Respuesta :
Answer:
a) [tex] P(t) =\frac{400}{3} e^{0.2027325541 t}[/tex]
b) [tex] P(t=7) =\frac{400}{3} e^{0.2027325541*7}= 551.135 \approx 552[/tex]
c) For this case we need to satisfy this equation:
[tex] 2P_o = P_o e^{kt}[/tex]
If we divide both sides by [tex]P_o[/tex] we got:
[tex] 2 = e^{kt}[/tex]
Now we can apply natural log on both sides and we got:
[tex] ln(2) = kt[/tex]
And the [tex] t = \frac{ln(2)}{0.2027325541}=3.419 hours[/tex]
Step-by-step explanation:
For this case the proportional model is given by the following differential equation:
[tex] \frac{dP}{dt} =kP[/tex]
Where P is the population, t the time and k a proportional constant.
IWe can rewrite the differential equation like this:
[tex] \frac{dP}{P} =kdt[/tex]
And if we integrate both sides we got:
[tex] ln |P| = kt +c [/tex]
Part a
And if we apply exponentiation on both sides we got:
[tex] P(t) = e^{kt} e^c = e^{kt} P_o = P_o e^{kt} [/tex]
So then our model is given by:
[tex] P(t) = P_o e^{kt}[/tex]
Where [tex] P_o[/tex] is the initial population for t =0.
For this case we have some conditions given:
[tex] P(2) = 200, P(4) = 300[/tex]
So then we have two equations:
[tex] 200 = P_o e^{2k}[/tex] (1)
[tex] 300 = P_o e^{4k}[/tex] (2)
From equation (1) we can solve for [tex]P_o[/tex] like this:
[tex] P_o =\frac{200}{e^{2k}}[/tex] (3)
And if we replace equation (3) into equation (2) we got:
[tex] 300 =\frac{200}{e^{2k}} e^{4k}= 200e^{2k}[/tex]
And we can divide both sides by 200 and we got:
[tex] \frac{3}{2} = e^{2k}[/tex]
Now we can apply natural logs on both sides and we got:
[tex] ln(\frac{3}{2}) = 2k[/tex]
And then the value for k is:
[tex] k = \frac{ln(\frac{3}{2})}{2}= 0.2027325541[/tex]
And since we have the value for k we can find the value or [tex] P_o[/tex] like this:
[tex] P_o =\frac{200}{e^{2* 0.2027325541}}= \frac{400}{3}[/tex]
And with that our model is given by:
[tex] P(t) =\frac{400}{3} e^{0.2027325541 t}[/tex]
Part b
For this case we just need to replace t=7 and see what we got:
[tex] P(t=7) =\frac{400}{3} e^{0.2027325541*7}= 551.135 \approx 552[/tex]
Part c
For this case we need to satisfy this equation:
[tex] 2P_o = P_o e^{kt}[/tex]
If we divide both sides by [tex]P_o[/tex] we got:
[tex] 2 = e^{kt}[/tex]
Now we can apply natural log on both sides and we got:
[tex] ln(2) = kt[/tex]
And the [tex] t = \frac{ln(2)}{0.2027325541}=3.419 hours[/tex]
