Problem 51 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

RADIOACTIVE DECAY A radioactive substance decays exponentially. If 500 grams of the substance were present initially and 400 grams are prescnt 50 years later, how many grams will be present after 200 years?

Step-by-Step Solution

Answer

After 200 years, approximately 204.5 grams will be present.

Step 1: Identify the exponential decay formula

The exponential decay formula is given by y=y_0e^(-kt), where \(y\) is the amount of substance at time \(t\), \(y_0\) is the initial amount of the substance, \(k\) is the decay constant, and \(t\) is time.

Step 2: Define the given values

Initial amount \(y_0 = 500\) gramsAmount after 50 years \(y(50) = 400\) gramsTime \(t = 50\) years

Step 3: Set up the equation

Using the formula \(y = y_0 e^{-kt}\), substitute the known values to find the decay constant \(k\).\(400 = 500 e^{-50k}\)

Step 4: Solve for the decay constant \(k\)

Divide by 500:\(\frac{400}{500} = e^{-50k} \ 0.8 = e^{-50k}\)Take the natural logarithm of both sides:\(\ln(0.8) = -50k \ -0.2231 ≈ -50k\)Solve for \(k\):\(k ≈ \frac{0.2231}{50} \approx 0.004462\)

Step 5: Find the amount after 200 years

Add \(t = 200\) years into the decay formula:\(y(200) = 500 e^{-0.004462 \times 200}\)Simplify the exponent:\(-0.004462 \times 200 ≈ -0.8924\)Calculate the exponential:\( y(200) = 500 e^{-0.8924} \approx 500 \times 0.409 \approx 204.5\text{ grams}\)

Key Concepts

Exponential Decay

Radioactive decay is a great example of exponential decay. In this process, the amount of the radioactive substance decreases at a rate proportional to its current value. This means that the larger the current amount, the faster it decreases. The general formula used to describe exponential decay is \( y = y_0 e^{-kt} \). Here, \( y \) is the remaining quantity of the substance at time \( t \), \( y_0 \) is the initial quantity, \( k \) is the decay constant, and \( t \) is time.
Understanding exponential decay is vital as it applies not only to radioactive substances but also to scenarios like population decline, cooling of objects, and even depreciation of assets.

Decay Constant

The decay constant, represented as \( k \) in the exponential decay formula, is a crucial factor. It determines how quickly the substance decays. The larger the decay constant, the faster the substance will decay. In our exercise, we solved for \( k \) using the relationship \( y = y_0 e^{-kt} \), and the given values \( y_0 = 500 \) grams, \( y(50) = 400 \) grams, and \( t = 50 \) years.
The computation involved:

• Setting up the equation \( 400 = 500 e^{-50k} \).
• Solving for \( e^{-50k} \), we get \( 0.8 = e^{-50k} \).
• Taking the natural logarithm of both sides, \( \text{ln}(0.8) = -50k \).
• Finally, \( k ≈ \frac{0.2231}{50} ≈ 0.004462 \).

This process underscores the importance of the decay constant in predicting the behavior of radioactive substances over time.

Natural Logarithm

Natural logarithms, denoted as \( \text{ln} \), are used extensively in exponential decay calculations. They are the inverse of the exponential function (\( e^x \)). This means \( e^{\text{ln}(x)} = x \) and \( \text{ln}(e^x) = x \).
In the context of our problem, using the natural logarithm helped us solve for the decay constant. We transformed the exponential equation \( e^{-50k} = 0.8 \) into a linear form. By taking the natural logarithm of both sides, \( \text{ln}(0.8) = -50k \), we simplified the process of isolating \( k \).
Mastering natural logarithms not only aids in solving exponential decay problems but also broadens understanding in fields like finance, biology, and engineering.

Half-Life

The half-life of a substance is the time it takes for half of the initial quantity to decay. Though not directly mentioned in our exercise, it's a closely related concept to exponential decay. If we know the decay constant, we can calculate the half-life using the formula \( t_{\frac{1}{2}} = \frac{\text{ln}(2)}{k} \).
In this exercise, with \( k ≈ 0.004462 \), if we were to find the half-life, it would be \[ t_{\frac{1}{2}} = \frac{\text{ln}(2)}{0.004462} \ \text{ln}(2) ≈ 0.693 \rightarrow t_{\frac{1}{2}} ≈ \frac{0.693}{0.004462} ≈ 155.35 \text{ years} \].
Understanding the half-life adds depth to our comprehension of how substances decay over time, helping to estimate remaining amounts or the duration until a certain level is reached.