basic rates question (1 Viewer)

[rainnwind]

1. a formula for the rate of change in population of a colony of bacteria is given by: P = 3200 + 400 e^kt. If the population double after 20hrs, how long it take to triple the original population.

2. In a certain bacteria culture, the rate of increase is proportioned to the number of bacteria present.
i) if the number doubles in 3 hrs. find the hourly growth rate. [ans: dN/dt = 0.23N]
got (i) but stuck on the rest...
ii) how many bacteria are there after 9hrs if the original population in 10^4? [ans: 8 X 10^4]
iii) after how many hours are there 4 X 10^4? [ans: 6]

 

[lyounamu]

1. a formula for the rate of change in population of a colony of bacteria is given by: P = 3200 + 400 e^kt. If the population double after 20hrs, how long it take to triple the original population.

2. In a certain bacteria culture, the rate of increase is proportioned to the number of bacteria present.
i) if the number doubles in 3 hrs. find the hourly growth rate. [ans: dN/dt = 0.23N]
got (i) but stuck on the rest...
ii) how many bacteria are there after 9hrs if the original population in 10^4? [ans: 8 X 10^4]
iii) after how many hours are there 4 X 10^4? [ans: 6]

1.

P = 3200 + 400 e^kt

When t=0, P = 3200 + 400 x 1 = 3600
When t = 20, P = 3600 x 2 = 7200

so 7200 = 3200 + 400e^20k
4000 = 400e^20k
e^20k = 10
20k = 2.39258509....
k = 0.115129254...


To triple the original population, you need P to be 3600 x 3 = 10800

so 10800 = 3200 + 400 e^(0.115129254... x t)
400 e^(0.115129254.. x t) = 7600
e^(0.115129254...x t) = 19
0.115129254... x t = 2.944...
t = 25.575072....
= 26 hours to nearest hours


2.

ii)

let N = Ae^kt

dN/dt = 0.23N
N = Ae^0.23t

A = 10^4 and t = 9
so N = 10^4 x e^(0.23 x 9) = 79248.23118... = 8 x 10^4 (roughly)

iii) after how many hours are there 4 X 10^4? [ans: 6]

N = 4x10^4
so
4 x 10^4 = 10^4 x e^(0.23 x t)
e^(0.23 x t) = 4
0.23 x t = 1.38629436...
t = 6.027366.... = 6 roughly

 

[rainnwind]

1.

P = 3200 + 400 e^kt

When t=0, P = 3200 + 400 x 1 = 3600
When t = 20, P = 3600 x 2 = 7200

so 7200 = 3200 + 400e^20k
4000 = 400e^20k
e^20k = 10
20k = 2.39258509....
k = 0.115129254...


To triple the original population, you need P to be 3600 x 3 = 10800

so 10800 = 3200 + 400 e^(0.115129254... x t)
400 e^(0.115129254.. x t) = 7600
e^(0.115129254...x t) = 19
0.115129254... x t = 2.944...
t = 25.575072....
= 26 hours to nearest hours


2.

ii)

let N = Ae^kt

dN/dt = 0.23N
N = Ae^0.23t

A = 10^4 and t = 9
so N = 10^4 x e^(0.23 x 9) = 79248.23118... = 8 x 10^4 (roughly)

iii) after how many hours are there 4 X 10^4? [ans: 6]

N = 4x10^4
so
4 x 10^4 = 10^4 x e^(0.23 x t)
e^(0.23 x t) = 4
0.23 x t = 1.38629436...
t = 6.027366.... = 6 roughly

i worked out q2 after i posted it... =o=
anyway, THANKS!!