Respuesta :
Answer:
a) [tex]P(t) =1000 e^{0.1732867951 t}[/tex]
b) [tex]P(t=10) =1000 e^{0.1732867951(10)}=5656.854[/tex]
c) [tex] rate= \frac{(5656.854-1000) cells}{(10-0) hours}=465.685 \frac{cells}{hour}[/tex]
Step-by-step explanation:
Part 1
For this case we assume an exponential growth model given by the following differential equation:
[tex]\frac{dP}{dt}= KP[/tex]
Where P represent the population and t the times in hours. And each 4 hours the population is doubled.
If we solve the differential equation we got:
[tex]\frac{dP}{P}= Kdt[/tex]
Integrating both sides we got:
[tex]ln P= Kt + C[/tex]
If we use exponential on both sides we got:
[tex]P = e^{kt} e^C = P_o e^{kt}[/tex]
And for this case [tex]P_o = 1000[/tex] and we have this:
[tex]1000= P_o e^{k(0)}= P_o[/tex]
Then after 4 hours we have that the population is doubled so then would be 2000 cells and we got:
[tex]2000=1000 e^{4k}[/tex]
And we can solve for k like this:
[tex]ln(2) = 4k[/tex]
[tex]k = \frac{ln2}{4}=0.1732867951[/tex]
So then our model is given by:
[tex]P(t) =1000 e^{0.1732867951 t}[/tex]
Part 2
For this case we just need to replace t= 10 hours into the model and we got:
[tex]P(t=10) =1000 e^{0.1732867951(10)}=5656.854[/tex]
Part 3
For this case we can find the rate like this:
[tex] rate= \frac{(5656.854-1000) cells}{(10-0) hours}=465.685 \frac{cells}{hour}[/tex]
