# Integrals and initial-value problems?

• January 14th 2010, 08:53 PM
oObutterfly-chaserOo
Integrals and initial-value problems?
I thought I understood this problem, until I solved it and got an answer of -4000 bacteria. Here is the problem:

A population of bacteria is changing at a rate of

dP/dt = 3000 / ( 1+ 0.25t)

where t is the time in days. The initial population is 1000. Write an equation that gives te population at any time t, and find the population when t = 3 days.

From this information I know that P(0) = 1000 and I went to integrate the function.

The equation I got was P = 3000 ln(1-0.25t) + k

But I don't think this is right because, well, -4158.5 just doesn't seem like a plausable answer to me. What should I have done differently?
• January 14th 2010, 10:32 PM
Quote:

Originally Posted by oObutterfly-chaserOo
I thought I understood this problem, until I solved it and got an answer of -4000 bacteria. Here is the problem:

A population of bacteria is changing at a rate of

dP/dt = 3000 / ( 1+ 0.25t)

where t is the time in days. The initial population is 1000. Write an equation that gives te population at any time t, and find the population when t = 3 days.

From this information I know that P(0) = 1000 and I went to integrate the function.

The equation I got was P = 3000 ln(1-0.25t) + k

But I don't think this is right because, well, -4158.5 just doesn't seem like a plausable answer to me. What should I have done differently?

You want to integrate like this:

$\int\frac{3,000}{1+0.25t}dt=3000\int\frac{1}{1+0.2 5t}dt$

You want this in the form $\int\frac{f'(t)}{f(t)}dt$, so you need a $.25$ in the numerator. So use the fact that $1=\frac{.25}{.25}$ to write the integral as $\frac{3000}{.25}\int\frac{.25}{1+0.25t}dt=12,000\l n(1+0.25t)+k$

Using your $P(0)=1,000$ gives:

$12,000\ln(1)+k=1,000$.

$k=1,000$

$P(t)=12,000\ln(1+0.25t)+1,000$