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April 24, 2014

April 24, 2014

Posted by **ariana`** on Tuesday, August 4, 2009 at 10:26am.

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 10:54amIs there a question here?

- calculus -
**ariana`**, Tuesday, August 4, 2009 at 11:04amyes,

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 11:32amTake 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.

- calculus -
**ariana`**, Tuesday, August 4, 2009 at 11:35amhuh?

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

- calculus -
**ariana`**, Tuesday, August 4, 2009 at 11:45amthank 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 11:51amYou 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 11:57ami get -17 is that remotely correct?

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

- calculus -
**ariana`**, Tuesday, August 4, 2009 at 12:04pmnever mind. please help me w/the question from the beginning. thanks

- calculus -
**MathMate**, Tuesday, August 4, 2009 at 12:18pmI 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 12:20pmThe 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.

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