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The rate of change
(dP)/(dt) of the number of deer on an island is modeled by a logistic differential equation. The maximum capacity of the island is 673 deer. At
8PM, the number of deer on the island is 172 and is increasing at a rate of 37 deer per day. Write a differential equation to describe the situation.

(dP)/(dt)=◻

The rate of change $\frac{d P}{d t}$ of the number of deer on an island is modeled by a logistic differential equation. The maximum capacity of the island is $673$ deer. At $8 \mathrm{PM}$ , the number of deer on the island is $172$ and is increasing at a rate of $37$ deer per day. Write a differential equation to describe the situation. $\newline$ $\frac{d P}{d t}=\square$

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Write exponential functions: word problems

Full solution

Q. The rate of change

\frac{d P}{d t}

of the number of deer on an island is modeled by a logistic differential equation. The maximum capacity of the island is

673

deer. At

8 \mathrm{PM}

, the number of deer on the island is

172

and is increasing at a rate of

37

deer per day. Write a differential equation to describe the situation.

\newline

\frac{d P}{d t}=\square

Logistic Growth Model: The logistic growth model can be represented by the differential equation $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$ , where $P$ is the population at time $t$ , $r$ is the intrinsic growth rate, and $K$ is the carrying capacity of the environment.
Given Carrying Capacity: We are given the carrying capacity $K = 673$ deer. This value will be used in our differential equation.
Population Data Given: We are also given that at a certain time ( $8$ PM), the population $P = 172$ deer and the rate of change of the population $\frac{dP}{dt} = 37$ deer per day. We can use these values to solve for the intrinsic growth rate $r$ .
Substitute Values: Substitute $P = 172$ and $\frac{dP}{dt} = 37$ into the logistic growth model equation to solve for $r$ : $\newline$ $37 = r \cdot 172 \left(1 - \frac{172}{673}\right)$ .
Calculate Value: Calculate the value inside the parentheses: $\newline$ $1 - \frac{172}{673} = \frac{673 - 172}{673} = \frac{501}{673}$ .
Solve for r: Now, solve for $r$ : $\newline$ $37 = r \cdot 172 \cdot \frac{501}{673}$ .
Isolate r: Divide both sides by $172 \cdot \frac{501}{673}$ to isolate $r$ : $\newline$ $r = \frac{37}{172 \cdot \frac{501}{673}}$ .
Perform Division: Perform the division to find $r$ : $\newline$ $r = \frac{37 \cdot 673}{172 \cdot 501}$ .
Calculate r: Calculate the value of $r$ : $\newline$ $r \approx \frac{24841}{86172} \approx 0.2883$ per day (rounded to four decimal places).
Write Differential Equation: Now that we have the value of $r$ , we can write the logistic differential equation for the deer population: $\newline$ $\frac{dP}{dt} = 0.2883P\left(1 - \frac{P}{673}\right)$ .

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