differential equation problem(i think)

**jeph** · December 16th 2006, 07:00 PM

I didn't know what to title this post

Suppose that the population of bacteria P(t) changes according to

P(t) = 4000(1 + (2t/1+t^2)) where t is given in hours and t is eual to or greater than 0

a) when does the population reach the maximum number?
b) What is the maximum population?

I been looking through this from a previous test and trying to figure out how I got the answer...According to my professor I did some things wrong and got marked down 3 out of 10 points, but I still ended up with the right answer.

**ThePerfectHacker** · December 16th 2006, 07:14 PM

Originally Posted by jeph

P(t) = 4000(1 + (2t/1+t^2)) where t is given in hours and t is eual to or greater than 0

a) when does the population reach the maximum number?
b) What is the maximum population?

First, use paranthesis

$P=4000 \left(1+\frac{2t}{1+t^2} \right)$
$P=4000+\frac{8000t}{1+t^2}$
$\frac{dP}{dt}=\frac{(8000t)'(1+t^2)-(8000t)(1+t^2)'}{(1+t^2)^2}$
$\frac{8000(1+t^2)-8000t(2t)}{(1+t^2)^2}=0$
$8000(1+t^2)-8000t(2t)=0$
$1+t^2-t(2t)=0$
$-2t^2+t^2+1=0$
$t^2=1$ and $t>0$
$t=1$

**jeph** · December 16th 2006, 08:27 PM

what happened to the 2 8000s here?

8000(1+t^2)-8000t(2t)=0

**ThePerfectHacker** · December 17th 2006, 05:54 AM

Originally Posted by jeph

what happened to the 2 8000s here?

8000(1+t^2)-8000t(2t)=0

I divided both sides by 8000.

The reason why I skipped explanations is because since you are in a Calculus class you need to be able to do this stuff well already. Thus, I assumed an explanation was not necessary.

**jeph** · December 17th 2006, 05:46 PM

Oh!

LOL

I didnt think of that....

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