Problem 15 - Applied Calculus

Question

In some chemical reactions, the rate at which the amount of a substance changes with time is proportional to the amount present. For example, this is the case as \(\delta\) glucono-lactone changes into gluconic acid. (a) Write a differential equation satisfied by \(y\), the quantity of \(\delta\) -glucono-lactone present at time \(t\). (b) If 100 grams of \(\delta\) -glucono-lactone is reduced to \(54.9\) grams in one hour, how many grams will remain after 10 hours?

Step-by-Step Solution

Answer

(a) Differential equation: \( \frac{dy}{dt} = ky \). (b) After 10 hours, 4.54 grams remain.

Step 1: Identify the Type of Differential Equation

The problem states that the rate of change of a substance is proportional to the amount present. This suggests a first-order linear differential equation of the form \( \frac{dy}{dt} = ky \), where \( y \) is the quantity of the substance, \( t \) is time, and \( k \) is the proportionality constant.

Step 2: Formulate the Differential Equation

Using the information provided, the differential equation is \( \frac{dy}{dt} = ky \). This mathematical expression captures that the rate at which \( \delta \)-glucono-lactone changes with time is proportional to its current amount.

Step 3: Solve the Differential Equation for General Solution

The equation \( \frac{dy}{dt} = ky \) can be solved by separation of variables. Integrate both sides: \( \int \frac{1}{y} dy = \int k \, dt \), which gives \( \ln|y| = kt + C \). Solving for \( y \), we get \( y = Ce^{kt} \), where \( C \) is a constant determined by initial conditions.

Step 4: Apply Initial Conditions

The problem states that initially (\( t=0 \)), the amount \( y = 100 \) grams. Thus, \( 100 = Ce^{0} \) implies \( C = 100 \). Therefore, the equation becomes \( y = 100e^{kt} \).

Step 5: Determine the Rate Constant \( k \)

Use the condition at \( t = 1 \text{ hour}\), \( y = 54.9 \) grams. Substituting into \( y = 100e^{kt} \), we get \( 54.9 = 100e^{k} \). Solving for \( e^{k} \), we find \( e^{k} = \frac{54.9}{100} \). So, \( k = \ln\left(\frac{54.9}{100}\right) \approx -0.5988 \).

Step 6: Calculate Remaining Amount after 10 hours

Now, find \( y \) when \( t = 10 \) hours using the equation \( y = 100e^{-0.5988 \times 10} \). Calculate \( y = 100e^{-5.988} \approx 4.54 \) grams. This is the amount of \( \delta \)-glucono-lactone remaining after 10 hours.

Key Concepts

First-Order Linear Differential Equation

A first-order linear differential equation is a type of differential equation that involves derivatives of a function. These equations model processes involving rates of change, where one quantity varies continuously with another. In this context, the equation is often expressed as \( \frac{dy}{dt} = ky \), where:

• \( y \) represents the quantity of a substance changing over time.
• \( t \) denotes time.
• \( k \) is a constant of proportionality reflecting how the rate of change is related to the current amount of the substance.

This specific problem in the exercise captures how the amount of a chemical, \( \delta\)-glucono-lactone, depletes over time because its transformation rate to gluconic acid is directly proportional to the quantity present.

Rate of Change

The rate of change refers to how quickly a quantity changes over a particular period. This is a crucial concept in differential equations, especially when examining real-world phenomena. In this exercise, the rate of change is illustrated by the equation \( \frac{dy}{dt} = ky \). Here's how it works:

• The left side, \( \frac{dy}{dt} \), represents the rate at which the quantity \( y \) changes with respect to time.
• The right side, \( ky \), indicates that the rate depends linearly on the current quantity of the substance.

In simple terms, more \( \delta\)-glucono-lactone you have, the faster it converts to gluconic acid. This is a common occurrence in chemical reactions, making the study of rate of change vital to understanding these processes.

Separation of Variables

Separation of variables is a mathematical method used to solve differential equations. It involves rearranging the equation to isolate the variables on either side, making it possible to integrate and find a solution. This method was used to solve the differential equation \( \frac{dy}{dt} = ky \) in the exercise:

• First, rewrite the equation to separate \( y \) and \( t \): \( \frac{1}{y} dy = k \, dt \).
• Integrate both sides: \( \int \frac{1}{y} dy = \int k \, dt \).
• This yields \( \ln|y| = kt + C \), where \( C \) is a constant of integration.

Finally, by solving for \( y \), we obtain the general solution \( y = Ce^{kt} \). This solution reflects how the quantity changes over time, facilitated by the initial conditions provided in the problem. Separation of variables is a key tool for deriving solutions for many first-order linear differential equations.