150 Basic Engineering Mathematics
a and b. Determine the cross-sectional area
needed for a resistance reading of 0.50 ohms.
7. Corresponding experimental values of two
quantities x and y are given below.
x 1.5 3.0 4.5 6.0 7.5
9.0
y 11.5 25.0 47.5 79.0 119.5 169.0
By plotting a suitable graph, verify that y
and x are connected by a law of the form
y = kx 2 + c, where k and c are constants.
Determine the law of the graph and hence find
the value of x when y is 60.0
8. Experimental results of the safe load L kN,
applied to girders of varying spans, d m, are
shown below.
Span, d (m)
2.0 2.8 3.6 4.2 4.8
Load, L (kN) 475 339 264 226 198
It is believed that the relationship between load
and span is L = c/d, where c is a constant.
Determine (a) the value of constant c and (b)
the safe load for a span of 3.0 m.
9. The following results give corresponding values of two quantities x and y which are
believed to be related by a law of the form
y = ax 2 + bx, where a and b are constants.
x 33.86 55.54 72.80 84.10 111.4 168.1
y 3.4
5.2
6.5
7.3
9.1 12.4
Verify the law and determine approximate values of a and b. Hence, determine (a) the value
of y when x is 8.0 and (b) the value of x when
y is 146.5
18.3 Revision of laws of logarithms
The laws of logarithms were stated in Chapter 15 as
follows:
log(A × B) = log A + log B
(1)
log
A
B
= log A − log B
(2)
logA
n
= n × logA
(3)
Also,
ln e = 1 and if, say, lg x = 1.5,
then x = 10
1.5
= 31.62
Further, if 3 x = 7 then lg 3 x = lg 7 and x lg 3 = lg 7,
from which x =
lg 7
lg 3
= 1.771
These laws and techniques are used whenever non-linear
laws of the form y = ax n , y = ab x and y = ae bx are
reduced to linear form with the values of a and b needing
to be calculated. This is demonstrated in the following
section.
18.4 Determination of laws involving
logarithms
Examples of the reduction of equations to linear form
involving logarithms include
(a) y = ax n
Taking logarithms to a base of 10 of both sides
gives
lg y = lg(ax
n
)
= lg a + lg x
n by law (1)
i.e.
lg y = n lg x + lg a by law (3)
which compares with Y = m X + c
and shows that lg y is plotted vertically against
lg x horizontally to produce a straight line graph
of gradient n and lg y-axis intercept lg a.
See worked Problems 4 and 5 to demonstrate how
this law is determined.
(b) y = ab x
Taking logarithms to a base of 10 of both sides
gives
lg y = lg(ab
x
)
i.e.
lg y = lg a + lg b
x by law (1)
lg y = lg a + x lg b by law (3)
i.e.
lg y = x lg b + lg a
or
lg y = (lg b)x + lg a
which compares with
Y = m X + c
a and b. Determine the cross-sectional area
needed for a resistance reading of 0.50 ohms.
7. Corresponding experimental values of two
quantities x and y are given below.
x 1.5 3.0 4.5 6.0 7.5
9.0
y 11.5 25.0 47.5 79.0 119.5 169.0
By plotting a suitable graph, verify that y
and x are connected by a law of the form
y = kx 2 + c, where k and c are constants.
Determine the law of the graph and hence find
the value of x when y is 60.0
8. Experimental results of the safe load L kN,
applied to girders of varying spans, d m, are
shown below.
Span, d (m)
2.0 2.8 3.6 4.2 4.8
Load, L (kN) 475 339 264 226 198
It is believed that the relationship between load
and span is L = c/d, where c is a constant.
Determine (a) the value of constant c and (b)
the safe load for a span of 3.0 m.
9. The following results give corresponding values of two quantities x and y which are
believed to be related by a law of the form
y = ax 2 + bx, where a and b are constants.
x 33.86 55.54 72.80 84.10 111.4 168.1
y 3.4
5.2
6.5
7.3
9.1 12.4
Verify the law and determine approximate values of a and b. Hence, determine (a) the value
of y when x is 8.0 and (b) the value of x when
y is 146.5
18.3 Revision of laws of logarithms
The laws of logarithms were stated in Chapter 15 as
follows:
log(A × B) = log A + log B
(1)
log
A
B
= log A − log B
(2)
logA
n
= n × logA
(3)
Also,
ln e = 1 and if, say, lg x = 1.5,
then x = 10
1.5
= 31.62
Further, if 3 x = 7 then lg 3 x = lg 7 and x lg 3 = lg 7,
from which x =
lg 7
lg 3
= 1.771
These laws and techniques are used whenever non-linear
laws of the form y = ax n , y = ab x and y = ae bx are
reduced to linear form with the values of a and b needing
to be calculated. This is demonstrated in the following
section.
18.4 Determination of laws involving
logarithms
Examples of the reduction of equations to linear form
involving logarithms include
(a) y = ax n
Taking logarithms to a base of 10 of both sides
gives
lg y = lg(ax
n
)
= lg a + lg x
n by law (1)
i.e.
lg y = n lg x + lg a by law (3)
which compares with Y = m X + c
and shows that lg y is plotted vertically against
lg x horizontally to produce a straight line graph
of gradient n and lg y-axis intercept lg a.
See worked Problems 4 and 5 to demonstrate how
this law is determined.
(b) y = ab x
Taking logarithms to a base of 10 of both sides
gives
lg y = lg(ab
x
)
i.e.
lg y = lg a + lg b
x by law (1)
lg y = lg a + x lg b by law (3)
i.e.
lg y = x lg b + lg a
or
lg y = (lg b)x + lg a
which compares with
Y = m X + c