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The rate of change
(dP)/(dt) of the number of wolves at a national park is modeled by a logistic differential equation. The maximum capacity of the park is 955 wolves. At 8 PM, the number of wolves at the national park is 218 and is increasing at a rate of 26 wolves 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 wolves at a national park is modeled by a logistic differential equation. The maximum capacity of the park is $955$ wolves. At $8$ PM, the number of wolves at the national park is $218$ and is increasing at a rate of $26$ wolves 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 wolves at a national park is modeled by a logistic differential equation. The maximum capacity of the park is

955

wolves. At

8

PM, the number of wolves at the national park is

218

and is increasing at a rate of

26

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

\newline

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

Calculate Fraction Occupied: First, calculate the fraction of the carrying capacity that is currently occupied by the wolves: $\newline$ $\frac{218}{955}$
Solve for r: Next, we solve for $r$ using the equation we derived: $\newline$ $26 = r \cdot 218 \left(1 - \frac{218}{955}\right)$ $\newline$ First, calculate the term inside the parentheses: $\newline$ $1 - \frac{218}{955} = 1 - 0.2283 = 0.7717$
Calculate Term Inside Parentheses: Finally, divide the rate of change of the population by this product to solve for $r$ : $\newline$ $r = \frac{26}{168.3506}$
Divide Rate of Change: Now we can write the logistic growth differential equation with the values we have found: $\newline$ $\frac{dP}{dt} = 0.1544 \cdot P \left(1 - \frac{P}{955}\right)$ $\newline$ This is the differential equation that models the logistic growth of the number of wolves in the national park.

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