Rate of increase/differential equation

[Gwilim]

The Problem:

For a species of bacterium, the rate of increase of the population is determined by the following information:

the rate of increase is proportional to the population P(t)

the rate of increase is proportional to the temperature T(t)

My question:

I initially wrote out two separate equations for this, namely dP/dt = aP(t) and dP/dt = bT(t), then using the initial conditions P(0)=1000 and P(10)=2000 and the information T(t)=40-2t solved each, one giving an exponential and the other a polynomial.

Should I have instead solved the equation dP/dt = kP(t)T(t)?

I'm pretty sure the answer is 'yes' but I'd like some confirmation anyway.

 

[Gwilim]

incidentally I ultimately have to find P(t) at t=20, and using the method I just described I get 1000 times the cube root of 16. Is that correct?

 

[Pinu7]

Yes, solve for dP/dt = kP(t)T(t). Think way back from algebra.

 

[Gwilim]

Thanks

incidentally I ultimately have to find P(t) at t=20, and using the method I just described I get 1000 times the cube root of 16. Is that correct?

I would like confirmation for this also

 

[Cyosis]

Giving you confirmation would mean that we would have to solve your entire problem first and then compare it to your answer. Secondly we don't even know to which method your solution belongs, the wrong one or the correct one?

If you want confirmation show us your work step by step so we can see where and if you made mistakes. We will then help you from thereon.

 

[Gwilim]

Yeah okay.

P'(t) = k P(t)T(t)

T=(40-2t)

P'(t)/P(t) = 40k - 2kt

P^(-1)dP = (40k -2kt)dt

lnP = 40kt - kt^2 + C

P(t) = Ce^(40kt - kt^2)

P(0) = 1000 => C=1000

P(t) = 1000e^(40kt - kt^2)

P(10) = 2000 = 1000e^300k => k = (ln2)/300

P(t) = 1000e^(((2/15)ln2)t - ((1/300)ln2)T^2)

P(20) = 1000e^((4/3)ln2)
P(20) = 1000*(16^(1/3))

 

[Cyosis]

Every step looks correct to me. A slightly nicer way to write it would perhaps be 2000 2^(1/3), put that's purely aesthetic.

 

[Gwilim]

thanks again. I'll probably be posting one or two more of these this weekend