Respuesta :
Answer:
a) r=-0.00012097
b) [tex]\large\ \boxed{Q(t)=Q_0e^{-0.00012097*t}}[/tex]
c) 13,304.65 years
Step-by-step explanation:
a)
Q(t) satisfies the differential equation Q'(t) = rQ(t), hence
[tex]\large \frac{Q'(t)}{Q(t)}=r[/tex]
Integrating on both sides
[tex]\int \frac{Q'(t)}{Q(t)}dt=\int rdt\rightarrow log(Q(t))=rt+C[/tex]
where C is a constant. Taking the exponential on both sides
[tex]\large e^{log(Q(t))}=e^{rt+C}=e^Ce^{rt}\rightarrow Q(t)=C_0e^{rt}[/tex]
In this case, [tex]\large C_0[/tex] is the initial value Q(0), the amount of the initial value of carbon-14 and we have
[tex]\large Q(t)=Q_0e^{rt}[/tex]
As we know that the half-life of carbon-14 is approximately 5,730 years we have
[tex]\large Q(5,730)=\frac{Q_0}{2}\rightarrow Q_0e^{5,730r}=\frac{Q_0}{2}\rightarrow e^{5,730r}=\frac{1}{2}[/tex]
Taking logarithm on both sides
[tex]\large log(e^{5,730r})=log(\frac{1}{2})\rightarrow 5,730r=-0.693147\rightarrow r=\frac{-0.693147}{5,730}[/tex]
and
[tex]\large \boxed{r=-0.00012097}[/tex]
b)
[tex]\large \boxed{Q(t)=Q_0e^{-0.00012097*t}}[/tex]
c)
We want to find a value of t for which
Q(t) = 20% of [tex]\large Q_0[/tex] = [tex]\large\ 0.2Q_0[/tex]
[tex]\large Q(t)=0.2Q_0\rightarrow Q_0e^{-0.00012097*t}=0.2Q_0\rightarrow e^{-0.00012097*t}=0.2\rightarrow\\-0.00012097*t=log(0.2)\rightarrow t=\frac{log(0.2)}{-0.00012097}[/tex]
and
[tex]\large \boxed{t\approx 13,304.65\;years}[/tex]