What is the age of the tree to the nearest year?

[karush]

$\tiny{242.7x.14}$
$\textsf{The charcoal from a tree killed in a volcanic eruption contained 62.8 percent of the carbon-14 found in living matter. }\\$
$\textsf{ How old is the tree to nearest year? Use 5700 years for the half-life of carbon-14}$
\begin{align*}\displaystyle
1&=2e^k(5700) \therefore k\approx 0.0001 \\
.628\cdot{11400} &= e^{-0.0001(t)}\\
t&=4652 \, \textit{years}
\end{align*}
$\textit{no sure about the 62.8% how to implement it}$

 

[MarkFL]

The percentage of $C$ carbon-14 present at time $t$ is:

\(\displaystyle C(t)=100\left(\frac{1}{2}\right)^{\frac{t}{5700}}\)

Hence:

\(\displaystyle t=\frac{5700\ln\left(\frac{100}{C}\right)}{\ln(2)}\)

Plug in $C=62.8$:

\(\displaystyle t=\frac{5700\ln\left(\frac{100}{62.8}\right)}{\ln(2)}\approx3826\)

Thus, we have found the tree is approximately 3,826 years old.

If we want to actually use calculus, we could set up an IVP from the given information:

\(\displaystyle \d{C}{t}=-kC\) where $C(0)=100$ and $C(5700)=50$

The ODE is separable, and using the boundaries, we obtain:

\(\displaystyle \int_{100}^{C}\frac{du}{u}=-k\int_0^t\,dv\)

\(\displaystyle \ln\left(\frac{100}{C}\right)=kt\)

\(\displaystyle t=\frac{1}{k}\ln\left(\frac{100}{C}\right)\)

Now, we can use the other given point to determine $k$:

\(\displaystyle 5700=\frac{1}{k}\ln\left(\frac{100}{50}\right)\implies k=\frac{\ln(2)}{5700}\)

And so we have:

\(\displaystyle t=\frac{5700\ln\left(\frac{100}{C}\right)}{\ln(2)}\)

As we found above. :D

 

[karush]

\begin{align*}\displaystyle
0.5&=e^k(5700) \therefore k\approx 0.0001 \\
1/0.628&= e^{k(t)}\\
t&\approx 3826 \, \textit{years}
\end{align*}
$\textit{my jury rigged version}$