1313 13.1direct variations 13.2inverse variations 13.3joint variations chapter summary case study...

13

13.1 Direct Variations

13.2 Inverse Variations

13.3 Joint Variations

Chapter Summary

Case Study

Variations

13.4 Partial Variations

P. 2

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.

Case StudyCase Study

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.

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?

P. 3

Variation describes a relation between two variables.

There are different kinds of variations. In this section, we will discuss direct variations first.

1133 .1 .1 Direct VariationsDirect Variations

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,

89

72

7

56

5

40

3

24

1

8

P. 4

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. 5


kx

y 1

1

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.

Direct VariationBoth 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.

This kind of relation is an example of direct variation.

Notes:

1. k is called the variation constant. Since y kx, if y1 is the

corresponding value of x1, then .

2. ‘’ means ‘varies directly as’.

P. 6

Example 13.1T

Solution:


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.

(a) Since y x 3, we havey k(x 3), where k 0.

Substituting y 32 and x 5 into the equation, we have

32 k(5 3)32 8kk 4

y 4(x 3)

(b) (i) When x 4,

y 4(4 3) 28

(ii) When y 48, 48 4(x 3)

12 x 3 x 9

P. 7

Example 13.2T

2

2

1

1 1212

x

y

x

y

2

1)5(2

4

1)14(2

x

3627 2 x

3

42 x

.3

4 ,5 When xy


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:

Since for any

pairs of the corresponding values of x any y, we can find x without finding the variation constant k.

kx

y

12

Substituting x1 4, y1 14 and y2 5 into the equation, we have

P. 8

Example 13.3T


h 7f

The height of David is 185.5 cm.

The height h cm of an average person varies directly with theirfoot 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 hf , we have h kf, where k 0.

Substituting h 168 and f 24 into the equation, we have

k 7

When f 26.5, h 7(26.5) 185.5

168 24k

P. 9

Example 13.4T


sv

.Then 11 skv

144.1 sk

Percentage change in v

%1001

12

v

vv

%1002.1

1

11 v

vv

%20

v increases by 20%.

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 . If s increases by 44%, find the percentage change in v.

Solution:Since , we have v , where k 0.sv sk

Let v1 and s1 be the original speed and the original distance respectively.

12.1 v

New distance s2 (1 44%)s1

1.44s1

New speed v2 2sk

12.1 sk

P. 10

1133 ..22 Inverse VariationsInverse 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 .w

l84

P. 11


The relation between the length and the width can also be represented graphically.

The relation between the length and the width is called an inverse variation.

Width (w) 2 4 6 8 10

Length (l) 42 21 14 10.5 8.4

Inverse VariationBoth statements ‘y varies inversely as x’ and ‘y is inversely

proportional to x’ mean that xy k or for a non-zero

constant k. Symbolically, we write .

x

ky

xy

1

P. 12


xy

24Suppose .

x 2 4 6 8 10

y 12 6 4 3 2.4

x

10.10.130.170.250.5

x

1

xky

1

x

ky

Remark:We observe that the graph of y against is

a straight line. In fact, from the relation of

, we have . This means y varies

directly as .x

1

P. 13

Example 13.5T

Solution:


2)13(8

k

32k

2)1(

32

xy

81

32

)110(

322

y

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.

0. where,)1(

have we,)1(

1 Since

22

k

x

ky

xy

Substituting y 8 and x 3 into the equation, we have

(a) When x 10,

P. 14

Example 13.5T


2)1(

322

x

2

32)1( 2 x

16)1( 2 x

Solution:

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.

(b) When y 2,

x 1 4x 4 1 or 4 1x 3 or 5

constant. a is)1(

)1(

1

2

2

xy

xy

Alternative Solution:

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. 15

Example 13.6T

Solution:


xy

3

.0 where,3

have we,3

Since kx

ky

xy

Let x and y be the new values of x and y respectively.

x

ky

3

x

k

15.1

3

15.1

y

%100y

yy

%10015.1

y

yy

%0.13 (cor. to 3 sig. fig.)

Suppose . Find the percentage change in y when x increases

by 15%. (Give the answer correct to 3 significant figures.)

Then x (1 15%)x 1.15x

Percentage change

y decreases by 13.0% when x increases by 15%.

P. 16

Example 13.7T


0. where,Let kV

kP

V

kP

V

k

8.0

8.0

P

Percentage change in the pressure of the gas

%100

P

PP

%1008.0

P

PP

%25 The pressure of the gas incr

eases by 25%.

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 V and P be the new volume and the new pressure respectively. Then V 0.8V.

P. 17

1133 ..33 Joint VariationsJoint 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 formulaV r2h.

