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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 dPdt \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 541541 deer. At 12PM 12 \mathrm{PM} , the number of deer on the island is 228228 and is increasing at a rate of 1515 deer per day. Write a differential equation to describe the situation.\newlinedPdt=□ \frac{d P}{d t}=\square

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Q. The rate of change dPdt \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 541541 deer. At 12PM 12 \mathrm{PM} , the number of deer on the island is 228228 and is increasing at a rate of 1515 deer per day. Write a differential equation to describe the situation.\newlinedPdt=□ \frac{d P}{d t}=\square
  1. Logistic Differential Equation: The logistic differential equation is generally given by the formula dPdt=rP(1−PK)\frac{dP}{dt} = rP(1 - \frac{P}{K}), where PP is the population at time tt, rr is the intrinsic growth rate, and KK is the carrying capacity of the environment. In this case, KK is given as 541541 deer, which is the maximum capacity of the island.
  2. Find Intrinsic Growth Rate: To find the intrinsic growth rate rr, we use the information that at 1212PM, the number of deer is 228228 and is increasing at a rate of 1515 deer per day. We can plug these values into the logistic growth equation dPdt=rP(1−PK)\frac{dP}{dt} = rP(1 - \frac{P}{K}) and solve for rr.
  3. Substitute Values and Solve: Substituting the given values into the logistic growth equation, we get:\newline15=r×228×(1−228/541)15 = r \times 228 \times (1 - 228/541)\newlineNow we need to solve for rr.
  4. Simplify Fraction: First, simplify the fraction inside the parentheses: 1−228541=541−228541=3135411 - \frac{228}{541} = \frac{541 - 228}{541} = \frac{313}{541}
  5. Divide to Solve for r: Now, substitute the simplified fraction back into the equation:\newline15=r×228×(313/541)15 = r \times 228 \times (313/541)\newlineNext, we solve for r by dividing both sides of the equation by (228×313/541)(228 \times 313/541).
  6. Calculate Intrinsic Growth Rate: r=15(228×313541)r = \frac{15}{(228 \times \frac{313}{541})}\newliner=15(228×313)×541r = \frac{15}{(228 \times 313)} \times 541\newliner=15×541(228×313)r = \frac{15 \times 541}{(228 \times 313)}\newliner≈0.0089r \approx 0.0089 deer per deer per day (rounded to four decimal places)
  7. Write Logistic Differential Equation: Now that we have the value of rr, we can write the logistic differential equation for this situation:\newlinedPdt=0.0089×P×(1−P541)\frac{dP}{dt} = 0.0089 \times P \times (1 - \frac{P}{541})

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