Chapter 8: Problem 28
(a) Suppose that a quantity \(y=y(t)\) changes in such a way that \(d y / d t=k \sqrt{y},\) where \(k>0 .\) Describe how \(y\) changes in words. (b) Suppose that a quantity \(y=y(t)\) changes in such a way that \(d y / d t=-k y^{3},\) where \(k>0 .\) Describe how \(y\) changes in words.
Short Answer
Expert verified
(a) As \( y \) increases, it accelerates; (b) \( y \) decreases more rapidly for larger values.
Step by step solution
01
Understand the differential equation for (a)
The equation given is \( \frac{d y}{d t} = k \sqrt{y} \). This describes a rate of change for \( y \) in relation to \( t \), which indicates how quickly \( y \) changes as time progresses.
02
Analyze the implication of the equation for (a)
Since \( k > 0 \) and the square root function \( \sqrt{y} \) is always non-negative, the rate of change \( \frac{d y}{d t} = k \sqrt{y} \) implies \( y \) is increasing over time as long as \( y > 0 \). The larger \( y \) gets, the faster the rate \( \frac{d y}{d t} \) because \( \sqrt{y} \) increases with \( y \).
03
Consolidate the behavior of y for (a)
\( y \) is increasing over time. The growth rate increases as the value of \( y \) increases, leading to accelerated growth.
04
Understand the differential equation for (b)
The equation given is \( \frac{d y}{d t} = -k y^3 \). This represents a different type of rate of change for \( y \). This equation describes a scenario where the rate of change of \( y \) is proportional to the negative cube of \( y \).
05
Analyze the implication of the equation for (b)
Since \( k > 0 \) and \( y^3 \) increases with \( y \), \( -k y^3 \) implies that \( y \) is decreasing over time whenever \( y > 0 \). The larger the value of \( y \), the faster \( y \) decreases since the negative impact grows as \( y \) increases.
06
Consolidate the behavior of y for (b)
\( y \) decreases over time, and as \( y \) becomes larger, the rate of decrease becomes more rapid, leading to decelerating quantities at lower values.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rate of Change
The "rate of change" in mathematics refers to how quickly a quantity evolves as time progresses. In the context of differential equations, this is often expressed with terms like \( \frac{dy}{dt} \), which signifies how a variable \( y \) changes with respect to another variable, typically time \( t \). Understanding the rate of change is crucial because it tells us whether a quantity is increasing or decreasing, and at what speed.
In the example from the exercise, we are given two differential equations: one for increasing values and another for decreasing. If \( \frac{dy}{dt} = k \sqrt{y} \), this indicates a positive rate of change. Therefore, as \( y \) grows, the rate increases, suggesting accelerating growth. Conversely, \( \frac{dy}{dt} = -k y^3 \) implies a negative rate of change. Here, as \( y \) becomes larger, the decrease becomes more rapid.
In the example from the exercise, we are given two differential equations: one for increasing values and another for decreasing. If \( \frac{dy}{dt} = k \sqrt{y} \), this indicates a positive rate of change. Therefore, as \( y \) grows, the rate increases, suggesting accelerating growth. Conversely, \( \frac{dy}{dt} = -k y^3 \) implies a negative rate of change. Here, as \( y \) becomes larger, the decrease becomes more rapid.
- Positive rate - the quantity increases over time.
- Negative rate - the quantity decreases over time.
Mathematical Modeling
Mathematical modeling involves using mathematical language and equations to represent real-world phenomena. It transforms a problem into an understandable format, often using differential equations, to explain how a system evolves over time.
In our exercise, two different scenarios are modeled via differential equations. For part (a), the model \( \frac{dy}{dt} = k \sqrt{y} \) can represent processes like population growth where more individuals lead to faster growth. For part (b), \( \frac{dy}{dt} = -k y^3 \) might illustrate phenomena like radioactive decay, where the material diminishes more quickly the larger it originally is.
To effectively create a mathematical model, one must:
In our exercise, two different scenarios are modeled via differential equations. For part (a), the model \( \frac{dy}{dt} = k \sqrt{y} \) can represent processes like population growth where more individuals lead to faster growth. For part (b), \( \frac{dy}{dt} = -k y^3 \) might illustrate phenomena like radioactive decay, where the material diminishes more quickly the larger it originally is.
To effectively create a mathematical model, one must:
- Identify the core variables involved.
- Understand the relationships and forces driving change.
- Translate these relationships into mathematical equations.
Growth and Decay
Growth and decay describe how a quantity increases (growth) or decreases (decay) over time. These concepts are fundamental in both natural and social sciences, illuminating how processes unfold.
In the exercise, the differential equation \( \frac{dy}{dt} = k \sqrt{y} \) represents a growth scenario. The rate of growth is dependent on the current size of \( y \). More \( y \) leads to faster growth, highlighting exponential properties in certain processes like populations or investments.
Alternatively, \( \frac{dy}{dt} = -k y^3 \) demonstrates decay. The term \( -k y^3 \) shows that the rate of decrease is hefty when \( y \) is large, similar to situations like carbon decay or cooling, where larger entities lose material or heat quickly initially.
In the exercise, the differential equation \( \frac{dy}{dt} = k \sqrt{y} \) represents a growth scenario. The rate of growth is dependent on the current size of \( y \). More \( y \) leads to faster growth, highlighting exponential properties in certain processes like populations or investments.
Alternatively, \( \frac{dy}{dt} = -k y^3 \) demonstrates decay. The term \( -k y^3 \) shows that the rate of decrease is hefty when \( y \) is large, similar to situations like carbon decay or cooling, where larger entities lose material or heat quickly initially.
- Growth - characterized by rates that increase as the quantity itself increases.
- Decay - represented by rates that become more negative as the quantity grows.