Chapter 8: Problem 38
(a) What effect does increasing \(y_{0}\) and keeping \(k\) fixed have on the doubling time or half-life of an exponential model? Justify your answer. (b) What effect does increasing \(k\) and keeping \(y_{0}\) fixed have on the doubling time and half-life of an exponential model? Justify your answer.
Short Answer
Expert verified
(a) Increasing \( y_0 \) does not affect doubling time or half-life. (b) Increasing \( k \) decreases both doubling time and half-life.
Step by step solution
01
Understanding Exponential Growth and Decay
An exponential model is described by the equation \( y(t) = y_0 e^{kt} \), where \( y_0 \) is the initial quantity, \( k \) is the growth (\( k > 0 \)) or decay (\( k < 0 \)) rate, and \( t \) is time.
02
Doubling Time in Exponential Growth
For a positive growth rate \( k \), the doubling time \( T_d \) can be calculated using \( T_d = \frac{ \,\ln(2)}{k} \). Thus, \( T_d \) depends only on \( k \) and is independent of \( y_0 \).
03
Effect of Increasing \( y_0 \) on Doubling Time: Part (a)
Since \( T_d = \frac{ \,\ln(2)}{k} \) does not involve \( y_0 \), increasing \( y_0 \) while keeping \( k \) fixed does not affect the doubling time. The doubling time remains the same.
04
Understanding Half-life in Exponential Decay
For a negative growth rate \( k \), half-life \( T_h \) is given by \( T_h = \frac{ \,\ln(2)}{\mid k \mid} \). Similar to doubling time, \( T_h \) depends on \( k \) and not on \( y_0 \).
05
Effect of Increasing \( y_0 \) on Half-life: Part (a)
Since half-life \( T_h = \frac{ \,\ln(2)}{\mid k \mid} \) does not depend on \( y_0 \), increasing \( y_0 \) does not change the half-life when \( k \) is fixed.
06
Effect of Increasing \( k \) on Doubling Time: Part (b)
Doubling time \( T_d = \frac{ \,\ln(2)}{k} \) decreases as \( k \) increases, because \( T_d \) is inversely proportional to \( k \). Thus, increasing \( k \) while keeping \( y_0 \) fixed results in a shorter doubling time.
07
Effect of Increasing \( k \) on Half-life: Part (b)
Half-life \( T_h = \frac{ \,\ln(2)}{\mid k \mid} \) decreases as \( \mid k \mid \) increases, similarly to doubling time. Therefore, increasing \( k \) (making it more negative) while \( y_0 \) is fixed also results in a shorter half-life.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Doubling Time
Doubling time refers to the time it takes for a quantity that is growing exponentially to double in size. In exponential growth models, the doubling time is a very useful concept because it provides a simple way to understand how fast something is growing. The formula for calculating doubling time in a scenario where the growth rate is positive is given by:
- \( T_d = \frac{\ln(2)}{k} \)
- \( T_d \) is the doubling time.
- \( k \) is the growth rate, which is a positive constant in this case.
- If \( y_0 \) is increased while \( k \) is fixed, the doubling time remains unchanged.
- If \( k \) increases, the doubling time becomes shorter, indicating faster growth.
Half-life
Half-life is a concept used in the context of exponential decay. It describes the time required for a quantity to reduce to half of its initial value. This concept is widely used in fields like physics and chemistry, particularly when discussing radioactive decay. The formula to calculate half-life when decay is exponential and the decay rate \( k \) is negative, is:
- \( T_h = \frac{\ln(2)}{\mid k \mid} \)
- \( T_h \) is the half-life.
- \( \mid k \mid \) is the absolute value of the decay rate, making it positive for calculation purposes.
- An increase in the magnitude of \( k \) (faster decay) results in a shorter half-life.
Growth and Decay Rates
The growth and decay rates are crucial components in exponential models and can greatly influence both doubling time and half-life. The growth rate \( k \) determines how quickly a quantity increases over time, while a negative growth rate (or decay rate) shows how fast a quantity decreases. When a system undergoes exponential growth or decay, the value of \( k \) plays a pivotal role in the speed of these processes.
- For a positive \( k \), the larger the value, the faster the exponential growth occurs.
- For \( k \) being negative, the larger its absolute value, the quicker the decay of the quantity.
- Increasing \( k \) leads to a decrease in doubling time, reflecting rapid growth.
- Increasing \( \mid k \mid \) (making \( k \) more negative) decreases the half-life, indicating faster decay.