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

Step-by-Step Solution

Answer

(a) Rate is nonnegative for \(0 \leq y \leq a\). (b) Rate is fastest at \( y = a \).

Step 1: Identify Constraints for Rate

The formula for the rate of reaction is given by \( k(a-y)(b-y) \). Since \( k \) is a positive constant, the expression \((a-y)(b-y)\) must be non-negative for the rate to be non-negative. Thus, the product of \((a-y)\) and \((b-y)\) should be non-negative.

Step 2: Determine Nonnegative Values of \(y\)

For the product \((a-y)(b-y)\) to be non-negative, consider the case when \((a-y)\) and \((b-y)\) have the same sign. That is, either:1. \( y \leq a \) and \( y \leq b \), or2. \( y > a \) and \( y > b \).Given that \( a < b \), only the first case is valid within the given constraints \( y \leq a \). Thus, \( 0 \leq y \leq a \).

Step 3: Graph Rate Against \(y\)

To graph \( k(a-y)(b-y) \) against \( y \), note that it is a quadratic equation that opens downward (since the leading coefficient is negative, \(-k\)). The parabola has roots at \( y = a \) and \( y = b \). Thus, it reaches its maximum at the midpoint of \( a \) and \( b \), which is at \( y = \frac{a+b}{2} \).

Step 4: Determine Fastest Rate of Reaction

The rate is fastest at the vertex of the parabola, which occurs at \( y = \frac{a+b}{2} \). Given that this value should be within the domain \( 0 \leq y \leq a \), and based on the constraint \( a < b \), the precise maximum rate within the valid interval \( 0 \leq y \leq a \) is at \( y = a \). Therefore, \( y = a \) is when the reaction rate is highest in the valid range.

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in mathematics, often appearing in the form \( ax^2 + bx + c = 0 \). They involve a squared variable, typically yielding two roots or solutions.
In our context, the rate of a chemical reaction is expressible as a quadratic function: \[Rate = k(a-y)(b-y) \] where \( k \) is a constant. This equation resembles the standard quadratic form \( ax^2 + bx + c = 0 \) but is expressed for the variable \( y \). The roots of this equation are \( y = a \) and \( y = b \).
Quadratic equations typically form a parabola when graphed. In this case, the parabola opens downward because the leading term's coefficient, \(-k\), is negative. This means the highest value of the rate, or the maximum point of the parabola, is at the vertex of this curve.
To find the maximum rate in the context of our chemical reaction, we look at where the parabola reaches its peak.
The vertex formula \( y = \frac{-b}{2a} \) or more practically \( y = \frac{a+b}{2} \) in our context, helps you determine this maximum point, providing the condition of optimal \( y \).

• The vertex is the spot where the rate is maximized within the feasible range of \( y \).
• For the chemical reaction, it reveals when the reaction progresses at its fastest.

Chemical Reactions

Chemical reactions represent processes where substances (reactants) convert into new substances (products). In the studied reaction, substance \( A \) combines with substance \( B \) to form substance \( Y \).
This transformation is quantified over time, indicated by the rate of the reaction, which measures how fast the reactants change.
The rate is expressed as: \[Rate = k(a-y)(b-y) \] Here:

• \( a-y \) denotes the remaining amount of substance \( A \).
• \( b-y \) signifies the remaining amount of substance \( B \).
• \( k \) is a rate constant, unique to the reaction conditions.

At each moment \( t \), \( y \) signifies the mass of the new substance \( Y \) formed, determining how fast \( A \) and \( B \) are reacting. It shows how the transformation progresses as more \( Y \) is formed.
The role of \( y \) is vital.As \( y \) approaches \( a \), the rate approaches maximum within defined constraints, since more \( Y \) has been formed in a shorter time.
This equation gives an insight into the balance of chemical equations at different phases of the reaction.

Graphing Functions

Graphing functions like our reaction rate equation helps visualize the relationship between quantities.
In this exercise, we graph the rate equation \( k(a-y)(b-y) \) to observe how the reaction evolves over time.
This specific function forms a parabola that descends, since the negative coefficient \(-k\) flips it downwards.
The roots of this graph occur at \( y = a \) and \( y = b \), meaning these are points where the rate is zero.Finding the maximum rate involves identifying the graph's vertex, providing insights into when the reaction accelerates the most before slowing down again.
When interpreting such graphs:

• The shape of the graph (parabola) indicates the function's behavior over \( y \).
• The maximum point, or vertex, tells us the peak rate when the reaction is fastest.
• Understanding these aspects can simplify complex reaction dynamics into manageable visual insights.

Graphing is indispensable in translating mathematical expressions into visual data, making it easier to understand function behaviors.