Basic Engineering Mathematics, Fifth Edition

Introduction to differentiation 323

34.10 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.

Here are some worked problems to demonstrate practi-
cal examples of rates of change.

Problem 26. The lengthLmetres of a certain

metal rod at temperaturet◦Cisgivenby

L= 1 + 0. 00003 t+ 0. 0000004 t^2. Determine the

rate of change of length, in mm/◦C, when the

temperature is (a) 100◦C (b) 250◦C

The rate of change of length means

dL

dt

Since lengthL= 1 + 0. 00003 t+ 0. 0000004 t^2 ,then

dL

dt

= 0. 00003 + 0. 0000008 t.

(a) Whent= 100 ◦C,

dL

dt

= 0. 00003 +( 0. 0000008 )( 100 )

= 0 .00011m/◦C=0.11mm/◦C.

(b) Whent= 250 ◦C,

dL

dt

= 0. 00003 +( 0. 0000008 )( 250 )

= 0 .00023m/◦C=0.23mm/◦C.

Problem 27. The luminous intensityIcandelas

of a lamp at varying voltageVis given by

I= 5 × 10 −^4 V^2. Determine the voltage at which

the light is increasing at a rate of 0.4 candelas

per volt

The rate of change of light with respect to voltage is

given by

dI
dV
SinceI= 5 × 10 −^4 V^2 ,
dI
dV

=( 5 × 10 −^4 )( 2 V)= 10 × 10 −^4 V= 10 −^3 V.

When thelightis increasing at 0.4 candelas per voltthen

+ 0. 4 = 10 −^3 V, from which

voltage,V=

0. 4

10 −^3

= 0. 4 × 10 +^3 =400volts

Problem 28. Newton’s 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

50s, given thatθ 0 = 15 ◦Candk=− 0. 02

The rate of change of temperature is

dθ

dt

Sinceθ=θ 0 e−kt,then

dθ

dt

=(θ 0 )(−ke−kt)

=−kθ 0 e−kt

Whenθ 0 = 15 ,k=− 0 .02 andt=50, then

dθ

dt

=−(− 0. 02 )( 15 )e−(−^0.^02 )(^50 )

= 0. 30 e^1 =0.815◦C/s

Problem 29. The pressurepof the atmosphere at

heighthabove ground level is given by

p=p 0 e−h/c,wherep 0 is the pressure at ground

level andcis a constant. Determine the rate

of change of pressure with height when

p 0 = 105 pascals andc= 6. 2 × 104 at 1550 metres

The rate of change of pressure with height is

dp

dh

Sincep=p 0 e−h/c,then

dp

dh

=(p 0 )

(

−

1

c

e−h/c

)

=−

p 0

c

e−h/c

Whenp 0 = 105 ,c= 6. 2 × 104 andh=1550, then

rate of change of pressure,

dp

dh

=−

105

6. 2 × 104

e−

(

1550 / 6. 2 × 104

)

=−

10

6. 2

e−^0.^025 =−1.573Pa/m