Higher Engineering Mathematics, Sixth Edition

300 Higher Engineering Mathematics

The rate of change of temperature is

dθ

dt

Since θ=θ 0 e−kt

then

dθ

dt

=(θ 0 )(−k)e−kt=−kθ 0 e−kt

When θ 0 = 16 ,k=− 0 .03 and t= 40

then

dθ

dt

=−(− 0. 03 )( 16 )e−(−^0.^03 )(^40 )

= 0 .48e^1.^2 = 1. 594 ◦C/s

Problem 4. The displacementscm of the end

of a stiff spring at timetseconds is given by

s=ae−ktsin2πft. Determine the velocity of the

end of the spring after 1s, ifa=2,k= 0 .9and

f=5.

Velocity, v=

ds

dt

where s=ae−ktsin2πft (i.e. a

product).

Using the product rule,

ds

dt

=(ae−kt)( 2 πfcos2πft)

+(sin 2πft)(−ake−kt)

Whena=2,k= 0 .9,f=5andt=1,

velocity,v=(2e−^0.^9 )( 2 π5cos2π 5 )

+(sin 2π 5 )(− 2 )( 0. 9 )e−^0.^9

= 25 .5455cos10π− 0 .7318sin10π

= 25. 5455 ( 1 )− 0. 7318 ( 0 )

=25.55cm/s

(Note that cos10πmeans ‘the cosine of 10πradians’,

notdegrees, and cos10π≡cos2π=1.)

Now try the following exercise

Exercise 120 Further problems on rates of

change

1. An alternating current,iamperes, is given by
   i=10sin2πft,wheref is the frequency in
   hertz andtthe time in seconds. Determine the

rate of change of current whent=20ms, given

thatf=150Hz. [3000πA/s]

2. The luminous intensity,Icandelas, of a lamp
   is given byI= 6 × 10 −^4 V^2 ,whereVis the
   voltage.Find(a)therateofchangeofluminous
   intensity withvoltagewhenV=200volts, and
   (b) the voltage at which the light is increasing
   at a rate of 0.3candelas per volt.
   [(a) 0.24cd/V (b) 250V]


3. The voltage across the plates of a capacitor at
   any timetseconds is given byv=Ve−t/CR,
   whereV,CandRare constants.
   GivenV=300volts,C= 0. 12 × 10 −^6 Fand
   R= 4 × 106 find (a) theinitialrate of change
   of voltage, and (b) the rate of change ofvoltage
   after 0.5s. [(a)−625V/s (b)−220.5V/s]


4. The pressurepof the atmosphere at heighth
   above ground level is given byp=p 0 e−h/c,
   where p 0 is the pressure at ground level
   and c is a constant. Determine the rate
   of change of pressure with height when
   p 0 = 1. 013 × 105 pascals andc= 6. 05 × 104 at
   1450metres. [− 1 .635Pa/m]

28.2 Velocity and acceleration

When acar moves a distancexmetresinatimetseconds

along a straight road, if the velocityvis constant then

v=

x

t

m/s, i.e. the gradient of the distance/time graph

shown in Fig. 28.1 is constant.

If, however, the velocity of the car is not constant then

the distance/timegraph willnotbe a straightline.It may

be as shown in Fig. 28.2.

Theaveragevelocity over asmall timeδtand distance

δx is given by the gradient of the chordAB,i.e.the

average velocity over timeδtis

δx

δt

.

Asδt→0, the chordABbecomes a tangent, such that

at pointA, the velocity is given by:

v=

dx

dt

Hence the velocity of the car at any instant is given by

the gradient of the distance/time graph. If an expression