Problem 20 - Calculus

Question

If \(y=100\) at \(t=4\) and \(y=10\) at \(t=8\) (exponential decay) when will \(y=1\) ? What was \(y_{0}\) ?

Step-by-Step Solution

Answer

\( y_0 = 1000 \) and \( y = 1 \) at \( t \approx 12 \).

Step 1: Understand Exponential Decay

Exponential decay can be represented by the formula: \( y = y_0 e^{-kt} \), where \( y_0 \) is the initial amount, \( k \) is the decay constant, and \( t \) is time. We'll use the given values to find \( y_0 \) and determine when \( y = 1 \).

Step 2: Set up Equations with Given Values

Using the information from the problem, we know \( y = 100 \) when \( t = 4 \) and \( y = 10 \) when \( t = 8 \). Set up the equations: 1. \( 100 = y_0 e^{-4k} \) 2. \( 10 = y_0 e^{-8k} \).

Step 3: Solve for the Decay Constant \(k\)

Divide the second equation by the first equation to eliminate \( y_0 \): \( \frac{10}{100} = \frac{y_0 e^{-8k}}{y_0 e^{-4k}} \). This simplifies to \( 0.1 = e^{-4k} \). Take the natural logarithm of both sides: \( \ln(0.1) = -4k \). Solve for \( k \): \( k = -\frac{\ln(0.1)}{4} \approx 0.5753 \).

Step 4: Find the Initial Value \(y_0\)

Substitute \( k = 0.5753 \) back into the first equation: \( 100 = y_0 e^{-4 \times 0.5753} \). Calculate the value: \( 100 = y_0 (0.1) \), hence \( y_0 = \frac{100}{0.1} = 1000 \).

Step 5: Determine When \(y = 1\)

Now use the equation \( 1 = 1000 e^{-kt} \) to solve for \( t \): \( \frac{1}{1000} = e^{-0.5753t} \). Take the natural logarithm: \( \ln(0.001) = -0.5753t \). Solve for \( t \): \( t = -\frac{\ln(0.001)}{0.5753} \approx 12 \).

Key Concepts

Decay Constant

The decay constant, often denoted as \( k \), is a crucial parameter in exponential decay equations. It tells us how quickly a quantity reduces over time. In the context of our problem, it shows how fast the values of \( y \) drop from one time point to another.

• The decay constant is typically a positive number when expressed in the formula \( y = y_0 e^{-kt} \).
• A larger decay constant means a faster rate of decay. A smaller decay constant means the quantity decays more slowly.

To find the decay constant, we often use given values for \( y \) at different time points. We set up equations using these values and solve for \( k \). Here, by dividing the known values and simplifying, we end up using the natural logarithm to find \( k \). This step is essential as it sets the pace for how our exponential function behaves over time.

Natural Logarithm

In solving exponential decay problems, the natural logarithm, denoted as \( \ln \), plays an essential role. The natural logarithm is the inverse function of the exponential function with base \( e \), where \( e \approx 2.71828 \).

• The natural logarithm helps to pull down the exponent and make the solution of equations easier when dealing with exponentials.
• In our context, when we have an equation of the form \( e^{-4k} = 0.1 \), taking the natural logarithm of both sides allows us to solve for \( k \).

By using \( \ln(0.1) \), we can transform the equation into a form that makes \( k \) easy to isolate and compute. This technique is not just limited to exponential decay but can be useful in a vast array of mathematical and scientific calculations involving growth and decay functions.

Initial Value

The initial value, denoted as \( y_0 \), is another key element in the study of exponential decay. It represents the starting quantity of our decaying process before any time has elapsed. This value is the baseline from which all decay occurs.

• The initial value can often be guessed if the conditions are simple enough, but sometimes, it needs to be calculated using known information from a specific time point.
• In exponential decay scenarios, this is the value of \( y \) when \( t = 0 \).

In our exercise, we identified \( y_0 \) by substituting the known value and the known decay constant into the equation \( y = y_0 e^{-kt} \). We calculated this by setting the equation using data where \( y = 100 \) when \( t = 4 \), transforming it to solve for \( y_0 \). Recognizing the initial value is a fundamental step in understanding the complete behavior of any decay process.