Reduction of non-linear laws to linear form
Exercise 60 Further problems on reducing non-linear laws to linear form
(Answers on page 277)
In Problems 1 to 5, x and y are two related variables and all other letters denote constants. For the stated laws to be verified it is necessary to plot graphs of the variables in a modified form. State for each (a) what should be plotted on the vertical axis, (b) what should be plotted on the horizontal axis, (c) the gradient and (d) the vertical axis intercept.
1. y= d + cx2 2. y− a = b√ x 3. y− e =f
x 4. y− cx = bx2
5. y=a x+ bx
6. In an experiment the resistance of wire is measured for wires of different diameters with the following results.
R ohms 1.64 1.14 0.89 0.76 0.63 d mm 1.10 1.42 1.75 2.04 2.56 It is thought that R is related to d by the law R= (a/d2)+ b, where a and b are constants. Verify this
and find the approximate values for a and b. Determine the cross-sectional area needed for a resistance reading of 0.50 ohms.
7. Corresponding experimental values of two quantities x and y are given below.
x 1.5 3.0 4.5 6.0 7.5 9.0
y 11.5 25.0 47.5 79.0 119.5 169.0 By plotting a suitable graph verify that y and x are con-nected by a law of the form y= kx2+ c, where k and c are constants. Determine the law of the graph and hence find the value of x when y is 60.0
8. Experimental results of the safe load L kN, applied to girders of varying spans, d m, are shown below.
Span, d m 2.0 2.8 3.6 4.2 4.8
Load, L kN 475 339 264 226 198
It is believed that the relationship between load and span is L= c/d, where c is a constant. Determine (a) the value of constant c and (b) the safe load for a span of 3.0 m.
9. The following results give corresponding values of two quantities x and y which are believed to be related by a law of the form y= ax2+ bx where a and b are constants.
y 33.86 55.54 72.80 84.10 111.4 168.1
x 3.4 5.2 6.5 7.3 9.1 12.4
Verify the law and determine approximate values of a and b.
Hence determine (i) the value of y when x is 8.0 and (ii) the value of x when y is 146.5.
16.2 Determination of law involving logarithms
Examples of reduction of equations to linear form involving logarithms include:
(i) y= axn
Taking logarithms to a base of 10 of both sides gives:
lg y= lg (axn)= lg a + lg xn
i.e. lg y= n lg x + lg a by the laws of logarithms which compares with Y= mX + c
and shows that lg y is plotted vertically against lg x hori-zontally to produce a straight line graph of gradient n and lg y-axis intercept lg a
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(ii) y= abx
Taking logarithms to a base of 10 of the both sides gives:
lg y= lg(abx) i.e. lg y= lg a + lg bx
i.e. lg y= x lg b + lg a by the laws of logarithms or lg y= ( lg b)x + lg a
which compares with Y= mX + c
and shows that lg y is plotted vertically against x horizontally to produce a straight line graph of gradient lg b and lg y-axis intercept lg a
(iii) y= aebx
Taking logarithms to a base of e of both sides gives:
ln y= ln(aebx) i.e. ln y= ln a + ln ebx i.e. ln y= ln a + bx ln e
i.e. ln y= bx + ln a since ln e = 1 which compares with Y= mX + c
and shows that ln y is plotted vertically against x horizontally to produce a straight line graph of gradient b and ln y-axis intercept ln a
Problem 4. The current flowing in, and the power dissi-pated by, a resistor are measured experimentally for various values and the results are as shown below.
Current, I amperes 2.2 3.6 4.1 5.6 6.8 Power, P watts 116 311 403 753 1110 Show that the law relating current and power is of the form P= RIn, where R and n are constants, and determine the law.
Taking logarithms to a base of 10 of both sides of P= RIngives:
lg P= lg(RIn)= lg R + lg In= lg R + n lg I by the laws of logarithms,
i.e. lg P= n lg I + lg R, which is of the form Y = mX + c, show-ing that lg P is to be plotted vertically against lg I horizontally.
A table of values for lg I and lg P is drawn up as shown below.
I 2.2 3.6 4.1 5.6 6.8
lg I 0.342 0.556 0.613 0.748 0.833
P 116 311 403 753 1110
lg P 2.064 2.493 2.605 2.877 3.045
A graph of lg P against lg I is shown in Fig. 16.4 and since a straight line results the law P= RInis verified.
3.0 2.98
2.78
2.5 lg P
2.18
2.0
0.30 0.40
C B
D A
0.50 0.60 lg l
0.80
0.70 0.90
Fig. 16.4
Gradient of straight line, n= AB
BC = 2.98− 2.18 0.8− 0.4 = 0.80
0.4 = 2
It is not possible to determine the vertical axis intercept on sight since the horizontal axis scale does not start at zero. Select-ing any point from the graph, say point D, where lg I= 0.70 and lg P= 2.78, and substituting values into lg P = n lg I + lg R gives
2.78= (2)(0.70) + lg R from which lg R= 2.78 − 1.40 = 1.38
Hence R= antilog 1.38 (= 101.38)= 24.0 Hence the law of the graph is P= 24.0 I2
Problem 5. The periodic time, T , of oscillation of a pen-dulum is believed to be related to its length, l, by a law of the form T= kln, where k and n are constants. Values of T were measured for various lengths of the pendulum and the results are as shown below.
