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The rate of change (dP)/(dt) of the number of bacteria in a tank is modeled by the following differential equation: (dP)/(dt)=(2)/(9849)P(598-P) At t=0, the number of bacteria in the tank is 196 and is increasing at a rate of 16 bacteria per minute. At what value of P does the graph of P(t) have an inflection point? Answer:

The rate of change dPdt \frac{d P}{d t} of the number of bacteria in a tank is modeled by the following differential equation:\newlinedPdt=29849P(598−P) \frac{d P}{d t}=\frac{2}{9849} P(598-P) \newlineAt t=0 t=0 , the number of bacteria in the tank is 196196 and is increasing at a rate of 1616 bacteria per minute. At what value of P P does the graph of P(t) P(t) have an inflection point?\newlineAnswer:

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Q. The rate of change dPdt \frac{d P}{d t} of the number of bacteria in a tank is modeled by the following differential equation:\newlinedPdt=29849P(598−P) \frac{d P}{d t}=\frac{2}{9849} P(598-P) \newlineAt t=0 t=0 , the number of bacteria in the tank is 196196 and is increasing at a rate of 1616 bacteria per minute. At what value of P P does the graph of P(t) P(t) have an inflection point?\newlineAnswer:
  1. Find First Derivative: To find the inflection point, we need to find the second derivative of PP with respect to tt and set it equal to zero. The second derivative will tell us where the concavity of the graph changes, which is the inflection point.\newlineFirst, let's find the first derivative, which is given by the differential equation:\newlinedPdt=29849P(598−P)\frac{dP}{dt} = \frac{2}{9849}P(598 - P)
  2. Find Second Derivative: Now, we need to find the second derivative d2Pdt2\frac{d^2P}{dt^2}. To do this, we will differentiate dPdt\frac{dP}{dt} with respect to PP and then multiply by dPdt\frac{dP}{dt} again due to the chain rule.\newlined2Pdt2=ddP(dPdt)×dPdt\frac{d^2P}{dt^2} = \frac{d}{dP}\left(\frac{dP}{dt}\right) \times \frac{dP}{dt}
  3. Differentiate First Derivative: Differentiating dPdt\frac{dP}{dt} with respect to PP gives us: ddP(29849P(598−P))=29849(598−2P)\frac{d}{dP}\left(\frac{2}{9849}P(598 - P)\right) = \frac{2}{9849}(598 - 2P)
  4. Multiply Derivatives: Now, we multiply this derivative by (dPdt)(\frac{dP}{dt}) to get the second derivative with respect to time: (d2Pdt2)=(29849)(598−2P)×(29849)P(598−P)(\frac{d^2P}{dt^2}) = (\frac{2}{9849})(598 - 2P) \times (\frac{2}{9849})P(598 - P)
  5. Set Second Derivative Equal to Zero: To find the inflection point, we set the second derivative equal to zero and solve for PP:0=(29849)(598−2P)∗(29849)P(598−P)0 = \left(\frac{2}{9849}\right)\left(598 - 2P\right) * \left(\frac{2}{9849}\right)P\left(598 - P\right)
  6. Simplify Equation: We can simplify the equation by setting the factors equal to zero:\newline598−2P=0598 - 2P = 0 or P(598−P)=0P(598 - P) = 0
  7. Solve for P: Solving the first equation for P gives us:\newline2P=5982P = 598\newlineP=299P = 299
  8. Disregard Endpoints: Solving the second equation for PP gives us two solutions:\newlineP=0P = 0 or 598−P=0598 - P = 0\newlineP=0P = 0 or P=598P = 598
  9. Identify Inflection Point: However, since we are looking for the inflection point where PP is changing, we disregard the solutions P=0P = 0 and P=598P = 598 because they represent the endpoints of the interval where the population exists. The inflection point occurs at P=299P = 299.

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