Problem 75 - Calculus with Concepts in Calculus

Question

The rate of a certain autocatalytic reaction is given by $$ v(x)=k x(a-x) \quad \text { for } 0 \leq x \leq a $$ where \(a\) is the original amount of the substance, \(x\) is the amount of the substance produced by the reaction, and \(k\) is positive. Determine where the rate of reaction is increasing and where it is decreasing.

Step-by-Step Solution

Answer

The reaction rate increases for \(0 \leq x < \frac{a}{2}\) and decreases for \(\frac{a}{2} < x \leq a\).

Step 1: Identify Functions to Analyze

We are given the rate of reaction function \(v(x)=k x(a-x)\). Our goal is to find where this function is increasing or decreasing. The amount \(x\) represents the product formed by the reaction, and \(a\) is the initial amount of reactant. The variable \(k\) is a positive constant.

Step 2: Find the First Derivative

To determine where the function \(v(x)\) is increasing or decreasing, we need to find its first derivative. The function is \(v(x)=k x(a-x)\). By using the product rule, the derivative is computed as \(v'(x) = k \cdot [(a-x)(1) + x(-1)] = k \cdot (a - 2x)\).

Step 3: Solve the Derivative for Critical Points

Set the first derivative \(v'(x) = k(a - 2x)\) equal to zero to find the critical points. Solve the equation \(k(a - 2x) = 0\). Since \(k\) is non-zero, this simplifies to \(a - 2x = 0\). Solve for \(x\) to get \(x = \frac{a}{2}\).

Step 4: Analyze the Sign of the Derivative

To determine the intervals of increase or decrease, analyze \(v'(x) = k(a - 2x)\) around the critical point \(x = \frac{a}{2}\). For \(0 \leq x < \frac{a}{2}\), \(a - 2x > 0\), so \(v'(x) > 0\) (increasing). For \(\frac{a}{2} < x \leq a\), \(a - 2x < 0\), so \(v'(x) < 0\) (decreasing).

Step 5: Conclusion on Increasing/Decreasing Intervals

The rate of reaction \(v(x)\) is increasing for \(0 \leq x < \frac{a}{2}\) and decreasing for \(\frac{a}{2} < x \leq a\). At \(x = \frac{a}{2}\), the rate has a maximum value as determined by the zero derivative.

Key Concepts

Rate of Reaction

The rate of reaction is a fundamental concept in chemistry, representing how quickly a chemical reaction occurs. For autocatalytic reactions, this rate can be influenced by several factors including the concentration of reactants and the presence of a catalyst that is part of the reaction. This catalyst increases the reaction rate by providing an alternative pathway with a lower activation energy.

In the given function of the rate of reaction, \(v(x) = k x(a-x)\), the rate depends on variables like the amount of substance produced \(x\) and initial reactant amount \(a\), with \(k\) acting as a positive constant. Understanding the rate of reaction helps in predicting how the concentration of products changes over time in a chemical process.

First Derivative

The first derivative of a function provides valuable information about the function's rate of change at any given point. For the rate of reaction function \(v(x) = k x(a-x)\), we employ the concept of the first derivative to determine how the rate changes as the reaction progresses.

By calculating the first derivative \(v'(x) = k(a - 2x)\), we determine whether the reaction rate is increasing or decreasing. The first derivative essentially measures the slope of the tangent line to the curve of the original function at any point \(x\). A positive derivative indicates an increasing function, while a negative derivative suggests a decreasing function.
Understanding how to derive and interpret the first derivative is crucial in analyzing and graphing functions.

Critical Points

Critical points occur where the first derivative is zero or undefined, indicating potential maxima, minima, or points of inflection on a graph. For the rate of reaction \(v(x) = k x(a-x)\), we find the critical points by setting the first derivative \(v'(x) = k(a - 2x)\) to zero.

This gives the critical point at \(x = \frac{a}{2}\). This point divides the graph into different sections where the behavior of the function may change. At this critical point, the slope of the tangent line is zero, suggesting a possible maximum rate of reaction.

• Critical points help in identifying significant changes in the behavior of the function.
• They are particularly useful in determining where functions reach their highest or lowest values.

Understanding critical points allows us to gain deep insights into the shape and dynamics of a function.

Increasing and Decreasing Intervals

An increasing interval on a graph represents a section where the function values rise as \(x\) increases, while a decreasing interval is where the function values fall. Identifying these intervals for a function like the reaction rate \(v(x) = k x(a-x)\) involves analyzing the sign of its first derivative.

By examining \(v'(x) = k(a - 2x)\), we determine:

• If \(v'(x) > 0\) for \(0 \leq x < \frac{a}{2}\), the rate is increasing.
• If \(v'(x) < 0\) for \(\frac{a}{2} < x \leq a\), the rate is decreasing.

These intervals are essential in understanding how and when the reaction rate changes, allowing chemists to optimize process conditions. Knowing where a function increases or decreases can guide predictions and control strategies in chemical reactions.