Respuesta :
Answer:
Find the amount Q1
[tex]P(T)=10000 e^{\frac{ln(2)}{5}20}=160000[/tex]
In what year will the population reach one million?
[tex]Q_2= \frac{ln(100)}{\frac{ln(2)}{5}}=33.219[/tex] years
What will the population of the city be at the beginning of the year 2010?Assuming that the starting point is th year 2000, we just need to replace T =10 in the model in order to find the population, like this:
[tex]P(T)=10000 e^{\frac{ln(2)}{5}10}=40000[/tex]
Step-by-step explanation:
Notation
T= represent the time in years
P(T)= represent the population at the time T
K= represent the proportional constant for the model
Solution to the problem
So for this case we need a proportional model given by:
[tex]\frac{DP}{DT}=KP[/tex]
And we can express the model like this
[tex]\frac{DP}{P}=K DT[/tex]
And integrating on both sides we got:
[tex]ln P= KT +C[/tex]
And we can exponentiate both sides:
[tex]P(T)=P_o e^{KT}[/tex]
We have two initial conditions on this case, given by:
[tex]P(0)=10000, P(5)=20000[/tex]
We can use the first condition in order to find [tex]P_o[/tex], like this:
[tex]10000=P_o e^{0}[/tex], so then we have that [tex]P_o =10000[/tex]
And we can use the second condition to find the constant K, like this:
[tex]20000=10000 e^{5K}[/tex],a dn solving for K we have:
[tex]ln(2)= 5K[/tex] and [tex]K=\frac{ln(2)}{5}[/tex]
So then our model would be given by:
[tex]P(T)=10000 e^{\frac{ln(2)}{5}T}[/tex]
Find the amount Q1
[tex]P(T)=10000 e^{\frac{ln(2)}{5}20}=160000[/tex]
In what year will the population reach one million?
We can find the time Q2 where the population is 1000000, like this:
[tex]1000000=10000 e^{\frac{ln(2)}{5}T}[/tex]
[tex]100 =e^{\frac{ln(2)}{5}Q_2}[/tex]
And using natural log on both sides we have:
[tex]ln(100)=\frac{ln(2)}{5}Q_2[/tex]
And [tex]Q_2= \frac{ln(100)}{\frac{ln(2)}{5}}=33.219[/tex] years
What will the population of the city be at the beginning of the year 2010?Assuming that the starting point is th year 2000, we just need to replace T =10 in the model in order to find the population, like this:
[tex]P(T)=10000 e^{\frac{ln(2)}{5}10}=40000[/tex]
