exp and log q (1 Viewer)

[pLuvia]

The natural rate of growth of a colony of micro-organism in a liquid is k₁ organisms per organism per minute. If the organisms are removed at the rate of k₂ organisms per minute, write down a differential equation to describe the variation in size of the colony.

Initially the colony contains N organisms. Find the number of organisms after t minutes and prove that the colony will be removed in a finite time provided that N < k₂ < k₁

 

[insert-username]

Just a note, this isn't exponentials and logs - it's exponential growth and decay, which is under Applications of Calculus to the Physical World.

Anyways, for the first part:

y = Ae^kt (formula for exponential/natural growth, k is a constant)

dy/dt = k.Ae^kt

= k₁y (given - k₁ is the growth rate, is the y is the number of organisms)

The organisms are being removed at a rate of k₂ organisms per minute, so our differential equation is:

dy/dt = k₁y - k₂

Initally, there are N organisms, so

y = Ne^k₁t

dy/dt = k₁.Ne^k₁t - k₂

And that's as far as I can get. Someone else can keep going/correct me. By the way, are you sure that you've posted the whole question? It seems to terminate after N₂₁...


I_F

 

[pLuvia]

Edited sorry, I don't know what happened the last bit didn't show up properly

 

[Mountain.Dew]

ummm this is how i would do it...

dy/dt = (k₁-k₂)N₁

so, using the general formulae that y =Ae^kt (formula for exponential/natural growth, k is a constant), then we get:

y = Ne^(k₁-k₂)t

now since k2 < k1, we know that k2 - k1 < 0, Or k1 - k2 > 0...

this doesnt seem to work...it will only work if k₁-k₂ < 0, also the number of bacteria will never be completely removed, its a feature of y= e^x that there is an asymptote at y=0.

or, we could simply do the method below, but it isnt 'natural' growth:

dy/dt = k₁-k₂

so y = (k₁-k₂)t + c, t = 0, y = N therefore c = N

so y = N + (k₁-k₂)t...