Question

If anyone has the time and patience to help me please do :)

The question is below:

Answer 1

1.) 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.

A. Find C(t) as a function of t only.
B. Calculate C(10).
C. Find the limit as t tends to infinity of C(t) , and explain its meaning.

Answer 2

For A I got C(t) = 3500 – 2500 e^((ln .6)/5 * T) bit I'm unsure of myself

Answer 3

but*

Answer 4

Show how you got it so we can check?

Answer 5

3500 . C(t) not sure what this is... 3500*C(t)?

If it is that \[\Large \frac{ dC }{ dt } = 3500 C\]

Answer 6

dC/dt = k ( 3500 – C(t))
dC/dt+ kC(t) = 3500k
e^(kt)(dC/dt + kC(t) ) = 3500k e^(kt)
(e^(kt)C(t))’ = 3500k e^(kt)
integrate
e^(kt)C(t) = 3500e^(kt) + A
C(t) = 3500 + Ae^(-kt)
when t = 0:
1000 = 3500 + A
A = -2500
t = 5:
2000 = 3500 – 2500 e^(-5k)
2500e^(-5k) = 1500
e^(-5k)= .6
-5k = ln .6
k = -(ln .6)/5
C(t) = 3500 – 2500 e^((ln .6)/5 * T)

Answer 7

Does 3500 . C(t). means 3500 - C(t) ?

Answer 8

If you have \[\large \frac{ dC }{ dt } = k ( 3500 – C(t)) \]
then it's a separable equation
\[\large \frac{ dC }{ 3500-C(t) }=k dt\]

Answer 9

Yes this method is much better and fast

Answer 10

\[\large \int\limits \frac{ dC }{ 3500-C(t) }= \int\limits k dt\] \[\Large -\ln(3500-C(t)) = kt+B\] I'm using B as the constant of integration, because C is taken.
Then just solve for C(t).

Answer 11

But your method worked fine.

Answer 12

She multiplied the equation by negative 1 afterwards

Answer 13

2000 = 3500 – 2500 e^(-5k)
-1500 = -2500 e^(-5k)
-1 x -1500 = -1 x -2500 e^(-5k)
1500 = 2500e^(-5k)
2500e^(-5k) = 1500

Answer 14

B. Calculate C(10).
C. Find the limit as t tends to infinity of C(t) , and explain its meaning.

These are easy now.
B. plug in t=10
C. is a little tougher but not hard... just find \[\large \lim_{t \rightarrow \infty} \left( 3500 – 2500 e^{t* \ln .6/5 }\right)\]

Answer 15

Remember that ln(0.6) is negative (because I didn't)

Answer 16

Thank you both

Answer 17

Just to clarify, B. would be 3470?

Answer 18

@agent0smith

Answer 19

Idk, I didn't work it out, just plug it in on wolframalpha or something to check, cos I gotta go

Answer 20

But no it's not 3470

Answer 21

I'm sorry Calculation error. I got 1025.41125 instead?

Answer 22

No... not close... show your work, but i gotta go anyway

all you have to do is plug in t=10 into: 3500 – 2500 e^((ln .6)/5 * T)

and type it exactly like that into a calculator or w/e

Answer 23

2600??? @agent0smith

Answer 24

Yes

Answer 25

Gosh dangit lol thanks. Now for the limit I got 3500. Is that correct?

Answer 26

Yes.

Answer 27

Now what does it mean by "explain it's meaning"? Like 3500 is the limit of the amount of cougars?

Answer 28

@agent0smith

Answer 29

Exactly!