Model radioactive decay using the notation t = time (independent variable), r(t) = amount of particular radioactive isotope present at time t (dependent variable), -lambda = decay rate (parameter). (a) Using this notation, write a model for the decay of a particular radioactive isotope. (b) If the amount of the isotope present at t = 0 is r_0, state the corresponding initial-value problem for the model in part (a).

Respuesta :

Answer:

a) [tex] \frac{dr}{dt}= -\lambda r[/tex]

[tex] r(t) = K e^{-\lambda t}[/tex]

b) [tex] r(t)= r_o e^{-\lambda t}[/tex]

Step-by-step explanation:

For this case we have the following info provided:

[tex] r(t) [/tex] represent the amount of particular radioactive isotope at time t

[tex] t[/tex] represent the time

[tex]-\lambda[/tex] represent the decay rate parameter.

Part a

We can use the following proportional model given by this differential equation:

[tex] \frac{dr}{dt}= -\lambda r[/tex]

If we reorder the expression we got:

[tex] \frac{dr}{r} = - \lambda dt[/tex]

If we integrate both sides we got:

[tex] ln|r| = -\lambda t +C[/tex]

And if we apply exponentials we got:

[tex] r = e^{-\lambda t} e^C[/tex]

So then if [tex] e^C =k[/tex]we can rewrite the model like this:

[tex] r(t) = K e^{-\lambda t}[/tex]

Part b

For this case since we know that [tex] t=0, r(0) = r_o[/tex] if we replace this condition in our formula we got:

[tex] r_o = K e^{-\lambda *0} =K[/tex]

So then [tex] K=r_o[/tex] and our model is given by:

[tex] r(t)= r_o e^{-\lambda t}[/tex]