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Chapter 10: Problem 13

Suppose that substance \(A\) is converted into substance \(B\) at a rate that, at any time \(t\), is proportional to the square of the amount of \(A .\) This situation occurs, for instance, when it is necessary for two molecules of \(A\) to collide to create one molecule of \(B\). Set up the differential equation that is satisfied by \(y=f(t)\), the amount of substance \(A\) at time \(t .\) Sketch a solution.

Short Answer

Expert verified
The differential equation is \( \frac{dy}{dt} = -ky^2 \), and its solution is \( y = \frac{1}{C + kt} \).

Step by step solution

01

Understand the Proportionality

The rate at which substance A is converted into substance B is proportional to the square of the amount of A. Mathematically, this means the derivative of the amount of A with respect to time, \(\frac{dy}{dt}\), is proportional to \(y^2\).
02

Set Up the Differential Equation

Since \( \frac{dy}{dt} \) is proportional to \(y^2\), we can write this as: \[ \frac{dy}{dt} = -ky^2 \] where \(k\) is the proportionality constant and the negative sign indicates that the amount of A is decreasing over time.
03

Sketch the Solution

To solve the differential equation \[ \frac{dy}{dt} = -ky^2 \], we separate the variables: \[ \frac{dy}{y^2} = -k \, dt \] Integrate both sides: \[ \begin{align*} \ \int \frac{1}{y^2} \, dy & = \ \int -k \, dt \ \end{align*} \] which gives \[ -\frac{1}{y} = -kt + C \]. We can solve for \(y\) to get the explicit solution: \[ y = \frac{1}{C + kt} \] where \(C\) is an integration constant. This solution shows that the amount of substance A decreases over time in an inverse relation.
04

Graph the Solution

The solution \( y = \frac{1}{C + kt} \) suggests that as time increases, the denominator grows, and hence the value of \( y \) (amount of substance A) decreases. It approaches zero as \( t \) goes to infinity, which aligns with our understanding that substance A is being converted into substance B over time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

proportionality
To understand the concept of proportionality in this context, consider how the rate of conversion from substance A to substance B depends on the amount of substance A. When we say that the rate at which A is converted to B is proportional to the square of the amount of A, we mean that the change in substance A, denoted \(\frac{dy}{dt}\), is directly related to some constant multiple of \(y^2\). Proportionality is a key concept in many types of differential equations, as it allows us to model real-world phenomena mathematically.
substance conversion
Substance conversion in this problem refers to the process where substance A is converted to substance B. The rate of this conversion is described by the differential equation we set up: \[ \frac{dy}{dt} = -ky^2 \]. This equation tells us how quickly substance A decreases over time. The negative sign indicates a decrease in substance A as it is converted into B. This conversion process is critical in many scientific scenarios, such as chemical reactions where reactants are turned into products.
rate of change
The rate of change in this problem is key to setting up and solving the differential equation. The rate of change of substance A with respect to time is depicted by the derivative \(\frac{dy}{dt}\). Here, this rate is proportional to \(y^2\), leading to the equation \[ \frac{dy}{dt} = -ky^2 \]. This tells us how the amount of substance A changes over time due to its conversion into substance B. Understanding the rate of change helps in predicting how the quantities involved will evolve over time.
separation of variables
Separation of variables is a technique used to solve differential equations. By rearranging the differential equation \(\frac{dy}{dt} = -ky^2 \) into the form \[ \frac{dy}{y^2} = -k \, dt \], we effectively separate the variables y and t on different sides of the equation. This allows us to integrate both sides independently: \[ \begin{align*} \int \frac{1}{y^2} \, dy \, = \, \int -k \, dt \end{align*} \]. After integrating, we obtain \[ -\frac{1}{y} = -kt + C \]. Solving this equation for y provides us with a function that explains how y varies over time, highlighting how separation of variables can simplify and solve differential equations.

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Most popular questions from this chapter

Solve the given equation using an integrating factor. Take \(t>0\). $$ y^{\prime}-2 y=e^{2 t} $$

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