Respuesta :
Answer:
a) [tex]\frac{dC}{dt} = rC[/tex]
And for this case we can rewrite the model like this:
[tex]\frac{dC}{C} = r dt[/tex]
If we integrate both sides we got:
[tex] ln C = rt + k[/tex]
If we use exponentials for both sides we got:
[tex] C = e^{rt} e^k = C_o e^{rt}[/tex]
For this case [tex] C_o = 100 mg[/tex] and r = -0.16
So then our model would be given by:
[tex] C(t) = 100 e^{-0.16 t}[/tex]
Where t represent the number of hours
b) [tex] C(5) = 100 e^{-0.16*5}= 44.9329[/tex]
Step-by-step explanation:
Part a
For this case we can assume the proportional model given by:
[tex]\frac{dC}{dt} = rC[/tex]
And for this case we can rewrite the model like this:
[tex]\frac{dC}{C} = r dt[/tex]
If we integrate both sides we got:
[tex] ln C = rt + k[/tex]
If we use exponentials for both sides we got:
[tex] C = e^{rt} e^k = C_o e^{rt}[/tex]
For this case [tex] C_o = 100 mg[/tex] and r = -0.16
So then our model would be given by:
[tex] C(t) = 100 e^{-0.16 t}[/tex]
Where t represent the number of hours
Part b
For this case we can replace the value t=5 into the model and we got:
[tex] C(5) = 100 e^{-0.16*5}= 44.9329[/tex]
