Assume that the rate of population growth of ants is given by the equation:

dP/dt=.1P(1 - P/730)-(129000/7300)

a. If P(0)=150, what happens to P as t gets very large?

b. If P(0)=800, what happens to P as t gets very large?

c. If P(0)=800, what happens to dP/dt as t gets very large?

d. What are the equilibrium solutions?

a. If P(0)=150, we can first find dP/dt when P=150:

dP/dt = 0.1(150)(1 - 150/730) - (129000/7300)
dP/dt = 15(1 - 0.2055) - 17.67
dP/dt ≈ 8.325

Since dP/dt is positive when P = 150, this means the population is increasing at this point in time. We can also analyze the equation further to determine what happens as t gets very large:

dP/dt = 0.1P(1 - P/730) - (129000/7300)
dP/dt = (0.1P - 0.1P^2/730) - 17.67
dP/dt = -0.1P^2/730 + 0.1P - 17.67

Looking at the equation, as t gets very large and P increases, the first term becomes more negative, causing dP/dt to decrease. Eventually, this will cause dP/dt to become negative, which means the population will start decreasing.

b. If P(0)=800, we can first find dP/dt when P=800:
dP/dt = 0.1(800)(1 - 800/730) - (129000/7300)
dP/dt = 80(-0.0959) - 17.67
dP/dt ≈ -24.142

Since dP/dt is negative when P = 800, this means the population is decreasing at this point in time. As the population continues to decrease, dP/dt will eventually approach the equilibrium solution.

c. If P(0)=800, as t gets very large, dP/dt will approach the equilibrium solution as well.

d. The equilibrium solutions are when dP/dt = 0:
0 = -0.1P^2/730 + 0.1P - 17.67

We can multiply through by 7300 to remove the fractions and simplify:
0 = -10P^2 + 730P - 128990

We can solve the quadratic equation for P using the quadratic formula:
P = (-b ± √(b²-4ac)) / 2a
P = (-730 ± √(730² - 4(-10)(-128990))) / (-20)

P ≈ 129 or P ≈ 10

Therefore, there are two equilibrium solutions, P = 129 and P = 10.

To answer these questions, we'll analyze the given differential equation in parts:

a. If P(0) = 150, we need to determine the long-term behavior of P as t gets very large. To do this, we can first look at the right-hand side of the equation:
dP/dt = 0.1P(1 - P/730) - (129000/7300)

As t gets very large, we consider the steady-state or equilibrium solutions, where dP/dt = 0. Setting dP/dt = 0 and solving for P gives us the equilibrium points:
0.1P(1 - P/730) - (129000/7300) = 0

Simplifying this equation, we get:
0.1P(1 - P/730) = (129000/7300)

Now, solve for P:
1 - P/730 = 1290/73

- P/730 = 1290/73 - 1
- P/730 = (1290 - 73)/73

P/730 = (1290 - 73)/73

P = (1290 - 73) * (730/73)
P = 14 * 10 = 140

The equilibrium solution is P = 140. If P(0) = 150, which is greater than the equilibrium value, P will decrease towards 140 as time approaches infinity.

b. If P(0) = 800, we can follow the same process as in part a to find the equilibrium solution. Setting dP/dt = 0 and solving for P gives us:
0.1P(1 - P/730) - (129000/7300) = 0

Simplifying this equation, we get:
0.1P(1 - P/730) = (129000/7300)

Now, solve for P:
1 - P/730 = 1290/73

- P/730 = 1290/73 - 1
- P/730 = (1290 - 73)/73

P/730 = (1290 - 73)/73

P = (1290 - 73) * (730/73)
P = 14 * 10 = 140

The equilibrium solution is P = 140. If P(0) = 800, which is greater than the equilibrium value, P will decrease towards 140 as time approaches infinity.

c. To determine what happens to dP/dt as t gets very large when P(0) = 800, we can substitute P = 140 into the equation for dP/dt:
dP/dt = 0.1P(1 - P/730) - (129000/7300)

dP/dt = 0.1 * 140(1 - 140/730) - (129000/7300)
dP/dt = 0.1 * 140(1 - 0.191780822) - (129000/7300)
dP/dt = 0.1 * 140 * 0.808219178 - (129000/7300)
dP/dt = 0.1 * 140 * 0.808219178 - 129000/7300
dP/dt = 11.35616438 - 17.67123288
dP/dt = -6.315068493

As t gets very large, dP/dt approaches -6.315068493. Therefore, dP/dt will approach and remain at -6.315068493.

d. The equilibrium solutions are the values of P where dP/dt = 0. Setting dP/dt = 0 and solving for P gives us the equilibrium points:
0.1P(1 - P/730) - (129000/7300) = 0

Simplifying this equation, we get:
0.1P(1 - P/730) = (129000/7300)

Now, solve for P:
1 - P/730 = 1290/73

- P/730 = 1290/73 - 1
- P/730 = (1290 - 73)/73

P/730 = (1290 - 73)/73

P = (1290 - 73) * (730/73)
P = 14 * 10 = 140

The equilibrium solution is P = 140.

To answer the given questions, we'll need to analyze the given equation and understand its behavior as t (time) gets very large. Let's break down the questions one by one:

a. If P(0) = 150, what happens to P as t gets very large?

To find out what happens to P as t approaches infinity, we need to consider the equilibrium behavior of the equation. In this case, we have the equation:

dP/dt = 0.1P(1 - P/730) - (129000/7300)

The equilibrium solutions occur when dP/dt = 0. This means that at equilibrium, the population growth rate is zero.

Setting dP/dt = 0 and solving for P:

0.1P(1 - P/730) - (129000/7300) = 0

Simplifying the equation:

0.1P - 0.1P^2/730 - (129000/7300) = 0

Rearranging the terms:

0.1P - (0.1P^2)/730 = 129000/7300

To find the equilibrium solutions, we can solve this equation either graphically or algebraically. The resulting equilibrium values for P will be the answers to question a.

b. If P(0) = 800, what happens to P as t gets very large?

To determine what happens to P as t approaches infinity when P(0) = 800, we again need to consider the equilibrium behavior of the equation. We can use the same equation as above:

dP/dt = 0.1P(1 - P/730) - (129000/7300)

Since P(0) = 800 is given, we already have an initial condition at t = 0. In this case, we can numerically integrate the equation using software or numerical methods to observe the behavior of P as t increases.

c. If P(0) = 800, what happens to dP/dt as t gets very large?

Similarly, to determine what happens to dP/dt as t approaches infinity when P(0) = 800, we can use the same equation:

dP/dt = 0.1P(1 - P/730) - (129000/7300)

By evaluating dP/dt numerically for large values of t, we can observe the behavior of the growth rate.

d. What are the equilibrium solutions?

As stated earlier, equilibrium solutions occur when dP/dt = 0. By solving the equation 0.1P(1 - P/730) - (129000/7300) = 0, we can find the values of P at equilibrium. The resulting equilibrium values will be the answers to question d.