Posted by ariana` on Tuesday, August 4, 2009 at 10:26am.
Bacterial Population: A bacterial colony is estimated to have a population of
P(t)=(24t+10)/(t^2+1)
Million t hours after the introduction or a toxin.
At what rate is the population changing in 1 hour after the toxin is introduced (t=1)? Is the population increasing or decreasing at this time?
At what time does the population begin to decline?

calculus  bobpursley, Tuesday, August 4, 2009 at 10:54am
Is there a question here?

calculus  ariana`, Tuesday, August 4, 2009 at 11:04am
yes,
A Bacterial colony is estimated to have a population of million t hours after the introduction of a toxin
(a) At what rate is the population changing during 1 hr after the toxin is introduced (t=1)? Is the population increasing or decreasing @ this time?
(b) At what time does the population begin to decline? 
calculus  bobpursley, Tuesday, August 4, 2009 at 11:32am
Take the derivative
P' = dP/dt=24/(t^2+1) + ((24t+1)*(1)(t^2+1)^2 (2t)
a) for rate,put t=1 and compute. I get about 1225=13 check that. It is reducing (negative)
b) when does it decline is the same question as when is it max?
set P'=0, solve for t. 
calculus  ariana`, Tuesday, August 4, 2009 at 11:35am
huh?

calculus  ariana`, Tuesday, August 4, 2009 at 11:36am
i'm new at this, please show me how to answer the problem. thanks

calculus  ariana`, Tuesday, August 4, 2009 at 11:45am
thank you for this, can you just please explain in a little more detail how i would go about getting the answer t question b? thanks

calculus  MathMate, Tuesday, August 4, 2009 at 11:51am
You need to start with question a.
Redo the differentiation of a quotient:
d(u/v) = (u dv  v du)/v²
using
u=24t+10
v=t^2+1
If you cannot do this part, backtrack on your notes to make sure you can do it. Otherwise you will have a difficulty doing other exercises. 
calculus  ariana`, Tuesday, August 4, 2009 at 11:57am
i get 17 is that remotely correct?

calculus  MathMate, Tuesday, August 4, 2009 at 11:59am
What is 17? It seems to form part of the answer. Please elaborate.

calculus  ariana`, Tuesday, August 4, 2009 at 12:04pm
never mind. please help me w/the question from the beginning. thanks

calculus  MathMate, Tuesday, August 4, 2009 at 12:18pm
I will resume what Mr. Bob suggested.
Mr. Bob provided you with the derivative of the Population P(x) as:
P'(x)=dP/dx
=24/(t²+1)  2t(24t+10)/(t²+1)²
after minor editorial corrections.
Substitute t=1 into the derivative (P'(x)) to find the rate of change.
At the maximum when the population starts to decline, the value of the derivative is zero, namely P'(x)=0.
You will note that P'(x) before 1 hour because at t=1, P'(x) is already negative.
I strongly suggest to work out the expression P'(x) from first principles and by yourself to fully understand your course content. Use the above expression as a check only.
Tell us what you get for P'(t) for t=1, i.e. P'(1)=? It happens to be a prime number under 10, but negative. 
calculus  MathMate, Tuesday, August 4, 2009 at 12:20pm
The time when the population starts to decline is a whole number of minutes under one hour.
I will be out for the next hour or so, but will be back.
Post your results for check when you can.