Problem 84 - A Graphical Approach to College Algebra

Question

Solve each problem. Suppose \(p\) varies directly with the square of \(z\) and inversely with \(r .\) If \(p=\frac{32}{5}\) when \(z=4\) and \(r=10,\) find \(p\) when \(z=2\) and \(r=16\)

Step-by-Step Solution

Answer

The value of \(p\) is 1.

Step 1: Understand the Variation Formula

The problem states that \(p\) varies directly with the square of \(z\) and inversely with \(r\). This can be expressed by the formula \(p = k \cdot \frac{z^2}{r}\), where \(k\) is a constant of proportionality.

Step 2: Determine the Constant of Proportionality

To find \(k\), use the given values \(p=\frac{32}{5}\), \(z=4\), and \(r=10\). Plug these values into the equation: \[\frac{32}{5} = k \cdot \frac{4^2}{10}\] Simplifying, \[\frac{32}{5} = k \cdot \frac{16}{10}\]\[\frac{32}{5} = k \cdot 1.6\]. Solving for \(k\), \(k = \frac{32}{5} \div 1.6\), or \(k = 4\).

Step 3: Use the Constant to Find New Value of p

With \(k = 4\), substitute \(z = 2\) and \(r = 16\) into the variation formula: \[p = 4 \cdot \frac{2^2}{16}\]. Compute \(p\): \[p = 4 \cdot \frac{4}{16}\] \[p = 4 \cdot \frac{1}{4}\]\(p = 1\).

Key Concepts

Proportionality Constant

In variation problems, the proportionality constant, often denoted by the letter \(k\), is key to understanding how variables relate to each other. This constant stems from the relationship outlined by the problem, connecting variables in either a direct or inverse manner. In our example, this connection is described by the equation:\[p = k \cdot \frac{z^2}{r}\]This equation signifies that the variable \(p\) changes directly with the square of \(z\) and inversely with \(r\). The role of \(k\) is to bridge these relationships into a singular equation. Using known values given in the problem, you can solve for \(k\). By substituting these values in, as shown in the solution, you determine \(k = 4\). With this constant, any variation in \(z\) or \(r\) can be computed to find the corresponding \(p\). The concept revolves around consistency of this constant across different scenarios, making it a powerful tool for solving such problems.

Algebraic Equations

Algebraic equations form the foundation of solving variation problems. Here, they are used to express the direct and inverse relationships in mathematical terms. The core equation used in this example is:\[p = k \cdot \frac{z^2}{r}\]These mathematical expressions provide a structured method by which different variables can be interrelated. When you input known values into the equation, it becomes simpler to isolate and solve for the unknown variables. This process involves basic algebraic manipulations like multiplication, division, and substitution. In this case, the given values for \(p\), \(z\), and \(r\) helped us establish \(k\), the missing link connecting the equation's terms. This approach simplifies complex scenarios into manageable mathematical tasks, highlighting the power of algebra to solve real-world problems using systematic steps and logic.

Variation Problems

Variation problems often involve understanding how one quantity changes based on the change of others under specified conditions. They are divided primarily into direct and inverse variations. In this exercise:- **Direct Variation**: This refers to how \(p\) changes directly with \(z^2\). If \(z\) increases, \(p\) increases proportionally, as long as \(r\) remains fixed.- **Inverse Variation**: This indicates how \(p\) decreases as \(r\) increases when \(z\) remains constant.Understanding these types is crucial, such as recognizing that when \(p\) varies directly with \(z^2\) and inversely with \(r\), an increase in \(r\) should logically decrease \(p\) if \(z\) stays the same.When solving variation problems, interpreting the relationships correctly ensures that when you substitute new values, the outcome will be logical and consistent with the established rules. Each variation problem demands translating a real-world relationship into a formula that can be manipulated, aiding in predicting outcomes with precision.