# 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 $2.22\times10^{-3}mg/minutes.mL$

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