Higher Engineering Mathematics, Sixth Edition

Exponential functions 31

decay curve over the first 6seconds. From the

graph, find (a) the voltage after 3.4s, and (b) the

time when the voltage is 150V.

A table of values is drawn up as shown below.

t 0 1 2 3

e

−t

(^3) 1.00 0.7165 0.5134 0.3679
v=250e
−t
(^3) 250.0 179.1 128.4 91.97
t 4 5 6
e
−t
(^3) 0.2636 0.1889 0.1353
v=250e
−t
(^3) 65.90 47.22 33.83
The natural decay curve ofv=250e
−t
(^3) is shown in
Fig. 4.4.
250
200
Voltage
v
(volts)
Time t(seconds)
150
100
80
50
0 1 1.5 2 3 3.4 4 5 6
y 5 250e
t
(^23)
Figure 4.4
From the graph:
(a) when timet=3.4s, voltagev=80Vand
(b) when voltagev=150V, timet=1.5s.
Now try the following exercise
Exercise 16 Further problemson
exponential graphs

1. Plot a graph of y=3e^0.^2 x over the range
   x=−3tox=3. Hence determine the value

ofywhenx= 1 .4 and the value ofxwhen

y= 4 .5. [3.95, 2.05]

2. Plot a graph of y=^12 e−^1.^5 x over a range
   x=− 1 .5tox= 1 .5 and hence determine the
   value ofywhenx=− 0 .8 and the value ofx
   wheny= 3 .5. [1.65,−1.30]


3. In a chemical reaction the amount of starting
   materialCcm^3 left aftertminutes is given by
   C=40e−^0.^006 t.PlotagraphofCagainsttand
   determine (a) the concentrationCafter 1 hour,
   and (b) the time taken for the concentration to
   decrease by half.
   [(a) 28cm^3 (b) 116min]


4. The rate at which a body cools is given by
   θ=250e−^0.^05 twhere the excess of tempera-
   ture of a body above its surroundings at
   timetminutes isθ◦C. Plot a graph showing
   the natural decay curve for the firsthour of
   cooling. Hence determine (a) the temperature
   after 25 minutes, and (b) the time when the
   temperature is 195◦C.
   [(a) 70◦C(b)5min]

4.4 Napierian logarithms

Logarithms having a base of ‘e’ are calledhyperbolic,

Napierianornatural logarithmsand the Napierian

logarithm ofxis written as logex, or more commonly

as lnx. Logarithms were invented by John Napier, a

Scotsman (1550–1617).

The most common method of evaluating a Napierian

logarithmisbyascientificnotationcalculator.Useyour

calculator to check the following values:

ln4. 328 = 1. 46510554 ...

= 1. 4651 ,correct to 4 decimal places

ln1. 812 = 0. 59443 ,correct to 5 significant figures

ln1= 0

ln527= 6. 2672 ,correct to 5 significant figures

ln0. 17 =− 1. 772 ,correct to 4 significant figures

ln0. 00042 =− 7. 77526 ,correct to 6 significant

figures

lne^3 = 3

lne^1 = 1