Revision Test 6 : Quadratics, logarithms and exponentials
18.4 Determination of laws involving logarithms
150 Basic Engineering Mathematics
a and b. Determine the cross-sectional area needed for a resistance reading of 0.50 ohms.
7. Corresponding experimental values of two quantitiesxandyare 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 connected 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 ofxwhenyis 60.0
8. Experimental results of the safe load LkN, applied to girders of varying spans,dm, 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, wherec is a constant.
Determine (a) the value of constantcand (b) the safe load for a span of 3.0 m.
9. The following results give corresponding val-ues of two quantities x and y which are believed to be related by a law of the form y=ax2+bx, whereaandbare constants.
x 33.86 55.54 72.80 84.10 111.4 168.1 y 3.4 5.2 6.5 7.3 9.1 12.4 Verify the law and determine approximate val-ues ofaandb. Hence, determine (a) the value ofywhenxis 8.0 and (b) the value ofxwhen yis 146.5
Graphs reducing non-linear laws to linear form 151
and shows thatlgy is plotted vertically against x horizontallyto producea straight line graph of gradient lgband lgy-axis intercept lg a.
See worked Problem 6 to demonstrate how this law is determined.
(c) y=aebx
Taking logarithms to a base of e of both sides gives
lny=ln(aebx)
i.e. lny=lna+lnebx by law (1) i.e. lny=lna+bxlne by law (3) i.e. lny=bx+lna since lne=1 which compares with
Y=m X+c
and shows thatlnyis plotted vertically against x horizontallyto producea straight line graph of gradientband lny-axis intercept ln a.
See worked Problem 7 to demonstrate how this law is determined.
Problem 4. The current flowing in, and the power dissipated 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 formP=R In, whereRandnare constants, and determine the law
Taking logarithms to a base of 10 of both sides of P=R Ingives
lgP=lg(R In)=lgR+lgIn=lgR+nlgI by the laws of logarithms i.e. lgP=nlgI+lgR
which is of the formY =m X+c, showing that lgPis to be plotted vertically against lgI horizontally.
A table of values for lgIand lgPis drawn up as shown below.
I 2.2 3.6 4.1 5.6 6.8
lgI 0.342 0.556 0.613 0.748 0.833
P 116 311 403 753 1110
lgP 2.064 2.493 2.605 2.877 3.045
A graph of lgP against lgI is shown in Figure 18.4 and,since a straight line results, the lawP=RInis verified.
C B
A D
2.0
0.30 0.40 0.50 0.60 0.70 0.80 0.90 2.18
2.5
lg P
lg L 2.78
2.983.0
Figure 18.4
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. Selecting any point from the graph, say point D, where lgI=0.70 and lgP=2.78 and substituting values into
lgP=nlgI+lgR gives 2.78=(2)(0.70)+lgR from which, lgR=2.78−1.40=1.38
Hence, R=antilog 1.38=101.38=24.0 Hence,the law of the graph isP=24.0I2
Problem 5. The periodic time,T, of oscillation of a pendulum is believed to be related to its length,L, by a law of the formT =kLn, wherekandnare constants. Values ofT 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 ofkandn. Hence find the periodic time when the length of the pendulum is 0.75 m
152 Basic Engineering Mathematics
From para (a), page 150, if T =kLn
then lgT =nlgL+lgk
and comparing with Y =m X+c
shows that lgT is plotted vertically against lgL hor-izontally, with gradient n and vertical-axis intercept lgk.
A table of values for lgTand lgLis drawn up as shown below.
T 1.0 1.3 1.5 1.8 2.0 2.3
lgT 0 0.114 0.176 0.255 0.301 0.362 L 0.25 0.42 0.56 0.81 1.0 1.32 lgL −0.602−0.377−0.252−0.092 0 0.121 A graph of lgTagainst lgLis shown in Figure 18.5 and thelawT =kLnis true since a straight line results.
A
C B 20.40 0 20.50
20.60 20.3020.20 0.20
0.10 0.20 0.30
0.25
0.05
lg L
lg T 0.40
20.10 0.10
Figure 18.5
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, lgk=0.30. Hence, k=antilog 0.30=100.30=2.0
Hence, the law of the graph is T =2.0L1/2 or T =2.0√
L.
When lengthL=0.75 m,T=2.0√
0.75=1.73 s
Problem 6. Quantitiesxandyare believed to be related by a law of the formy=abx, whereaandb are constants. The values ofxand corresponding values ofyare
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 ofaandb. Hence determine (a) the value of ywhenxis 2.1 and (b) the value ofxwhenyis 100 From para (b), page 150, if y=abx
then lgy=(lgb)x+lga
and comparing with Y =m X+c
shows that lgyis plotted vertically andx horizontally, with gradient lgband vertical-axis intercept lga.
