1. ## Growth/Decay problem

I have the following question from a past paper which I seemed to be rusty on and just wanted to go through it:

A virus spreads slowly through a population of animals. Initially no animals are infected, but after 10 days 25% of the population in infected. Let $P(t)$ denote the proportion of the population that is infected. Then

$\frac{dP}{dt} = k(1-P)$

(i) Solve the differential equation to get an expression for P(t).
(ii) Find the value of the constant k.
(iii) How long will it take for 75% of the population to become infected?

My attempt:

(i)

$\integral \frac{dP}{dt} = \integral k(1-P)$

$dP=k(1-P)~dt$

$\int \frac{1}{1-P}~dP = \int k~dt$

$-ln|1-P| = kt~dt$

$-1+P = e^{kt}$

$P(t) = e^{kt} +1$

This seems wrong as at $P(0)$, no animals should be infected, but the equation I derived gives a value of 2. What have I done wrong?

Also, why don't I have a coefficient in front of the $e^{kt}$ term? I have also never seen a constant next to the normal $P_0e^{kt}$ term; is that normal?

I will wait till we have figured out part (i) before I attempt the next two parts.

Thanks.

2. ## Re: Growth/Decay problem

Originally Posted by terrorsquid
I have the following question from a past paper which I seemed to be rusty on and just wanted to go through it:

A virus spreads slowly through a population of animals. Initially no animals are infected, but after 10 days 25% of the population in infected. Let $P(t)$ denote the proportion of the population that is infected. Then

$\frac{dP}{dt} = k(1-P)$

(i) Solve the differential equation to get an expression for P(t).
(ii) Find the value of the constant k.
(iii) How long will it take for 75% of the population to become infected?

My attempt:

(i)

$\integral \frac{dP}{dt} = \integral k(1-P)$

$dP=k(1-P)~dt$

$\int \frac{1}{1-P}~dP = \int k~dt$

$-ln|1-P| = kt~dt$

$-1+P = e^{kt}$

$P(t) = e^{kt} +1$

This seems wrong as at $P(0)$, no animals should be infected, but the equation I derived gives a value of 2. What have I done wrong?

Also, why don't I have a coefficient in front of the $e^{kt}$ term? I have also never seen a constant next to the normal $P_0e^{kt}$ term; is that normal?

I will wait till we have figured out part (i) before I attempt the next two parts.

Thanks.
When integrating you have to include the arbitrary constant of integration '+ C'. Get its value from the initial condition.

3. ## Re: Growth/Decay problem

Is this what you mean?

$-ln|1 -P| = kt + C$

$-1+P=e^{kt}e^{C}$

$let~e^C = P_0$

$P(t) = P_0e^{kt}+1$

$P(0) = P_0 + 1$

Not sure how to proceed... Initially 0 are infected, so that means $P_0 = -1 ?$

4. ## Re: Growth/Decay problem

$\int \frac{dP}{1-P} = \int k \, dt$

$\ln|1-P| = -kt + C$

$1 - P = e^{-kt+C}$ ... let $A = e^C$

$1 - P = Ae^{-kt}$

$P = 1 - Ae^{-kt}$

when $t = 0$ , $P = 0$ (which is very unrealistic for a problem like this)

$A = 1$

$P = 1 - e^{-kt}$

at $t = 10$, $P = 0.25$

$0.25 = 1 - e^{-10k}$

solve for $k$ and finish it.

5. ## Re: Growth/Decay problem

Just out of curiosity, how did you spot to switch the sides of kt and ln|1-P|? After considering the situation presented by the question, it seems we are dealing with an exponential growth problem which would suggest I need to find a formula with a +k value. Typically I would try to get it into the form of +k for growth and -k for decay. I can see now with the final answer that it makes sense, but how would you spot it before getting to the end?

6. ## Re: Growth/Decay problem

Originally Posted by terrorsquid
Just out of curiosity, how did you spot to switch the sides of kt and ln|1-P|? After considering the situation presented by the question, it seems we are dealing with an exponential growth problem which would suggest I need to find a formula with a +k value. Typically I would try to get it into the form of +k for growth and -k for decay. I can see now with the final answer that it makes sense, but how would you spot it before getting to the end?
$\int \frac{1}{1-P} \, dP = \int k \, dt$

$\int \frac{-1}{1-P} \, dP = \int -k \, dt$

... since the derivative of (1-P) is -1

7. ## Re: Growth/Decay problem

So, whenever I have a problem where the term being multiplied by the constant k in the instantaneous growth/decay rate equation: $\frac{dP}{dt}=k(term)$
is negative, I look to find a P(t) term with a -k exponent and vice-versa, regardless of whether the question is about growth or decay?

Continuing:

(i)

$0.25 = 1-e^{-10k}$

$-0.75 = -e^{-10k}$

$e^{-10k}=0.75$

$-10k = ln(0.75)$

$k = 0.02876820725$

(ii)

$0.75 = 1-e^{-0.02876820725t}$

$t = \frac{ln(0.25)}{-0.0287682072}$

$=48.19~days$

8. ## Re: Growth/Decay problem

you're asking me "do I always?" ... understand that I cannot say yes. each problem is different. I just took advantage of being able to multiply both sides by -1 to make the antiderivative easier to find.