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)=◻

Question

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)=◻

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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 $r$: $37 = r \cdot 172 \left(1 - \frac{172}{673}\right)$.Calculate Value: Calculate the value inside the parentheses: $1 - \frac{172}{673} = \frac{673 - 172}{673} = \frac{501}{673}$.Solve for r: Now, solve for $r$: $37 = r \cdot 172 \cdot \frac{501}{673}$.Isolate r: Divide both sides by $172 \cdot \frac{501}{673}$ to isolate $r$: $r = \frac{37}{172 \cdot \frac{501}{673}}$.Perform Division: Perform the division to find $r$: $r = \frac{37 \cdot 673}{172 \cdot 501}$.Calculate r: Calculate the value of $r$: $r \approx \frac{24841}{86172} \approx 0.2883$ 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: $\frac{dP}{dt} = 0.2883P\left(1 - \frac{P}{673}\right)$.

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