Differential equations help, modelling population

In summary, a simple model of population growth was used to study the size of a population at different times in a scientific study. The initial population was 560 and after 6 hours, it was found to be 1218. By finding the proportionality constant k, the size of the population after 24 hours was predicted to be 12532. However, it was calculated incorrectly due to a mistake in finding the k value. The correct answer is 1218e^(\frac{1}{6}ln(\frac{1218}{560})24). The model can also be used to determine how long it will take for the population to increase tenfold, by using the equation p = p_0e^(kt)
  • #1
tweety1234
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.
 
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  • #2
tweety1234 said:
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
rl.bhat said:
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 that's what I have dont, is that not the same as multiplying the denominator by 6?
 
  • #4
tweety1234 said:
Thank you,

But that's 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
Okay, thank you.

I get it now
 
  • #6
tweety1234 said:
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.
 

FAQ: Differential equations help, modelling population

What are differential equations?

Differential equations are mathematical equations that describe how a quantity changes over time. They involve derivatives and are often used to model various real-world phenomena, such as population growth.

Why are differential equations important in modelling population?

Differential equations are important in modelling population because they allow us to understand and predict how a population will change over time. By using differential equations, we can take into account factors such as birth rates, death rates, and migration to create accurate models of population growth.

What is the difference between discrete and continuous population models?

Discrete population models use whole numbers to represent the size of a population, while continuous population models use real numbers. Discrete models are often used for smaller populations, while continuous models are more accurate for larger populations.

Can differential equations accurately model population growth?

Yes, differential equations can accurately model population growth as long as the necessary parameters, such as birth and death rates, are known and the model is continuously updated with new data. However, they are not perfect and may not accurately predict sudden changes or unexpected events.

Are there any limitations to using differential equations for population modelling?

Yes, there are some limitations to using differential equations for population modelling. These models assume that the population is homogenous and does not take into account individual characteristics or behaviors. They also rely on accurate data and assumptions, which may not always reflect real-world situations.

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