Chapter 10: Problem 14
In each part, find an exponential decay model \(y=y_{0} e^{-k t}\) that satisfies the stated conditions. (a) \(y_{0}=10 ;\) half-life \(T=5\) (b) \(y(0)=10:\) decay rate \(1.5 \%\) (c) \(y(1)=100 ; \quad y(10)=1\) (d) \(y(1)=10:\) half-life \(T=5\)
Short Answer
Expert verified
(a) \(k = -\frac{\ln(0.5)}{5}\); (b) \(k=0.015\); (c) \(k=-\frac{\ln(0.01)}{9}, y_0=100e^k\); (d) \(y_0 = 20, k=-\frac{\ln(0.5)}{5}\).
Step by step solution
01
Understand the Exponential Decay Model
The general form of an exponential decay model is given by the equation \(y = y_0 e^{-kt}\), where \(y_0\) is the initial amount, \(k\) is the decay constant, and \(t\) is time.
02
Solve Part (a) for Half-life
Given \(y_0 = 10\) and half-life \(T = 5\), we know that after time \(T\), the remaining amount is \(\frac{y_0}{2}\). Therefore, \(10 e^{-5k} = 5\). Solving for \(k\), we get \(e^{-5k} = 0.5\). Taking the natural logarithm of both sides, \(-5k = \ln(0.5)\), which gives \(k = -\frac{\ln(0.5)}{5}\).
03
Solve Part (b) Using Decay Rate
Given \(y(0) = 10\) and decay rate is \(1.5\%\), convert the percentage into a decimal for \(k\), i.e., \(k = 0.015\). The model becomes \(y = 10 e^{-0.015t}\).
04
Solve Part (c) with Two Conditions
For \(y(1) = 100\) and \(y(10) = 1\), use the equations \(100 = y_0 e^{-k}\) and \(1 = y_0 e^{-10k}\). Divide the second equation by the first: \(\frac{1}{100} = e^{-9k}\). Taking the natural logarithm gives \(-9k = \ln(0.01)\) so \(k = -\frac{\ln(0.01)}{9}\). Substitute \(k\) back to find \(y_0\) using \(y_0 e^{-k} = 100\).
05
Solve Part (d) for Half-life
Given \(y(1) = 10\) and half-life \(T = 5\). As before, \(e^{-5k} = 0.5\) so \(k = -\frac{\ln(0.5)}{5}\). Use \(10 = y_0 e^{-k}\) to find \(y_0\): \(y_0 = 10e^{k}\), substituting \(k\) with \( -\frac{\ln(0.5)}{5}\) and using \(e^{k} = 2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Half-life
The concept of half-life is essential in understanding exponential decay. It refers to the time it takes for a substance to reduce to half its initial amount through a decay process. This process is common in radioactive materials, but can be applied to any process involving exponential decay. In the equation for exponential decay, given as \( y = y_0 e^{-kt} \), the half-life (\( T \)) helps determine the value of the decay constant \( k \).
To find \( k \) using the half-life, we use the relation \( y_0 e^{-kT} = \frac{y_0}{2} \). Solving, we find \( e^{-kT} = 0.5 \), leading to \( k = -\frac{\ln(0.5)}{T} \).
The half-life allows us to predict how quickly a substance will decrease, which is useful in fields like chemistry and physics.
To find \( k \) using the half-life, we use the relation \( y_0 e^{-kT} = \frac{y_0}{2} \). Solving, we find \( e^{-kT} = 0.5 \), leading to \( k = -\frac{\ln(0.5)}{T} \).
The half-life allows us to predict how quickly a substance will decrease, which is useful in fields like chemistry and physics.
Decay Rate
The decay rate in exponential decay represents how quickly a substance is reducing over time. It is often expressed as a percentage and gives the rate at which the quantity decreases. Understanding the decay rate involves converting the percentage to a decimal to use in the exponential decay formula.
For instance, a decay rate of 1.5% means that every unit of time sees the substance decrease by 1.5% of its current value. In mathematical terms, if the decay rate is \( 1.5\% \), it translates to \( k = 0.015 \).
For instance, a decay rate of 1.5% means that every unit of time sees the substance decrease by 1.5% of its current value. In mathematical terms, if the decay rate is \( 1.5\% \), it translates to \( k = 0.015 \).
- This affects the function as \( y = y_0 e^{-0.015t} \), dictating the slope of the exponential decay curve.
- Determining the decay rate can help forecast the future concentration of a substance or population.
Exponential Functions
Exponential functions are a type of mathematical function that describe processes involving constant multiplicative rates of change, such as growth or decay.
In an exponential decay function, expressed as \( y = y_0 e^{-kt} \), the following components are critical:
In an exponential decay function, expressed as \( y = y_0 e^{-kt} \), the following components are critical:
- \( y_0 \): The initial quantity or starting value.
- \( k \): The decay constant, a positive number dictating the rate of decay.
- \( t \): Time or the variable over which the decay occurs.
Math Problem Solving
Solving math problems involving exponential decay requires a strategic approach and understanding of related concepts. First, recognize the type of problem and what is given, such as initial amount, half-life, or decay rate. In each specific scenario, the approach may require:
- Identifying known values like \( y_0 \) or given decay conditions.
- Using logarithms to solve for the decay constant \( k \) when needed, especially when time or percentage rates are involved.
- Substituting values back into the exponential decay formula \( y = y_0 e^{-kt} \) to complete calculations or to find unknown variables.