Problem 36 - Algebra for College Students

Question

The number of grams \(Q\) of a certain radioactive substance present after \(t\) seconds is given by the equation \(Q=1500 e^{-0.4 t}\). How many grams remain after \(5 \mathrm{sec}-\) onds? 10 seconds? 20 seconds? \(203 ; 27 ; 1\)

Step-by-Step Solution

Answer

After 5 seconds: ~203g; after 10 seconds: ~27g; after 20 seconds: ~1g.

Step 1: Understanding the Problem

We need to find out the amount of a radioactive substance remaining after certain time intervals (5, 10, and 20 seconds). The decay of the substance is described by the equation \(Q = 1500 e^{-0.4 t}\), where \(Q\) is the amount in grams and \(t\) is the time in seconds.

Step 2: Substitute for 5 seconds

Substitute \(t = 5\) into the equation to find \(Q\) after 5 seconds: \[Q = 1500 e^{-0.4 imes 5}\] Simplify the exponent: \[Q = 1500 e^{-2}\].

Step 3: Calculate Q for 5 seconds

Compute \(Q\) using a calculator:\[Q = 1500 \times e^{-2} \approx 1500 \times 0.1353 \approx 202.95\] So, after 5 seconds, the approximate quantity is 202.95 grams.

Step 4: Substitute for 10 seconds

Substitute \(t = 10\) into the equation to find \(Q\) after 10 seconds:\[Q = 1500 e^{-0.4 imes 10}\] Simplify the exponent:\[Q = 1500 e^{-4}\].

Step 5: Calculate Q for 10 seconds

Compute \(Q\) using a calculator:\[Q = 1500 \times e^{-4} \approx 1500 \times 0.0183 \approx 27.45\] So, after 10 seconds, the approximate quantity is 27.45 grams.

Step 6: Substitute for 20 seconds

Substitute \(t = 20\) into the equation to find \(Q\) after 20 seconds:\[Q = 1500 e^{-0.4 imes 20}\] Simplify the exponent:\[Q = 1500 e^{-8}\].

Step 7: Calculate Q for 20 seconds

Compute \(Q\) using a calculator:\[Q = 1500 \times e^{-8} \approx 1500 \times 0.000335 \approx 0.5025\] So, after 20 seconds, the approximate quantity is 0.5025 grams.

Key Concepts

Radioactive Decay

Radioactive decay is a natural process where unstable atomic nuclei lose energy by emitting radiation. This phenomenon happens at a predictable rate for each type of radioactive material.
In our example, the problem revolves around understanding how much of a radioactive substance is left over time. The equation presented, \( Q = 1500 e^{-0.4 t} \), helps us calculate the remaining quantity. Here, the material's half-life plays a critical role since it determines the rate at which the substance decays.

• Radioactive substances continuously decay into lighter substances, releasing energy.
• The rate of decay is exponential, meaning it decreases rapidly at first and then more slowly over time.
• The term \( e^{−0.4t} \) indicates the portion of the initial quantity that remains as time passes.

These principles provide a foundation for calculating how much of the substance is left at any given moment. Understanding radioactive decay is crucial for fields like nuclear physics and medicine.

Exponential Functions

Exponential functions describe processes that grow or decay at a constant percentage rate over time. In our equation, \( Q = 1500 e^{-0.4 t} \), the variable \( e \) represents Euler's number, approximately equal to 2.71828.
These functions are essential in capturing dynamics of processes that change at rates proportional to their current value, like population growth or radioactive decay.

• The negative sign in the exponent \(-0.4t\) indicates a decrease, aligning with the nature of radioactive decay.
• \(Q\) represents the remaining quantity, while \(t\) is the time in seconds.
• The initial quantity, 1500 grams in this case, shows what we start with before any decay occurs.

Exponential functions allow us to create models that predict how systems evolve over time, providing valuable insights into various scientific fields.

Mathematical Modeling

Mathematical modeling uses equations or formulas to represent real-world situations, enabling predictions and problem-solving. The equation \( Q = 1500 e^{-0.4 t} \) serves as a mathematical model for radioactive decay.
It allows us to predict the amount of radioactive material remaining at any point.

• This model offers a simplified representation of a complex natural occurrence.
• By substituting different time values, it predicts behavior under different conditions.
• Mathematical models are crucial for understanding phenomena too complex or large to observe directly.

Using such models helps students and scientists alike in visualizing and estimating outcomes, making them indispensable tools in scientific research and education.

Problem-solving in Mathematics

Problem-solving is at the heart of mathematics, requiring analytical thinking and creativity. Solving the problem step-by-step, as demonstrated, involves using the given exponential decay equation effectively.
The key aspects of solving such problems include:

• Understanding the problem statement and what is being asked.
• Identifying known variables and substituting them into the equation.
• Performing calculations accurately to find the desired results.
• Verifying if the results make sense in a real-world context.

This process not only aids in finding solutions but also strengthens mathematical reasoning skills. Mastery of such skills is critical for tackling various mathematical challenges and applying mathematics in everyday scenarios.