Respuesta :
Answer:
1) [tex] P(t) = e^{kt +C}= P_o e^{kt}[/tex]
For this case we know that the population is doubled after 5 years so then we can assume that the initial population is Po and we have this:
[tex] 2P_o = P_o e^{5k}[/tex]
[tex] 2 = e^{5k}[/tex]
We can apply natural log on both sides and we got:
[tex] ln (2) = 5k[/tex]
[tex] k= \frac{ln(2)}{5}= 0.138629436[/tex]
We know also that the population after 3 years is 9000, so we can use this condition:
[tex] 9000 = P_o e^{3*0.138629436}[/tex]
And we can solve for Po and we got:
[tex] P_o = \frac{9000}{e^{3*0.138629436}} = 5937.786[/tex]
2) [tex] P(t) = 5937.786 e^{0.138629436 t}[/tex]
If we replace the value of t =10 we got:
[tex] P(10) = 5937.786 e^{0.138629436 *10}= 23751.144[/tex]
3) [tex] \frac{dP}{dt} = kP[/tex]
Since we know the value of the population at t=10 , P= 23751.144 and k = 0.138629436 we can replace and we got:
[tex] \frac{dP}{dt} = kP =0.138629436 *23751.144 = 3292.607[/tex]
Explanation:
Part 1
For this case we know that we can use the proportional model given by:
[tex]\frac{dP}{dt}= kP[/tex]
We can rewrite the expression like this:
[tex]\frac{dP}{P}= kdt[/tex]
If we integrate both sides we got:
[tex] ln P = kt +C[/tex]
After apply exponentials in both sides we got:
[tex] P(t) = e^{kt +C}= P_o e^{kt}[/tex]
For this case we know that the population is doubled after 5 years so then we can assume that the initial population is Po and we have this:
[tex] 2P_o = P_o e^{5k}[/tex]
[tex] 2 = e^{5k}[/tex]
We can apply natural log on both sides and we got:
[tex] ln (2) = 5k[/tex]
[tex] k= \frac{ln(2)}{5}= 0.138629436[/tex]
We know also that the population after 3 years is 9000, so we can use this condition:
[tex] 9000 = P_o e^{3*0.138629436}[/tex]
And we can solve for Po and we got:
[tex] P_o = \frac{9000}{e^{3*0.138629436}} = 5937.786[/tex]
Part 2
For this case we have the following model:
[tex] P(t) = 5937.786 e^{0.138629436 t}[/tex]
If we replace the value of t =10 we got:
[tex] P(10) = 5937.786 e^{0.138629436 *10}= 23751.144[/tex]
Part 3
For this case we know that:
[tex] \frac{dP}{dt} = kP[/tex]
Since we know the value of the population at t=10 , P= 23751.144 and k = 0.138629436 we can replace and we got:
[tex] \frac{dP}{dt} = kP =0.138629436 *23751.144 = 3292.607[/tex]
