Problem 11 - Calculus

Question

A stream flowing into a lake brings with it a pollutant at a rate of 8 metric tons per year. The river leaving the lake removes the pollutant at a rate proportional to the quantity in the lake, with constant of proportionality -0.16 if time is measured in years. (a) Is the quantity of pollutant in the lake increasing or decreasing at a moment at which the quantity is 45 metric tons? At which the quantity is 55 metric tons? (b) What is the quantity of pollutant in the lake after a long time?

Step-by-Step Solution

Answer

Quantity is increasing at 45 metric tons and decreasing at 55 metric tons. Long-term quantity is 50 metric tons.

Step 1: Determine Rate of Change Equation

To find whether the pollutant quantity increases or decreases, we need to establish the differential equation of change. The rate of change of pollutant quantity \( Q \) in the lake can be written as:\[\frac{dQ}{dt} = \, \text{Rate in} \, - \, \text{Rate out}\]The rate at which the pollutant enters is 8 metric tons per year. The rate at which it leaves is proportional to the quantity \( Q \) with a proportionality constant of -0.16. Therefore, the equation becomes:\[\frac{dQ}{dt} = 8 - 0.16Q\]

Step 2: Analyze the Rate of Change at 45 metric tons

At \( Q = 45 \) metric tons, substitute \( Q \) into the differential equation to find \( \frac{dQ}{dt} \):\[\frac{dQ}{dt} = 8 - 0.16 \times 45\]\[\frac{dQ}{dt} = 8 - 7.2 = 0.8\]Since \( \frac{dQ}{dt} > 0 \), the quantity is **increasing** at 45 metric tons.

Step 3: Analyze the Rate of Change at 55 metric tons

At \( Q = 55 \) metric tons, substitute \( Q \) into the differential equation:\[\frac{dQ}{dt} = 8 - 0.16 \times 55\]\[\frac{dQ}{dt} = 8 - 8.8 = -0.8\]Since \( \frac{dQ}{dt} < 0 \), the quantity is **decreasing** at 55 metric tons.

Step 4: Determine the Long-Term Quantity of Pollutant

To find the equilibrium quantity (long-term quantity), set \( \frac{dQ}{dt} = 0 \) and solve for \( Q \):\[0 = 8 - 0.16Q\]Rearrange to find \( Q \):\[0.16Q = 8 \quad \Rightarrow \quad Q = \frac{8}{0.16} = 50\]After a long time, the quantity of pollutant will be **50 metric tons**.

Key Concepts

Rate of Change

In differential equations, the rate of change is used to describe how a particular quantity varies over time. In our example, we are dealing with the amount of pollutant in a lake. The rate at which this amount changes is given by the equation: \[ \frac{dQ}{dt} = \text{rate in} - \text{rate out} \]. This means that the rate of change of the pollutant is calculated by subtracting the outflow rate from the inflow rate.
For this problem, the inflow rate of the pollutant is 8 metric tons per year. The outflow rate is 0.16 times the current quantity of the pollutant in the lake, written as \( 0.16Q \). By establishing this equation, we can predict whether the quantity is increasing or decreasing over time simply by evaluating the sign of \( \frac{dQ}{dt} \). - **Positive Rate of Change:** Indicates that the quantity is increasing. - **Negative Rate of Change:** Indicates that the quantity is decreasing.By plugging in specific values of \( Q \) such as 45 and 55 metric tons, we can determine the behavior of the system at those points.

Equilibrium Quantity

The equilibrium quantity refers to the state where the rate of change becomes zero, meaning the inputs and outputs are balanced. In our differential equation, we find the equilibrium by setting \( \frac{dQ}{dt} = 0 \). This implies that there is no net change in the pollutant level. The equation \( 0 = 8 - 0.16Q \) represents this balance, where the inflow of pollutants precisely matches the outflow, driven by the quantity present in the lake.

To find this equilibrium quantity, solve the equation for \( Q \). Initially, rearrange the equation to give \( 0.16Q = 8 \), leading us to \( Q = \frac{8}{0.16} = 50 \). Thus, after a long time, the pollutant settles at 50 metric tons, establishing a steady state whereby inflow equals outflow. Understanding the concept of equilibrium helps us know what happens over a long time, where temporary increases or decreases in pollutant levels ultimately level out.

Proportionality Constant

The proportionality constant is a crucial component in differential equations that helps express how a specific rate is related to another variable. In this case, it connects the outflow rate of the pollutant to the quantity of the pollutant in the lake. The rate of change equation \( \frac{dQ}{dt} = 8 - 0.16Q \) features - 0.16: The proportionality constant.This constant indicates how intensely the quantity within the lake impacts the rate at which pollutants are washed away by the river leaving the lake. Essentially, the larger the quantity of pollutants present, the faster they leave, thanks to this constant. However, it's important to note that this relationship is proportional, meaning that an increase in the quantity directly increases the outflow at a rate governed by this constant.

Recognizing the role of the proportionality constant helps in understanding and predicting how systems react to change, specifically how the lake system will self-correct or adjust its pollutant level based on the current quantity.