91Ó°ÊÓ

Exponential functions are crucial in various fields, such as science, finance, and engineering. When working with exponential functions, the general form is typically expressed as: $$ A(t) = A_0 e^{kt} $$ Here, \( A(t) \) is the amount at time \( t \), \( A_0 \) is the initial amount, \( e \) is the base of the natural logarithm (approximately 2.71828), and \( k \) is a constant that represents the growth (or decay) rate. If \( k \) is positive, the function models exponential growth. Conversely, if \( k \) is negative, it represents exponential decay. To understand this better, let’s consider the equation in our problem: $$ A(t) = 2.00 e^{-0.053t} $$ Here, the negative exponent indicates the exponential decay of Plutonium 241 over time. Exponential decay is an essential concept for understanding how radioactive substances decrease over time.

Radioactive decay is a process by which an unstable atomic nucleus loses energy by emitting radiation. This decay is typically random in nature but can be modeled statistically. The amount of a remaining radioactive substance decreases over time, following an exponential decay pattern. In the context of our exercise, we have the radioactive decay of Plutonium 241, described by the equation: $$ A(t) = 2.00 e^{-0.053t} $$ This means that the quantity of Plutonium 241 reduces exponentially as time passes. The decay constant \( -0.053 \) signifies the rate at which decay occurs. Understanding radioactive decay is essential for areas such as nuclear physics, medicine (e.g., radiotherapy), and archaeology (carbon dating).

Evaluating a function means determining the outcome or output value for a given input. For the exercise, we are given the function: $$ A(t) = 2.00 e^{-0.053t} $$ To evaluate it at specific points, we substitute the value of \( t \) (in years) into the function. Here are the precise steps:
1. Substitute the value of \( t \) into the function.
2. Compute the exponent \( -0.053t \).
3. Calculate \( e \) raised to the power of the computed exponent.
4. Multiply the result by 2.00 to get the final amount. Let’s take an example from the exercise: For \( t = 10 \) years, the steps are: Substitute \( t \): \( A(10) = 2.00 e^{-0.053 \times 10} \)
Compute: \( -0.053 \times 10 = -0.53 \)
Calculate \( e^{-0.53} \approx 0.5880 \)
Multiply: \( A(10) = 2.00 \times 0.5880 = 1.18 \text{ grams} \).

Algebraic substitution involves replacing a variable in an equation with a given value. This method is essential for solving specific types of functions and equations. In our exercise, we use algebraic substitution to determine the amount of plutonium left after a certain number of years: The function provided is: $$ A(t) = 2.00 e^{-0.053t} $$ Let's walk through the steps for \( t = 20 \) years:
1. **Substitute \( t \) with 20**: $$ A(20) = 2.00 e^{-0.053 \times 20} $$ 2. **Compute the exponent**: $$ -0.053 \times 20 = -1.06 $$ 3. **Calculate \( e^{-1.06} \)** using a calculator to get approximately 0.3470.
4. **Multiply the result by 2.00**: $$ A(20) = 2.00 \times 0.3470 = 0.69 \text{ grams} $$ These steps simplify finding the amount of a substance at different times by substituting the values into the given function.