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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?

Short Answer

Expert verified
Approximately 4.88% of the ozone will decay in 20 years.

Step by step solution

01

Set Up the Differential Equation

The problem states that the rate of change of ozone, \( Q \), is proportional to the amount of ozone present. Mathematically, we express this as: \( \frac{dQ}{dt} = kQ \). Given, the constant of proportionality is \( k = -0.0025 \). Thus, the differential equation becomes \( \frac{dQ}{dt} = -0.0025Q \).
02

Solve the Differential Equation

To solve the differential equation \( \frac{dQ}{dt} = -0.0025Q \), we separate variables and integrate: \( \frac{1}{Q} \, dQ = -0.0025 \, dt \). Integrating both sides, we get \( \ln Q = -0.0025t + C \), where \( C \) is the constant of integration. Exponentiating both sides, we arrive at \( Q = e^C e^{-0.0025t} \). Let \( Q_0 = e^C \), the initial amount of ozone, giving us the general solution: \( Q = Q_0 e^{-0.0025t} \).
03

Calculate the Percent Decay Over 20 Years

To find the percent of ozone that will decay in the next 20 years, we compute the ratio of the ozone remaining to the initial amount. Using the solution \( Q = Q_0 e^{-0.0025t} \), substitute \( t = 20 \) to get \( Q = Q_0 e^{-0.0025 \times 20} = Q_0 e^{-0.05} \). The fraction of remaining ozone is \( e^{-0.05} \), so the decayed percentage is \( (1 - e^{-0.05}) \times 100\% \). Calculating, this gives approximately 4.88%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ozone Depletion
Ozone depletion refers to the gradual thinning of the Earth's ozone layer in the upper atmosphere. This is a significant environmental concern because the ozone layer plays a crucial role in protecting life on Earth by absorbing most of the Sun's harmful ultraviolet radiation.
Without sufficient ozone, increased levels of UV radiation can reach the Earth's surface, potentially causing a range of effects such as:
  • Increased skin cancers and cataracts in humans
  • Negative impacts on wildlife, especially marine life like plankton
  • Damage to crops and terrestrial ecosystems
The differential equation derived in the exercise helps model the depletion by predicting how the quantity of ozone, denoted as \( Q \), decreases over time. This model is essential for understanding long-term environmental changes and planning interventions.
Exponential Decay
Exponential decay describes a process where the quantity of something decreases at a rate proportional to its current value. This aspect is crucial in modeling the thinning of the ozone layer.
Mathematically, we indicate this by using the formula for exponential decay:
  • \( Q = Q_0 e^{-kt} \)
  • Where \( Q_0 \) is the initial amount
  • \( k \) is the positive rate constant, with \( -0.0025 \) in our case indicating decay
  • \( t \) is time
The concept of exponential decay helps illustrate how processes like ozone depletion have initially large impacts, which slow down as the levels decrease. This is often visualized as an exponential curve, highlighting the quick loss of ozone initially, followed by a leveling off over time.
Separation of Variables
Separation of variables is a powerful method used to solve ordinary differential equations (ODEs). It involves rearranging the equation so that each variable and its differential are on separate sides, and then integrating both sides.
In the given problem, we start with the differential equation:\[ \frac{dQ}{dt} = kQ \]To separate the variables, we rearrange the terms:
  • \( \frac{1}{Q} \, dQ = -0.0025 \, dt \)
By integrating both sides, we isolate the natural logarithm of the ozone amount on one side and the linear function of time on the other, leading to:
  • \( \ln Q = -0.0025t + C \)
Subsequently solving for \( Q \) gives us the exponential model \( Q = Q_0 e^{-0.0025t} \), which is key to predicting how much ozone will remain over a period like 20 years.
This method is widely used in problems where rates of change depend directly on the current state, simplifying the process and enabling deeper insights into dynamical systems.

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Most popular questions from this chapter

A bank account earns \(7 \%\) annual interest compounded continuously. You deposit \(\$ 10,000\) in the account, and withdraw money continuously from the account at a rate of \(\$ 1000\) per year. (a) Write a differential equation for the balance, \(B\), in the account after \(t\) years. (b) What is the equilibrium solution to the differential equation? (This is the amount that must be deposited now for the balance to stay the same over the years.) (c) Find the solution to the differential equation. (d) How much is in the account after 5 years? (e) Graph the solution. What happens to the balance in the long run?

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