2 Linear Law

Chapter 2

Linear Law

2.1 Lines of Best Fit A. Drawing lines of best fit Example 1

V 4.0

3.96 – 1.08 = 2.88

3.5 3.0 2.5 2.0 1.5 1.0 0.5 Touch at y-axis 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

I

B. Equations of lines of best fit Example 2

y 28 26 24 22

Not a line of best fit cause

20 18

-Number of points above

16

and below line not balance

14 12 10 8 6 4 2 0

x 1

2

3

4

5

6

7

8

9

10

11

12

13

y

Any 2 points on the line,

28 26 24 22 20 18 16

(6,16)

14 12 10 8 6

(0,4)

4 2 0

1

2

x 3

4

5

6

7

8

9

10

11

From the graph,

c = 4,

16 − 4 m= 6−0

12 = 6

=2

12

y=mx+c

13

m = gradient c = y - intercept

y=mx+c

y=2x+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 fit Example 3

y=4.9

(a) y = 3 (b) y = 0.2 (c) x = 2.4

X=1.4

(b) From the equations of lines of best fit Example 4

(b) From the equations of lines of best fit Example 4

y=mx+c 16.5 − 9 m= 0.2 − 0.8

7.5 = − 0.6

= −12.5

When m=-12.5 and (0.8, 9.0)

9 = -12.5(0.8) + c

9 = -10 + c 19= c

y= m x+c y = -12.5 x + 19

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

A. Reducing non-linear relations to the linear form B. Values of constants of non-linear relations

6

log y = log 2 x 6 log y = log 2 + log x log y = log 2 + 6 log x

Y = log y

log y = 6 log x + log 2

m=6

Y

=

m X

+ c

X = log x

c = log 2

Exercise 2.2 1e

y =ax

b

log y = log a + log x

b

log y = log a + b log x

log y = b log x + log a Y

=

m X

+ c

Y = log y X = log x m=b c = log a

Exercise 2.2 1e

xy =m

log x + log y = log m

Y = log y

log y = log m − log x log y = − log x + log m Y

=

m X

+ c

X = log x m=-1 c = log m

Exercise 2.2 1e

y =

a x

b x log y = log a − log b log y = log a − x log b Y

=

m X

+ c

Y = log y X= x

log y = −(log b) x + log a m = - log b c = log a

(a)Given lines of best fit Example 6

Only unknown so, each term ÷ b

By comparing, From equation,

x a y = + b bx X x

xx ax xy = + b bx

From graph,

50 −35 m= 4 −1

=5

x2 a xy = + b b

Y = mX + c

1 2 a xy = x + b b

35 = 5(1) + c

Y

=

m X

1 m= b

+

c

a ,c = b

c = 30

1 m= =5 b 1 b= 5 a c = = 30 b

a =30b 1 = 30( ) 5

=6

From equation,

From graph,

b y =ax + x

7 −3 m= 3 −1

b=2

m =2

a=1

y b =a + 2 x x y 1 =b( 2 ) +a x x Y

= m = b,

m X c=a

+

c

Y= m X + c 3 = 2(1) + c c=1

By comparing,

Y

= m= −

From equation,

px +qy =xy px qy xy + = y y y

px +q =x y ÷ px 1 q x + = y px px 1 q 1 1 + ( ) = y p x p

m X + c

q p

1 c= p

From graph,

0.85 −0.6 m= 0.3 −0.8

By comparing,

1 1 = 0 .2 = p 5

m = 0.5

p=5

Y= m X + c

q − = 0 .5 p q − = 0.5 5

0.6 = 0.5(0.8) + c c = 0.2

1 q 1 1 =− ( ) + y p x p

q = − 2.5

v =

v t

h t

+ k

t

From equation, By comparing,

v t =

h t t

+k

v

t =h +kt

v

t =kt +h

Y

=

t

m X + c

m = k,

c=h

t

From graph,

m=

8 52 −1 54 4 −1

m = 2.2 Y= m X + c 4 1 = 2.2(1) +C 5 c = - 0.4

k = 2.2 h = -0.4

From equation,

log y = log nx m

From graph,

log y = log n + log x m

0.98 − 0.73 m= 1 − 0.3

log y = log n + m log x

m = 0.357

log y = m log x + log n

m ≈ 0.36

Y= m X + c

gradient = m Y − int ercept = log n

Y= m X + c 0.98 = (0.357)(1) + c c = 0.623

By comparing,

m = 0.36 log n = 0.623

n = 10 0.623 n = 4.198

•From data Example 7

y =h

x +

k x

y =h

x +

k x

(a)

y =h x +

k x

× x

y x =h x x+ y

x =hx +k

Y =y x X =x gradient = h Y-intercept = k

k x x

(b)

x =hx +k

y x y



x

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

1

2

3

4

5

1

4.002

6.599

10.000

13.193

Label on1 both 1 = 1 axes. 2.83 2 3.81 3 5.9 5 5 4 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.

ax 2 +by 2 =x

2

2

ax + by = x

÷x

2

y ax + b( ) = 1 x 2

÷b

y b( ) = −ax +1 x 2 y2 a 1 =− x + x b b

Example 8