Problem 24 - Calculus Early Transcendentals

Question

The water in a polluted lake initially contains 1 lb of mercury salts per \(100,000\) gal of water. The lake is circular with diameter \(30 \mathrm{m}\) and uniform depth \(3 \mathrm{m}\). Polluted water is pumped from the lake at a rate of \(1000 \mathrm{gal} / \mathrm{h}\) and is replaced with fresh water at the same rate. Construct a table that shows the amount of mercury in the lake (in \(1 \mathrm{b})\) at the end of each hour over a 12 -hour period. Discuss any assumptions you made. [Note: Use \(1 \mathrm{m}^{3}=264\) gal.]

Step-by-Step Solution

Answer

The mercury decreases gradually, starting at 5.61 lbs and reduces to 5.48 lbs after 12 hours.

Step 1: Calculate the Total Volume of the Lake

The lake is circular with a diameter of 30 m and a uniform depth of 3 m. The volume of a cylinder (which describes the lake) is given by \( V = \pi r^2 h \) where \( r \) is the radius and \( h \) is the height (or depth in this case). Here, \( r = \frac{30}{2} = 15 \) meters and \( h = 3 \) meters. Thus:\[V = \pi (15^2)(3) = 675\pi \text{ cubic meters}\]Then, convert this to gallons using the conversion \(1 \text{ m}^3 = 264 \text{ gal}\):\[V = 675\pi \times 264 \approx 561,349 \text{ gallons}\]

Step 2: Initial Mercury Calculation

Initially, there is 1 lb of mercury per 100,000 gallons of water. To find the initial amount of mercury in the lake, calculate:\[\text{Initial Mercury} = \frac{561,349}{100,000} = 5.61349 \text{ lbs}\]This rounds to approximately 5.61 lbs.

Step 3: Set Up Mercury Reduction Function

As water is pumped out and replaced by fresh water at the same rate, the concentration of mercury decreases exponentially over time. The amount of mercury left after each hour can be found using:\[y = y_0 \times \left(1 - \frac{r}{V}\right)^t\]where \(y_0 = 5.61\) lbs (initial mercury), \(r = 1000\) gal/hour (rate of pumping), \(V = 561,349\) gal (total volume), and \(t\) is the time in hours.

Step 4: Calculate Mercury for Each Hour

Fill in the table using the reduction formula from Step 3 to calculate the mercury for each hour until 12 hours:\[y_t = 5.61 \times \left(1 - \frac{1000}{561349}\right)^t\]Calculate \(y_t\) for \(t = 1, 2, 3, \ldots, 12\).

Step 5: Construct the Table

Compile the results of Step 4 into a table:\[\begin{array}{|c|c|}\hline\text{Hour} & \text{Mercury (lbs)} \\hline0 & 5.61 \1 & 5.60 \2 & 5.59 \3 & 5.57 \4 & 5.56 \5 & 5.55 \6 & 5.54 \7 & 5.53 \8 & 5.52 \9 & 5.51 \10 & 5.50 \11 & 5.49 \12 & 5.48 \\hline\end{array}\]

Key Concepts

Exponential Decay

Exponential decay is a fundamental concept in understanding how quantities decrease over time. It's commonly modeled with a formula that describes how the amount of a substance decreases, where the rate of reduction is proportional to the current quantity.
In the case of mercury in the polluted lake, we have an initial quantity that reduces every hour due to the replacement of polluted water with fresh water at a constant rate. This continuous process is where exponential decay shines.

• Exponential decay is represented mathematically as \(y = y_0 \times \left(1 - \frac{r}{V}\right)^t\).
• Here, \(y_0\) is the initial amount (5.61 lbs of mercury), \(r\) is the rate at which the water is pumped (1000 gal/hour), \(V\) is the lake's total volume (561,349 gallons), and \(t\) is time in hours.

This model assumes that the process continues indefinitely, meaning that as long as water is being pumped out and fresh water is being added, the concentration will keep decreasing exponentially. Each hour, the quantity of mercury diminishes slightly, showing a classic exponential decay pattern.

Pollution Modeling

Pollution modeling in a body of water involves understanding how pollutants are distributed and reduced over time. The mercury in the lake illustrates a real-world scenario of pollution modeling.

• We start with an initial concentration of mercury in the water, which was 1 lb per 100,000 gallons initially.
• To model the reduction in pollution, we consider the process of removing and replacing water at the continuous rate of 1000 gallons per hour with fresh water, thereby diluting the concentration of mercury over time.

In pollution modeling, assumptions play a crucial role. We assumed the mixing of water and mercury is uniform and immediate, meaning the mercury evenly spreads across the lake each hour. This helps in simplifying the model to use exponential decay to predict mercury levels.
Such modeling can be extended to other pollutants and environments, offering crucial insights into environmental management and policymaking.

Volume Calculation

Calculating volume is a critical skill and here, we addressed the volume of the lake, which affects the pollution modeling and exponential decay.

• The lake is modeled as a cylinder with a diameter of 30 meters and depth of 3 meters.
• The formula for the volume of a cylinder, \( V = \pi r^2 h \), helps us determine the lake's volume where \( r \) (radius) is half the diameter (15 meters) and \( h \) (height) is 3 meters.
• The calculated volume in cubic meters is \(675\pi\).

This volume was then converted to gallons using the conversion factor of 1 cubic meter being equal to 264 gallons, giving about 561,349 gallons. Knowing the volume allowed us to set up the differential equation for mercury reduction and to track how pollution changes over time. Understanding these conversion processes and volume calculations is vital in many areas of science and engineering, especially those involving environmental studies and natural resource management.