Question

Three thousand 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?

Answer 1

@TuringTest @AccessDenied Any help??

Answer 2

Any ideas on how to approach the problem? Or are you completely uncertain of where to start?

Answer 3

I am uncertain of where to start. :0

Answer 4

Read this statement carefully from the problem excerpt:
''What is the rate of population decrease when the number of trout in the pond reaches 2000?''
We want to find a _rate_ of population decrease. What does this translate to finding of our function N(t) = 3000(0.9)^t ?

Answer 5

\(\bf \textit{rate of change}\implies \cfrac{rise}{run}\implies ?\)

Answer 6

Would it become N(t) = 2000(0.9)^2 ??

Answer 7

Where did you get that equation?
N(t) = 3000 (0.9)^t represents the number of trout in the pond, N, after some years, t.
The rate at which the population of trout is changing would be the rate at which N(t) changes with respect to time. Have you learned about derivatives?

Answer 8

Yea I have. Then we would have to take the derivative?

Answer 9

Yes. The derivative of N(t) with respect to t gives us that rate of change we need. So we find the derivative first. Then we just need to find it "when the population reaches 2000."

Answer 10

So once we find the derivative, then we set it equal to 2000? Or no?

Answer 11

Not quite. 2000 is a population, not a rate of change. N(t) = 2000 is what we should look at, where we solve for t. N'(t) = 2000 doesn't make practical sense because N'(t) is trout/year, and 2000 is the number of trout. :)

Answer 12

Okay, give me a second to process everything you just said. Sorry, I am slow when it comes to math. `__`

Answer 13

It is fine, take your time! :)

Answer 14

2000 = 3000(0.9)^t
t = 3.85

Plug t into derivative:
N'(t) = -316.08 * (0.9^t)
= -210.68

So the answer will be B, or approximately 211 trouts per year? @AccessDenied

Answer 15

Yep, that looks good to me! :)