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.
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?
1 answer
1 answer