Question

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

A.Find the limit as t tends to infinity of C(t) , and explain its meaning.
B.On the axes provided, draw a graph showing the number of cougars as a function of time:

Answer 2

@perl may you help me?

Answer 3

I think you have a typo in your question.
The number of cougars is increasing at a rate directly proportional to `3500 - C(t).`
The directions can be translated into the following differential equation.
$$ \Large \rm {
\frac{dC}{dt} = k \cdot (3500 -C(t))
}$$

Answer 4

@perl may you help me?

Answer 5

we can solve that differential equation

Answer 6

it is seperable

Answer 7

How do we find the limit from there?

Answer 8

$$ \Large \rm {
\frac{dC}{dt} = k \cdot (3500 -C(t))\\~\\
dC = k \cdot (3500 -C(t))\cdot dt \\ ~\\

\frac{dC}{3500 -C(t)} = k \cdot dt \\ ~\\
\int \frac{dC}{3500 -C(t)} = \int k \cdot dt \\ ~\\
-\ln | 3500 - C(t) | = k\cdot t + h
\\ ~ \\ solve~ for~constants~ k~ and~ h

} $$

Answer 9

does that make sense up to here?

Answer 10

I understand how to get there, but I'm a little confused where to go from there.

Answer 11

now we plug in our initial conditions

Answer 12

Is that 1000 and 2000?

Answer 13

@perl

Answer 14

right plug those in

Answer 15

I got -log(2500) for h

Answer 16

k = 1.02

Answer 17

h= -7.82
h= 1.02

Answer 18

this is an easier expression to plug in ,

Answer 19

Well are my solutions still correct?

Answer 20

$$
\Large \rm {
\frac{dC}{dt} = k \cdot (3500 -C(t))\\~\\
dC = k \cdot (3500 -C(t))\cdot dt \\ ~\\

\frac{dC}{3500 -C(t)} = k \cdot dt \\ ~\\
\int \frac{dC}{3500 -C(t)} = \int k \cdot dt \\ ~\\
-\ln | 3500 - C(t) | = k\cdot t + h\\
\ln | 3500 - C(t) | = -k\cdot t + h'\\~\\
e^{\ln | 3500 - C(t) |} =e^{ -k\cdot t + h'}\\
\cancel e^{\cancel \ln | 3500 - C(t) |} =e^{ -k\cdot t + h'}\\
3500 - C(t) =e^{ -k\cdot t + h'\\}\\
3500 - C(t) =e^{ -k\cdot t} \cdot e^{ h'}\\
3500 - C(t) =Ae^{ -k\cdot t} \\
C(t)= 3500 - Ae^{ -k\cdot t} \\~\\
1000 = C(0)= 3500 - Ae^{ -k\cdot 0} \\
\\ \therefore \\
1000 = 3500 - A\cdot 1 \\
A = 3500 - 1000 =2500
}


$$

Answer 21

yes they are correct, but you are introducing decimal error , which i want to avoid

Answer 22

Should I exclude decimals or go farther past the decimal?

Answer 23

I guess it doesn't matter, just keep going with what you were doing.

Answer 24

Is that the limit?

Answer 25

that is k

Answer 26

So now we need to find h?

Answer 27

$$
\large \rm {
\frac{dC}{dt} = k \cdot (3500 -C(t))\\~\\
dC = k \cdot (3500 -C(t))\cdot dt \\ ~\\

\frac{dC}{3500 -C(t)} = k \cdot dt \\ ~\\
\int \frac{dC}{3500 -C(t)} = \int k \cdot dt \\ ~\\
-\ln | 3500 - C(t) | = k\cdot t + h\\
\ln | 3500 - C(t) | = -k\cdot t + h'\\~\\
e^{\ln | 3500 - C(t) |} =e^{ -k\cdot t + h'}\\
\cancel e^{\cancel \ln | 3500 - C(t) |} =e^{ -k\cdot t + h'}\\
3500 - C(t) =e^{ -k\cdot t + h'\\}\\
3500 - C(t) =e^{ -k\cdot t} \cdot e^{ h'}\\
3500 - C(t) =Ae^{ -k\cdot t} \\
C(t)= 3500 - Ae^{ -k\cdot t} \\~\\
1000 = C(0)= 3500 - Ae^{ -k\cdot 0} \\
\\ \therefore \\
1000 = 3500 - A\cdot 1 \\
A = 3500 - 1000 =2500 \\~\\
C(t)= 3500 - 2500e^{ -k\cdot t} \\~\\
2000 = C(5) = 3500 - 2500e^{ -k\cdot 5} \\~\\
\\ \therefore \\~\\
2000 = 3500 - 2500e^{ -k\cdot 5} \\~\\
\frac{2000-3500}{-2500} = e^{ -k\cdot 5} \\~\\
\frac 3 5 = e^{ -k\cdot 5} \\~\\
\ln (\frac 3 5) = \ln (e^{ -k\cdot 5}) \\~\\
\ln (\frac 3 5) = -k\cdot 5\\~\\
\frac{\ln ( \frac {3}{5}) }{-5} = k\\~\\

k \approx 0.102165\\~\\
C(t)= 3500 - 2500e^{ -.1022 t} \\~\\
}
$$

Answer 28

the limit you plug in infinity, and you should get

Answer 29

t = -3.29229

Answer 30

$$
\large \rm {
\frac{dC}{dt} = k \cdot (3500 -C(t))\\~\\
dC = k \cdot (3500 -C(t))\cdot dt \\ ~\\

