Q. 67 - Calculus

Question

The amount of the radioactive isotope carbon-$14$ present in small quantities can be measured with a Geiger counter. Carbon-$14$ is replenished in live organisms, and after an organism dies the carbon-$14$ in it decays at a rate proportional to the amount of carbon-$14$ present in the body. Suppose $C\left(t\right)$ is the amount of carbon-$14$ in a dead organism $t$ years after it dies.

(a) Set up a differential equation describing $\frac{dC}{dt}$, and solve it to get a formula for $C\left(t\right)$. Your answer will involve two constants.

(b) The half-life of carbon-$14$ is $5730$ years. (See part (b) of the previous problem for the definition of half-life.) Use this half-life to find the value of the proportionality constant for the model you found in part (a).

(c) Suppose you find a bone fossil that has $10\%$of its carbon-$14$ left. How old would you estimate the fossil to be?

Step-by-Step Solution

Answer

Part $\left(a\right)$: A formula for $C\left(t\right)$ is, $C\left(t\right)=A{e}^{-kt}$.

Part $\left(b\right)$: The value of the proportionality constant for the model is, $0.00012$.

Part $\left(c\right)$: The fossil is approximately $19188$ years old.

Part a Step 1 . Given information

Carbon-$14$ decays at a rate proportional to the amount of carbon-$14$ present in the body.

$C\left(t\right)$ is the amount of carbon-$14$ in a dead organism $t$ years after it dies.

Part a Step 2 . Note that the amount of carbon - 14 is continuously decreasing with time.

This means that the rate of change of Carbon$-14$ is negative. Hence, the mathematical equation representing this model will follow the model of negative growth rate.

From the given information, if $C\left(t\right)$ is the amount of carbon$-14$ remaining after $t$ years, then the decay rate $\frac{dC}{dt}$ of carbon$-14$ is proportional to $C\left(t\right)$.Hence, the model representing the decay rate of carbon$-14$ is,

$$\begin{gathered} \frac{dC}{dt}\alpha C \\ =-kC \end{gathered}$$

Part a Step 3 . In the above relation k is the decay constant.

Now integrate the obtained equation on both sides,

$$\begin{gathered} \int \frac{dC}{C}=-k\int dt \\ \ln \left|C\right|=-kt+C_1 \\ C=e^{-kt+C_1} \\ =A{e}^{-kt } (e^{C_1}=A) \end{gathered}$$

Not that the solution contains two constants namely the decay constant $k$ and the constant $A$.

If the initial amount of carbon$-14$ present in the body is taken as $C_0$,then the constant $A$ becomes $C_0$ and the obtained model becomes,

$C=C_0{e}^{-kt}$.

Part b Step 1 . Suppose that the quantity of carbon - 14 present initially is C 0 ,

Then the half life of $5730$ implies that the amount of carbon$-14$ remaining in the body after $5730$ years will be $\frac{C_0}{2}$. So, substitute $t=5730,C\left(t\right)=\frac{C_0}{2}$ in the obtained model .

$$\begin{gathered} \frac{C_0}{2}=C_0{e}^{-5730k} \\ {e}^{-5730k}=2 \\ k=\frac{1}{5730}\ln 2 \\ \approx 0.00012 \end{gathered}$$

Part c Step 1 . Substitute the value of k obtained in part b to get the solution of the decay model as,

$C\left(t\right)=C_0{e}^{-0.00012t}$

From the given information,

The fossil found has $10\%$ of its carbon$-14$ left. So, $C\left(t\right)=0.1C_0$.

Equate both the obtained equation.

$$\begin{gathered} 0.1C_0=C_0{e}^{-0.00012t} \\ {e}^{0.00012t}=10 \\ 0.00012t=\ln 10 \\ t=\frac{1}{0.00012}\ln 10 \\ t=19188.2 \\ t\approx 19188 \end{gathered}$$