Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises - Page 206: 85


(a) After 10 minutes, BAC increases by $$7.519\times10^{-3}(mg/minutes.mL)$$ (b) After half an hour later, BAC decreases by $$2.22\times10^{-3}(mg/minutes.mL)$$

Work Step by Step

$$C(t)=0.0225te^{-0.0467t}$$ 1) Here the question asks how rapidly BAC was increasing and decreasing, so it concerns the rate of change of BAC values. Therefore, to find out how rapidly BAC increases or decreases, first we need to find the derivative of the BAC formula, or $C'(t)$. $$C'(t)=(0.0225te^{-0.0467t})'$$ $$C'(t)=0.0225(t'e^{-0.0467t}+t(e^{-0.0467t})')$$ $$C'(t)=0.0225(e^{-0.0467t}+t(e^{-0.0467t})')$$ *Consider $(e^{-0.0467t})'$ $$(e^{-0.0467t})'=\frac{d(e^{-0.0467t})}{d(-0.0467t)}\frac{(-0.0467)dt}{dt}$$ $$(e^{-0.0467t})'=-0.0467e^{-0.0467t}$$ Therefore, $$C'(t)=0.0225[e^{-0.0467t}-0.0467te^{-0.0467t}]$$ $$C'(t)=0.0225e^{-0.0467t}(1-0.0467t)(mg/minutes.mL)$$ (a) After 10 minutes $(t=10)$, the rate of increase of BAC is $$C'(10)=0.0225e^{-0.0467\times10}(1-0.0467\times10)$$ $$C'(10)=0.0225e^{-0.467}(1-0.467)$$ $$C'(10)=0.0225\times0.627\times0.533\approx7.519\times10^{-3}(mg/minutes.mL)$$ (b) In the second part, the question asks for the rate of decrease of BAC half an hour later $(t=30)$. Again we would substitute $t=30$ into the formula of $C'(t)$ $$C'(30)=0.0225e^{-0.0467\times30}(1-0.0467\times30)$$ $$C'(30)=0.0225e^{-1.401}[1-1.401]$$ $$C'(30)=0.0225\times0.246\times(-0.401)\approx-2.22\times10^{-3}(mg/minutes.mL)$$So, after 30 minutes, the rate of decrease of BAC is $2.22\times10^{-3}mg/minutes.mL$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.