Problem 21 - Algebra and Trigonometry

Question

Radioactive Decay A radioactive substance decays in such a way that the amount of mass remaining after \(t\) days is given by the function $$ m(t)=13 e^{-0.015 t} $$ where \(m(t)\) is measured in kilograms. (a) Find the mass at time \(t=0\) . (b) How much of the mass remains after 45 days?

Step-by-Step Solution

Answer

(a) 13 kg; (b) approximately 6.622 kg.

Step 1: Understand the problem

This problem involves determining the remaining mass of a radioactive substance that decays according to the given function \(m(t) = 13 e^{-0.015 t}\). We are asked to find the initial mass (when \(t=0\)) and the mass after 45 days.

Step 2: Find mass at time t=0

To find the initial mass, substitute \(t=0\) into the function \(m(t) = 13 e^{-0.015 t}\). This gives us \(m(0) = 13 e^{0}\). Since \(e^{0} = 1\), the mass \(m(0) = 13 \times 1 = 13\) kilograms.

Step 3: Calculate mass after 45 days

Substitute \(t=45\) into the function. We calculate \(m(45) = 13 e^{-0.015 \, \times \, 45}\). First, calculate the exponent: \(-0.015 \, \times \, 45 = -0.675\). Now, compute \(m(45) = 13 e^{-0.675}\). Using a calculator, \(e^{-0.675} \approx 0.5094\), thus \(m(45) \approx 13 \, \times \, 0.5094 \approx 6.6222\) kilograms.

Key Concepts

Exponential Decay

Exponential decay describes the process where a quantity decreases at a rate proportional to its current value. In simpler terms, it means that as time progresses, the loss or reduction happens quickly at first and then slows down. It's like climbing a steep hill where the beginning is very steep, but it gradually flattens out.

For radioactive substances, this concept is particularly helpful because radioactive materials lose mass over time in a predictable pattern. When talking about the mass of a radioactive substance, we can represent this decay mathematically as a function involving the base of natural logarithms, which is denoted by the constant \(e\).

In our example function \(m(t) = 13 e^{-0.015 t}\), the term \(e^{-0.015 t}\) is key to understanding how the mass decreases. The exponent '-0.015t' shows how fast this decay is occurring. The negative sign indicates a decrease, while the number 0.015 dictates the rate of decay. Simply put, the larger this number, the quicker the substance decays.

Mathematical Modeling

Mathematical modeling allows us to use mathematical formulas to represent real-world systems. It's like creating a mini simulation of a problem so we can calculate outcomes easily and efficiently.

In this context, we use mathematical modeling to represent radioactive decay. The function \(m(t) = 13 e^{-0.015 t}\) is an example of such a model. It gives us a clear rule to compute how much mass is left at any moment \(t\).

This model is incredibly useful as it allows scientists and engineers to predict decay over time without the need for continuous measurement. Just by plugging different values of \(t\) (time in days) into the equation, they can determine how much of the substance remains. This is efficient and saves a lot of practical effort when dealing with substances that are hazardous or costly to measure directly.

Half-Life Calculation

The half-life of a radioactive substance is the time it takes for half of its mass to decay. It's a central concept because it provides a way to understand the speed of radioactive decay without complex calculations every time you need to know how much of the substance has decayed.

To calculate the half-life using the decay function, we set up the equation where the remaining mass \(m(t)\) is half of the initial mass. For our decay equation, the initial mass is 13 kg. So, we solve \(13 e^{-0.015 t} = 6.5\).

Simplifying gives us \(e^{-0.015 t} = 0.5\). To find \(t\), we apply the natural logarithm: \(-0.015 t = \ln(0.5)\). Solving for \(t\), we get \(t = -\frac{\ln(0.5)}{0.015}\). Using this calculation, the half-life tells us precisely when half of the substance's mass will have decayed. Understanding half-life is critical for fields like nuclear medicine, archaeology, and even environmental science.