\frac{dC}{3500 -C(t)} = k \cdot dt \\ ~\\
\int \frac{dC}{3500 -C(t)} = \int k \cdot dt \\ ~\\
-\ln | 3500 - C(t) | = k\cdot t + h\\
\ln | 3500 - C(t) | = -k\cdot t + h'\\~\\
e^{\ln | 3500 - C(t) |} =e^{ -k\cdot t + h'}\\
\cancel e^{\cancel \ln | 3500 - C(t) |} =e^{ -k\cdot t + h'}\\
3500 - C(t) =e^{ -k\cdot t + h'\\}\\
3500 - C(t) =e^{ -k\cdot t} \cdot e^{ h'}\\
3500 - C(t) =Ae^{ -k\cdot t} \\
C(t)= 3500 - Ae^{ -k\cdot t} \\~\\
1000 = C(0)= 3500 - Ae^{ -k\cdot 0} \\
\\ \therefore \\
1000 = 3500 - A\cdot 1 \\
A = 3500 - 1000 =2500 \\~\\
C(t)= 3500 - 2500e^{ -k\cdot t} \\~\\
2000 = C(5) = 3500 - 2500e^{ -k\cdot 5} \\~\\
\\ \therefore \\~\\
2000 = 3500 - 2500e^{ -k\cdot 5} \\~\\
\frac{2000-3500}{-2500} = e^{ -k\cdot 5} \\~\\
\frac 3 5 = e^{ -k\cdot 5} \\~\\
\ln (\frac 3 5) = \ln (e^{ -k\cdot 5}) \\~\\
\ln (\frac 3 5) = -k\cdot 5\\~\\
\frac{\ln ( \frac {3}{5}) }{-5} = k\\~\\

k \approx .102165
C(t)= 3500 - 2500e^{ -.1022\cdot t} \\~\\
\lim_{t \to \infty} C(t) = 3500 - 2500e^{ -.1022\cdot \infty} \\~\\

}
$$

Answer 31

what is the limit of e^{-infinity} ?

Answer 32

0

Answer 33

right

Answer 34

3500 cougars is the maximum capacity or maximum population this island can reach

Answer 35

Wow, that was harder than I thought it was going to be.

Answer 36

I'm pretty sure I understand all that though.

Answer 37

$$
\large \rm {
\frac{dC}{dt} = k \cdot (3500 -C(t))\\~\\
dC = k \cdot (3500 -C(t))\cdot dt \\ ~\\

\frac{dC}{3500 -C(t)} = k \cdot dt \\ ~\\
\int \frac{dC}{3500 -C(t)} = \int k \cdot dt \\ ~\\
-\ln | 3500 - C(t) | = k\cdot t + h\\
\ln | 3500 - C(t) | = -k\cdot t + h'\\~\\
e^{\ln | 3500 - C(t) |} =e^{ -k\cdot t + h'}\\
\cancel e^{\cancel \ln | 3500 - C(t) |} =e^{ -k\cdot t + h'}\\
3500 - C(t) =e^{ -k\cdot t + h'\\}\\
3500 - C(t) =e^{ -k\cdot t} \cdot e^{ h'}\\
3500 - C(t) =Ae^{ -k\cdot t} \\
C(t)= 3500 - Ae^{ -k\cdot t} \\~\\
1000 = C(0)= 3500 - Ae^{ -k\cdot 0} \\
\\ \therefore \\
1000 = 3500 - A\cdot 1 \\
A = 3500 - 1000 =2500 \\~\\
C(t)= 3500 - 2500e^{ -k\cdot t} \\~\\
2000 = C(5) = 3500 - 2500e^{ -k\cdot 5} \\~\\
\\ \therefore \\~\\
2000 = 3500 - 2500e^{ -k\cdot 5} \\~\\
\frac{2000-3500}{-2500} = e^{ -k\cdot 5} \\~\\
\frac 3 5 = e^{ -k\cdot 5} \\~\\
\ln (\frac 3 5) = \ln (e^{ -k\cdot 5}) \\~\\
\ln (\frac 3 5) = -k\cdot 5\\~\\
\frac{\ln ( \frac {3}{5}) }{-5} = k\\~\\

k \approx .102165\\~\\
C(t)= 3500 - 2500e^{ -.1022\cdot t} \\~\\
\lim_{t \to \infty} C(t) = 3500 - 2500e^{ -.1022\cdot \infty} \\
= 3500 - 2500\cdot 0 = 3500

}
$$

Answer 38

I'll be sure to look over all that work.

Answer 39

However, can you help me with B right now?

Answer 40

the dashed line is not part of the curve, that is an asymptote , as \(\bf t \to \infty \)

Answer 41

Yeah, I understand that.

Answer 42

You've been a real help. Thanks so much!