Higher Engineering Mathematics

SOME APPLICATIONS OF DIFFERENTIATION 299

G

Problem 4. The displacementscm of the end

of a stiff spring at timetseconds is given by

s=ae−ktsin 2πft. Determine the velocity of the

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

f=5.

Velocity, v=

ds

dt

where s=ae−ktsin 2πft (i.e. a

product).
Using the product rule,

ds

dt

=(ae−kt)(2πfcos 2πft)

+(sin 2πft)(−ake−kt)

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

velocity,v=(2e−^0.^9 )(2π5 cos 2π5)

+(sin 2π5)(−2)(0.9)e−^0.^9

= 25 .5455 cos 10π− 0 .7318 sin 10π

= 25 .5455(1)− 0 .7318(0)

=25.55 cm/s

(Note that cos 10πmeans ‘the cosine of 10πradians’,
notdegrees, and cos 10π≡cos 2π=1).

Now try the following exercise.

Exercise 122 Further problems on rates of

change

1. An alternating current,iamperes, is given by
   i=10 sin 2πft, wheref is the frequency in
   hertz andtthe time in seconds. Determine
   the rate of change of current whent=20 ms,
   given thatf=150 Hz. [3000πA/s]


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


3. The voltage across the plates of a capacitor at
   any timetseconds is given byv=Ve−t/CR,
   whereV,CandRare constants.

GivenV=300 volts,C= 0. 12 × 10 −^6 F and

R= 4 × 106  find (a) the initial rate of

change of voltage, and (b) the rate of change

of voltage after 0.5 s.

[(a)−625 V/s (b)−220.5 V/s]

4. The pressurepof the atmosphere at heighth
   above ground level is given byp=p 0 e−h/c,
   wherep 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 1450 metres. [− 1 .635 Pa/m]

28.2 Velocity and acceleration

When a car moves a distancexmetres in a timetsec-

onds along a straight road, if the velocityvis constant

thenv=

x

t

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

graph shown in Fig. 28.1 is constant.

Figure 28.1

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

then the distance/time graph will not be a straight

line. It may be as shown in Fig. 28.2.

The average velocity over a small timeδtand dis-

tanceδxis 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