Problem 55 - Calculus and its applications

Question

The following formula is used by psychologists and educators to predict the reading ease, \(E,\) of a passage of words: $$E=206.835-0.846 w-1.015 s$$ where \(w\) is the number of syllables in a 100 -word section and s is the average number of words per sentence. $$\text { Find } \frac{\partial E}{\partial w}$$

Step-by-Step Solution

Answer

The partial derivative \(\frac{\partial E}{\partial w}\) is \(-0.846\).

Step 1: Understanding the Problem

We are given the formula for reading ease \(E\) as a function of two variables, \(w\) and \(s\). We need to find the partial derivative of \(E\) with respect to \(w\). This means we will assume \(s\) is a constant and differentiate \(E\) with respect to \(w\).

Step 2: Identify Terms Involving Variable

In the formula \(E = 206.835 - 0.846w - 1.015s\), identify the term that involves the variable \(w\), which is \(-0.846w\). The rest terms, \(206.835\) and \(-1.015s\) will remain constant during differentiation with respect to \(w\).

Step 3: Differentiate with Respect to w

To find the partial derivative \(\frac{\partial E}{\partial w}\), differentiate the term \(-0.846w\) with respect to \(w\). The derivative of \(-0.846w\) with respect to \(w\) is \(-0.846\) while the derivatives of the constant terms are zero.

Step 4: Write Out the Partial Derivative

Since the derivative of the constant terms are zero, the partial derivative of \(E\) with respect to \(w\) is simply \(-0.846\). Therefore, \(\frac{\partial E}{\partial w} = -0.846\).

Key Concepts

Understanding the Reading Ease Formula

The reading ease formula is a popular tool among psychologists and educators. It helps predict how easy or difficult a passage of text might be for readers. This formula involves some simple mathematical operations based on two variables:

• Syllables in a 100-word section, represented by \(w\)
• Average number of words per sentence, represented by \(s\)

The formula itself is:\[ E = 206.835 - 0.846w - 1.015s \]This equation gives the reading ease score \(E\), where higher scores indicate easier reading passages. Each component of the formula plays a role:

• The constant \(206.835\) provides a base level of ease.
• The term \(-0.846w\) suggests that an increase in the number of syllables makes the text harder to read.
• Similarly, \(-1.015s\) indicates that longer sentences also decrease reading ease.

Summarizing, this formula allows experts to quantify readability, offering a valuable metric for assessing text complexity.

Demystifying Differentiation

Differentiation is a fundamental concept in calculus and is indispensable when dealing with changes in multivariable functions. When we differentiate, we find how a function changes as its variables change. In the context of multivariable functions, like our reading ease equation, we are often interested in how the function changes with respect to one specific variable, keeping the others constant.In our exercise, we perform partial differentiation of the function \(E = 206.835 - 0.846w - 1.015s\) with respect to \(w\). Partial differentiation with respect to \(w\) means that we treat \(s\) as a constant. Thus, we focus only on the terms involving \(w\):

• For \(-0.846w\), the derivative with respect to \(w\) is \(-0.846\).
• The constant term \(206.835\) and the term \(-1.015s\) do not change with \(w\), so their derivatives are zero.

Consequently, the partial derivative \(\frac{\partial E}{\partial w}\) simplifies to \(-0.846\), indicating how much the reading ease score changes with a unit increase in syllables within a 100-word section.

Exploring Multivariable Functions

Multivariable functions are essential in mathematics and its applications, allowing us to explore relationships between more than one variable. These functions can model complex systems more precisely compared to single-variable functions.In our given formula, \(E = 206.835 - 0.846w - 1.015s\), we deal with a function of two variables, \(w\) (syllables in a 100-word section) and \(s\) (average number of words per sentence). Here’s what makes multivariable functions exciting:

• Each variable can independently influence the output, offering a richer understanding of relationships.
• They provide a framework for partial derivatives, which show how each variable affects the function when others are kept constant.
• Multivariable functions are crucial in fields like economics, physics, and statistics where multiple factors simultaneously impact outcomes.

Understanding these functions is key to tackling real-world problems that depend on multiple changing conditions, making them an invaluable tool in analysis and prediction.