38 Higher Engineering Mathematics
determine approximate values of a and b. Also
determine the value of y when x is 3.8 and the value
of x when y is 85.
x −1.2 0.38 1.2 2.5 3.4 4.2 5.3
y 9.3 22.2 34.8 71.2 117 181 332
Since y = a e kx then ln y = kx + ln a (from above),
which is of the form Y = m X + c, showing that to produce a straight line graph ln y is plotted vertically against
x horizontally. The value of y ranges from 9.3 to 332
hence ‘log 3 cycle × linear’ graph paper is used. The
plotted co-ordinates are shown in Fig. 4.8 and since a
straight line passes through the points the law y = a e kx
is verified.
Gradient of straight line,
k =
AB
BC
=
ln 100 − ln 10
3.12 − (−1.08)
=
2.3026
4.20
= 0.55, correct to 2 significant figures.
Since ln y = kx + ln a, when x = 0, ln y = ln a, i.e. y = a
The vertical axis intercept value at x = 0 is 18, hence
a = 18
1000
y
100
10
1
22
21
0
1
2
3
4
5
6
x
A
B
C
y 5a e kx
Figure 4.8
The law of the graph is thus y = 18 e
0.55x
When x is 3.8, y = 18 e 0.55(3.8) = 18 e 2.09
= 18(8.0849) = 146
When y is 85, 85 = 18 e 0.55x
Hence,
e 0.55x =
85
18
= 4.7222
and
0.55x = ln 4.7222 = 1.5523
Hence
x =
1.5523
0.55
= 2.82
Problem 22. The voltage, v volts, across an
inductor is believed to be related to time, t ms, by
the law v = V e
t
T , where V and T are constants.
Experimental results obtained are:
v volts 883 347 90 55.5 18.6 5.2
t ms 10.4 21.6 37.8 43.6 56.7 72.0
Show that the law relating voltage and time is as
stated and determine the approximate values of V
and T . Find also the value of voltage after 25 ms
and the time when the voltage is 30.0 V.
Since v = V e
t
T then ln v =
1
T t + ln V which is of the
form Y = m X + c.
Using ‘log3 cycle × linear’ graph paper, the points
are plotted as shown in Fig. 4.9.
Since the points are joined by a straight line the law
v = Ve
t
T is verified.
Gradient of straight line,
1
T
=
AB
BC
=
ln 100 − ln 10
36.5 − 64.2
=
2.3026
−27.7
Hence T =
−27.7
2.3026
= −12.0, correct to 3 significant figures.
Since the straight line does not cross the vertical axis
at t = 0 in Fig. 4.9, the value of V is determined
by selecting any point, say A, having co-ordinates
(36.5,100) and substituting these values into v = V e
t
T .
determine approximate values of a and b. Also
determine the value of y when x is 3.8 and the value
of x when y is 85.
x −1.2 0.38 1.2 2.5 3.4 4.2 5.3
y 9.3 22.2 34.8 71.2 117 181 332
Since y = a e kx then ln y = kx + ln a (from above),
which is of the form Y = m X + c, showing that to produce a straight line graph ln y is plotted vertically against
x horizontally. The value of y ranges from 9.3 to 332
hence ‘log 3 cycle × linear’ graph paper is used. The
plotted co-ordinates are shown in Fig. 4.8 and since a
straight line passes through the points the law y = a e kx
is verified.
Gradient of straight line,
k =
AB
BC
=
ln 100 − ln 10
3.12 − (−1.08)
=
2.3026
4.20
= 0.55, correct to 2 significant figures.
Since ln y = kx + ln a, when x = 0, ln y = ln a, i.e. y = a
The vertical axis intercept value at x = 0 is 18, hence
a = 18
1000
y
100
10
1
22
21
0
1
2
3
4
5
6
x
A
B
C
y 5a e kx
Figure 4.8
The law of the graph is thus y = 18 e
0.55x
When x is 3.8, y = 18 e 0.55(3.8) = 18 e 2.09
= 18(8.0849) = 146
When y is 85, 85 = 18 e 0.55x
Hence,
e 0.55x =
85
18
= 4.7222
and
0.55x = ln 4.7222 = 1.5523
Hence
x =
1.5523
0.55
= 2.82
Problem 22. The voltage, v volts, across an
inductor is believed to be related to time, t ms, by
the law v = V e
t
T , where V and T are constants.
Experimental results obtained are:
v volts 883 347 90 55.5 18.6 5.2
t ms 10.4 21.6 37.8 43.6 56.7 72.0
Show that the law relating voltage and time is as
stated and determine the approximate values of V
and T . Find also the value of voltage after 25 ms
and the time when the voltage is 30.0 V.
Since v = V e
t
T then ln v =
1
T t + ln V which is of the
form Y = m X + c.
Using ‘log3 cycle × linear’ graph paper, the points
are plotted as shown in Fig. 4.9.
Since the points are joined by a straight line the law
v = Ve
t
T is verified.
Gradient of straight line,
1
T
=
AB
BC
=
ln 100 − ln 10
36.5 − 64.2
=
2.3026
−27.7
Hence T =
−27.7
2.3026
= −12.0, correct to 3 significant figures.
Since the straight line does not cross the vertical axis
at t = 0 in Fig. 4.9, the value of V is determined
by selecting any point, say A, having co-ordinates
(36.5,100) and substituting these values into v = V e
t
T .