Higher Engineering Mathematics, Sixth Edition

Exponential functions 39

1000

100

Voltage,

v volts

10

1

0 102030405060708090

Time, t ms

A

B C

(36.5, 100)

v 5 Ve

Tt

Figure 4.9

Thus 100=Ve

36. 5

− 12. 0

i.e. V=

100

e

− 36. 5

12. 0

=2090volts,

correct to 3 significant figures.

Hence the law of the graph isv=2090e

−t

12.0.

When time t=25ms,

voltage v=2090e

− 25

12. 
    

    (^0) =260V
    When the voltage is 30.0volts, 30. 0 =2090e
    −t

    
    


13. 
    

    (^0) ,
    hence e
    −t

    
    


14. 
    

    (^0) =

    
    


15. 
    

    0
    2090
    and e
    t

    
    


16. 
    

    (^0) =
    2090

    
    


17. 
    

    0
    = 69. 67
    Taking Napierian logarithms gives:
    t

    
    


18. 
    

    0
    =ln69. 67 = 4. 2438
    from which, timet=( 12. 0 )( 4. 2438 )=50.9ms
    Now try the following exercise
    Exercise 19 Further problemson reducing
    exponential laws to linear form

    
    


19. 
    

    Atmospheric pressurepis measured at vary-
    ing altitudeshand the results are as shown
    below:

    
    

Altitude,hm pressure,pcm

500 73.39

1500 68.42

3000 61.60

5000 53.56

8000 43.41

Show that the quantities are related by the

lawp=aekh,whereaandkare constants.

Determine the values ofaandk and state

the law. Find also the atmospheric pressure at

10000m.

[

a= 76 ,k=− 7 × 10 −^5 ,

p=76e−^7 ×^10

− (^5) h
, 37 .74cm
]

2. At particular times,tminutes, measurements
   are made of the temperature, θ◦C, of a
   cooling liquid and the following results are
   obtained:

Temperatureθ◦C Timetminutes

92.2 10

55.9 20

33.9 30

20.6 40

12.5 50

Prove that the quantities follow a law of the

formθ=θ 0 ekt,whereθ 0 andkare constants,

and determine the approximate value ofθ 0

andk.

[θ 0 =152,k=− 0 .05]