Differential equations help, modelling population

  • Thread starter tweety1234
  • Start date
  • #1
112
0
in a scientific study, the size , p of a population at t hours is being studied. Initially p = 560 and after 6 hours, p is found to be 1218. In a simple model of population growth, the rate of the population is taken to be proportional to the population at that time. Using this model predict.

a) the size of the population 24 hours after the start of the experiment.
b) how long it will take for the population to increase tenfold.

I keep getting the wrong answer, here's my working any help appreciated.


[tex] \frac{dp}{dt} = kt [/tex]

[tex] \int \frac{1}{p} dp = \int k dt [/tex]

[tex] ln (p) = kt +c [/tex]

[tex] p = p_{0}e^{kt} [/tex] p_{0} = 560 , t= 0

[tex] p = 560e^{kt} [/tex]

[tex] 1218 = 560e^{k6} [/tex]

[tex] k = ln(\frac{1218}{3360}) [/tex]

so to find after 24 hours I would just put t = 42 into my formula, and get ;

[tex] p = 560e^{ln\frac{1218}{3360}24} [/tex]

But when I solve this I get 4872, and the correct answer is 12532

What am I doing wrong?

Thank you.
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
8
in a scientific study, the size , p of a population at t hours is being studied. Initially p = 560 and after 6 hours, p is found to be 1218. In a simple model of population growth, the rate of the population is taken to be proportional to the population at that time. Using this model predict.

a) the size of the population 24 hours after the start of the experiment.
b) how long it will take for the population to increase tenfold.

I keep getting the wrong answer, here's my working any help appreciated.


[tex] \frac{dp}{dt} = kt [/tex]

[tex] \int \frac{1}{p} dp = \int k dt [/tex]

[tex] ln (p) = kt +c [/tex]

[tex] p = p_{0}e^{kt} [/tex] p_{0} = 560 , t= 0

[tex] p = 560e^{kt} [/tex]

Thank you.

Your calculation of k value is wrong.

[tex]6k = ln(\frac{1218}{560})[/tex]

[tex]k = \frac{1}{6}ln(\frac{1218}{560})[/tex]

Now find k.
 
  • #3
112
0
Your calculation of k value is wrong.

[tex]6k = ln(\frac{1218}{560})[/tex]

[tex]k = \frac{1}{6}ln(\frac{1218}{560})[/tex]

Now find k.

Thank you,

But thats what I have dont, is that not the same as multiplying the denominator by 6?
 
  • #4
rl.bhat
Homework Helper
4,433
8
Thank you,

But thats what I have dont, is that not the same as multiplying the denominator by 6?

No. Find the logarithm and divide it by 6 to get k.
 
  • #5
112
0
Okay, thank you.

I get it now
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,847
966
in a scientific study, the size , p of a population at t hours is being studied. Initially p = 560 and after 6 hours, p is found to be 1218. In a simple model of population growth, the rate of the population is taken to be proportional to the population at that time. Using this model predict.

a) the size of the population 24 hours after the start of the experiment.
b) how long it will take for the population to increase tenfold.

I keep getting the wrong answer, here's my working any help appreciated.


[tex] \frac{dp}{dt} = kt [/tex]
Though not important there is a typo in your original equation. You clearly intended the equation to be
[tex]\frac{dp}{dt}= kp[/tex]
not "= kt".

[tex] \int \frac{1}{p} dp = \int k dt [/tex]

[tex] ln (p) = kt +c [/tex]

[tex] p = p_{0}e^{kt} [/tex] p_{0} = 560 , t= 0

[tex] p = 560e^{kt} [/tex]

[tex] 1218 = 560e^{k6} [/tex]

[tex] k = ln(\frac{1218}{3360}) [/tex]

so to find after 24 hours I would just put t = 42 into my formula, and get ;

[tex] p = 560e^{ln\frac{1218}{3360}24} [/tex]

But when I solve this I get 4872, and the correct answer is 12532

What am I doing wrong?

Thank you.
 

Related Threads on Differential equations help, modelling population

Replies
9
Views
15K
Replies
5
Views
1K
  • Last Post
Replies
2
Views
4K
Replies
1
Views
3K
Replies
4
Views
3K
Replies
2
Views
2K
Replies
0
Views
4K
Replies
1
Views
1K
  • Last Post
2
Replies
28
Views
5K
  • Last Post
Replies
16
Views
2K
Top