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P3 Question on rate of change

This Question 6 from the Heinemann P3 book Review exercise 1.
I’m having problems applying the chain rule :s-smilie:

The population, p, of insects on an island, t hours after midday, is given by

p = 1000e^(kt)

where k is a constant

Given that when t = 0, the rate of change of the population with respect to time is 100 per hour,

a) find k.
b) Find also the population of insects on the island when t = 6. Give your answer to 3 sig figs.

Thanks.
Reply 1
a)dp/dt = 1000ke^kt

t = 0, dp/dt = 100

100=1000k
k=1/10

b) p = 1000e^0.1t
t = 6
p = 1000e^0.6 = 1820 3sf
Reply 2
awengraf
This Question 6 from the Heinemann P3 book Review exercise 1.
I’m having problems applying the chain rule :s-smilie:

The population, p, of insects on an island, t hours after midday, is given by

p = 1000e^(kt)

where k is a constant

Given that when t = 0, the rate of change of the population with respect to time is 100 per hour,

a) find k.
b) Find also the population of insects on the island when t = 6. Give your answer to 3 sig figs.

Thanks.


ok so we know that dp/dt =100 when t=0
dp/dt = 1000ke^(kt)
100=1000k
k=0.1


t=6
p=1000e^(.1*6)
p=1000e^(0.6)
p=1820

not sure why you need the chain rule. im pretty sure what i have done is correct(sorry if it is not)
Reply 3
thanks guys
awengraf
The population, p, of insects on an island, t hours after midday, is given by

p = 1000e^(kt)

where k is a constant

Given that when t = 0, the rate of change of the population with respect to time is 100 per hour,

a) find k.
b) Find also the population of insects on the island when t = 6. Give your answer to 3 sig figs.

a.) p = 1000e^(kt) ---> dp/dt = k1000e^(kt)
When t = 0: dp/dt = 100
---> 100 = k1000e^kt
---> 100 = k1000
---> k = 0.1

b.) When t = 6:
p = 1000e^(0.1*6)
= 1000e^(0.6)
= 1820 (3.S.F)
Reply 5
thanks again, i can only give 1 rating per 24 hours :frown: