Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 6 - Differential Equations - 6.2 Exercises - Page 413: 29

Answer

Amount after 1000 years: 12.96g. Amount after 10,000 years: 0.26g.

Work Step by Step

Using Decay formula : $y=Ce^{kt}$ First solve for rate, k, to find amount in grams, after set amount of time. Knowing, half life is 1599 years, divide initial amount by 2, and set equal to decay formula and solve for k. $10=20e^{k1599}$ $\frac{1}{2}=e^{k1599}$ $\ln\frac{1}{2}=k1599$ $k=\frac{\ln\frac{1}{2}}{1599}$ $k\approx−4.33x10−4$ Using known k value, solve for t = 1000 and t = 10,000. $t=1000$ $y=20e^{(−4.33x10−4)(1000)}$ $y\approx12.96$ $t=10,000$ $y=20e^{(−4.33x10−4)(10000)}$ $y≈0.26$
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