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The rate of change (dP)/(dt) of the number of algae in a test tube is modeled by a logistic differential equation. The maximum capacity of the tube is 580 algae. At 8AM, the number of algae in the test tube is 155 and is increasing at a rate of 31 algae per minute. Write a differential equation to describe the situation. (dP)/(dt)=◻

The rate of change dPdt \frac{d P}{d t} of the number of algae in a test tube is modeled by a logistic differential equation. The maximum capacity of the tube is 580580 algae. At 8AM 8 \mathrm{AM} , the number of algae in the test tube is 155155 and is increasing at a rate of 3131 algae per minute. 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 algae in a test tube is modeled by a logistic differential equation. The maximum capacity of the tube is 580580 algae. At 8AM 8 \mathrm{AM} , the number of algae in the test tube is 155155 and is increasing at a rate of 3131 algae per minute. Write a differential equation to describe the situation.\newlinedPdt=□ \frac{d P}{d t}=\square
  1. Logistic Growth Model Equation: The logistic growth model is given by the differential equation:\newlinedPdt=rP(1−PK)\frac{dP}{dt} = rP(1 - \frac{P}{K})\newlinewhere:\newline- PP is the population at time tt,\newline- rr is the growth rate,\newline- KK is the carrying capacity of the environment (maximum population size).\newlineWe need to find the value of rr and use the given KK value to write the differential equation.
  2. Given Carrying Capacity: We are given the carrying capacity KK as 580580 algae. This value will be used in our differential equation.
  3. Initial Algae Population Data: We are given that at 8AM8\text{AM}, the number of algae is 155155 and is increasing at a rate of 3131 algae per minute. This rate of increase is the value of (dP)/(dt)(dP)/(dt) when P=155P = 155.
  4. Calculate Growth Rate: To find the growth rate rr, we can use the given rate of change when P=155P = 155. Plugging these values into the logistic growth model, we get:\newline31=r×155×(1−155/580)31 = r \times 155 \times (1 - 155/580)\newlineNow we need to solve for rr.
  5. Simplify Fraction: First, simplify the fraction inside the parentheses: 1−155580=1−0.26724≈0.732761 - \frac{155}{580} = 1 - 0.26724 \approx 0.73276
  6. Solve for Growth Rate: Now, plug this value back into the equation and solve for rr:31=r×155×0.7327631 = r \times 155 \times 0.73276r≈31/(155×0.73276)r \approx 31 / (155 \times 0.73276)r≈31/113.5778r \approx 31 / 113.5778r≈0.273r \approx 0.273
  7. Write Differential Equation: Now that we have the value of rr, we can write the logistic differential equation as:\newlinedPdt=0.273P(1−P580)\frac{dP}{dt} = 0.273P(1 - \frac{P}{580})\newlineThis equation models the rate of change of the number of algae in the test tube.

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