Chapter 2: Problem 361
For the following exercises, use \(y=y_{0} e^{\Lambda t}\). If \(y=100\) at \(t=4\) and \(y=10\) at \(t=8\), when does \(y=1 ?\)
Short Answer
Expert verified
The value of \( t \) when \( y = 1 \) is 12.
Step by step solution
01
Set Up the Equation for Given Conditions
We are given the equation \( y = y_0 e^{\Lambda t} \). Using the first condition \( y = 100 \) when \( t = 4 \), we have: \( 100 = y_0 e^{4\Lambda} \). Similarly, using the condition \( y = 10 \) when \( t = 8 \), we have: \( 10 = y_0 e^{8\Lambda} \).
02
Solve for Initial Value \( y_0 \) and \( \Lambda \)
Divide the two equations: \( \frac{100}{10} = \frac{y_0 e^{4\Lambda}}{y_0 e^{8\Lambda}} \). This simplifies to \( 10 = e^{-4\Lambda} \). Solving for \( \Lambda \), \( \Lambda = -\frac{1}{4} \ln(10) \).
03
Determine \( y_0 \) Using \( \Lambda \) Value
Substitute \( \Lambda = -\frac{1}{4} \ln(10) \) back into the first condition equation \( 100 = y_0 e^{4\Lambda} \). Since \( e^{4\Lambda} = \frac{1}{10} \), \( 100 = y_0 \times \frac{1}{10} \) implies \( y_0 = 1000 \).
04
Set Up the Equation for Unknown \( t \)
Now use the equation \( y = y_0 e^{\Lambda t} \) for \( y = 1 \) to find \( t \). Substitute \( y_0 = 1000 \) and \( \Lambda = -\frac{1}{4} \ln(10) \): \( 1 = 1000 e^{-\frac{1}{4} \ln(10)t} \).
05
Solve for \( t \) when \( y = 1 \)
Simplify the equation: \( \frac{1}{1000} = e^{-\frac{1}{4} \ln(10)t} \). Taking the natural logarithm of both sides gives \( \ln(\frac{1}{1000}) = -\frac{1}{4} \ln(10) t \). Solve for \( t \) to find \( t = 12 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Value Problems
An initial value problem is a type of differential equation problem that requires determining a specific function that satisfies both a differential equation and initial conditions. In these problems, we often start with a general equation that describes some form of change or process. This could represent physical phenomena like radioactive decay or population growth. In our exercise, the equation takes the form:
- \( y = y_0 e^{\Lambda t} \)
Natural Logarithm
Natural logarithms are a mathematical concept deeply connected to exponential functions and growth. The natural logarithm is the power to which \( e \) (approximately 2.71828) must be raised to produce a number. It is commonly denoted by \( \ln \). In exponential growth and decay equations, natural logarithms play a crucial role in solving for variables when the exponent is unknown. They adhere to various properties that make calculations easier:
- \( \ln(e^x) = x \)
- \( \ln(AB) = \ln(A) + \ln(B) \)
- \( \ln(A/B) = \ln(A) - \ln(B) \)
Exponential Functions
Exponential functions are essential tools in mathematics, particularly in modeling growth and decay phenomena. These functions are expressed in the form \( y = y_0 e^{\Lambda t} \), where \( y_0 \) is the initial value, \( \Lambda \) the rate of change, and \( e \) is the base of natural logarithms. They are characterized by their constant rate of proportional growth or decay, adapting to various real-world contexts like population or radioactive decay over time. The parameter \( \Lambda \) determines whether the function grows (if \( \Lambda > 0 \)) or decays (if \( \Lambda < 0 \)). These functions exhibit properties such as:
- Constant percentage growth or decay per time unit.
- The shape is a smooth curve that never touches the zero line, either approaching infinity (growth) or zero (decay).