Problem 39 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

NUCLEAR WASTE After \(t\) years of operation, a certain nuclear power plant produces radioactive waste at the rate of \(R(t)\) pounds per year, where $$ R(t)=300-200 e^{-0.03 t} $$ The waste decays exponentially at the rate of \(2 \%\) per year. How much radioactive waste from the plant will be present in the long run?

Step-by-Step Solution

Answer

The amount of radioactive waste in the long run is 15000 pounds.

Step 1 - Understand the Waste Production Rate

Identify that the waste production rate is given by the function: \[ R(t) = 300 - 200e^{-0.03t} \] This tells the rate at which waste is produced in pounds per year.

Step 2 - Concept of Long Run

In the long run, we need to find the total amount of waste that will be present. This involves integrating the production rate and accounting for the decay of the waste.

Step 3 - Recognize Decay Rate

The radioactive waste decays at a rate of \(2\%\) per year, which can be expressed as an exponential decay function with a decay constant of \(0.02\).

Step 4 - Setting Up the Integral

To find the amount of radioactive waste in the long run, set up the integral of \(R(t)\) over time: \[ \text{Amount of Waste} = \frac{R(t)}{\text{decay rate}} \]

Step 5 - Integral and Solution

Substitute \(R(t)\) and the decay rate: \[ \text{Amount of Waste} = \frac{300 - 200e^{-0.03t}}{0.02} \] As \(t\) approaches infinity, the term \(200e^{-0.03t}\) tends to zero, so: \[ \text{Amount of Waste} = \frac{300}{0.02} = 15000 \text{ pounds} \]

Key Concepts

exponential decay

Exponential decay describes the process by which a quantity decreases at a rate proportional to its current value. In the context of radioactive waste, this means that the amount of waste decreases over time at a rate proportional to how much waste is present. This can be mathematically represented by the equation: \[ N(t) = N_0 e^{-kt} \] where

• \(N(t)\) is the amount of waste at time \(t\)
• \(N_0\) is the initial amount of waste
• \( k \) is the decay constant
• \( e \) is the base of the natural logarithm (approximately equal to 2.71828).

For the given exercise, the decay rate is 2% per year. This translates to a decay constant \(k\) of 0.02. Over time, the waste will decrease exponentially, resulting in less and less waste remaining. This is an important concept in understanding how radioactive waste diminishes over time.

integral calculus

Integral calculus is a branch of mathematics that deals with finding total quantities when given a rate of change. In the context of our exercise, we have the waste production rate given by the function \[ R(t) = 300 - 200e^{-0.03t} \] To find the total amount of radioactive waste produced over time, calculus helps us integrate this rate function. Integration can be thought of as the reverse process of differentiation. In simple terms, it allows us to add up small quantities to find a total amount. The integral of the waste production rate over time will give us the total amount of waste produced. However, because the waste also decays over time, we need to consider this decay in our calculations. The decay is accounted for by dividing the rate function by the decay rate.This gives us:\[ \text{Amount of Waste} = \frac{R(t)}{\text{decay rate}} \]. In our exercise, when \(t\) approaches infinity, the term \(200e^{-0.03t}\) tends to zero. Therefore, the integral solution simplifies to \( \frac{300}{0.02} = 15000 \text{ pounds} \) of waste in the long run.

long-term equilibrium analysis

Long-term equilibrium analysis involves finding the steady state condition where the system has reached a balance and the values no longer change over time. In the context of radioactive waste management, we use this analysis to determine how much waste will ultimately be present in the facility after many years. Initially, the power plant produces waste at a certain rate, which then decays over time. The rate at which new waste is generated is given by \[ R(t) = 300 - 200e^{-0.03t} \] while the decay rate is a constant 2% per year. In the long-term (as \(t\) approaches infinity), we can analyze the behavior of the waste production and decay to find a balance point. The equilibrium amount of waste can be calculated by considering the ongoing production and the continuous decay. With our given functions, we found that the amount of radioactive waste in the long run is: \[ \text{Amount of Waste} = \frac{300}{0.02} = 15000 \text{ pounds} \] This means that the system will eventually stabilize, and there will be 15,000 pounds of radioactive waste remaining in the long-term equilibrium state.