Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Review - Test - Page 391: 12

Answer

(a) $m(t)=3\cdot 2^{-t/10}$ (b) $m(t)=3e^{-0.0693t}$ (c) $0.047$g (d) $3.6$ minutes

Work Step by Step

(a) Given $h=10s, m_0=3g$, the function becomes $m(t)=3\cdot 2^{-t/10}$ (b) In this case $m(t)=3e^{-rt}$ and half life $10s$ gives $m(10)=m_0/2=3/2=1.5$, so we have $1.5=3e^{-10r}$, solve for $r$ we get $r=-ln(1.5/3)/10=0.0693$ The function becomes $m(t)=3e^{-0.0693t}$ (c) Let $t=1min=60s$, use the formula in (a), we have $m(60)=3\times2^{-60/10}\approx0.047$g (d) Let $m(t)=1\mu g=10^{-6}g$, use the formula in (b), we have $3e^{-0.0693t}=10^{-6}$, solve for $t$ to get $t=-ln(10^{-6}/3)/0.0693\approx215.2s\approx3.6$ minutes
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