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 displacement s cm of the end
of a stiff spring at time t seconds is given by
s = ae −kt sin 2π f t. Determine the velocity of the
end of the spring after 1 s, if a = 2, k = 0.9 and
f = 5.
Velocity, v =
ds
dt
where s = ae −kt sin 2π f t (i.e. a
product).
Using the product rule,
ds
dt
= (ae
−kt
)(2π f cos 2π f t)
+ (sin 2π f t)(−ake
−kt
)
When a = 2, k = 0.9, f = 5 and t = 1,
velocity, v = (2e
−0.9
)(2π5 cos 2π5)
+ (sin 2π5)(−2)(0.9)e
−0.9
= 25.5455 cos10π − 0.7318 sin 10π
= 25.5455(1) − 0.7318(0)
= 25.55 cm/s
(Note that cos10π means ‘the cosine of 10π radians’,
not degrees, and cos 10π ≡ cos 2π = 1.)
Now try the following exercise
Exercise 120 Further problems on rates of
change
1. An alternating current, i amperes, is given by
i = 10 sin 2πf t, where f is the frequency in
hertz and t the time in seconds. Determine the
rate of change of current when t = 20 ms, given
that f = 150 Hz.
[3000π A/s]
2. The luminous intensity, I candelas, 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 when V = 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 time t seconds is given by v = V e −t /C R ,
where V , C and R are constants.
Given V = 300 volts, C = 0.12 × 10 −6 F and
R = 4 ×10 6 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 pressure p of the atmosphere at height h
above ground level is given by p = 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 × 10 5 pascals and c = 6.05 × 10 4 at
1450 metres.
[−1.635 Pa/m]
28.2 Velocity and acceleration
When a car moves a distance x metres in a time t seconds
along a straight road, if the velocity v is 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/time graph will not be a straight line. It may
be as shown in Fig. 28.2.
The average velocity over a small time δt and distance
δx is given by the gradient of the chord AB, i.e. the
average velocity over time δt is
δx
δt
.
As δt → 0, the chord AB becomes a tangent, such that
at point A, 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
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 displacement s cm of the end
of a stiff spring at time t seconds is given by
s = ae −kt sin 2π f t. Determine the velocity of the
end of the spring after 1 s, if a = 2, k = 0.9 and
f = 5.
Velocity, v =
ds
dt
where s = ae −kt sin 2π f t (i.e. a
product).
Using the product rule,
ds
dt
= (ae
−kt
)(2π f cos 2π f t)
+ (sin 2π f t)(−ake
−kt
)
When a = 2, k = 0.9, f = 5 and t = 1,
velocity, v = (2e
−0.9
)(2π5 cos 2π5)
+ (sin 2π5)(−2)(0.9)e
−0.9
= 25.5455 cos10π − 0.7318 sin 10π
= 25.5455(1) − 0.7318(0)
= 25.55 cm/s
(Note that cos10π means ‘the cosine of 10π radians’,
not degrees, and cos 10π ≡ cos 2π = 1.)
Now try the following exercise
Exercise 120 Further problems on rates of
change
1. An alternating current, i amperes, is given by
i = 10 sin 2πf t, where f is the frequency in
hertz and t the time in seconds. Determine the
rate of change of current when t = 20 ms, given
that f = 150 Hz.
[3000π A/s]
2. The luminous intensity, I candelas, 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 when V = 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 time t seconds is given by v = V e −t /C R ,
where V , C and R are constants.
Given V = 300 volts, C = 0.12 × 10 −6 F and
R = 4 ×10 6 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 pressure p of the atmosphere at height h
above ground level is given by p = 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 × 10 5 pascals and c = 6.05 × 10 4 at
1450 metres.
[−1.635 Pa/m]
28.2 Velocity and acceleration
When a car moves a distance x metres in a time t seconds
along a straight road, if the velocity v is 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/time graph will not be a straight line. It may
be as shown in Fig. 28.2.
The average velocity over a small time δt and distance
δx is given by the gradient of the chord AB, i.e. the
average velocity over time δt is
δx
δt
.
As δt → 0, the chord AB becomes a tangent, such that
at point A, 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