Post on 20-Jan-2016
Linear Law
“Transformation” of non-linear relationships to linear
relationships
How it works
y x 2 32Quadratic Curve: non-linear!
Graph of y vs x
Transforming to linear relationship
y x 2 32
cmXY
Linear ?
General equation of linear relationship:
y x 2 323
y
x2
Plot y vs x2
yx
2
3 31
2
xy
3
y
1/x
ya bx
x
bxaxy bx
ay
Plot xy vs x Plot y vs
Grad = b, y-intercept = a Grad = a, y-intercept = b
x
1
y ab x 3
bbxay
bxay
bay
abyx
x
lg3lglglg
lg)3(lglg
lglglg
)lg(lg3
3
cmXY
Plot lg y vs x m = lg b, c = lg a + 3 lg b
y ab x 4 cmXY
bxay
aby
abyx
x
lglg)4lg(
)lg()4lg(
4
Plot lg (y – 4) vs x m = lg b, c = lg a
baxy
2
1
baxy
bax
y
2
2
1
1
1
cmXY
Plot (1/y) vs x2 m = a, c = b
Q1a
cmXY
Plot xy vs x2 m = a, c = b
Q1b
x
baxy
baxxy 2
cmXY
m = b, c = a
Q1c
xbx
ay
xbxaxy
Plot xy vs xx
cmXY
Plot lg y vs x m = lg b, c = lg a
Q1d
xaby
bxay
bay
abyx
x
lglglg
lglglg
)lg(lg
cmXY
Plot lg y vs lg x m = b, c = - lg a
Q1e
bxay
axby
xbya
xay b
lglglg
lglglg
)lg()lg(
cmXY
m = p, c = - q
Q1f
qxpxe y 2
qpxx
e y
Plot vs xx
e y
Express y in terms of x? y = ??x
(0,1)
(4,9)
a) y
x2
cxmy )( 2
204
19
12
12
xx
yym
121
)0(21
)1,0(
)(2
2
2
xyc
c
At
cxy
12
)0(21
)(
)(
2
2
1
2
1
11
xy
xy
xxmyy
xxmyy
Express y in terms of x?
cx
my
11
2
1
40
02
12
12
xx
yym
22
11
21
2
112
)0(2
12
)2,0(
1
2
11
xy
xyc
c
At
cxy
22
11
)01
(2
12
1
)1
(1
)(
11
11
xy
xy
xx
myy
xxmyy
(4,0)
(0,2)
b)
y
1
x
1
(5, 9)
(2, 3)
x + 1
lg y
225
39
12
12
xx
yym
cxmy 1lg
12
10
11
11
10
12log
12lg
)1(23lg
)21(23lg
)1(2lg
)(
xy
xy
xy
xy
xy
xxyy
xxmyy
Q2a
313
39
12
12
xx
yym
cxmy )]1[ln(ln
3
3
11
11
)1(
)1ln(ln
)1ln(3ln
3)1ln(33ln
]1)1[ln(33ln
])1[ln(3ln
)(
xy
xy
xy
xy
xy
xxyy
xxmyy
Q (3, 9)
P (1, 3)ln (x – 1)
ln y
Q2b
The following table gives values of y corresponding to some value of x.
x 1 2 3 4 5
y 1 1.6 2 2.28 2.5
It is known that x and y are related by the equation 1a b
y x
.
(i)Explain how a straight-line graph of 1
y against
1
x
can be drawn to represent the given equation and draw it for the given data. Use this graph to estimate the value of a and of b.
(ii) Express the given equation in another form suitable for a straight-line graph to be drawn. State the variables whose values should be plotted.
.
(i)Explain how a straight-line graph of 1
y against
1
x
can be drawn to represent the given equation and draw it for the given data. Use this graph to estimate the value of a and of b.
(i) 1
1 1( ) ( ) 1
1 1 1( )
a b
y x
a by x
b
y a x a
In order to plot 1/y against 1/x, we need to arrange the equation into (1). b/a represents the gradient and 1/a represents the y-intercept.
