Higher Engineering Mathematics, Sixth Edition

Exponential functions 35

Hence α=

1

θ

ln

R

R 0

=

1

1500

ln

(

6 × 103

5 × 103

)

=

1

1500

( 0. 1823215 ...)

= 1. 215477 ···× 10 −^4

Hence α=1.215× 10 −^4 ,

correct to 4 significant figures.

From above, ln

R

R 0

=αθ

hence θ=

1

α

ln

R

R 0

When R= 5. 4 × 103 ,α= 1. 215477 ...× 10 −^4 and
R 0 = 5 × 103

θ=

1

1. 215477 ...× 10 −^4

ln

(

5. 4 × 103

5 × 103

)

=

104

1. 215477 ...

( 7. 696104 ...× 10 −^2 )

= 633 ◦C,correct to the nearest degree.

Problem 18. In an experiment involving

Newton’s law of cooling, the temperatureθ(◦C) is

given byθ=θ 0 e−kt. Find the value of constantk

whenθ 0 = 56. 6 ◦C,θ= 16. 5 ◦Candt= 83 .0seconds.

Transposing θ=θ 0 e−ktgives

θ

θ 0

=e−kt

from which

θ 0

θ

=

1

e−kt

=ekt

Taking Napierian logarithms of both sides gives:

ln

θ 0

θ

=kt

from which,

k=

1

t

ln

θ 0

θ

=

1

83. 0

ln

(

56. 6

16. 5

)

=

1

83. 0

( 1. 2326486 ...)

Hencek=1.485× 10 −^2

Problem 19. The currentiamperes flowing in a

capacitor at timetseconds is given by

i= 8. 0 ( 1 −e

−t

CR), where the circuit resistanceRis

25 × 103 ohms and capacitanceCis

16 × 10 −^6 farads. Determine (a) the currentiafter

0.5seconds and (b) the time, to the nearest

millisecond, for the current to reach 6.0A. Sketch

the graph of current against time.

(a) Currenti= 8. 0 ( 1 −e

−t

CR)

= 8 .0[1−e

− 0. 5

( 16 × 10 −^6 )( 25 × 103 )]= 8. 0 ( 1 −e−^1.^25 )

= 8. 0 ( 1 − 0. 2865047 ...)= 8. 0 ( 0. 7134952 ...)

=5.71amperes

(b) Transposingi= 8. 0 ( 1 −e

−t

CR)

gives

i

8. 0

= 1 −e

−t

CR

from which, e

−t

CR= 1 −

i

8. 0

=

8. 0 −i

8. 0

Taking the reciprocal of both sides gives:

e

t

CR=^8.^0

8. 0 −i

Taking Napierian logarithms of both sides gives:

t

CR

=ln

(

8. 0

8. 0 −i

)

Hence

t=CRln

(

8. 0

8. 0 −i

)

=( 16 × 10 −^6 )( 25 × 103 )ln

(

8. 0

8. 0 − 6. 0

)

wheni= 6 .0 amperes,

i.e. t=

400

103

ln

(

8. 0

2. 0

)

= 0 .4ln4. 0

= 0. 4 ( 1. 3862943 ...)= 0 .5545s

=555ms, to the nearest millisecond.

Agraphofcurrent against timeisshowninFig.4.6.