Exercise 1.4.32

Question

After alcohol is fully absorbed into the body, it is metabolized. Suppose that after consuming several alcoholic drinks earlier in the evening, your blood alcohol concentration (BAC) at midnight is $0.14 \mathrm{~g} / \mathrm{dL}$. After 1.5 hours your BAC is half this amount.

\begin{enumerate}[label=(\alph*)]
\item Find an exponential model for your BAC $t$ hours after midnight.

\item  Graph your BAC and use the graph to determine when your $\mathrm{BAC}$ reaches the legal limit of $0.08 \mathrm{~g} / \mathrm{dL}$.
\end{enumerate}

Khagan Gulusoy · Accepted Answer

\begin{enumerate}[label=(\alph*)]
\item There are a lot of exponential functions that fit the points $(0,0.14$ ) and $(1.5,0.07$ ) (the second coordinate is 0.07 since $0.14: 2=0.07$). Suppose that we want to find an exponential function that it is of type:

$$\forall t \in \mathbb{R}: \quad f(t) =C b^{t}$$

for some $C, b \in \mathbb{R}$. If we substitute the values of time $t$ to the formula above, we get:

$$
\begin{aligned}
& t=0: \quad C b^{0}=C \cdot 1=C=0.14 . \
& t=1.5: \quad C b^{1.5}=0.07 .
\end{aligned}
$$

If we substitute the first equation into the second one, we obtain the following:

$$
C b^{1.5}=C \cdot b^{1.5}=0.14 \cdot b^{1.5}=0.07 	ext {. }
$$

Thus, we have:

$$
\begin{aligned}
& b^{1.5}=0.07: 0.14 \
& b^{1.5}=0.5 \
& b=\sqrt[1.5]{0.5}=\sqrt[3]{(0.5)^{2}}=\sqrt[3]{0.25}
\end{aligned}
$$

Substituting $b$ and $C$ in the general formula above, we obtain the desired formula.

$$
\forall t \in[0,+\infty): f(t)=0.14 \cdot(\sqrt[3]{0.25})^{t} 	ext {. }
$$

\item

\begin{figure}[ht]
\centering
\begin{tikzpicture}
\begin{axis}[
    axis lines=middle,
    xlabel={$t$},
    ylabel={$f(t)$},
    domain=0:2,
    ymax=0.2,
    samples=200,
    legend style={
        at={(0.95,0.95)},
        anchor=north east,
    },
    ytick={0,0.07,0.08,0.14,0.17,0.2},
    yticklabel style={
        /pgf/number format/.cd,
        fixed,
        fixed zerofill,
        precision=2,
    },
]

% Plot the function
\addplot[mark=none, color=blue, ultra thick] {0.14 * (0.25^(1/3))^x};
\addlegendentry{$f(t) = 0.14 \cdot \left(\sqrt[3]{0.25}ight)^t$};

% Add dashed lines
\draw[dashed] (axis cs:0,0.07) -- (axis cs:1.5,0.07);
\draw[dashed] (axis cs:1.5,0.07) -- (axis cs:1.5,0);

\draw[dashed] (axis cs:0,0.08) -- (axis cs:1.2,0.08);
\draw[dashed] (axis cs:1.2,0.08) -- (axis cs:1.2,0);

\end{axis}
\end{tikzpicture}
\end{figure}
From the graph, we can conclude that $f$ reaches $0.08$ at $t \approx 1.2$.
\end{enumerate}

Exercise 1.4.32

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Exercise 1.4.32

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