# Thread: Growth/ Decay of bacteria popoulation

1. ## Growth/ Decay of bacteria popoulation

Hi everyone,

I got a question on a math quest book.

The questions said in a lake, the population of bacteria will grow at a rate that is proportional to their current population and that in the absence of any outside factors the population will double every 10 days. It is estimates that, on average, (it is an outside factor), on any given day 8% of them will die. The current population is 100.

Will this population survive or die out eventually?

How could I use differential equation to model this problem?

If I let P(t) be the population changes with time t (in days),
I know that without outside factor, P(t) = 100*2(t/10)
and with that outside factor, P(t) = P(t-1)*0.92
and P(0) = 100.

Then is it true to say that dP/dt = P*2(t/10) ?
Or how could I set up the equation of dP/dt ?
Hope you could give me some inspiration.

Thank you
Ken

2. Originally Posted by ken0605040
Hi everyone,

I got a question on a math quest book.

The questions said in a lake, the population of bacteria will grow at a rate that is proportional to their current population and that in the absence of any outside factors the population will double every 10 days. It is estimates that, on average, (it is an outside factor), on any given day 8% of them will die. The current population is 100.

Will this population survive or die out eventually?

How could I use differential equation to model this problem?

If I let P(t) be the population changes with time t (in days),
I know that without outside factor, P(t) = 100*2(t/10)
and with that outside factor, P(t) = P(t-1)*0.92
and P(0) = 100.

Then is it true to say that dP/dt = P*2(t/10) ?
Or how could I set up the equation of dP/dt ?
Hope you could give me some inspiration.

Thank you
Ken
To get P(t) = 100*2(t/10) , which ODE did you solve?

3. Should it be dP/dt = P*2(t/10)?

But it doesn't consider the factor that 8% of population decrease.

4. Originally Posted by ken0605040
Should it be dP/dt = P*2(t/10)?

No....

How did you get the function without getting the ODE?

The initial ODE (w/o) external factor is:

$\dfrac{dP}{dt} = kP$

Now continue.....

But it doesn't consider the factor that 8% of population decrease.

You'll have to add the population decrease term into the ODE.

You cannot solve the problem, if you do not know how to set up the ODE without the external factor.
,

5. Thanks, it makes more sense now.

If I combine the external factor into ODE, can I say

dP/dt = kP*0.92^t

if so, then I can find k by using
P(0) = 100

Am I right? But how can I treat 'the population will double every 10 days'?
Can I say P(10) = 200 ?

It is still a bit confusing.

6. Originally Posted by ken0605040
Thanks, it makes more sense now.

If I combine the external factor into ODE, can I say

dP/dt = kP*0.92^t

if so, then I can find k by using
P(0) = 100

Am I right? But how can I treat 'the population will double every 10 days'?
Can I say P(10) = 200 ?

It is still a bit confusing.
Before combining the external effect, calculate the valuue of 'kold' - then write the ODE for the external effect.

7. dP dt Do you mean after finding Kold by using
1. dP/dt = kP
2. P(0) = 100

then, I should work on the ODE for the external effect(i.e. the 8% decreasing population per day)?
And the ODE become dP/dt = koldP*0.92^t ?

Sorry that I am not sure if it is what you mean.

8. Originally Posted by ken0605040
dP dt Do you mean after finding Kold by using
1. dP/dt = kP
2. P(0) = 100

yes ..... and P(10) = 200

then, I should work on the ODE for the external effect(i.e. the 8% decreasing population per day)?
And the ODE become dP/dt = koldP*0.92^t ?

No... how are you getting the "^t" term included in ODE?

Please pay attention to what you are writing!

Sorry that I am not sure if it is what you mean.

.

9. Yes, it is not correct at all.

I want to ask if the result is something like

P = P0 e(koldknew*t)

and by using P(1) = 92 to find knew ?

10. Originally Posted by ken0605040
Yes, it is not correct at all.

I want to ask if the result is something like

P = P0 e(koldknew*t)

and by using P(1) = 92 to find knew ?
N0....

Before the introduction of external factor:

$\dfrac{dP}{dt} \ = \ k_{old}*P$

After the introduction of external factor:

$\dfrac{dP}{dt} \ = \ k_{old}*P - 0.08*P \ = \ (k_{old} - 0.08) * P$

Now continue......

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