Chapter 28

Some applications of

differentiation

28.1 Rates of change

If a quantityydepends on and varies with a quantity

xthen the rate of change ofywith respect toxis

dy

dx

.

Thus, for example, the rate of change of pressurepwith

heighthis

dp

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 lengthlmetres of a certain

metal rod at temperatureθ◦Cisgivenby

l= 1 + 0. 00005 θ+ 0. 0000004 θ^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. 00005 θ+ 0. 0000004 θ^2 ,

then

dl

dθ

= 0. 00005 + 0. 0000008 θ

(a) Whenθ= 100 ◦C,

dl

dθ

= 0. 00005 +( 0. 0000008 )( 100 )

= 0 .00013m/◦C

= 0 .13mm/◦C

(b) Whenθ= 400 ◦C,

dl

dθ

= 0. 00005 +( 0. 0000008 )( 400 )

= 0 .00037m/◦C

= 0 .37mm/◦C

Problem 2. The luminous intensityIcandelas

of a lamp at varying voltageVis 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

dI

dV

.

Since I= 4 × 10 −^4 V^2 ,

dI

dV

=( 4 × 10 −^4 )( 2 )V= 8 × 10 −^4 V

When thelightis increasing at 0.6candelas per voltthen

+ 0. 6 = 8 × 10 −^4 V, from which, voltage

V=

0. 6

8 × 10 −^4

= 0. 075 × 10 +^4

=750volts

Problem 3. Newtons law of cooling is given by

θ=θ 0 e−kt, where the excess of temperature at zero

time isθ 0 ◦C and at timetseconds isθ◦C. Determine

the rate of change of temperature after 40s, given

thatθ 0 = 16 ◦Candk=− 0. 03