Exponential functions 39
1000
100
Voltage, v volts
10
1
0
1 0
2 0
3 0
4 0
5 0
6 0
7 0
8 0
9 0
Time, t ms
A
B
C
(36.5, 100)
v 5Ve
t
T
Figure 4.9
Thus 100 = V e
36.5
−12.0
i.e.
V =
100
e
−36.5
12.0
= 2090 volts,
correct to 3 significant figures.
Hence the law of the graph is v = 2090 e
−t
12.0 .
When time t = 25 ms,
voltage
v = 2090 e
−25
12.0 = 260 V
When the voltage is 30.0 volts, 30.0 = 2090 e
−t
12.0 ,
hence e
−t
12.0 =
30.0
2090
and
e
t
12.0 =
2090
30.0
= 69.67
Taking Napierian logarithms gives:
t
12.0
= ln 69.67 = 4.2438
from which, time t = (12.0)(4.2438) = 50.9 ms
Now try the following exercise
Exercise 19 Further problems on reducing
exponential laws to linear form
1. Atmospheric pressure p is measured at varying altitudes h and the results are as shown
below:
Altitude, h m pressure, p cm
500
73.39
1500
68.42
3000
61.60
5000
53.56
8000
43.41
Show that the quantities are related by the
law p =a e kh , where a and k are constants.
Determine the values of a and k and state
the law. Find also the atmospheric pressure at
10 000 m.
a = 76, k = −7 × 10 −5 ,
p = 76 e −7×10 −5 h , 37.74 cm
2. At particular times, t minutes, measurements
are made of the temperature, θ ◦ C, of a
cooling liquid and the following results are
obtained:
Temperature θ ◦ C Time t minutes
92.2
10
55.9
20
33.9
30
20.6
40
12.5
50
Prove that the quantities follow a law of the
form θ = θ 0 e kt , where θ 0 and k are constants,
and determine the approximate value of θ 0
and k.
[θ 0 = 152, k = −0.05]
1000
100
Voltage, v volts
10
1
0
1 0
2 0
3 0
4 0
5 0
6 0
7 0
8 0
9 0
Time, t ms
A
B
C
(36.5, 100)
v 5Ve
t
T
Figure 4.9
Thus 100 = V e
36.5
−12.0
i.e.
V =
100
e
−36.5
12.0
= 2090 volts,
correct to 3 significant figures.
Hence the law of the graph is v = 2090 e
−t
12.0 .
When time t = 25 ms,
voltage
v = 2090 e
−25
12.0 = 260 V
When the voltage is 30.0 volts, 30.0 = 2090 e
−t
12.0 ,
hence e
−t
12.0 =
30.0
2090
and
e
t
12.0 =
2090
30.0
= 69.67
Taking Napierian logarithms gives:
t
12.0
= ln 69.67 = 4.2438
from which, time t = (12.0)(4.2438) = 50.9 ms
Now try the following exercise
Exercise 19 Further problems on reducing
exponential laws to linear form
1. Atmospheric pressure p is measured at varying altitudes h and the results are as shown
below:
Altitude, h m pressure, p cm
500
73.39
1500
68.42
3000
61.60
5000
53.56
8000
43.41
Show that the quantities are related by the
law p =a e kh , where a and k are constants.
Determine the values of a and k and state
the law. Find also the atmospheric pressure at
10 000 m.
a = 76, k = −7 × 10 −5 ,
p = 76 e −7×10 −5 h , 37.74 cm
2. At particular times, t minutes, measurements
are made of the temperature, θ ◦ C, of a
cooling liquid and the following results are
obtained:
Temperature θ ◦ C Time t minutes
92.2
10
55.9
20
33.9
30
20.6
40
12.5
50
Prove that the quantities follow a law of the
form θ = θ 0 e kt , where θ 0 and k are constants,
and determine the approximate value of θ 0
and k.
[θ 0 = 152, k = −0.05]