#### Answer

The concentration of pollutants (in grams per liter) in the east fork of the Big Weasel River is approximated by
$$
\begin{aligned}
P(x)&=0.04 e^{-4x}
\end{aligned}
$$
(a)
The concentration of pollutants $0.5$ mile downstream is :
$$
\begin{aligned}
P(0.5)&=0.04 e^{-4(0.5)}\\
& =0.04 e^{-2}\\
& \approx 0.005
\end{aligned}
$$
(b)
The concentration of pollutants $1$ mile downstream is:
$$
\begin{aligned}
P(1)&=0.04 e^{-4(1)}\\
& =0.04 e^{-4}\\
& \approx 0.0007
\end{aligned}
$$
(c)
The concentration of pollutants $2$ mile downstream is:
$$
\begin{aligned}
P(2)&=0.04 e^{-4(2)}\\
& =0.04 e^{-8}\\
& \approx 0.000013
\end{aligned}
$$
The rate of change of concentration with respect to distance is given by
$$
\begin{aligned} P^{\prime}(x) &=0.04 (-4)e^{-4x} \\
&=-0.16 e^{-4x} \\
\end{aligned}
$$
Now, the rate of change of concentration with respect to distance for the following distances.
(d) 0.5 mile
$$
\begin{aligned} P^{\prime}(0.5) &=0.04 (-4)e^{-4(0.5)} \\
&=-0.16 e^{-4(0.5)} \\
&=-0.16 e^{-2} \\
& \approx -0.022
\end{aligned}
$$
(e) 1 mile
$$
\begin{aligned} P^{\prime}(1) &=0.04 (-4)e^{-4(1)} \\
&=-0.16 e^{-4(1)} \\
&=-0.16 e^{-4} \\
& \approx -0.0029
\end{aligned}
$$
(f) 2 mile
$$
\begin{aligned} P^{\prime}(2) &=0.04 (-4)e^{-4(2)} \\
&=-0.16 e^{-8} \\
& \approx -0.000054
\end{aligned}
$$