Problem 47 - Applied Calculus

Question

In a chemical reaction, substance \(A\) combines with substance \(B\) to form substance \(Y .\) At the start of the reaction, the quantity of \(A\) present is \(a\) grams, and the quantity of \(B\) present is \(b\) grams. At time \(t\) seconds after the start of the reaction, the quantity of \(Y\) present is \(y\) grams. Assume \(a< b\) and \(y \leq a .\) For certain types of reactions, the rate of the reaction, in grams/sec, is given by Rate \(=k(a-y)(b-y), \quad k\) is a positive constant. (a) For what values of \(y\) is the rate nonnegative? Graph the rate against \(y\) (b) Use your graph to find the value of \(y\) at which the rate of the reaction is fastest.

Step-by-Step Solution

Answer

The rate is nonnegative for \( y \leq a \) and fastest near \( y = a \).

Step 1: Understanding the Problem

We have a chemical reaction where the rate of reaction is defined by the equation \( \text{Rate} = k(a-y)(b-y) \). We need to find when this rate is nonnegative and when it is at its maximum.

Step 1: Solve for Nonnegative Rates

The rate will be nonnegative as long as the expression \( (a-y)(b-y) \geq 0 \). This product is nonnegative if both terms are nonnegative or both are nonpositive.

Step 2: Analyze Inequalities

For nonnegative: \( a-y \geq 0 \) and \( b-y \geq 0 \), giving the range \( y \leq a \). For nonpositive: if \( a-y \leq 0 \) and \( b-y \leq 0 \), leading to \( y \geq b \). Since \( a < b \), only \( y \leq a \) holds. Thus, the rate is nonnegative for \( y \leq a \).

Step 3: Understand the Shape of the Parabola

The expression \( (a-y)(b-y) \) is a quadratic equation in \( y \) with roots at \( y = a \) and \( y = b \). It is an inverted parabola because the coefficient of \( y^2 \) is negative.

Step 4: Find the Vertex of the Parabola

The maximum point for \( f(y) = (a-y)(b-y) \) occurs at the vertex of the parabola. The vertex of a parabola in standard form is given by \( y = \frac{-b}{2a} \). Here, the vertex \( y \) is at the midpoint of \( a \) and \( b \), which is \( y = \frac{a+b}{2} \).

Step 5: Check Maximum Rate Condition

Since \( y = \frac{a+b}{2} \) is the midpoint and \( a < b \), the midpoint \( y \) is within the interval \( a < b \). Hence, for this reaction, considering \( y \leq a \), the fastest rate (maximum) is approached as \( y \to a^- \).

Step 6: Graph the Rate Against \( y \)

Plotting \( f(y) = k(a-y)(b-y) \) shows a downward-facing parabola with the maximum rate just before \( y = a \), since \( y \leq a \) constrains us from reaching the actual vertex.

Key Concepts

Chemical Reaction

Understanding a chemical reaction involves observing how substances transform into new products. In our exercise, substances \( A \) and \( B \) react to form substance \( Y \). One common way to gauge the progress of a chemical reaction is by monitoring the rate of reaction. The rate reflects how quickly reactants are consumed or products are formed over time. In our scenario, this rate is expressed as \( \text{Rate} = k(a-y)(b-y) \), where \( k \) is a constant, and \( y \) is the quantity of substance \( Y \) formed at any time \( t \).
Let's break down what this means:

• The term \( (a-y) \) signifies the remaining amount of \( A \) after \( y \) of it has reacted to form \( Y \).
• The term \( (b-y) \) represents the remaining amount of \( B \).

The equation indicates that the reaction rate depends on the unreacted portions of both \( A \) and \( B \). When the quantities of \( A \) and \( B \) decrease, so does the rate of reaction.
The rate will be nonnegative provided the expression \((a-y)(b-y) \geq 0\). This is a crucial part because it defines when the reaction is making progress by producing \( Y \). In essence, since \( a < b \), the rate remains nonnegative as long as \( y \leq a \). This insight helps us visualize and predict behavior changes in the reaction.

Quadratic Equation

A quadratic equation is a fundamental concept in algebra, representing equations of the form \( ax^2 + bx + c = 0 \). In the exercise, the quadratic play comes from the expression \( (a-y)(b-y) \). It multiplies to form \( y^2-(a+b)y+ab \). This is a classic form of a quadratic equation, explaining how the reaction rate varies with \( y \), the product formed.
Key properties of quadratic equations include:

• The leading term here is negative, implying that the graph of this equation is a parabola opening downward.
• The roots, or solutions, occur where the quadratic equals zero, specifically at \( y=a \) and \( y=b \).

In this case, understanding these roots means recognizing the points where the rate of the chemical reaction stops progressing. Mathematically, these points are crucial as they inform us about the interval of \( y \) which keeps the rate positive or nonnegative.

Parabola Analysis

Analyzing parabola behavior helps illustrate the changing rate of a chemical reaction. The quadratic equation derived, \( f(y) = (a-y)(b-y) \), forms a parabola when plotted against \( y \). This parabola's characteristics are vital for identifying how the reaction proceeds. Specifically, it is crucial to find the vertex—the highest point—of this inverted parabola because it represents the fastest rate of the reaction.

• The axis of symmetry for this parabola is at the midpoint \( y = \frac{a+b}{2} \).
• This midpoint denotes where the reaction reaches its maximum speed in theoretical terms, although practically constrained by \( y \leq a \).

For real-world implications, since \( a < b \), we see that \( y \) cannot actually reach the vertex without exceeding \( a \). Instead, observing that \( y \) is very close to \( a \), approaching it from the left (\( y \to a^- \)), indicates the reaction is running at its fastest permissible rate. This analysis uncovers how mathematical concepts apply to real-world chemical phenomena, simplifying complex dynamics into a clear visual understanding.