Question

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?

Answer 1

Is there a question here?

Answer 2

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?

Answer 3

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 12-25=-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.

Answer 4

huh?

Answer 5

i'm new at this, please show me how to answer the problem. thanks

Answer 6

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

Answer 7

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.

Answer 8

i get -17 is that remotely correct?

Answer 9

What is -17? It seems to form part of the answer. Please elaborate.

Answer 10

never mind. please help me w/the question from the beginning. thanks

Answer 11

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/(t&sup2+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.

Answer 12

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.