(1)
x 1 2 3 4 5
y 1 1.6 2 2.28 2.5
1/x 1 0.5 0.33 0.25 0.2
1/y 1 0.625 0.5 0.44 0.4
Choose appropriate scales1
y
1
x0.2 0.4 0.6 0.8 1.0
0.2
0.40.6
0.8
1.0
1
y
1
x0.2 0.4 0.6 0.8 1.0
0.2
0.40.6
0.8
1.0
1
1 1( ) ( ) 1
1 1 1( )
a b
y x
a by x
b
y a x a
, int 0.2
10.2
5
Fr graph y ercept
aa
,
0.7 0.20.83333
0.6 0
0.83333
5 0.83333
4.166 4.17(3 )
Fr graph
m
b
ab
sf
(0.6, 0.7)
(0,0.2)
(ii) Express the given equation in another form suitable for a straight-line graph to be drawn. State the variables whose values should be plotted.
1a b
y x
y ya b y y b a
x xy
plot y vsx
Q1 The data for x and y given in the table below are related by a law of the form y px x q 2
, where p and q are constants.
x 1 2 3 4 5
y 41.5 38.0 31.5 22.0 9.5
By drawing a suitable straight line, find estimates for p and q.
qxpxy 2
qpxxy 2
Plot (y ─ x) against x2, p represents the gradient and q represents the y-intercept.
qpxxy 2 x 1 2 3 4 5
y 41.5 38.0 31.5 22.0 9.5
x2 1 4 9 16 25
y ─ x 40.5 36.0 28.5 18.0 4.5
43q
5 10 15 20 25
5
1015
20
25
2x
xy
303540
45
53.1150
2043
p
)43,0(
)20,15(
Q2 The table below shows experimental values of two variables, x and y. One value of y has been recorded incorrectly.
x 1 2 3 4 5
y 5.71 6.38 9.10 14.20 20.49
It is believed that x and y are related in the form y = x 2 – ax + b, where a and b are constants. Draw a suitable straight-line graph to represent the given data. Use your graph to estimate (i) the value of a and of b, (ii) a value of y to replace the incorrect value.
baxxy 2
Plot (y ─ x2) against x, ─ a represents the gradient and b represents the y-intercept.
baxxy 2 x 1 2 3 4 5
y 5.71 6.38 9.10 14.20 20.49
x 1 2 3 4 5
y ─ x2 4.71 2.38 0.10 -1.80 -4.51
1 2 3 4 5
1
23
4
5
2xy
x
-5
-4-3
-2
-1
5.7b
5.2
5.230
05.7
a
a
)5.7,0(
)0,3(
8.1342.24
readingcorrect 2.2
readingincorrect 8.1
2
2
2
yx
xy
xy
Q3 The table shows the experimental values of two variables x and y which are known to be related by an equation of the form p(x + y – q) = qx 3, where p and q are constants.
x 0.5 1.0 1.5 2.0 2.5
y 1.06 1.00 1.69 3.50 6.81
Draw a suitable straight-line graph to represent the above data. Use your graph to estimate (i) the value of p and of q, (ii)the value of y when x = 2.2.
qxp
qyx
p
qxqyx
3
3
Plot (x + y) against x3, (q/p) represents the gradient and q represents the y-intercept.
qxp
qyx
3 x 0.5 1.0 1.5 2.0 2.5
y 1.06 1.00 1.69 3.50 6.81
x3 0.125 1 3.375 8 15.625
x+y 1.56 2 3.19 5.5 9.31
5 10 15 20
2
46
8
10
yx
3x
)10,20(
)6,10(
8.1q
5.44.0
8.1
4.08.1
4.01020
610
p
p
p
q
42.22.62.6,graphFr
648.102.2 3
yyx
xx
Identify the incorrect readings/ outliers!!1
y
1
xy x
2x
x 1 2 3 4 5
y 2.65 3.00 3.32 3.71 3.87
x+2 3 4 5 6 7
y2 7.02 9.00 11.02 13.76 14.98
2 ( 2)y m x c
2y
2x 1 2 3 4 5
2
4
6
8
10
6 7
12
14
16
? One of the values of y is subject to an abnormally large error
Identify the abnormal reading and estimate its correct value.
abnormal reading: y = 3.71Correct value should be
2 12.8
3.58
y
y
2y
2x 1 2 3 4 5
2
4
6
8
10
6 7
12
14
16
Estimate the value of x when y = 2
22 4y y
2When 4, 2 1.5y x
0.5x
2 1.99( 2) 1y x 22 1.99( 2) 1
3( 2)
1.990.492
x
x
x
2 ( 2)y m x c
Q4 The table below shows the experimental values of two variables x and y. It is known that one value of y has been recorded incorrectly x 0.5 1 1.5 2.0 2.5
y 1.20 1.00 0.86 0.70 0.66It is known that x and y are related by an equation of the form
ay
x b
, where a and b are constants. By plotting 1
y
against x, obtain a straight-line graph to represent the above data. Use your graph to estimate the value of a and of b, giving your answer to the nearest integers.
