Respuesta :
Answer:
a) [tex] D(t) = 500 e^{-0.3837641822t}[/tex]
b) For this case we just need to replace t=4 and see what we got:
[tex] D(t=4) =500 e^{-0.3837641822*4}= 107.721 \approx 102[/tex]
Step-by-step explanation:
For this case the proportional model is given by the following differential equation:
[tex] \frac{dD}{dt} =kD[/tex]
Where D is the concentration, t the time and k a proportional constant.
We can rewrite the differential equation like this:
[tex] \frac{dD}{D} =kdt[/tex]
And if we integrate both sides we got:
[tex] ln |D| = kt +c [/tex]
Part a
And if we apply exponentiation on both sides we got:
[tex] D(t) = e^{kt} e^c = e^{kt} D_o = D_o e^{kt} [/tex]
So then our model is given by:
[tex] D(t) = D_o e^{kt}[/tex]
Where [tex] D_o[/tex] is the initial amount for t =0.
For this case we have the initial condition assumed:
[tex] D(0) = 500[/tex]
So then our model is given by:
[tex] D(t) = 500 e^{kt}[/tex]
We have another condition given:
[tex] D=50 , t = 6[/tex]
[tex] 50 = 500 e^{6k}[/tex]
If we divide both sides by 500 we got:
[tex] \frac{1}{10} =e^{6k}[/tex]
And if we apply natutral log on both sides we got:
[tex] ln(\frac{1}{10}) = 6k[/tex]
[tex] k = \frac{\frac{1}{10}}{6}= -0.3837641822[/tex]
And our model then is given by:
[tex] D(t) = 500 e^{-0.3837641822t}[/tex]
Part b
For this case we just need to replace t=4 and see what we got:
[tex] D(t=4) =500 e^{-0.3837641822*4}= 107.721 \approx 102[/tex]
