3000 young trout are introduced into a large fishpond. The number of trout still alive after t years is modeled by the formula N(t) = 3000(.9)^t. What is the rate of population decrease when the number of trout in the pond reaches 2000?
A. Approximately 520 trout per year
B. Approximately 211 trout per year (I put this)
C. Approximately 2000 trout per year
D. Approximately 494 trout per year
E. None of the above
Please tell what is the correct answer.
dN/dt = 3000 ln(0.9) * 0.9^t = ln(0.9) N(t)
so, dN/dt = 2000 ln(0.9) = -211 fish/tr
So, B?
Yes, @Miller, the answer would be B.
To find the rate of population decrease when the number of trout in the pond reaches 2000, we need to find the rate of change of the population with respect to time (t).
Given the formula for the number of trout still alive after t years: N(t) = 3000(.9)^t, we can differentiate it with respect to t to find the rate of change.
dN(t)/dt = 3000 ln(0.9) * (0.9)^t
To find the rate of population decrease when N(t) reaches 2000, we substitute N(t) = 2000 into the formula and solve for dt:
2000 = 3000(.9)^t
Dividing both sides by 3000:
0.67 = 0.9^t
Take the natural logarithm of both sides:
ln(0.67) = ln(0.9^t)
Using properties of logarithms:
ln(0.67) = t * ln(0.9)
Solve for t:
t = ln(0.67) / ln(0.9) ≈ 4.445
Now that we have the value of t, we can find the rate of population decrease by substituting t into the formula for dN(t)/dt:
dN(t)/dt ≈ 3000 * ln(0.9) * (0.9)^t = 3000 * ln(0.9) * (0.9)^4.445
Calculating this value, we get:
dN(t)/dt ≈ 211
Therefore, the correct answer is B. Approximately 211 trout per year.