Chapter 28
Some applications of
differentiation
28.1 Rates of change
If a quantity y depends on and varies with a quantity
x then the rate of change of y with respect to x is
d y
dx
.
Thus, for example, the rate of change of pressure p with
height h is
d p
dh
.
A rate of change with respect to time is usually just
called ‘the rate of change’, the ‘with respect to time’
being assumed. Thus, for example, a rate of change of
current, i, is
di
dt
and a rate of change of temperature,
θ, is
dθ
dt
, and so on.
Problem 1. The length l metres of a certain
metal rod at temperature θ ◦ C is given by
l = 1 + 0.00005θ + 0.0000004θ 2 . Determine the
rate of change of length, in mm/ ◦ C, when the
temperature is (a) 100 ◦ C and (b) 400 ◦ C.
The rate of change of length means
dl
dθ
.
Since length
l = 1 +0.00005θ + 0.0000004θ 2 ,
then
dl
dθ
= 0.00005 + 0.0000008θ
(a) When θ = 100 ◦ C,
dl
dθ
= 0.00005 + (0.0000008)(100)
= 0.00013 m/
◦ C
= 0.13 mm/
◦ C
(b) When θ = 400 ◦ C,
dl
dθ
= 0.00005 + (0.0000008)(400)
= 0.00037 m/
◦ C
= 0.37 mm/
◦ C
Problem 2. The luminous intensity I candelas
of a lamp at varying voltage V is given by
I = 4 ×10 −4 V 2 . Determine the voltage at which the
light is increasing at a rate of 0.6 candelas per volt.
The rate of change of light with respect to voltage is
given by
d I
dV
.
Since
I = 4 × 10
−4 V
2
,
d I
dV
= (4 × 10
−4
)(2)V = 8 × 10
−4 V
When the light is increasing at 0.6 candelas per volt then
+0.6 = 8 × 10 −4 V , from which, voltage
V =
0.6
8 × 10 −4 = 0.075 × 10
+4
= 750 volts
Problem 3. Newtons law of cooling is given by
θ = θ 0 e −kt , where the excess of temperature at zero
time is θ ◦
0 C and at time t seconds is θ ◦ C. Determine
the rate of change of temperature after 40 s, given
that θ 0 = 16 ◦ C and k = −0.03