Question

CALCULUS HELP PLZ? Let C(t) be the number of cougars on an island at time t years (where t > 0). The number of cougars is increasing at a rate directly proportional to 3500 - C(t). Also, C(0) = 1000, and C(5) = 2000.

Answer 1

sorry. havent taken calc or even pre-calc yet.

Answer 2

Wld u pls check on "the # of cougars is increasing at a rate proportional to .... what ??

Answer 3

Too bad we have to guess. I have little choice but to do that myself in this case.

But anyway: Let's assume, for now, that the differential equation is \[\frac{ dy }{ dt }=3500*c*e ^{t}\].... how much sense does this make to you? If you're not comfortable with this, how would you change it? This model says that the rate of growth of the cougar population is directly proportional to 3500c and that the growth is exponential. Can you either accept that or refute it?

Answer 4

Excuse me: ignore that final e^t and replace dy/dt by dc/dt. Please write this out. Is this form of a d. e. familiar to you?

Answer 5

Having gone back to read what you typed in initially I don't see any question there at all, nor have I seen the first part of the question.

what is the goal of the problem you're now dealing with?

Answer 6

Unless I'm sorely mistaken, if the growth rate of the cougar population is proportional to 3500 C, with nothing to slow it down, then pretty soon we'd have a cougar carpet...wall to wall cougars. In reality, food would run out and starvation would keep the cougar pop'n in check.

Answer 7

@prodemis: Please save this for another place, another time.. IDKwut and I really need to solve this math problem. Thank you.

Answer 8

Prodemis, I would appreciate if you would not joke around. Please leave if you are not going to help.

Answer 9

So: if the growth rate is indeed proportional to 3500 C, then the cougar population would increase without bound. So something is wrong with our model.

Could you possibly take a screen shot of the original problem and upload it here to this conversation?

Answer 10

This is what the document looks like.

Answer 11

thanks for sharing it. What a shame that we're losing time to what may be a misprint.
All I can tell you for sure at this point is that the growth rate of the cougar pop'n is not 3500 C, because, if it were, the pop'n would grow without limit.

So it's back to the drawing board... should we say that the growth rate is prop. to

3500-C and determine whether the result make sense or not?

Answer 12

\[\frac{ dC }{ dt }=3500-C\] looks valid to me as a differential equation.

Answer 13

this equation could be re-arranged to appear as\[\frac{ dC }{ 3500-C }=dt\]

No guarantee that this is correct, but it's worth a try (that is, worth solving).

Answer 14

Hint: \[\int\limits_{}^{}\frac{ dC }{ C }=\ln C + some~constant\]

Answer 15

Can we instead say that is proportional to 3500 - C(t) ? I feel a little more comfortable just following what it says in the post.

Answer 16

Certainly we can do that. Strictly speaking, then, we'd have \[\frac{ dC }{ dt }=k(3500-C)\] this equation says that "the rate of change of the population C of cougars is proportional to 3500-C." The k would be a proportionality constant.

Answer 17

let's stop and assess where we are just now. What do you still need to know / find?

Answer 18

I think I have letter a and b completed. The equation I found for letter a was C(t) = 3500 - 2500 e^((ln .6)/5 * t). You plug in 10 for t in the equation and you get the solution to question b. Now I am working on question c but don't know where to start.

Answer 19

In that case let's by all means focus on your equation:\[C(t) = 3500 - 2500 e^(\ln .6)/5 * t)\] (I copied your equation and pasted it into Equation Editor). If you'd please clear up the latter half of this equation, then we'll take the limit as t goes to infinity.

Answer 20

\[e^(\ln .6)/5 * t)\] isn't clear for me.

Answer 21

\[C(t) = 3500 - 2500 e ^{(\ln .6)/(5 * t)}\]

Answer 22

[(ln .6)/(5*t)] is the exponent of e.

Answer 23

That's better. But the result still stirkes me as being odd: \[C(t) = 3500 - 2500 (0.6)^{1/(5t)}\] .... odd because of the 1/(5t).

Answer 24

Is it possible that you mean\[C(t) = 3500 - 2500 (0.6)^{-5t}?\]

Answer 25

OK...let 's take this version: C(t) = 3500 - 2500e^[ln .6/5) * T] and simplify it a bit.

Answer 26

\[e^[\ln .6/5)]~simplifies \to (.6/5)\]

Answer 27

so we end up with C(t)=3500-2500(.6/5)^t.

Answer 28

\[C(t)=3500-2500(.6/5)^t \rightarrow 3500-2500(.12)^t\]

Answer 29

Do you realize that 0.12^t will go to zero as t goes to infinity, leaving you with a limit of 3500?

Answer 30

So that will be the solution to C??

Answer 31

I can't say for certain, but among the things we've tried so far, this tentative solution makes the most sense to me by far.

Answer 32

We know the cougar population can't grow indefinitely, so learning that the upper limit is 3500 is practically a relief.

Answer 33

Ah I see. So applying the limit to the cougar problem, what does it ultimately mean?

Answer 34

You asked what the limit of c(t) would be as t approaches infinity. Our result says that as time goes on and on. the cougar population would reach an upper limit of 3,500.

Answer 35

Okay, just clarifying with you. This is great! Thank you for all of your help and precious time. :) Very much appreciated.

Answer 36

Really happy to be of help; hope we can meet again here on OpenStudy! Best to you. Good night!