   Chapter 4.6, Problem 20E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Carbon Dating Repeat Exercise 19 for a piece of charcoal that contains 30% as much radioactive carbon as a modem piece.

To determine

To calculate: The age of the charcoal using carbon dating that assumes the carbon dioxide on the earth today has the same radioactive content as it did centuries ago, provided that a piece of charcoal contains only 30% as much of the radioactive carbon as a modern charcoal. The half-life of C14 is 5715.

Explanation

Given Information:

The provided information is that a piece of charcoal contains only 30% as much of the radioactive carbon as a modern charcoal and the half-life of C14 is 5715.

Formula used:

Exponential growth and decay:

If the rate of change of a positive quantity y with respect to time is proportional to the amount of quantity present at any time t, that is dydt=ky, then y is given by the equation, y=Cekt, where C is the value of the quantity at time t=0 and k is the constant of proportionality.

If k>0 then there is exponential growth and when k<0 then there is exponential decay.

Calculation:

Consider the provided information that the half-life of C14 is 5715 years and a piece of charcoal contains only 30% as much of the radioactive carbon as a modern charcoal.

Let the C be the amount of radioactive carbon present in modern charcoal.

Then the amount of carbon present in the ancient charcoal is 0.30C.

Let t be the age of ancient charcoal, then at time t amount of carbon y=0.30C.

Substitute y=0.30C in the equation y=Cekt,

0.30C=Cektekt=0.30CCekt=0.30

Also, the half-life of C14 is 5715 years, that is, at t=5715 the initial amount reduces to half so,

y=C2

Substitute t=5715 and y=C2 in the equation y=5ekt

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