Worlds population in 1960 was approximately 3 billion; in 2000 it was approximately 6 billion. Model this using a differential equation of your choice. You should include your reasons for your choice of models, the solution to the equation, and a graph spanning the given data and some time into the future. Please help me get this finished!!! Medal when done! :)

Question

ghazi · Answer

i am sorry i could have solved but its a lengthy process , all i want to say is ..total stochastic calculus failed in developing and framing expansion of population so i am sure this equation wont give solution either

unklerhaukus · Answer

i think more than two data point would be helpful

unklerhaukus · Answer

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anonymous · Answer

@UnkleRhaukus Two points is enough it you assume it is some exponential function.

unklerhaukus · Answer

which ones can we assume?

anonymous · Answer

$$
P(t) = P_0e^{kt}
$$$$
P(1960) = 3 = P_0e^{k(1960)}\
P(2000) = 6 = P_0e^{k(2000)}
$$Two equations, two variables $P_0$ and $k$.

anonymous · Answer

You could also think of it as, the double life of the function is 40 years, and suppose the function starts at year 1960.  In that case: $$
P(t) = 3\cdot 2^{t/40}
$$

unklerhaukus · Answer

but these models are so inaccurate

anonymous · Answer

What do you think the the carrying capacity as a function of time?

anonymous · Answer

Okay how can I make them accurate then?

anonymous · Answer

Because Im lost...One person says this and the other says something else...

zehanz · Answer

Accuracy is not the point here.
When so little data is given, the only thing you can reasonably do is assume there is exponential growth.
The data support @wio's formula: $P(t)=3 \cdot 2^{\frac{1}{40}t}=3 \cdot e^{\ln2^{\frac{1}{40}t}}=3e^{\frac{\ln 2}{40} \cdot t} \approx 3e^{0.01733t}$.

If we take the derivative of P, we get: $P'(t)=0.01733P(t)$.

The differential equation that fits the situation is therefore:

$\dfrac{dx}{dt}=0.01733x$.

Solving can be done with separation of variables:

$\dfrac{dx}{x}=0.01733dt$
$\int \dfrac{dx}{x}=\int 0.01733dt +C$; C is a real constant.

$ln|x|=0.01733t+C$

$|x|=e^{0.01733t+C}=e^{C}e^{0.01733t}=C_{1}e^{0.01733t};~C_{1}$  is a positive constant.

Because x is the population, it is always positive, so we don't need the absolute value:

$x=C_{1}e^{0.01733t}$.

To find the constant, we set x(0)=3, so the constant is 3.

Solution: $x(t)=3e^{0.01733t}$, of course the same as the population function P.

BTW: if we forget about the practical situation this is about, the general solution of the differential equation is $x(t)=Ce^{0.01733t}$, where C can be any real constant.

Here is also a graph with prediction for 2020:

anonymous · Answer

Okay thanks so much! So may I ask...What would be different if you would have made the choice of a different model?

anonymous · Answer

Please help me with this integration!

anonymous · Answer

If you wanted to change the model, you'd probably want the change of population (derivative of population) to tend toward $0$ as the the population reached some carrying capacity.