Calculus with Applications (10th Edition)

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}
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}