In this formula, 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 VariationIf 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. 18

Example 13.8T

Solution:


0. where, have we, Since)(a 22 kvkuPvuP

)16()3(54 2kk3654

2

3k

vuP 2

2

3

)64()6(2

3 2P

432

Suppose P varies jointly as u2 and . 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.

v

Substituting u 3, v 16 and P 54 into the equation, we have

(b) When u 6 and v 64,

8362

3

P. 19

Example 13.9T

Solution:


y

4

1

0. where, have we, Since(a)22

kx

ykz

x

yz

Substituting x 4, y 25 andinto the above equation, 4

1z

24

25

4

1 k5

4k

25

4

x

yz

(b) When x 6 and y 64,

2)6(5

644z

180

32

45

8

Suppose z varies inversely as x2 and directly as .

When x 4 and y 25, z .

(a) Express z in terms of x and y. (b) Find the value of z when x 6 and y 64.

P. 20

Example 13.10T


L

kWM

15

3300

k

1500k

L

WM

1500

When W 4 and L 30,

20030

)4(1500

M

The bar can support a mass of 200 kg.

The mass M kg that a wooden bar of a certain thickness cansupport 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 , where k 0.

Substituting W 3, L 15 and M 300 into the equation,

P. 21


R

kVI

VVV 1.1%)101( RRR 9.0%)101(

R

VkI

9.0

)1.1(

9.0

1.1 I

Percentage change in I

%100I

II

%1009.0

1.1

I

II

(cor. to 1 d. p.)

Example 13.11TThe current I flowing through a resistor varies directly as thevoltage 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 , where k 0.

(b) Let I, V and R be the new current, voltage and resistance respectively.

%22

P. 22

1133 ..44 Partial VariationsPartial 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. 23

33π3

4srV


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:

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. 24

Example 13.12T

ykcx , where c and k are non-zero constants. Substituting y 16 and x 116 into the equation,

16116 kc )1.....(..........4116 kc

Substituting y 64 and x 132 into the equation,

64132 kc )2....(..........8132 kc

(2) (1):

Substituting k 4 into (1), )4(4116 c

100cyx 4100


ySuppose x is partly constant and partly varies directly as . 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 , we havey

16 4k k 4

(b) When y 25, 120

254100 x

P. 25

Example 13.13T

Solution:


)1....(..........2621 kk

263

)3( 221 k

k

)2......(7827 21 kk

262 2 k242 k

xxQ

242 2

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.

Substituting x 1 and Q 26 into the equation,

Substituting x 3 and Q 26 into the equation,

(2) (1):

Substituting k1 2 into (1),

(a) Since Q partly varies directly as x2 and partly varies inversely as x,

we have , where k1 and k2 are non-zero constants.

26k1 52 k1 2

x

kxkQ 22

1

P. 26

Example 13.13T


6

24)6(2 2 Q

76

2024

2 2 x

x

xx 20242 3 012103 xx

x 2 is a factor of P(x).

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,

(c) When Q 20,

Let P(x) x3 10x 12.

P(2) 23 10(2) 12 0

0)62)(2( 2 xxx

By long division,

71or 2 x

62

126

21 6

4 2

21 102

2

21 1002

2

2

2

23

23

xx

x

x

xx

xx

xx

xxxx

P. 27

Example 13.14T


Substituting N 500 and C 15.5 500 into the equation,

)1........(5007750 21 kk

The total cost $C of printing a magazine is partly constant andpartly 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 1200 and C 8.5 1200 into the equation,

21 5005005.15 kk

)2......(1200200 10 21 kk 21 120012005.8 kk

P. 28

Example 13.14T


(b) When N 1000, )1000(5.36000 C

The total cost is $9500. Cost of printing per copy

1000

9500$

5.9$

The total cost $C of printing a magazine is partly constant andpartly 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:

)2........(120020010

)1...(..........5007750

21

21

kk

kk

(2) (1): 2450 700k2

k2 3.5Substituting k2 3.5 into (1),

7750 k1 500(3.5)k1 6000

C 6000 3.5N

9500

P. 29

Example 13.15T

4.18)4( 221 kk

6.31)7( 221 kk


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,

Substituting s 7 and C 31.6 into the equation,

)1......(4.1816 21 kk

)2......(6.3149 21 kk

(2) (1): 33k2 13.2 k2 0.4Substituting k2 0.4 into (1),

24.012 sC

18.4 k1 16(0.4) k1 12

P. 30

Example 13.15T

cost )64.012$( 24.26$

Selling price $[26.4 (1 25%)]


Solution:(b) For a cube with sides of 6 cm,

$33

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?

P. 31

13.1 Direct Variations

Variation describes a relation between two changing quantities and we can use an equation to express the relation.

Chapter Chapter SummarySummary

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. 32

13.2 Inverse Variations


xy

1

x

ky

Both statements ‘y varies inversely as x’ and ‘y is inversely

proportional to x’ mean that xy k or , where k is a non-zero

constant. Symbolically, we write .

P. 33

13.3 Joint Variations


If one variable z varies jointly as two (say x and y) or moreother variables (either directly or inversely), it is a joint variation. Symbolically, if z varies jointly as x and y, we write z xy.

P. 34

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.

1313 13.1direct variations 13.2inverse variations 13.3joint variations chapter summary case study...

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