Basic Engineering Mathematics, Fifth Edition

Graphs reducing non-linear laws to linear form 151

and shows thatlgyis plotted vertically against

xhorizontallyto 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 toabaseofeof 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=mX+c

and shows thatlnyis plotted vertically against

xhorizontallyto 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=RIn,whereRandnare constants,

and determine the law

Taking logarithms to a base of 10 of both sides of
P=RIngives
lgP=lg(RIn)=lgR+lgIn=lgR+nlgI
by the laws of logarithms
i.e. lgP=nlgI+lgR
which is of the formY=mX+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 lgPagainst lgIis 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

lg 2.5

P

lg L

2.78

2.98

3.0

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,wherelgI= 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. 0 I^2

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 ofTwere 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. 250. 420. 560. 811. 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

0.75m