question about growth and decay (1 Viewer)

[sasquatch]

In the formula N = N₀e^kt, does k stand for the "growth/decay rate". Because i thought the rate was dN/dt = kN. But questions like the following:

The population of a town has its rate of growth proportional to the polulation. If the anual growth rate is 0.02, and the original poluation is 2400, what would be the population in.. whatever.

So to establish the growth/decay formula (growth in this case because of the proportionality), dP/dt = kP, P₀ = 2400, and k = 0.02?

It wouldnt make sense to say dP/dt = 0.02, and hence solve for k, kP = 0.02, because P is a variable. So im confused...

 

[Mountain.Dew]

In the formula N = N₀e^kt, does k stand for the "growth/decay rate". Because i thought the rate was dN/dt = kN. But questions like the following:

The population of a town has its rate of growth proportional to the polulation. If the anual growth rate is 0.02, and the original poluation is 2400, what would be the population in.. whatever.

So to establish the growth/decay formula (growth in this case because of the proportionality), dP/dt = kP, P₀ = 2400, and k = 0.02?

It wouldnt make sense to say dP/dt = 0.02, and hence solve for k, kP = 0.02, because P is a variable. So im confused...

small mistake there. dP/dt = kP = 0.02P, not 0.02.

 

[sasquatch]

I think you missed the point of my question i said

It wouldnt make sense to say dP/dt = 0.02

But a rate, is the change in something with respect to time (i.e. dy/dt). And by saying that dP/dt = 0.02P, that means that k is the "anual growth rate is 0.02".

So yeah you didnt read what i said properly..

 

[Mountain.Dew]

But a rate, is the change in something with respect to time (i.e. dy/dt).

agreed.

And by saying that dP/dt = 0.02P, that means that k is the "anual growth rate is 0.02".

true, but what dP/dt represents is the change in population size over time, not the change in RATE over time.

please read this: The population of a town has its rate of growth proportional to the population.

this means the rate of growth = dP/dt = kP=0.02P, not simply 0.02.

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i think we are going in circles. i will start again.

query: So to establish the growth/decay formula (growth in this case because of the proportionality), dP/dt = kP, P0 = 2400, and k = 0.02?

answer: thats right.

query: It wouldnt make sense to say dP/dt = 0.02, and hence solve for k, kP = 0.02

answer: wholeheartedly in agreement.

 

[sasquatch]

true, but what dP/dt represents is the change in population size over time, not the change in RATE over time.

hehe cool i think i get it now thanks!

 

[sasquatch]

For questions involving growth and decay, can you just quote the formulae

N = N₀e^kt, N = P + Ae^kt and etc. or is it neccessary to establish them using the expressions, dN/dt = k(N-P)?

Cuz the proof is kinda confusing..

 

[Mountain.Dew]

hehe cool i think i get it now thanks!

*phew* hope it helped

For questions involving growth and decay, can you just quote the formulae

N = N₀e^kt, N = P + Ae^kt and etc. or is it neccessary to establish them using the expressions, dN/dt = k(N-P)?

Cuz the proof is kinda confusing..

in the 3U exam they should give u the form that u will use, either the N = N₀e^kt form, or the N = P + Ae^kt form. if they dont (which is unusual), see if u can derive it yourself.

query: is it neccessary to establish them using the expressions, dN/dt = k(N-P)?

answer: actually, quite the opposite. usually a (i) would be to go from N = P + Ae^kt ==> dN/dt = k(N-P), not the other way around. advised not to go from dN/dt --> N.

 

[Riviet]

advised not to go from dN/dt --> N.

This method is definitely more time-consuming and challenging than simply differentiating, I believe this used to be in the syllabus and students were expected to know how to derive it by integration. Luckily for us, we are encouraged to use differentiation, the much easier way.