The rate of change (dP)/(dt) of the number of algae in a tank is modeled by a logistic differential equation. The maximum capacity of the tank is 630 algae. At 1PM, the number of algae in the tank is 227 and is increasing at a rate of 39 algae per minute. Write a differential equation to describe the situation. (dP)/(dt)=◻

Question

The rate of change  (dP)/(dt) of the number of algae in a tank is modeled by a logistic differential equation. The maximum capacity of the tank is 630 algae. At  1PM, the number of algae in the tank is 227 and is increasing at a rate of 39 algae per minute. Write a differential equation to describe the situation.  (dP)/(dt)=◻

Bytelearn · Accepted Answer

Logistic Differential Equation: The logistic differential equation is generally given by the formula: $$ \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) $$ where $ P $ is the population at time $ t $, $ r $ is the intrinsic growth rate, and $ K $ is the carrying capacity of the environment. In this context, $ K $ is the maximum capacity of the tank, which is 630 algae.Calculate Intrinsic Growth Rate: We are given that at 1 PM, the number of algae is 227 and is increasing at a rate of 39 algae per minute. This gives us the value of $ \frac{dP}{dt} $ when $ P = 227 $. So, we can plug these values into the logistic equation to solve for $ r $: $$ 39 = r \cdot 227 \left(1 - \frac{227}{630}\right) $$Solve for r: Now we solve for $ r $: $$ 39 = r \cdot 227 \left(\frac{630 - 227}{630}\right) $$ $$ 39 = r \cdot 227 \left(\frac{403}{630}\right) $$ $$ r = \frac{39}{227 \cdot \frac{403}{630}} $$ $$ r = \frac{39 \cdot 630}{227 \cdot 403} $$ $$ r \approx \frac{24570}{91361} $$ $$ r \approx 0.269 $$Write Logistic Differential Equation: Now that we have the value of $ r $, we can write the logistic differential equation for this situation: $$ \frac{dP}{dt} = 0.269P\left(1 - \frac{P}{630}\right) $$

Full solution

More problems from Interpreting Linear Expressions

Related topics