Linear Law “Transformation” of non- linear relationships to linear relationships
Mar 16, 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
3 312
xy
3
y
1/x
y a bxx
bxaxy bxay
Plot xy vs x Plot y vs
Grad = b, y-intercept = a Grad = a, y-intercept = b
x1
y ab x 3
bbxaybxay
bayaby
x
x
lg3lglglglg)3(lglg
lglglg)lg(lg
3
3
cmXY
Plot lg y vs x m = lg b, c = lg a + 3 lg b
y ab x 4 cmXY
bxayaby
abyx
x
lglg)4lg()lg()4lg(
4
Plot lg (y – 4) vs x m = lg b, c = lg a
baxy
21
baxy
baxy
2
2
11
1
cmXY
Plot (1/y) vs x2 m = a, c = b
Q1a
cmXY
Plot xy vs x2 m = a, c = b
Q1b
xbaxy
baxxy 2
cmXY
m = b, c = a
Q1c
xbxay
xbxaxy
Plot xy vs xx
cmXY
Plot lg y vs x m = lg b, c = lg a
Q1d
xaby
bxaybay
abyx
x
lglglglglglg)lg(lg
cmXY
Plot lg y vs lg x m = b, c = - lg a
Q1e
bxay
axbyxbya
xay b
lglglglglglg)lg()lg(
cmXY
m = p, c = - q
Q1f
qxpxe y 2
qpxxe y
Plot vs xxe y
Express y in terms of x? y = ??x
(0,1)
(4,9)
a) y
x2
cxmy )( 2
20419
12
12
xxyym
121)0(21)1,0(
)(2
2
2
xycc
Atcxy
12)0(21
)()(
2
2
12
1
11
xyxy
xxmyyxxmyy
Express y in terms of x?
cx
my
11
21
4002
12
12
xxyym
2211
212112
)0(212
)2,0(
1211
xy
xyc
c
At
cxy
2211
)01(2121
)1(1)(
11
11
xy
xy
xx
myy
xxmyy
(4,0)
(0,2)
b)
y1
x1
(5, 9)
(2, 3)
x + 1
lg y
22539
12
12
xxyym
cxmy 1lg
12
10
11
11
1012log
12lg)1(23lg
)21(23lg)1(2lg
)(
xyxy
xyxyxy
xxyyxxmyy
Q2a
31339
12
12
xxyym
cxmy )]1[ln(ln
3
3
11
11
)1()1ln(ln
)1ln(3ln3)1ln(33ln
]1)1[ln(33ln])1[ln(3ln
)(
xyxy
xyxyxy
xxyyxxmyy
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 5y 1 1.6 2 2.28 2.5
It is known that x and y are related by the equation 1a by x
.
(i)Explain how a straight-line graph of 1y
against 1x
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 1y
against 1x
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 by x
a by x
by 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 5y 1 1.6 2 2.28 2.5
1/x 1 0.5 0.33 0.25 0.21/y 1 0.625 0.5 0.44 0.4
Choose appropriate scales1y
1x0.2 0.4 0.6 0.8 1.0
0.2
0.40.60.8
1.0
1y
1x0.2 0.4 0.6 0.8 1.0
0.2
0.40.60.8
1.0
1
1 1( ) ( ) 1
1 1 1( )
a by x
a by x
by a x a
, int 0.21 0.2
5
Fr graph y ercept
aa
,0.7 0.2 0.833330.6 0
0.83333
5 0.833334.166 4.17(3 )
Fr graph
m
bab
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 by x
y ya b y y b ax x
yplot 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 5y 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 5y 41.5 38.0 31.5 22.0 9.5
x2 1 4 9 16 25y ─ x 40.5 36.0 28.5 18.0 4.5
43q
5 10 15 20 25
5
101520
25
2x
xy
30354045
53.11502043
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 5y 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 5y 5.71 6.38 9.10 14.20 20.49
x 1 2 3 4 5y ─ x2 4.71 2.38 0.10 -1.80 -4.51
1 2 3 4 5
1
234
5
2xy
x
-5
-4-3-2
-1
5.7b
5.2
5.23005.7
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 2.5y 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.
qxpqyx
pqxqyx
3
3Plot (x + y) against x3, (q/p) represents the gradient and q represents the y-intercept.
qxpqyx
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 15.625x+y 1.56 2 3.19 5.5 9.31
5 10 15 20
2
468
10
yx
3x
)10,20(
)6,10(
8.1q
5.44.08.1
4.08.1
4.01020610
p
p
pq
42.22.62.6,graphFr648.102.2 3
yyxxx
Identify the incorrect readings/ outliers!!1y
1x
y x
2x
x 1 2 3 4 5y 2.65 3.00 3.32 3.71 3.87
x+2 3 4 5 6 7y2 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 12.83.58
yy
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)1.99
0.492
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
ayx b
, where a and b are constants. By plotting
1y
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 =
109.
(iii) By inserting another straight line to your graph, find the value of x and of y which satisfy the simultaneous equations
ayx b
and 1015 12
yx
x 0.5 1 1.5 2.0 2.5y 1.20 1.00 0.86 0.70 0.66
x 0.5 1 1.5 2.0 2.51/y 0.83 1 1.16 1.43 1.52
ayx b
1
1 1
x by a
bxy a a
1y
x0.5 1.0 1.5 2.0 2.5
0.2
0.40.60.81.01.21.4
1.6
13.2
68.0
bab
)68.0,0(
)4.1,25.2(
13.3025.268.04.11
aa
1y
x0.5 1.0 1.5 2.0 2.5
0.2
0.40.60.81.01.21.4
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 2.5y 1.20 1.00 0.86 0.70 0.66
x 0.5 1 1.5 2.0 2.51/y 0.83 1 1.16 1.43 1.52
1 1.35
0.741yy
1y
x0.5 1.0 1.5 2.0 2.5
0.2
0.40.60.81.01.21.4
1.6
Estimate the value of x when y = 109
10 1 0.99
yy
1When 0.9, 0.75xy
1 0.319 0.68xy
0.9 0.319 0.680.690
xx
ayx b
and 1015 12
yx
1 15 1210
1 1.5 1.2
xy
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
1 1.5 1.2
y y
xy
1.20 1.5 1.2 0.81.5
x x
1y
x0.5 1.0 1.5 2.0 2.5
0.2
0.40.60.81.01.21.4
1.6
Vertical intercept (0, -1.2)
Horizontal intercept (0.8, 0)
ayx b
-0.2-0.4-0.6-0.8-1.0
-1.2
1015 12
yx
At point of intersection,1 1.3 0.769
1.8
yyx
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 7y 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
axCyCay x
lglglglglg
x 1 2 3 4 5 6 7y 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
0.40.60.81.01.21.4
1.6
6 7
1.8
2.0
(i) abnormal reading: y = 25.10 Correct value should be
lg 1.2819.05
yy
(ii) lg 2.0100
CC
2.0 0.4(ii) lg0 6.5
1.76
a
a
)0.2,0(
)4.0,5.6(
(iii) 1lg 0
8.3
yy
x
lg y
x1 2 3 4 5
0.2
0.40.60.81.01.21.4
1.6
6 7
1.8
2.0
8 9
estimate the value of x when y = 1.