2 Linear Law
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Transcript of 2 Linear Law
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