Periodic time, T s 1.0 1.3 1.5 1.8 2.0 2.3 Length, l m 0.25 0.42 0.56 0.81 1.0 1.32 Show that the law is true and determine the approximate values of k and n. Hence find the periodic time when the length of the pendulum is 0.75 m.
From para (i), if T= klnthen lg T= n lg l + lg k and comparing with
Y = mX + c Property of Mr. Alphonce Kimutai Kirui
Reduction of non-linear laws to linear form 121
shows that lg T is plotted vertically against lg l horizontally.
A table of values for lg T and lg l is drawn up as shown below.
T 1.0 1.3 1.5 1.8 2.0 2.3
lg T 0 0.114 0.176 0.255 0.301 0.362
l 0.25 0.42 0.56 0.81 1.0 1.32
lg l −0.602 −0.377 −0.252 −0.092 0 0.121 A graph of lg T against lg l is shown in Fig. 16.5 and the law T= klnis true since a straight line results.
0.40
lg T
lg I
0.30
0.20
0.10
⫺0.10 0 0.10 0.20
⫺0.20
⫺0.30
⫺0.40
⫺0.50
⫺0.60
0.25
0.05 A
B C
Fig. 16.5
From the graph, gradient of straight line,
n= AB
BC = 0.25− 0.05
−0.10 − (−0.50) = 0.20 0.40= 1
2 Vertical axis intercept, lg k= 0.30
Hence k= antilog 0.30 (= 100.30)= 2.0
Hence the law of the graph is T= 2.0 l1/2 or T= 2.0√ l When length l= 0.75 m then T = 2.0√
0.75= 1.73 s
Problem 6. Quantities x and y are believed to be related by a law of the form y= abx, where a and b are constants.
Values of x and corresponding values of y are:
x 0 0.6 1.2 1.8 2.4 3.0
y 5.0 9.67 18.7 36.1 69.8 135.0 Verify the law and determine the approximate values of a and b. Hence determine (a) the value of y when x is 2.1 and (b) the value of x when y is 100
From para (ii), if y= abxthen lg y= (lgb)x + lg a and comparing with
Y = mX + c
shows that lg y is plotted vertically and x horizontally.
Another table is drawn up as shown below.
x 0 0.6 1.2 1.8 2.4 3.0
y 5.0 9.67 18.7 36.1 69.8 135.0
lg y 0.70 0.99 1.27 1.56 1.84 2.13
A graph of lg y against x is shown in Fig. 16.6 and since a straight line results, the law y= abxis verified.
2.50
2.13 2.00
1.50
lgy
1.17 1.00
0.70 0.50
0 1.0 2.0 3.0 x
A
B C
Fig. 16.6
Gradient of straight line,
lg b= AB
BC = 2.13− 1.17 3.0− 1.0 = 0.96
2.0 = 0.48
Hence b= antilog 0.48 (= 100.48)= 3.0, correct to 2 significant figures.
Vertical axis intercept, lg a= 0.70, from which a= antilog 0.70 (= 100.70)
= 5.0, correct to 2 significant figures Hence the law of the graph is y= 5.0(3.0)x Telephone:+254728801352
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(a) When x= 2.1, y = 5.0(3.0)2.1= 50.2 (b) When y= 100, 100 = 5.0(3.0)x,
from which 100/5.0= (3.0)x,
i.e. 20= (3.0)x
Taking logarithms of both sides gives lg 20= lg(3.0)x= x lg 3.0 Hence x= lg 20
lg 3.0=1.3010 0.4771= 2.73
Problem 7. The current i mA flowing in a capacitor which is being discharged varies with time t ms as shown below.
i mA 203 61.14 22.49 6.13 2.49 0.615
t ms 100 160 210 275 320 390
Show that these results are related by a law of the form i= Iet/T, where I and T are constants. Determine the approximate values of I and T .
Taking Napierian logarithms of both sides of i= Iet/Tgives ln i= ln (Iet/T)= ln I + ln et/T
i.e. ln i= ln I + t
T (since ln e= 1) or ln i=
1 T
t+ ln I
which compares with y= mx + c, showing that ln i is plotted vertically against t horizontally. (For methods of evaluating Napierian logarithms see Chapter 15.) Another table of values is drawn up as shown below.
t 100 160 210 275 320 390
i 203 61.14 22.49 6.13 2.49 0.615
ln i 5.31 4.11 3.11 1.81 0.91 −0.49 A graph of ln i against t is shown in Fig. 16.7 and since a straight line results the law i= Iet/Tis verified.
Gradient of straight line, 1
T = AB
BC = 5.30− 1.30 100− 300 = 4.0
−200= −0.02 Hence T= 1
−0.02= −50
Selecting any point on the graph, say point D, where t= 200 and ln i= 3.31, and substituting into ln i =
1 T
t+ ln I gives
3.31= −1
50(200)+ ln I from which ln I= 3.31 + 4.0 = 7.31
5.0 A
D (200, 3.31)
B C
4.0 3.31
lni
3.0
2.0 1.30 1.0
0
⫺1.0
100 200 300 400 t (ms)
Fig. 16.7
and I= antilog 7.31 (= e7.31)= 1495 or 1500 correct to 3 signifi-cant figures.
Hence the law of the graph is i= 1500e−t/50
Now try the following exercise
Exercise 61 Further problems on reducing non-linear