Question

Ötzi the Iceman In September 1991 of two German tourists found the well-preserved body of a man partially frozen in a glacier in the Ötztal Alps on the border between Austria and Italy. See Figure 3.R.1. Through the carbon-dating technique it was found that the body of Ötzi the iceman—as he came to be called—contained 53% as c-14 much as found in a living person. Assume that the iceman was carbon dated in 1991 . Use the method illustrated in Example 3 of Section 3.1 to ­find the approximate date of his death

Short answer

The approximate date of his death is $3260BCE$.

Step-by-step solution

Concept of modelling with first order differential equations:

A first-order differential equation is defined by an equation: $\frac{dy}{dx}=f(x,y)$ of two variables x and y with its function$f(x,y)$ defined on a region in the xy plane.

Find the equation.

We have an iceman is carbon dated in 1991 then we found that it contains 0.53 of c-14 in a living person, has the differential equation.

$\frac{dC}{dt}=kc$ equation 1

Where C is the amount of carbon.

The differential equation is linear and separable, then we solve it as

$$\begin{gathered} \int \frac{1}{kC}dC=\int dt \\ \frac{1}{k}ln\left(C\right)=t+h_1 \\ {e}^{ln\left(C\right)}=e^{(kt+h_2)} \end{gathered}$$

Then we have

$$\begin{gathered} C=e^{h_2}{e}^{\frac{k}{t}} \\ =h{e}^{kt}equation\left(2\right) \end{gathered}$$

Now, to find the value of constant h , we have to apply the point of condition $(C,t)=(C_0,t)$ into equation 2 as

$$\begin{gathered} C_0=h{e}^0 \\ h=C_0 \end{gathered}$$

After that, substitute with the value of constant h into equation (2) , then we have

$C=C_0{e}^{kt}$ equation (3)

Find the k value.

Now, to find the value of constant of proportionality k , we have to apply the point of condition$(C,t)=(0.5C_0,5730years)$ into equation (2) as

$$\begin{gathered} 0.5C_0=C_0{e}^{5730k} \\ ln\left(0.5\right)=1991k \end{gathered}$$

Then we have

$$\begin{gathered} k=\frac{-0.693}{5730} \\ =-0.0001209 \end{gathered}$$

find the approximate death.

After that, substitute with the value of constant k into equation (3) then we have

$C\left(t\right)=C_0{e}^{-0.0001209t}$ equation (4)

Now, we have to obtain the approximate date of iceman’s death by substituting with $C=0.53C_0$into equation (4) as

$$\begin{gathered} 0.53C_0=C_0{e}^{-0.0001209t} \\ 0.53=C\left(t\right)=e^{-0.0001209t} \\ ln\left(0.53\right)=-0.0001209t \end{gathered}$$

Then we have

$$\begin{gathered} t=\left(\ln \right(0.53\left)\right)/(-0.0001209t) \\ \approx 5251years \end{gathered}$$

Then we can obtain the approximate date of death as

$$\begin{gathered} t_{app}=5251-1991 \\ =3260BCE \end{gathered}$$

Thus, the approximate date of his death is$=3260BCE$.