Higher Engineering Mathematics

LOGARITHMS AND EXPONENTIAL FUNCTIONS 29

A

(b) 5

(

e^0.^25 −e−^0.^25

e^0.^25 +e−^0.^25

)

= 5

(

1. 28402541 ...− 0. 77880078 ...

1. 28402541 ...+ 0. 77880078 ...

)

= 5

(

0. 5052246 ...

2. 0628261 ...

)

=1.2246, correct to 4 decimal places

Problem 13. The instantaneous voltagevin

a capacitive circuit is related to time t by

the equationv = Ve

−t

CR whereV,CandR

are constants. Determinev, correct to 4 sig-

nificant figures, whent= 30 × 10 −^3 seconds,

C= 10 × 10 −^6 farads,R= 47 × 103 ohms and

V=200 V.

v=Ve

−t

CR=200 e

(− 30 × 10 −^3 )

(10× 10 −^6 × 47 × 103 )

Using a calculator,

v=200 e−^0.^0638297 ...=200(0. 9381646 ...)

=187.6 V

Now try the following exercise.

Exercise 18 Further problems on evaluat-

ing exponential functions

1. Evaluate, correct to 5 significant figures:

(a) 3.5e^2.^8 (b)−

6

5

e−^1.^5 (c) 2.16 e^5.^7

⎡

⎣

(a) 57. 556

(b)− 0. 26776

(c) 645. 55

⎤

⎦

In Problems 2 and 3, evaluate correct to 5

decimal places.

2. (a)

1

7

e^3.^4629 (b) 8.52 e−^1.^2651

(c)

5e^2.^6921

3e^1.^1171 ⎡

⎣

(a) 4. 55848

(b) 2. 40444

(c) 8. 05124

⎤

⎦

3. (a)

5. 6823

e−^2.^1347

(b)

e^2.^1127 −e−^2.^1127

2

(c)

4(e−^1.^7295 −1)

e^3.^6817 [

(a) 48. 04106

(b) 4. 07482

(c)− 0. 08286

]

4. The length of a bar,l, at a temperatureθ
   is given byl=l 0 eαθ, wherel 0 andαare
   constants. Evaluatel, correct to 4 signifi-
   cant figures, whenl 0 = 2 .587,θ= 321 .7 and
   α= 1. 771 × 10 −^4. [2.739]

4.6 The power series for ex

The value of excan be calculated to any required

degree of accuracy since it is defined in terms of the

followingpower series:

ex= 1 +x+

x^2

2!

+

x^3

3!

+

x^4

4!

+···

(where 3!= 3 × 2 ×1 and is called ‘factorial 3’)

The series is valid for all values ofx.

The series is said toconverge, i.e. if all the terms

are added, an actual value for ex(wherexis a real

number) is obtained. The more terms that are taken,

the closer will be the value of exto its actual value.

The value of the exponent e, correct to say 4 decimal

places, may be determined by substitutingx=1in

the power series of equation (1). Thus,

e^1 = 1 + 1 +

(1)^2

2!

+

(1)^3

3!

+

(1)^4

4!

+

(1)^5

5!

+

(1)^6

6!

+

(1)^7

7!

+

(1)^8

8!

+···

= 1 + 1 + 0. 5 + 0. 16667 + 0. 04167

+ 0. 00833 + 0. 00139 + 0. 00020

+ 0. 00002 +···

i.e. e= 2. 71828 = 2 .7183,

correct to 4 decimal places

The value of e^0.^05 , correct to say 8 significant figures,

is found by substitutingx= 0 .05 in the power series