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The rate of change 
(dP)/(dt) of the number of fox at a national park is modeled by a logistic differential equation. The maximum capacity of the park is 888 fox. At 8 PM, the number of fox at the national park is 210 and is increasing at a rate of 36 fox 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 fox at a national park is modeled by a logistic differential equation. The maximum capacity of the park is 888888 fox. At 88 PM, the number of fox at the national park is 210210 and is increasing at a rate of 3636 fox 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 fox at a national park is modeled by a logistic differential equation. The maximum capacity of the park is 888888 fox. At 88 PM, the number of fox at the national park is 210210 and is increasing at a rate of 3636 fox 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: \newline(dPdt)=rP(1PK)(\frac{dP}{dt}) = rP(1 - \frac{P}{K})\newlinewhere:\newline- (dPdt)(\frac{dP}{dt}) is the rate of change of the population PP with respect to time tt,\newline- rr is the intrinsic growth rate of the population,\newline- PP is the current population size,\newline- KK is the carrying capacity of the environment (the maximum population size that the environment can sustain indefinitely).\newlineIn this case, we need to find the value of rr since we have the carrying capacity KK and the current rate of change (dPdt)(\frac{dP}{dt}) at a specific population size PP.
  2. Given Parameters: We are given:\newline- The carrying capacity K=888K = 888 foxes,\newline- The current population size P=210P = 210 foxes,\newline- The current rate of change dPdt=36\frac{dP}{dt} = 36 foxes per day.\newlineWe can use the given rate of change to find the intrinsic growth rate rr by rearranging the logistic equation:\newline36=r×210×(1210888)36 = r \times 210 \times (1 - \frac{210}{888})
  3. Calculate Intrinsic Growth Rate: Now we solve for rr:36=r×210×(1210/888)36 = r \times 210 \times (1 - 210/888)36=r×210×(678/888)36 = r \times 210 \times (678/888)36=r×210×(0.7635)36 = r \times 210 \times (0.7635)36=r×160.33536 = r \times 160.335r=36/160.335r = 36 / 160.335r0.2245r \approx 0.2245 per day
  4. Write Logistic Differential Equation: Now that we have the value of rr, we can write the logistic differential equation for this situation:\newlinedPdt=0.2245×P×(1P888)\frac{dP}{dt} = 0.2245 \times P \times (1 - \frac{P}{888})

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