Question

Suppose the rate of growth of population is given by:
dP/dT=0.010P(100-P), where time is in months and P(0) = 2

A) Find formula for the population in terms of T

B) How long will it take the population to reach 60 individuals??

Answer 1

isolate dp
take integral both sides to get P = ...... + C
replace t = 0, P =2 to get C
plug back to get the whole answer

Answer 2

Not sure if you are familiar with the logistic differential equation:
http://en.wikipedia.org/wiki/Logistic_function

In particular,
\[\frac{dP}{dt}=rP\left(1 - \frac{P}{K}\right)\]
where the constant r defines the growth rate and K is the carrying capacity.

This has the solution
\[P(t) = \frac{K P_0 e^{rt}}{K + P_0 \left( e^{rt} - 1\right)} \]
with \(P_0\) being the initial population.

Answer 3

cam you shoew me step by step how to get the answer

Answer 4

@cruffo @Loser66

Answer 5

Using the formula above, or actually separating variables and integrating?

Answer 6

using the formula

Answer 7

Oh good, the integration would require use to find a partial fraction decomposition :)

To use the formula we have to rewrite the right hand side to find out what r and K will be:

Concentrate on the (100 - P) part first. We want this to look like (1 - P/K).

Factor out 100 from the (100 - P) to get \(\;100(1 - \frac{P}{100})\). You can check the factoring by redistributing the 100.

ok so far?

Answer 8

Okay

Answer 9

Next step?

Answer 10

Great, becaue that was the tricky part.
To continue, we have
\[0.01P(100 - P) = 0.01\cdot P \cdot 100 \cdot \left( 1 - \frac{p}{100}\right)\]

Answer 11

Ok next?

Answer 12

\[\frac{dP}{dt}=rP\left(1 - \frac{P}{K}\right)\]


If you simplify the expression in the previous step, what do you have for r and K?

Answer 13

multiple to cancel out?

Answer 14

multiply

Answer 15

right,
\[ 0.01P(100 - P) = 0.01\cdot P \cdot 100 \cdot \left( 1 - \frac{p}{100}\right) = 1 P\left( 1 - \frac{p}{100}\right)\]
so r = 1 and k = 100.

Answer 16

next step???

Answer 17

\[P(t) = \frac{K P_0 e^{rt}}{K + P_0 \left( e^{rt} - 1\right)}\]

Replace K with 100, r with 1, and the initial population \(P_0 = P(0) = 2\).

Answer 18

what about part B

Answer 19

How long will it take for the population to reach 60 individuals?

Answer 20

Here we have to solve the equation:
\[\frac{200e^t}{100 + 2(e^t-1)} = 60\]

Answer 21

multiply 60 with the den?

Answer 22

Yep. That's what I would do :)

Answer 23

6000+120e^t-120=200e^t

Answer 24

5880=80e^t

Answer 25

right!

Answer 26

e^t=73.5

now what?

Answer 27

take the natural log of both sides :)

Answer 28

show me please!

Answer 29

i wanna make sure i did it correctly

Answer 30

\[e^t = 73.5\]
\[\ln(e^t) = \ln(73.5)\]
then use power rule :
\[t \cdot \ln(e) = \ln(73.5)\]
and ln(e) is just 1 so
\[t = \ln(73.5)\]

Answer 31

you my friend earned yourself a best response point!

Answer 32

can you help me with one more question?

Answer 33

yah, but make it a new post.