Problem 13 - Applied Calculus

Question

Filters at a water treatment plant become less effective over time. The rate at which pollution passes through the filters into a nearby lake is given in the following table. (a) Estimate the total quantity of pollution entering the lake during the 30 -day period. (b) Your answer to part (a) is only an estimate. Give bounds (lower and upper estimates) between which the true quantity of pollution must lie. (Assume the rate of pollution is continually increasing.) $$\begin{array}{c|c|c|c|c|c|c} \hline \text { Day } & 0 & 6 & 12 & 18 & 24 & 30 \\ \hline \text { Ratc (kg/day) } & 7 & 8 & 10 & 13 & 18 & 35 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

Estimated pollution: 420 kg. Bounds: 336 kg to 504 kg.

Step 1: Understanding the Problem

We need to estimate the total quantity of pollution entering the lake over a 30-day period, using the rate data provided for specific days. Then, calculate lower and upper estimates assuming the pollution rate is continually increasing.

Step 2: Estimate Total Quantity Using the Trapezoidal Rule

To estimate the total pollution, apply the Trapezoidal Rule over each interval [0, 6], [6, 12], etc. The formula for each interval is \( \frac{b-a}{2} \times (f(a) + f(b)) \), where \( a \) and \( b \) are consecutive days, and \( f(a) \) and \( f(b) \) are the corresponding rates. Calculate for each segment, sum them up.

Step 3: Calculate the Pollution for Each Interval

- For [0, 6]: \( \frac{6-0}{2} \times (7 + 8) = 6 \times 7.5 = 45 \) kg- For [6, 12]: \( \frac{12-6}{2} \times (8 + 10) = 6 \times 9 = 54 \) kg- For [12, 18]: \( \frac{18-12}{2} \times (10 + 13) = 6 \times 11.5 = 69 \) kg- For [18, 24]: \( \frac{24-18}{2} \times (13 + 18) = 6 \times 15.5 = 93 \) kg- For [24, 30]: \( \frac{30-24}{2} \times (18 + 35) = 6 \times 26.5 = 159 \) kg

Step 4: Sum Up the Estimated Pollution

Add up the pollution from each interval: \( 45 + 54 + 69 + 93 + 159 = 420 \) kg. The total estimated pollution over 30 days is 420 kg.

Step 5: Determine Lower and Upper Bounds

For lower bound: Assume pollution rate is constant at the lower value of each interval. Calculate as follows:- \( [0, 6]: 7 \times 6 = 42 \)- \( [6, 12]: 8 \times 6 = 48 \)- \( [12, 18]: 10 \times 6 = 60 \)- \( [18, 24]: 13 \times 6 = 78 \)- \( [24, 30]: 18 \times 6 = 108 \)Sum is \( 42 + 48 + 60 + 78 + 108 = 336 \) kg.For upper bound: Assume pollution rate jumps to the upper value of each interval.- \( [0, 6]: 8 \times 6 = 48 \)- \( [6, 12]: 10 \times 6 = 60 \)- \( [12, 18]: 13 \times 6 = 78 \)- \( [18, 24]: 18 \times 6 = 108 \)- \( [24, 30]: 35 \times 6 = 210 \)Sum is \( 48 + 60 + 78 + 108 + 210 = 504 \) kg.

Step 6: Conclude with Bounds

The true quantity of pollution lies between the lower bound of 336 kg and the upper bound of 504 kg.

Key Concepts

Pollution Estimation

When dealing with environmental concerns, it's crucial to estimate the amount of pollutants entering ecosystems, particularly for preventive measures and regulation. In the context of the given exercise, we're estimating pollution transferred into a lake through water treatment plant filters over a specific period.
At its core, pollution estimation involves understanding rates and applying mathematical methods to approximate total quantities. The core idea is to take given values from intervals (in this exercise, the rate of pollution per day) and project an estimated total. Using numerical integration techniques, especially when dealing with varying rates, such as the Trapezoidal Rule, allows us to achieve better accuracy over simply summing values.
Moreover, acknowledging that filter efficiency decreases over time, it becomes more significant to estimate accurately how much pollution enters the surrounding environment due to increased rates. It's not just about an approximate figure but also understanding limitations and deriving bounds, such as assuming extremes, which help refine the estimate's accuracy.

Rate of Change

The rate of change in pollution levels is a crucial factor in this problem. Rates provide a snapshot of how fast pollution is passing through the filters at any given moment. An understanding of rate of change helps in analyzing how these rates evolve as days go by.
This can be seen clearly in our problem scenario: as time progresses, the rate at which pollution enters the lake increases. Such an increase implies that over time, as filters wear out, they let more pollution through.
Recognizing a trend like this is important. Not only for this estimation problem but also for forming strategies in real-world pollution control. When plant management knows the rate of change, they can implement timely maintenance or upgrades to the filters, minimizing ecological impact and ensuring the longevity of environmental resources.

Integration

Integration is the mathematical process used to compute the total quantity of pollutants entering the lake over several days. It is a fundamental concept in calculus that allows the calculation of area under a curve, in this case, representing the total pollution.
In practice, integration can be quite complex with continuous functions. However, in our specific case, we use the Trapezoidal Rule, a straightforward numerical method. This technique is beneficial when dealing with discrete sets of data, as it approximates the area under the curve by splitting it into trapezoids.
The formula used in the Trapezoidal Rule is \[\frac{b-a}{2} \times (f(a) + f(b))\]Here, \( a \) and \( b \) represent interval limits, while \( f(a) \) and \( f(b) \) are the respective rates of pollution on those days. Integration, in this context, provides an accurate approach for calculating an estimated total amount of pollution given variable rates over time.