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 lgy 0.70 0.99 1.27 1.56 1.84 2.13 A graph of lgy against x is shown in Figure 18.6 and,since a straight line results, the lawy=abx is verified.
C
B A
lg y
x 2.50
2.13 2.00
1.50
1.00 1.17
0.50
0 1.0 2.0 3.0
0.70
Figure 18.6
Graphs reducing non-linear laws to linear form 153
Gradient of straight line, lgb= 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, lga=0.70,from which a=antilog 0.70
=100.70=5.0, correct to 2 significant figures.
Hence,the law of the graph isy=5.0(3.0)x (a) Whenx=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=xlg 3.0 Hence,x= lg 20
lg 3.0=1.3010 0.4771=2.73 Problem 7. The currentimA flowing in a capacitor which is being discharged varies with timetms, 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 formi=I et/T, whereI andT are constants.
Determine the approximate values ofI andT Taking Napierian logarithms of both sides of
i=I et/T
gives lni=ln(I et/T)=lnI+lnet/T
=lnI+ t T lne
i.e. lni=lnI+ t
T since lne=1
or lni=
1 T
t+lnI which compares with y=mx+c
showing that lni is plotted vertically against t hor-izontally, with gradient 1T and vertical-axis intercept lnI.
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 lni 5.31 4.11 3.11 1.81 0.91 −0.49 A graph of lni against t is shown in Figure 18.7 and, since a straight line results,the lawi=I et /T is verified.
A
B C
ln i
5.0
4.0
3.0 3.31
2.0
1.0
100 200 300 400 t (ms)
D (200, 3.31)
21.0 1.30
0
Figure 18.7
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 pointD, where t=200 and lni=3.31, and substituting
into lni=
1 T
t+lnI
gives 3.31= − 1
50(200)+lnI from which, lnI =3.31+4.0=7.31 and I=antilog 7.31=e7.31=1495 or1500 correct to 3 significant figures.
Hence,the law of the graph isi=1500e−t /50
154 Basic Engineering Mathematics
Now try the following Practice Exercise
Practice Exercise 71 Determination of law involving logarithms (answers on page 348) In problems 1 to 3,x andyare two related vari-ables 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 horizon-tal axis, (c) the gradient and (d) the vertical axis intercept.
1. y=bax 2. y=kxL 3. y m =enx 4. The luminosity I of a lamp varies with
the applied voltage V and the relationship between I and V is thought to be I =kVn. Experimental results obtained are
I(candelas) 1.92 4.32 9.72
V(volts) 40 60 90
I(candelas) 15.87 23.52 30.72 V(volts) 115 140 160 Verify that the law is true and determine the law of the graph. Also determine the luminosity when 75 V is applied across the lamp.
5. The head of pressure h and the flow veloc-ity v are measured and are believed to be connected by the lawv=ahb, wherea and b are constants. The results are as shown below.
h 10.6 13.4 17.2 24.6 29.3 v 9.77 11.0 12.44 14.88 16.24 Verify that the law is true and determine values ofaandb.
6. Experimental values ofxandyare measured as follows.
x 0.4 0.9 1.2 2.3 3.8
y 8.35 13.47 17.94 51.32 215.20 The law relatingx andyis believed to be of the formy=abx, whereaandbare constants.
Determine the approximate values ofaandb.
Hence, find the value ofywhenx is 2.0 and the value ofxwhenyis 100.
7. The activity of a mixture of radioactive iso-topes is believed to vary according to the law R=R0t−c, where R0 and c are constants.
Experimental results are shown below.
R 9.72 2.65 1.15 0.47 0.32 0.23
t 2 5 9 17 22 28
Verify that the law is true and determine approximate values ofR0andc.
8. Determine the law of the formy=aekxwhich relates the following values.
y 0.0306 0.285 0.841 5.21 173.2 1181 x –4.0 5.3 9.8 17.4 32.0 40.0 9. The tensionT in a belt passing round a
pul-ley wheel and in contact with the pulpul-ley over an angle ofθ radians is given byT =T0eμθ, whereT0 andμare constants. Experimental results obtained are
T(newtons) 47.9 52.8 60.3 70.1 80.9 θ(radians) 1.12 1.48 1.97 2.53 3.06 Determine approximate values of T0 andμ.
Hence, find the tension whenθis 2.25 radians and the value ofθ when the tension is 50.0 newtons.