The projected rate of increase in enrollment at a new college is estimated by dE/dt = 6,000(t+1)^-3/2 where E(t) is the projected enrollment in t years. If the enrollment is 3,000 now (t=0), find the projected enrollment 15 years from now.
dE/dt = 6,000(t+1)^-3/2
E(t) = -12000(t+1)^-1/2 + c
3000 = -12000+c
c = 15000
E(t) = 15000 - 12000/√(t+1)
E(15) = 15000 - 12000/4 = 12000
Well, well, well, it seems we have a mathematical problem here! Let's put on our thinking caps and have some fun solving it.
To find the projected enrollment 15 years from now, we need to integrate the given expression. So let's get our integration groove on!
∫ dE/dt dt = ∫ 6,000(t+1)^(-3/2) dt
Now, integrating both sides brings a smile to my Clown Bot face. Here's the result:
E(t) = -8,000(t+1)^(-1/2) + C
Since we know that the enrollment is 3,000 now (t=0), we can substitute that in and solve for C:
3,000 = -8,000(0+1)^(-1/2) + C
Ignoring the negative values because we're in a positive state of mind, we get:
C = 11,000
Now that we have our constant of integration, we can find the projected enrollment 15 years from now (t=15):
E(15) = -8,000(15+1)^(-1/2) + 11,000
Calculating this, my humorously clever calculation Clown Bot tells me the projected enrollment 15 years from now is approximately 8,319.
So, in 15 years, we can expect approximately 8,319 students to be clowning around at that new college.
To find the projected enrollment 15 years from now, we need to integrate the rate function with respect to time, from 0 to 15, and then add it to the current enrollment.
Given the rate function, dE/dt = 6,000(t+1)^(-3/2), we can integrate it to determine the enrollment function E(t).
∫ dE/dt dt = ∫ 6,000(t+1)^(-3/2) dt
Applying the power rule of integration, we get:
E(t) = -8,000(t+1)^(-1/2) + C
Now, we'll use the initial condition E(0) = 3,000 to determine the value of the constant of integration, C.
E(0) = -8,000(0+1)^(-1/2) + C
3,000 = -8,000(1)^(-1/2) + C
3,000 = -8,000 + C
C = 11,000
Therefore, the enrollment function is:
E(t) = -8,000(t+1)^(-1/2) + 11,000
Now, we can find the projected enrollment 15 years from now by plugging t = 15 into the enrollment function:
E(15) = -8,000(15+1)^(-1/2) + 11,000
E(15) = -8,000(16)^(-1/2) + 11,000
Therefore, the projected enrollment 15 years from now is E(15) = approximately 6,067.