 ( 4 ) 18 . 4
Substituting k2 0.4 into (1),
k1 16 k 2 18 . 4 ......( 1)
Substituting s 7 and C 31.6
18.4 k1 16(0.4)
into the equation,
k1 12
2
k 1 k 2 ( 7 ) 31 . 6
C 12 0 . 4 s
2
k1 49 k 2 31 . 6 ......( 2 )
P. 29
Slide 30
13.4 Partial Variations
Example 13.15T
The cost $C of making a wooden cube is partly constant and partly varies
directly as the surface area of the cube. When the side s cm of the cube is
4 cm, the cost is $18.4; when the side of the cube is 7 cm, the cost is $31.6.
(a) Express C in terms of s.
(b) In order to make a profit percentage of 25%, what is the selling price
of a cube with sides of 6 cm?
Solution:
(b) For a cube with sides of 6 cm,
cost $( 12 0 . 4 6 2 )
$ 26 . 4
Selling price $[26.4 (1 25%)]
$33
P. 30
Slide 31
Chapter Summary
13.1 Direct Variations
Variation describes a relation between two changing quantities
and we can use an equation to express the relation.
Both statements ‘y varies directly as x’ and ‘y is directly proportional
to x’ mean that y kx, where k is a non-zero constant. Symbolically,
we write y x.
P. 31
Slide 32
Chapter Summary
13.2 Inverse Variations
Both statements ‘y varies inversely as x’ and ‘y is inversely
proportional to x’ mean that xy k or y
constant. Symbolically, we write y
1
x
k
x
, where k is a non-zero
.
P. 32
Slide 33
Chapter Summary
13.3 Joint Variations
If one variable z varies jointly as two (say x and y) or more
other variables (either directly or inversely), it is a joint variation.
Symbolically, if z varies jointly as x and y, we write z xy.
P. 33
Slide 34
Chapter Summary
13.4 Partial Variations
1. If z is partly constant and partly varies directly as x, then
z c kx,
where c is a constant and k is the variation constant.
2. If z partly varies directly as x and partly varies directly as y, then
z k1x k2y,
where k1 and k2 are variation constants.
P. 34