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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 541 deer. At
12PM, the number of deer on the island is 228 and is increasing at a rate of 15 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 $541$ deer. At $12 \mathrm{PM}$ , the number of deer on the island is $228$ and is increasing at a rate of $15$ 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

541

deer. At

12 \mathrm{PM}

, the number of deer on the island is

228

and is increasing at a rate of

15

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

\newline

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

Logistic Differential Equation: The logistic differential equation is generally given by the formula $\frac{dP}{dt} = rP(1 - \frac{P}{K})$ , where $P$ is the population at time $t$ , $r$ is the intrinsic growth rate, and $K$ is the carrying capacity of the environment. In this case, $K$ is given as $541$ deer, which is the maximum capacity of the island.
Find Intrinsic Growth Rate: To find the intrinsic growth rate $r$ , we use the information that at $12$ PM, the number of deer is $228$ and is increasing at a rate of $15$ deer per day. We can plug these values into the logistic growth equation $\frac{dP}{dt} = rP(1 - \frac{P}{K})$ and solve for $r$ .
Substitute Values and Solve: Substituting the given values into the logistic growth equation, we get: $\newline$ $15 = r \times 228 \times (1 - 228/541)$ $\newline$ Now we need to solve for $r$ .
Simplify Fraction: First, simplify the fraction inside the parentheses: $1 - \frac{228}{541} = \frac{541 - 228}{541} = \frac{313}{541}$
Divide to Solve for r: Now, substitute the simplified fraction back into the equation: $\newline$ $15 = r \times 228 \times (313/541)$ $\newline$ Next, we solve for r by dividing both sides of the equation by $(228 \times 313/541)$ .
Calculate Intrinsic Growth Rate: $r = \frac{15}{(228 \times \frac{313}{541})}$ $\newline$ $r = \frac{15}{(228 \times 313)} \times 541$ $\newline$ $r = \frac{15 \times 541}{(228 \times 313)}$ $\newline$ $r \approx 0.0089$ deer per deer per day (rounded to four decimal places)
Write Logistic Differential Equation: Now that we have the value of $r$ , we can write the logistic differential equation for this situation: $\newline$ $\frac{dP}{dt} = 0.0089 \times P \times (1 - \frac{P}{541})$

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