Chapter 3.1, Problem 58E

Solution

a.

To find: the initial amount of carbon-14 present.

20g

Given information:

The amount C in grams of carbon-14 present in a certain substance after t years is given by:

  $C=20e^{-0.0001216t}$

Explanation:

The initial amount of carbon-14 present is the C-14 that was present at $t=0$ yearss. So, substitute $t=0$ in the given model,

  $\begin{array}{c}C\left(0\right)=20e^{-0.0001216\left(0\right)}\\ =20e^0\\ =20\left(1\right)\\ =20\end{array}$

Initial amount of carbon-14 present was 20g.

b.

To find: the amount of carbon-14 left after 10,400 years. And when will the amount left be 10 g.

After 10,400 years the amount of carbon-14 left is 5.65 g.

Nearly 5700 years

Given information:

The amount C in grams of carbon-14 present in a certain substance after t years is given by:

  $C=20e^{-0.0001216t}$

Explanation:

First, find the amount of carbon-14 left after 10,400 years.

Substitute $t=10,400$ in the given model,

  $\begin{array}{c}C\left(10,400\right)=20e^{-0.0001216\left(10,400\right)}\\ =20e^{1.26464}\\ \approx 20\cdot 0.282341\\ \approx 5.65g\end{array}$

After 10,400 years the amount of carbon-14 left is 5.65 g.

Now, to find when will the amount left be 10 g. Equate the given model to 10 and solve for t .

  $\begin{array}{c}20e^{-0.0001216t}=10\\ e^{-0.0001216t}=\frac{10}{20}\\ e^{-0.0001216t}=0.5\end{array}$

Take natural logarithm on both sides,

  $\begin{array}{c}\ln \left(e^{-0.0001216t}\right)=\ln 0.5\\ -0.0001216t=\ln 0.5\\ t=-\frac{\ln 0.5}{0.0001216}\\ t\approx 5700\text{ years}\end{array}$