# Two ODE problems not sure about

• clynne21
In summary, the conversation discussed modifying the logistic model to account for fishing in a lake stocked with walleye pike. The modified model is P'=.1P(1-P/10)-.1, taking into account the removal of 100 fish daily. The second question involved proving that the rate at which the population is increasing is maximized when the population is at one-half of its carrying capacity, which can be found by setting the second derivative of the logistic equation equal to 0.
clynne21

## Homework Statement

consider a lake that is stocked with walleye pike and that the pike population is governed by P'=.1P(1-P/10) where time is measured in days and P is thousands of fish. Suppose that fishing is started in this lake and that 100 fish are removed daily. modify the logistic model to account for the fishing

P'=.1P(1-P/10)

## The Attempt at a Solution

I am thinking it's just P'=.1P(1-P/10)-.1 but that seems too easy LOL. any thoughts?

## Homework Statement

Suppose a population is growing according to the logistic eqn

dP/dt=rP(1-P/K)

Prove that the rate at which the population is increasing is at its greatest when the population is at one-half of it's carrying capacity. Hint: Consider the second derivative of P

dP/dt=rP(1-P/K)

## The Attempt at a Solution

Absolutely no idea where to start with this one :-(

clynne21 said:

## Homework Statement

consider a lake that is stocked with walleye pike and that the pike population is governed by P'=.1P(1-P/10) where time is measured in days and P is thousands of fish. Suppose that fishing is started in this lake and that 100 fish are removed daily. modify the logistic model to account for the fishing

P'=.1P(1-P/10)

## The Attempt at a Solution

I am thinking it's just P'=.1P(1-P/10)-.1 but that seems too easy LOL. any thoughts?
Looks good to me.

## Homework Statement

Suppose a population is growing according to the logistic eqn

dP/dt=rP(1-P/K)

Prove that the rate at which the population is increasing is at its greatest when the population is at one-half of it's carrying capacity. Hint: Consider the second derivative of P

dP/dt=rP(1-P/K)

## The Attempt at a Solution

Absolutely no idea where to start with this one :-(
the rate at which the population is increasing = dP/dt
is at its greatest = is maximized

You should recall from calculus that when a function attains a local maximum, its derivative is equal to 0. In this problem, the function is dP/dt, and its derivative is therefore d2P/dt2. You want to set that equal to 0 and solve for P.

Next, you want to find the carrying capacity of the system in terms of r and K. Do you know how to find this?

Finally, you just need to show that your first answer is twice your second answer.

vela said:
Looks good to me.

the rate at which the population is increasing = dP/dt
is at its greatest = is maximized

You should recall from calculus that when a function attains a local maximum, its derivative is equal to 0. In this problem, the function is dP/dt, and its derivative is therefore d2P/dt2. You want to set that equal to 0 and solve for P.

Next, you want to find the carrying capacity of the system in terms of r and K. Do you know how to find this?

Finally, you just need to show that your first answer is twice your second answer.

Got through the first question fine and did take the second derivative of the equation, but once I do that all variables are codependent on each other so setting it equal to zero makes everything zero.

d2P/dt2= -2Pr/K

so I'm kind of at a loss of how to make that a maximum. I must be missing something. Thank you for the help! I do know how to find the carrying capacity.

You calculated the second derivative incorrectly. You have to use the product rule, or you can just multiply dP/dt out first:

$$\frac{dP}{dt} = rP - \frac{r}{K} P^2$$

and then differentiate each term separately. Don't forget you're differentiating with respect to t, not P.

vela said:
and then differentiate each term separately. Don't forget you're differentiating with respect to t, not P.

That's where I messed up! Was very tired last night LOL. Thanks for straightening me out!

## 1. What is an ODE?

An ODE, or ordinary differential equation, is a mathematical equation that describes the relationship between a function and its derivatives. It is used to model many physical and natural phenomena in science and engineering.

## 2. What is the difference between a linear and a nonlinear ODE?

A linear ODE is an equation where the dependent variable and its derivatives appear in a linear form, meaning they are raised to the first power and are not multiplied together. A nonlinear ODE, on the other hand, contains terms where the dependent variable and its derivatives are raised to powers other than one and/or are multiplied together.

## 3. What are initial conditions in the context of ODEs?

Initial conditions are values given to the dependent variable and its derivatives at a specific point in the domain of the ODE. They are required to uniquely determine a solution to the ODE.

## 4. How are ODEs solved?

ODEs can be solved using various analytical and numerical methods. Analytical methods involve finding an exact solution to the equation, while numerical methods use algorithms to approximate the solution. The choice of method depends on the complexity of the ODE and the desired accuracy.

## 5. What are some real-world applications of ODEs?

ODEs are used in many fields of science and engineering, including physics, chemistry, biology, economics, and engineering. They can be used to model phenomena such as population growth, chemical reactions, motion of objects, and electrical circuits.

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