Chapter 2 Linear Law

2.1 Lines of Best FitA. Drawing lines of best fit

Example 1

V

I 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1

3.96 – 1.08 = 2.88

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

Touch at y-axis

B. Equations of lines of best fitExample 2

1 2 3 4 5 6 7 8 9 10 11 12 130

2

4

6

8

10

28

26

24

22

2018

16

14

12

y

x

Not a line of best fit cause

-Number of points above

and below line not balance

1 2 3 4 5 6 7 8 9 10 11 12 130

24

6

810

28

262422

2018

16

14

12

y

x y = m x + c

m = gradient

c = y - interceptFrom the graph,

c = 4,

Any 2 points on the line,

(0,4)

(6,16)

06

416

m

6

12

2

y = m x + cy = 2 x + 4

C. Determining the values of variables

(a)From lines of best fit

(b) From the equations of lines of best fit

(a)From lines of best fitExample 3

X=1.4

(a) y = 3

(b) y = 0.2

y=4.9

(c) x = 2.4

(b) From the equations of lines of best fit

Example 4

(b) From the equations of lines of best fit

Example 4

y = m x + c

8.02.0

95.16

m

6.0

5.7

5.12

y = m x + c

y = -12.5 x + 19

When m=-12.5 and (0.8, 9.0)

9 = -12.5(0.8) + c

9 = -10 + c19= c

y = -12.5 x + 19

(i) y = -12.5 (0.7) + 19

= 10.25

(ii) 22 = -12.5 x + 19

12.5 x = 19 – 22 = -3 x = -0.24

2.2 Applications of Linear Law

to Non-linear Relation

B. Values of constants of non-linear relations

A. Reducing non-linear relations to the linear form

62loglog xy 6log2loglog xy xy log62loglog

2loglog6log xy

Y = log y

X = log x

m = 6

c = log 2 Y = m X + c

baxy bxay logloglog

xbay logloglog

axby logloglog

Y = log y

X = log x

m = b

c = log a Y = m X + c

Exercise 2.2

1e

mxy

myx logloglog Y = log y

X = log x

m = - 1

c = log m

Y = m X + c

Exercise 2.2

1e

xmy logloglog

mxy logloglog

xb

ay

xbay logloglog

bxay logloglog

axby log)(loglog

Y = log y

X = x

m = - log b

c = log a

Y = m X + c

Exercise 2.2

1e

(a)Given lines of best fitExample 6

14

3550

m

bm

1

cmXY

bx

a

b

xy

bx

ax

b

xxxy

b

a

b

xxy

2

b

ax

bxy 21

5

c )1(535

30c

b

ac ,

From equation,From graph,

By comparing,

51

b

m

5

1b

ba 30

)5

1(30

6

Only unknown

so, each term ÷ b

X x

Y = m X + c

30b

ac

x

baxy

2x

ba

x

y

ax

bx

y )

1(

2

Y = m X + c

m = b, c = a

From equation, From graph, By comparing,

13

37

m

2m

Y = m X + c

3 = 2(1) + c

c = 1

b = 2

a = 1

xyqypx

y

xy

y

qy

y

px

xqy

px

Y = m X + c

From equation, From graph, By comparing,

8.03.0

6.085.0

m

5.0m

Y = m X + c

0.6 = 0.5(0.8) + c

c = 0.2

px

x

px

q

y

1÷ px

pxp

q

y

1)

1(

1

pxp

q

y

1)

1(

1

p

qm

pc

1

5

12.0

1

p

5p

5.0p

q

5.05

q

5.2q

tv tkt

hv

ttkt

thtv

kthtv

hkttv

Y = m X + c

m = k, c = h

From equation,

From graph,

By comparing,

14

18 54

52

m

2.2m

Y = m X + c

c = - 0.4

k = 2.2

h = -0.4

C )1(2.25

41

tt

t

mnxy loglog mxny logloglog

xmny logloglog

From equation, From graph, By comparing,

3.01

73.098.0

m

357.0m

Y = m X + c

c = 0.623

mgradient

36.0m

623.0log n

623.010n

198.4n

nxmy logloglog

Y = m X + c

nerceptY logint

36.0m

0.98 = (0.357)(1) + c

(a)From dataExample 7

x

kxhy

x

kxhy

x

kxhy

x

xkxxhxy

(a)

x

khxxy

xXxyY

gradient = h

Y-intercept = k

(b) khxxy

x 1 2 3 4 5

xy 1 4.002 6.599 10.000 13.193

111 283.2 381.3 45 59.5 Point (2, 4.002) cannot be seem accurate on (2, 4).

4.002 must be a little bit above 4.Mark “x” on every point when plotting the graph.

They should be clear and big enough for examining.Use a long ruler which is luminous to construct a smooth continuous straight line.

The intercept on the vertical axis (y-axis) must be shown.

Get the value of the gradient and the intercept from the graph, and not from the table directly.

Label on both axes.

Round off to 4 sf or 4 dp in the table.

xbyax 22

xbyax 22

x1)(

2

x

ybax

1)(2

axx

yb

bb

xb

a

x

y 12

x

y2x

Example 8