Im stuck on this question. Could someone answer it. Id appreciate it.
The projected rate of increase in enrollment at a new college is estimated by:
dE/dt = 9000(t + 1)^-3/2
where E(t) is the projected enrollment in t years. If the enrollment is 5000 now (t = 0), find the projected enrollment 35 years from now. Round to the nearest integer as needed.
DEPAUL
To find the projected enrollment 35 years from now, we need to use the given differential equation and initial condition. Here's what you need to do step by step:
Step 1: Set up the differential equation.
Given: dE/dt = 9000(t + 1)^(-3/2)
Step 2: Integrate both sides of the equation with respect to t.
∫ dE = ∫ 9000(t + 1)^(-3/2) dt
Step 3: Integrate the right side of the equation.
E = ∫ 9000(t + 1)^(-3/2) dt
To evaluate this integral, we will use the substitution method. Let u = t + 1, then du = dt.
E = ∫ 9000(u)^(-3/2) du
Step 4: Evaluate the integral.
E = 9000 * ∫ (u)^(-3/2) du
E = 9000 * (2u^(-1/2))
Step 5: Substitute back for u.
E = 9000 * 2(u)^(-1/2)
E = 18000(u)^(-1/2)
Step 6: Evaluate the integral.
E = 18000 * 2(u)^(-1/2)
E = 36000(u)^(-1/2)
Step 7: Substitute back for u with t + 1.
E = 36000(t + 1)^(-1/2)
Step 8: Apply the initial condition.
At t = 0, E = 5000.
5000 = 36000(0 + 1)^(-1/2)
5000 = 36000(1)^(-1/2)
5000 = 36000
Step 9: Find the projected enrollment 35 years from now (t = 35).
E = 36000(35 + 1)^(-1/2)
Step 10: Calculate the value of E.
E = 36000(36)^(-1/2)
E = 36000/6
E = 6000
Therefore, the projected enrollment 35 years from now is 6000.