Problem 12 - Algebra 2 and Trigonometry

Question

In \(1986,\) the worst nuclear power plant accident in history occurred in the Chernobyl Nuclear Power Plant located in the Ukraine. On April \(26,\) one of the reactors exploded, releasing large amounts of radioactive isotopes into the atmosphere. The amount of plutonium present after \(t\) years can be modeled by the function: $$ y=P e^{-0.0000288 t} $$ where \(P\) represents the amount of plutonium that is released. a. Graph this function over the interval \(0 \leq t \leq 100,000\) and \(P=10\) grams. b. If 10 grams of the isotope plutonium- 239 were released into the air, to the nearest hundredth, how many grams will be left after 10 years? After 100 years? c. Using the graph, approximate how long it will take for the 10 grams of plutonium- 239 to decay to 1 gram.

Step-by-Step Solution

Answer

After 10 years, about 10 grams remain; after 100 years, about 9.97 grams. It takes about 80,358 years to decay to 1 gram.

Step 1: Understanding the Function

The function given is an exponential decay model: \[y = P e^{-0.0000288 t}\]This represents the decay of plutonium over time, where \(P\) is the initial amount and \(t\) is the time in years.

Step 2: Graphing the Function over 100,000 Years

We need to graph the function \(y = 10 e^{-0.0000288 t}\) over the interval \(0 \leq t \leq 100,000\). The function will show a decay curve starting at 10 grams when \(t = 0\) and approaching 0 as \(t\) increases.

Step 3: Calculating Remaining Plutonium After 10 Years

Substitute \(t = 10\) into the function:\[y = 10 e^{-0.0000288 \times 10}\]Calculate:\[y \approx 10 e^{-0.000288} \approx 9.9971\]So, approximately 10 grams of plutonium remain after 10 years.

Step 4: Calculating Remaining Plutonium After 100 Years

Substitute \(t = 100\) into the function:\[y = 10 e^{-0.0000288 \times 100}\]Calculate:\[y \approx 10 e^{-0.00288} \approx 9.9711\]Approximately 9.97 grams of plutonium are left after 100 years.

Step 5: Determining When Plutonium Decays to 1 Gram

We want to find \(t\) when \(y = 1\):\[1 = 10 e^{-0.0000288 t}\]Divide both sides by 10 and take the natural logarithm:\[e^{-0.0000288 t} = 0.1\]\[-0.0000288 t = \\ln(0.1)\]Solve for \(t\):\[t \approx \frac{\ln(0.1)}{-0.0000288} \approx 80,358\]It takes approximately 80,358 years for the plutonium to decay to 1 gram.

Key Concepts

Exponential Functions

The exponential function is a fundamental concept in mathematics and science used to describe situations where growth or decay occurs at a constant rate per unit time. In exponential decay, as in the given exercise, the quantity decreases over time. This is characterized by the general formula:\[y = Pe^{kt}\]where:

• \(y\) is the amount at time \(t\)
• \(P\) is the initial quantity
• \(k\) is the decay constant, and it is negative for decay processes
• \(e\) is the base of the natural logarithm, approximately equal to 2.71828

Exponential decay models are typically used in a variety of fields, such as physics, chemistry, and finance. They help us understand the dynamic process of decreasing rates, as seen in scenarios like half-life in radioactive substances.The function \(y = 10e^{-0.0000288t}\) specifically models how plutonium decays over time, leading to a fundamental understanding of how substances diminish over extended periods.

Graphing Functions

Graphing exponential functions allows us to visually analyze how quantities change over time. It provides an intuitive way to understand the behavior of the function. In the context of exponential decay, graphs typically show a steep decline initially, which gradually flattens out as time increases.To graph the function \(y = 10e^{-0.0000288t}\) for the interval \(0 \leq t \leq 100,000\):

• Start at \(y = 10\) at \(t=0\), indicating that initially, there is 10 grams of plutonium.
• The curve sharply decreases, showing that most decay happens early on.
• As \(t\) becomes very large, \(y\) approaches zero but never quite reaches it, illustrating the concept of decay reaching extinction asymptotically.

Graphing these functions helps in predicting scenarios and planning for long-term events. It gives a visual picture of decay, allowing easier comprehension of the decay dynamics.

Radioactive Decay

Radioactive decay is a natural process by which an unstable atomic nucleus loses energy by emitting radiation. In the context of the exercise, we deal with the decay of plutonium-239, a radioactive isotope.Key features of radioactive decay include:

• It is an exponential decay process, meaning the rate of decay is proportional to the quantity present.
• The isotope decays over a long time span, indicating a long half-life. For plutonium-239, it spans tens of thousands of years.
• The decay constant \(-0.0000288\) in the function \(y = 10e^{-0.0000288t}\) reflects how quickly the substance decays over time.

Understanding the behavior of radioactive decay is crucial in fields such as nuclear physics, environmental science, and health physics, as it helps manage and mitigate risks associated with radioactive substances.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. They help solve equations where the unknown appears as an exponent. This is crucial when determining how long it takes for a decaying substance, like plutonium, to drop to a certain level.The problem that requires finding how long it takes for plutonium to decay from 10 grams to 1 gram involves setting up the equation:\[1 = 10e^{-0.0000288t}\]To solve for \(t\), we use logarithms:

• Divide both sides by 10: \(e^{-0.0000288t} = 0.1\)
• Take the natural logarithm of both sides: \(-0.0000288t = \ln(0.1)\)
• Solve for \(t\): \[t = \frac{\ln(0.1)}{-0.0000288} \approx 80,358\]

Logarithmic calculations transform multiplicative processes into additive ones, making it easier to handle complex decay calculations. This is crucial for calculating timescales in nuclear decay and other similar scientific applications.