linear law “transformation” of non-linear relationships into linear relationships
TRANSCRIPT
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Linear Law
“Transformation” of non-linear relationships into linear
relationships
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How it works22 3y x Quadratic Curve: non-linear!
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Transforming to linear relationship
22 3y x
cmXY
Linear ?
General equation of linear relationship:
22 3y x −3
y
x2
Plot y vs x2
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yx
2
3 31
2
xy
3
y
1/x
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baxy
2
1
baxy
bax
y
2
2
1
1
1
cmXY
Plot (1/y) vs x2 m = a, c = b
Example 1
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cmXY
Plot xy vs x2 m = a, c = b
Example 2
x
baxy
baxxy 2
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cmXY
m = b, c = a
Example 3
xbx
ay
xbxaxy
Plot xy vs xx
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cmXY
Plot lg y vs x m = lg b, c = lg a
Q9
xaby
bxay
bay
abyx
x
lglglg
lglglg
)lg(lg
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ya bx
x
bxaxy bx
ay
Plot xy vs x Plot y vs
Grad = b, xy-intercept = a Grad = a, y-intercept = b
x
1
Q16
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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
Q17
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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
Q18
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cmXY
Plot lg y vs lg x m = b, c = - lg a
Q19
bxay
axby
xbya
xay b
lglglg
lglglg
)lg()lg(
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cmXY
m = p, c = - q
Q20
qxpxe y 2
qpxx
e y
Plot vs xx
e y
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Express y in terms of x? y = ??
(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
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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)y
1
x
1
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(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
Q1
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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
Q2
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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.
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.
(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 vertical intercept.
(1)
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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
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1
1 1( ) ( ) 1
1 1 1( )
a b
y x
a by x
b
y a x a
1, intercept 0.25
10.25
4
Fr graphy
aa
,
0.7 0.250.75
0.6 0
0.75
3
Fr graph
m
b
ab
1
y
1
x0.2 0.4 0.6 0.8 1.0
0.2
0.40.6
0.8
1.0
(0.6, 0.7)
(0,0.25)
1 1 1b
y a x a
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(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.
1
plot vs with gradient and vertical intercept
a b
y x
y ya b y y b a
x xy
y b ax
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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-x) -intercept.
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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
42.5q
42.5 20
0 151.5
p
5 10 15 20 25
5
1015
20
25
2x
xy
303540
45 (0, 42.5)
)20,15(
qpxxy 2
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Q2 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 (x + y) - intercept.
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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
qxp
qyx
3
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Q3 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 ─ x2) -intercept.
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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
baxxy 2
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Identify the incorrect readings/ outliers!!1
y
1
xy x
2x
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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
2 ( 2)y m x c
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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
2 ( 2)y m x c
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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 0.66
It is known that x and y are related by an equation of the form a
yx 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.
(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
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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 1 bx
y a a
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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 1 bx
y a a
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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
1 1 bx
y a a
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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
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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
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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.
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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(
lg lg lgy C x a
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(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.
lg lg lgy C x a