ABSTRACT
Problem 1. The length l metres of a certain metal rod at temperature θ◦C is given by l=1+0.00005θ +0.0000004θ2. Determine the rate of change of length, in mm/◦C, when the temperature is (a) 100◦C and (b) 400◦C
The of
Since length l = 1+ 0.00005θ + 0.0000004θ2, then
dl dθ
= 0.00005+ 0.0000008θ
(a) When θ = 100◦C, dl dθ
= 0.00005+ (0.0000008)(100) = 0.00013m/◦C = 0.13mm/◦C
(b) When θ = 400◦C, dl dθ
= 0.00005+ (0.0000008)(400) = 0.00037m/◦C = 0.37mm/◦C
Problem 2. The luminous intensity I candelas of a lamp at varying voltage V is given by: I =4× 10−4 V2. 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
d I dV
Since I = 4× 10−4V2, d I dV
= (4× 10−4)(2)V
= 8× 10−4V When the light is increasing at 0.6 candelas per volt then +0.6=8×10−4 V, from which, voltage
V = 0.6 8×10−4 =0.075×10
Problem 3. Newton’s law of cooling is given by: θ =θ0e−kt , where the excess of temperature at zero time is θ0◦C and at time t seconds is θ◦C. Determine the rate of change of temperature after 40 s, given that θ0=16◦C and k=−0.03
The rate of change of temperture is dθ dt
Since θ = θ0e−kt then dθdt = (θ0)(−k)e −kt
= −kθ0e−kt
When θ0=16, k=−0.03 and t=40 then dθ dt
= −(−0.03)(16)e−(−0.03)(40)
= 0.48e1.2 = 1.594◦C/s
Problem s cm of of a stiff spring at time t seconds is given by: s= ae−kt sin2π 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 sin2π f t (i.e. a product) Using the product rule,
ds dt
= (ae−kt )(2π f cos2π f t) +(sin 2π f t)(−ake−kt )
When a=2, k=0.9, f =5 and t =1,
velocity, v = (2e−0.9)(2π5cos2π5) +(sin2π5)(−2)(0.9)e−0.9
= 25.5455cos10π − 0.7318sin10π = 25.5455(1)− 0.7318(0) = 25.55 cm/s
(Note that cos10π means ‘the cosine of 10π radians’, not degrees, and cos10π ≡ cos2π =1)
Now try the following Practice Exercise
Practice Exercise 230 Further problems on rates of change (answers on page 1136)
1. An alternating current, i amperes, is given by i =10sin2π f t , where f is the frequency in hertz and t the time in seconds. Determine the rate of change of currentwhen t=20ms, given that f =150Hz.
