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Chapter 10: Problem 38

Rate of Decomposition When a certain liquid substance \(A\) is heated in a flask, it decomposes into a substance \(B\) at such a rate (measured in units of \(A\) per hour) that at any time \(t\) is proportional to the square of the amount of substance \(A\) present. Let \(y=f(t)\) be the amount of substance \(A\) present at time \(t .\) Construct and solve a differential equation that is satisfied by \(f(t).\)

Short Answer

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

Step by step solution

01

Formulate the proportional relationship

The rate of decomposition of substance A is proportional to the square of the amount of substance A present. Mathematically, this relationship can be expressed as: \[ \frac{dy}{dt} = -k y^2 \] where \( k \) is the constant of proportionality.
02

Separate the variables

To solve the differential equation, first separate the variables. This involves rearranging the equation so that all terms involving \( y \) are on one side, and all terms involving \( t \) are on the other side: \[ \frac{dy}{y^2} = -k \, dt \].
03

Integrate both sides

Integrate both sides of the equation to find \( y \) as a function of \( t \). For the left side: \[ \int \frac{dy}{y^2} = \int y^{-2} \, dy = -y^{-1} \] For the right side: \[ \int -k \, dt = -kt + C \], where \( C \) is the constant of integration.
04

Combine the results from integration

Combine the results from integrating both sides: \[ -\frac{1}{y} = -kt + C \] We can simplify this to: \[ \frac{1}{y} = kt - C \].
05

Solve for y

Rearrange the equation to solve for \( y \): \[ y = \frac{1}{kt - C} \]. For better clarity, we can rename the constant terms to \( C' = \frac{1}{C} \): \[ y = \frac{1}{C' + kt} \].

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Key Concepts

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

Differential Equation
A differential equation is a mathematical equation that relates a function with its derivatives. If you're dealing with rates of change, differential equations are extremely useful. For this exercise, the rate at which substance A decomposes is proportional to the square of the amount of substance A present. This relationship is represented as:
\[ \frac{dy}{dt} = -k y^2 \]
Here, \( y \) represents the amount of substance A, \( t \) is time, and \( k \) is the constant of proportionality.
This equation is a first-order, non-linear differential equation, meaning it involves the first derivative of \( y \) and the function \( y \) is raised to a power other than one. Understanding how to form such an equation is crucial for modeling real-world phenomena, like chemical reactions or populations dynamics.
Separation of Variables
Separation of variables is a technique for solving differential equations. The method involves rearranging the equation so that each variable and its differential are on opposite sides of the equation.
In this exercise, we start with the differential equation:
\[ \frac{dy}{dt} = -k y^2 \]
To separate the variables, we arrange the equation such that all terms involving \( y \) are on one side, and all terms involving \( t \) are on the other side:
\[ \frac{dy}{y^2} = -k \, dt \]
By doing so, we can now integrate both sides separately. This technique is extremely useful when you can express the equation in a form that allows straightforward integration.
Integration
Integration is the process of finding the integral of a function, which is essentially the reverse operation of differentiation. It helps us find the original function given its derivative.
For our equation, we need to integrate both sides:
\[ \text{Left side: } \int \frac{dy}{y^2} = \int y^{-2} \, dy = -y^{-1} \]
\[ \text{Right side: } \int -k \, dt = -kt + C \]
Here, \( C \) is the constant of integration. After integrating, we combine the results:
\[ -\frac{1}{y} = -kt + C \]
Solving for \( y \), we rearrange and rename the constants for simplicity:
\[ y = \frac{1}{kt - C'} \]
Integration allows us to convert the rate of change given by the differential equation into an explicit function describing the amount of substance A over time.

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Most popular questions from this chapter

A certain drug is administered intravenously to a patient at the continuous rate of \(r\) milligrams per hour. The paticnt's body removes the drug from the bloodstream at a rate proportional to the amount of the drug in the blood, with constant of proportionality \(k=.5\) (a) Write a differential equation that is satisfied by the amount \(f(t)\) of the drug in the blood at time \(t\) (in hours). (b) Find \(f(t)\) assuming that \(f(0)=0 .\) (Give your answer in terms of \(r .)\) (c) In a therapeutic 2 -hour infusion, the amount of drug in the body should reach 1 milligram within 1 hour of administration and stay above this level for another hour. However, to avoid toxicity, the amount of drug in the body should not exceed 2 milligrams at any time. Plot the graph of \(f(t)\) on the interval \(1 \leq t \leq 2,\) as \(r\) varies between 1 and 2 by increments of \(.1 .\) That is, plot \(f(t)\) for \(r=1,1.1,1.2,1.3, \ldots . .2 .\) By looking at the graphs, pick the values of \(r\) that yield a therapeutic and nontoxic 2-hour infusion.

Suppose that \(f(t)\) satisfies the initial-value problem \(y^{\prime}=y^{2}+t y-7, y(0)=3 .\) Is \(f(t)\) increasing or decreasing at \(t=0 ?\)

In an autocatalytic reaction, one substance is converted into a second substance in such a way that the second substance catalyzes its own formation. This is the process by which trypsinogen is converted into the enzyme trypsin. The reaction starts only in the presence of some trypsin, and each molecule of trypsinogen yields 1 molecule of trypsin. The rate of formation of trypsin is proportional to the product of the amounts of the two substances present. Set up the differential equation that is satisfied by \(y=f(t)\) the amount (number of molecules) of trypsin present at time \(t\) Sketch the solution. For what value of \(y\) is the reaction proceeding the fastest? [Note: Letting \(M\) be the total amount of the two substances, the amount of trypsinogen present at time \(t \text { is } M-f(t) .\)]

Use Euler's method with \(n=2\) on the interval \(2 \leq t \leq 3\) to approximate the solution \(f(t)\) to \(y^{\prime}=t-2 y, y(2)=3\) Estimate \(f(3)\).

Review concepts that are important in this section. In each exercise, sketch the graph of a function with the stated properties. Domain: \(0 \leq t \leq 5 ;(0,3)\) is on the graph; the slope is always negative, and the slope becomes less negative.
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