(i) Use your graph to estimate a value of y to replace the incorrect value.(ii) Find the value of x when y =
10
9.
(iii) By inserting another straight line to your graph, find the value of x and of y which satisfy the simultaneous equations
ay
x b
and
10
15 12y
x
x 0.5 1 1.5 2.0 2.5
y 1.20 1.00 0.86 0.70 0.66
x 0.5 1 1.5 2.0 2.5
1/y 0.83 1 1.16 1.43 1.52
ay
x b
1
1 1
x b
y a
bx
y a a
1
y
x0.5 1.0 1.5 2.0 2.5
0.2
0.40.6
0.8
1.01.2
1.4
1.6
13.2
68.0
ba
b
)68.0,0(
)4.1,25.2(
13.3025.2
68.04.11
aa
1
y
x0.5 1.0 1.5 2.0 2.5
0.2
0.40.6
0.8
1.01.2
1.4
1.6
)68.0,0(
)4.1,25.2(
abnormal reading: y = 0.70Correct value should be
x 0.5 1 1.5 2.0 2.5
y 1.20 1.00 0.86 0.70 0.66
x 0.5 1 1.5 2.0 2.5
1/y 0.83 1 1.16 1.43 1.52
11.35
0.741
y
y
1
y
x0.5 1.0 1.5 2.0 2.5
0.2
0.40.6
0.8
1.01.2
1.4
1.6
Estimate the value of x when y = 10
910 1
0.99
yy
1When 0.9, 0.75x
y
10.319 0.68x
y
0.9 0.319 0.68
0.690
x
x
ay
x b
and
10
15 12y
x
1 15 12
10
11.5 1.2
x
y
xy
Need to draw this and find the point of intersection of the 2 lines
Bear in mind: need to use the same axes as first line!
Vertical intercept (0, -1.2)
Horizontal intercept (0.8, 0)
1 11.5(0) 1.2 1.2
11.5 1.2
y y
xy
1.20 1.5 1.2 0.8
1.5x x
1
y
x0.5 1.0 1.5 2.0 2.5
0.2
0.40.6
0.8
1.01.2
1.4
1.6
Vertical intercept (0, -1.2)
Horizontal intercept (0.8, 0)
ay
x b
-0.2-0.4
-0.6
-0.8
-1.0
-1.2
10
15 12y
x
At point of intersection,
11.3 0.769
1.8
yy
x
Q5 The variables x and y are known to be connected by the equation
xCay
An experiment gave pairs of values of x and y as shown in the table.One of the values of y is subject to an abnormally large error.
x 1 2 3 4 5 6 7
y 56.20 29.90 25.10 8.91 6.31 3.35 1.78
Plot lg y against x and use the graph to
(i) identify the abnormal reading and estimate its correct value.(ii) estimate the value of C and of a.(iii) estimate the value of x when y = 1.
xCay
axCy
Cay x
lglglg
lglg
x 1 2 3 4 5 6 7
y 56.20 29.90 25.10 8.91 6.31 3.35 1.78
lg y 1.75 1.48 1.40 0.95 0.80 0.53 0.25
lg y
x1 2 3 4 5
0.2
0.40.6
0.8
1.0
1.2
1.4
1.6
6 7
1.8
2.0
(i) abnormal reading: y = 25.10 Correct value should be
lg 1.28
19.05
y
y
(ii) lg 2.0
100
C
C
2.0 0.4(ii) lg
0 6.51.76
a
a
)0.2,0(
)4.0,5.6(
(iii) 1
lg 0
8.3
y
y
x
lg y
x1 2 3 4 5
0.2
0.40.6
0.8
1.0
1.2
1.4
1.6
6 7
1.8
2.0
8 9
estimate the value of x when y = 1.