Problem 14 - 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

Approximately 3 grams will remain after 10 hours.

Step 1: Write the Differential Equation

Since the rate of change of the substance is proportional to the amount present, we can express this using the differential equation \( \frac{dy}{dt} = -ky \), where \( y \) is the amount of the substance at time \( t \), and \( k \) is the proportionality constant.

Step 2: Solve the Differential Equation

The differential equation \( \frac{dy}{dt} = -ky \) is a first-order linear differential equation, which can be solved by separation of variables. We separate the variables to get \( \frac{1}{y} dy = -k dt \). Integrating both sides, we find \( \ln|y| = -kt + C \), where \( C \) is the integration constant. Exponentiating both sides gives the solution \( y = Ce^{-kt} \).

Step 3: Determine the Constant C

We know at \( t = 0 \), \( y = 100 \) grams. Substituting these values into the equation \( y = Ce^{-kt} \) gives us \( 100 = Ce^0 \), hence \( C = 100 \). So the solution becomes \( y = 100e^{-kt} \).

Step 4: Calculate the Constant k

We are given that after 1 hour, the amount is reduced to 54.9 grams, so \( y(1) = 54.9 \). Substituting into the equation \( y = 100e^{-kt} \), we get \( 54.9 = 100e^{-k(1)} \). Solving for \( k \), we have \( e^{-k} = \frac{54.9}{100} \) or \( -k = \ln\left(\frac{54.9}{100}\right) \). Thus, \( k = -\ln\left(\frac{54.9}{100}\right) \).

Step 5: Calculate the Amount After 10 Hours

Substitute \( t = 10 \) into the equation \( y = 100e^{-kt} \) to find \( y(10) \). We have \( y(10) = 100e^{-10k} \). Using \( k = -\ln\left(\frac{54.9}{100}\right) \), substitute this into the equation and calculate \( y(10) = 100 \left(\frac{54.9}{100}\right)^{10} \). Solving this gives the amount of \( \delta \)-glucono-lactone after 10 hours.

Step 6: Final Calculation

Evaluate the expression \( 100 \left(\frac{54.9}{100}\right)^{10} \) using the value calculated for \( k \) to find the amount of \( \delta \)-glucono-lactone remaining after 10 hours. This will give us approximately 3 grams.

Key Concepts

Differential Equation

A differential equation is an equation that involves one or more functions and their derivatives. These equations describe how a particular quantity changes with respect to another quantity, which in this context is often time. In the exercise, the challenge was to model the change in quantity of \( \delta \)-glucono-lactone over time.
The differential equation provided was \( \frac{dy}{dt} = -ky \), representing how the quantity \( y \) changes with time \( t \).

• \( y \) is the quantity of the substance at time \( t \).
• \( \frac{dy}{dt} \) is the rate of change of this substance.
• \( k \) is a positive constant, representing the rate constant of the reaction.

This equation implies that the rate of change is proportional to the current amount of substance, which is typical in processes showing exponential decay, such as many chemical reactions.

First-order Differential Equation

First-order differential equations involve only the first derivative of the function and no higher derivatives. These types of equations are the simplest to solve and are very common in applications like physics, engineering, and finance. Here, our equation \( \frac{dy}{dt} = -ky \) is a first-order differential equation because the highest order of the derivative present is 1 (involving \( \frac{dy}{dt} \)).

• "First-order" means the equation involves only the first derivative.
• These equations often describe exponential growth or decay processes.

Solving this type of equation involves standard techniques, and in this case, we used a method called separation of variables, which allows us to integrate both sides easily.

Separation of Variables

Separation of variables is a classic technique used to solve differential equations where variables can be segregated onto separate sides of the equation. This is particularly useful for first-order equations like \( \frac{dy}{dt} = -ky \). Here’s how it works: by rearranging terms, we aim to isolate \( y \) and its derivative on one side, and time \( t \) on the other, like so: \( \frac{1}{y} dy = -k dt \).
This allows us to:

• Integrate the left side with respect to \( y \), giving \( \ln|y| \).
• Integrate the right side with respect to \( t \), which yields \( -kt + C \), where \( C \) is the constant of integration.

Upon integrating and exponentiating, we obtain \( y = Ce^{-kt} \), providing the solution to our initial differential equation.

Proportionality Constant

The proportionality constant \( k \) is a key component in the differential equation \( \frac{dy}{dt} = -ky \). It determines the speed of the reaction or process described by the equation. When solving the problem in the exercise, we calculate \( k \) using initial conditions given for the problem.

• At \( t = 0 \), \( y = 100 \).
• At \( t = 1 \), \( y = 54.9 \).

By substituting these into our solution \( y = 100e^{-kt} \), we solved for \( k \) and found it to be \(-\ln\left(\frac{54.9}{100}\right)\).
This constant shows how rapidly the substance diminishes over time. A larger \( k \) value would indicate a quicker decay, demonstrating the significance of understanding the proportionality constant's role in such equations.