calculus

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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 -

    Is there a question here?

  • calculus -

    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 -

    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.

  • calculus -

    huh?

  • calculus -

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

  • calculus -

    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 -

    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 -

    i get -17 is that remotely correct?

  • calculus -

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

  • calculus -

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

  • calculus -

    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 -

    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.

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