   Chapter 4.6, Problem 12E ### 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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# Modeling Radioactive Decay In Exercises 11-16, complete the table for each radioactive isotope. See Example 1. Isotope Half-life (in years) Initialquantity Amountafter1000years Amountafter10,000years 12. 14 C 5715 3 grams

To determine

To calculate: The amount of C14 left after 1000 years and 10,000 years, if the half-life of C14 is 5715 years and the initial quantity is 3 grams.

Explanation

Given Information:

The provided half-life of C14 is 5715 years and the initial quantity is 3 grams.

Formula used:

The equation for a positive quantity y whose rate of change with respect to time is proportional to the amount of quantity is,

y=Cekt

Where, C is the amount of quantity, t is the any time and k is the proportionality constant.

The exponential growth when k>0 and the exponential growth when k<0.

The property of natural logarithm,

lnxn=nlnx

Calculation:

Consider the provided information that the half-life of C14 is 5715 years and the initial quantity is 3 grams.

Now, at t=0, the initial amount y=3.

Therefore, the value of C is 3.

Substitute C=3 in the equation y=Cekt.

y=3ekt

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

y=32=1.5

Substitute t=5715 and y=1.5 in the equation y=3ekt.

1.5=3ek(5715)e5715k=1.53

Take natural log on both the sides and use the property of logarithm, lnxn=nlnx.

ln(e5715k)=ln(1.53)5715k(lne)=ln(12)5715k=ln(12)k=ln(12)5715

Therefore, the value of k is ln(12)5715

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