Problem 10 - Applied Calculus

Question

The amount of ozone, \(Q,\) in the atmosphere is decreasing at a rate proportional to the amount of ozone present. If time \(t\) is measured in years, the constant of proportionality is \(-0.0025 .\) Write a differential equation for \(Q\) as a function of \(t,\) and give the general solution for the differential equation. If this rate continues, approximately what percent of the ozone in the atmosphere now will decay in the next 20 years?

Step-by-Step Solution

Answer

The ozone decays by approximately 4.88% over 20 years.

Step 1: Setting up the Differential Equation

The problem states that the amount of ozone, \( Q \), decreases at a rate proportional to the amount present, which can be expressed as \( \frac{dQ}{dt} = kQ \), where \( k \) is the constant of proportionality. Given that the constant is \(-0.0025\), we have \( \frac{dQ}{dt} = -0.0025Q \).

Step 2: Solving the Differential Equation

The differential equation \( \frac{dQ}{dt} = -0.0025Q \) is a separable differential equation. Separate variables to get \( \frac{1}{Q}dQ = -0.0025 \, dt \). Integrate both sides to obtain \( \ln|Q| = -0.0025t + C \), where \( C \) is the constant of integration.

Step 3: Exponentiating to Solve for Q

To solve for \( Q \), exponentiate both sides: \( e^{\ln|Q|} = e^{-0.0025t + C} \). This simplifies to \( |Q| = e^C e^{-0.0025t} \). Let \( A = e^C \), then \( Q(t) = Ae^{-0.0025t} \).

Step 4: Determining the Decay Over 20 Years

To find out the percentage decay in 20 years, use the solution \( Q(t) = Ae^{-0.0025t} \). Calculate \( Q(20) = Ae^{-0.0025 imes 20} = Ae^{-0.05} \). The percentage remaining after 20 years is \( e^{-0.05} \times 100\% \).

Step 5: Calculating the Decay Percentage

Calculate \( e^{-0.05} \approx 0.9512 \), which means approximately 95.12% of the ozone remains. Therefore, the ozone has decayed by approximately \( 100\% - 95.12\% = 4.88\% \).

Key Concepts

Rate of Change

In mathematics and science, the rate of change is a fundamental concept. It describes how a quantity changes over time. When we talk about differential equations, the rate of change is often represented by a derivative. For example, in the problem we're dealing with, the rate of change of ozone concentration is given as \( \frac{dQ}{dt} \). This notation indicates how the amount of ozone, \(Q\), changes with respect to time, \(t\).

Here, the rate is not constant but instead varies depending on how much ozone is currently present. This is a typical scenario in many natural processes such as radioactive decay or population growth. Understanding the rate of change helps predict future outcomes by acknowledging current conditions.

Proportionality Constant

The proportionality constant is a special type of constant found in equations where two quantities are directly proportional to one another. In the scenario of our ozone problem, this constant is denoted by \(k\) and is the value that relates the rate of change of ozone to the amount currently present.

Because the constant of proportionality in this problem is negative, \(k = -0.0025\), it indicates a decrease over time, which is an integral part of exponential decay. This constant ensures that the differential equation accurately reflects how changes occur. To put it simply, if \(Q\) were to double, the rate at which it decreases would also double, with the rate being directly proportional to \(Q\). This helps maintain a consistent mathematical description of the decay process.

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. This can be seen in our solution formula, \(Q(t) = Ae^{-0.0025t}\).

Here, the function is exponential because the variable \(t\) is in the exponent. Exponential decay is characterized by a rapid decrease initially, which slows over time as the quantity reduces. The negative sign in the exponent reflects this decay. The decay constant, \(-0.0025\), affects how quickly or slowly this process occurs. Over 20 years, we calculated a decrease of about 4.88%.

Exponential decay applies to many real-life situations such as radioactive decay, depreciation of asset value, and cooling of warm objects.

Separable Differential Equations

Separable differential equations are among the simplest types of differential equations to solve. They involve separating the variables to opposite sides of the equation so each variable is integrated separately.

In our ozone scenario, the differential equation is written as \(\frac{dQ}{dt} = -0.0025Q\). We can rearrange it to \(\frac{1}{Q}dQ = -0.0025 \, dt\). Here, \(Q\) and \(t\) have been separated to opposite sides. By integrating both sides, we find a relation between \(Q\) and \(t\).

This separation makes it more straightforward to solve the equation analytically. It's a powerful method because it transforms a complex differential problem into a more manageable form, allowing us to solve for the quantity of interest.