Problem 21 - Algebra A Combined Function

Question

Practice using the exponential decay formula with half-lives by completing the table below. The first row has been completed for you. $$ \begin{array}{|c|c|c|c|c|c|} \hline \begin{array}{c} \text { Original } \\ \text { Amount } \end{array} & \begin{array}{c} \text { Half-Life } \\ \text { (in years) } \end{array} & \begin{array}{c} \text { Number } \\ \text { of Years } \end{array} & \begin{array}{c} \text { Time Intervals, } \boldsymbol{x}\left(\frac{\text { Years }}{\text { Half-Life }}\right) \\ \text { Rounded to Tenths if Needed } \end{array} & \begin{array}{c} \text { Final Amount after } \boldsymbol{x} \text { Time } \\ \text { Intervals (rounded to tenths) } \end{array} & \begin{array}{c} \text { Is Your Final Amount } \\ \text { Reasonable? } \end{array} \\ \hline 60 & 8 & 10 & \frac{10}{8}=1.25 & 25.2 & \text { yes } \\ \hline \text { a. } 40 & 7 & 14 & & & \\ \hline \text { b. } 40 & 7 & 11 & & & \\ \hline \end{array} $$

Step-by-Step Solution

Answer

Final amounts are 10 (a) and 13.4 (b), both are reasonable.

Step 1: Calculate the Time Intervals (a)

For the first problem (a), given the \( \text{Original Amount} = 40 \), \( \text{Half-Life} = 7 \), and \( \text{Number of Years} = 14 \). Calculate the Time Intervals (\( x \)) using the formula \( x = \frac{\text{Number of Years}}{\text{Half-Life}} \). Substitute the values: \[ x = \frac{14}{7} = 2 \].

Step 2: Calculate the Final Amount (a)

Use the exponential decay formula: \[ A = A_0 \times \left(\frac{1}{2}\right)^x \]. Where \( A_0 = 40 \) is the original amount and \( x = 2 \) is the time intervals. Calculate the final amount \( A \): \[ A = 40 \times \left(\frac{1}{2}\right)^2 = 40 \times \frac{1}{4} = 10 \].

Step 3: Reasonability Check (a)

Compare the calculated final amount to the original amount. Since 10 is less than 40 by a factor of four, and a time interval of 2 represents two half-lives (where each half-life reduces the amount by half), the result is reasonable. Thus, \( \text{Final Amount = 10} \) is reasonable.

Step 4: Calculate the Time Intervals (b)

For the second problem (b), given the \( \text{Original Amount} = 40 \), \( \text{Half-Life} = 7 \), and \( \text{Number of Years} = 11 \). Calculate the Time Intervals (\( x \)) using the formula \( x = \frac{\text{Number of Years}}{\text{Half-Life}} \). Substitute the values: \[ x = \frac{11}{7} \approx 1.57 \].

Step 5: Calculate the Final Amount (b)

Use the exponential decay formula: \[ A = A_0 \times \left(\frac{1}{2}\right)^x \]. Where \( A_0 = 40 \) is the original amount and \( x \approx 1.57 \) is the time intervals. Calculate the final amount \( A \): \[ A = 40 \times \left(\frac{1}{2}\right)^{1.57} \approx 40 \times 0.336 \approx 13.4 \].

Step 6: Reasonability Check (b)

Verify the reasonableness of the final amount. After about 1.57 half-lives, the amount should be reduced more than a half (20) but not quartered (10). Thus, \( \text{Final Amount} \approx 13.4 \) is reasonable.

Key Concepts

half-life

Half-life is a key concept in the study of exponential decay and is fundamental to understanding how certain quantities, like radioactive substances, decrease over time. The half-life of a substance is the time it takes for half of the original amount to decay or transform into another substance. This property is inherent to the material and remains constant regardless of the quantity present. For example, if you start with 40 grams of a substance, after one half-life, you will have 20 grams remaining. After another half-life, only 10 grams will be left. This predictable pattern makes calculating substance quantities over time manageable and reliable.

exponential functions

Exponential functions are mathematical expressions used to model scenarios where quantities decrease (or increase) at a rate proportional to their current size. In the context of decay, an exponential function is used to describe how the amount of a substance decreases over time. The general form of an exponential decay function is:

• \( A = A_0 \times \left(\frac{1}{2}\right)^x \)

Here, \( A_0 \) represents the initial amount, \( x \) is the number of half-life intervals that have passed, and \( A \) is the remaining amount. This formula is key to calculating how much of a substance remains after a given time, leveraging the consistent behavior of exponential decay.

mathematical modeling

Mathematical modeling involves using mathematical formulas and concepts to represent and analyze real-world scenarios. When dealing with decay processes, modeling can quantify how substances change over time. By using the exponential decay formula, scientists can predict and assess the time it takes for a substance to decrease to a specific amount. This modeling not only helps in fields like chemistry and physics but is also crucial in environmental studies, medicine, and even finance where such decay-like processes are observed.

• It enables precise forecasts and adjustments based on the substance's behavior.
• Helps in planning and decision-making through accurate predictions.

calculating time intervals

Calculating time intervals is essential when employing the exponential decay formula, as it determines how long the substance has been decaying. In practice, the time intervals are calculated using the formula:

• \( x = \frac{\text{Number of Years}}{\text{Half-Life}} \)

This equation finds the number of half-life periods that have elapsed. For instance, if a substance has a half-life of 7 years and it has been decaying for 14 years, there have been 2 half-life intervals. Knowing this, you can use these intervals in the decay formula to calculate the remaining amount of substance. This simple yet powerful calculation bridges theoretical decay processes with real-world applications, making it indispensable for accurate modeling.