If the population was 200,000 bacteria when observations began, what will the population be 12 hours later?

Question

Answer the given problem.

The population P(t) of a bacterial colony t hours after observation begins is

found to be changing at the rate dp/dt=200e^0.1t+150e^-0.03t. If the population

was 200,000 bacteria when observations began, what will the population be

12 hours later?

Include:

Important Notations

Useful Given

Solution:

Initial function

The population

function

value of C

Final Answer

Interpretation

Answer 1

We can use integration to get p(t), the population of bacteria at any time t. Just integrate the entire RHS and you find

p(t)=(1/0.1)*200*e^0.1t+(1/-0.03)*150*e^-0.03t+C

p(t)=2000e^0.1t-5000e^-0.03t+C

C is the usual constant of integration. We can find C by using the initial condition p(0)=200000.

p(0)=2000e^0.1*0-5000e^-0.03*0+C=2000-5000+C=C-3000=200000

C=203000

So now we know

p(t)=2000e^0.1t-5000e^-0.03t+203000

Here we have a problem (unless part of the problem got omitted), which is that we don't know the units of t in this expression. If t is in hours, then great, we just set t=12 and calculate. If t is in days, though, then we'd need to set t=0.5 (12 hours is half a day). So on for if t is in weeks, or months, or years. And obviously this can have a large effect on the answer we get.

I will assume that t is in fact in hours and get

p(12)=2000e^0.1*12-5000e^-0.03*12+203000

p(12)=206151.85 bacteria