Respuesta :
Answer:
[tex] \frac{dA}{dt} \propto M-A[/tex]
[tex] \frac{dA}{dt}= k (M-A)[/tex]
And the solution for this equation is:
[tex]\frac{dA}{M-A} = k dt[/tex]
Using the substitution u = M-A, we have that du = -dA and after integrate we got:
[tex] -ln |M-A|= kt +C[/tex]
C is a constant.
Now we can apply exponential in both sides and we got:
[tex]-(M-A)= e^{kt +C}= e^{kt} e^c = e^{kt} K_0[/tex]
[tex] A-M = K_0 e^{kt}[/tex]
[tex] A(t) = M+ K_0 e^{kt}[/tex]
Step-by-step explanation:
Let's define some notation first :
[tex] M[/tex] represent the total amount of the subject to be memorized
[tex] A(t)[/tex] represent the amount memorized at time t
t represent the time
k is a constant of proportionality
For this case we know that the rate of memorizing is proportional to the amount left to be memorized.
So then we can create the following relationship:
[tex] \frac{dA}{dt} \propto M-A[/tex]
[tex] \frac{dA}{dt}= k (M-A)[/tex]
Where k>0. We can solve this differential equation like this:
[tex]\frac{dA}{M-A} = k dt[/tex]
Using the substitution u = M-A, we have that du = -dA and after integrate we got:
[tex] -ln |M-A|= kt +C[/tex]
C is a constant.
Now we can apply exponential in both sides and we got:
[tex]-(M-A)= e^{kt +C}= e^{kt} e^c = e^{kt} K_0[/tex]
[tex] A-M = K_0 e^{kt}[/tex]
[tex] A(t) = M+ K_0 e^{kt}[/tex]
