Respuesta :
Answer:
a) [tex] \frac{1}{2} x_o = x_o e^{-r(5730}[/tex]
We can cancel [tex] x_o[/tex] and we got:
[tex] \frac{1}{2}= e^{-r(5730)}[/tex]
We apply natural log and we got:
[tex]ln(\frac{1}{2}) = -5730 r[/tex]
And [tex] r =\frac{ln(\frac{1}{2})}{-5730}=0.000121[/tex]
b) [tex] 0.2 x_o = x_o e^{-0.000121 t}[/tex]
We can cancel [tex]x_o[/tex] in both sides and we got:
[tex] 0.2 = e^{-0.000121 t}[/tex]
Now we can apply natural log on both sides and we got:
[tex] ln(0.2) = -0.00121 t [/tex]
And if we solve for t we got:
[tex] t =\frac{ln(0.2)}{-0.000121}=13301.139 yars[/tex]
So in order to have 20% of the original amount [tex]x_o[/tex] the total time is 13301.14 years approximately.
Step-by-step explanation:
Part a
For this case we have the following model given by the differential equation:
[tex] x' = \frac{dx}{dt}= -rt[/tex]
The solution for this model is given by:
[tex] X(t) = x_o e^{-rt}[/tex]
Using the half life we know that for 5730 years we need to have 1/2 of the initial amount [tex]x_o[/tex] so if we replace we have this:
[tex] \frac{1}{2} x_o = x_o e^{-r(5730)}[/tex]
We can cancel [tex] x_o[/tex] and we got:
[tex] \frac{1}{2}= e^{-r(5730)}[/tex]
We apply natural log and we got:
[tex]ln(\frac{1}{2}) = -5730 r[/tex]
And [tex] r =\frac{ln(\frac{1}{2})}{-5730}=0.000121[/tex]
Where [tex] x_o[/tex] represent the initial amount. For this case we know the value for the rate of decay [tex] r = 0.000121 \frac{1}{year}[/tex]
Part b
And the half like is 5730 years. And we want to find the time in years in order to have 20% of the original amount. So we can write the following expression:
[tex] 0.2 x_o = x_o e^{-0.000121 t}[/tex]
We can cancel [tex]x_o[/tex] in both sides and we got:
[tex] 0.2 = e^{-0.000121 t}[/tex]
Now we can apply natural log on both sides and we got:
[tex] ln(0.2) = -0.00121 t [/tex]
And if we solve for t we got:
[tex] t =\frac{ln(0.2)}{-0.000121}=13301.139 yars[/tex]
So in order to have 20% of the original amount [tex]x_o[/tex] the total time is 13301.14 years approximately.
