Problem 54 - Elementary Algebra

Question

For exercises 53-56, the formula \(F=\frac{100 S_{u} C_{p}}{S_{p} C_{u}}\) describes the fractional excretion of sodium, \(F\). Is the relationship of the given variables a direct variation or an inverse variation? $$ C_{p}, S_{p} \text {, and } C_{u} \text { are constant; the relationship of } F \text { and } S_{u} $$

Step-by-Step Solution

Answer

Direct variation because \(F\) increases proportionally with \(S_{u}\).

Step 1: Understand the Given Equation

Given the formula: \( F = \frac{100 S_{u} C_{p}}{S_{p} C_{u}} \) The variables \(C_{p}\), \(S_{p}\), and \(C_{u}\) are constants.

Step 2: Isolate the Relationship between \(F\) and \(S_{u}\)

Observe that the only variable in the numerator is \(S_{u}\). This suggests that \(F\) and \(S_{u}\) are directly related by this formula.

Step 3: Identify the Direct Variation Form

The formula can be rewritten as: \( F = k S_{u} \) where \( k \) is a constant: \( k = \frac{100 C_{p}}{S_{p} C_{u}} \). This demonstrates that as \(S_{u}\) increases, \(F\) increases proportionally, indicating a direct variation.

Key Concepts

fractional excretion of sodium

Fractional excretion of sodium (FENa) helps healthcare professionals understand how well the kidneys are functioning by measuring the amount of sodium excreted in the urine compared to the amount filtered by the kidneys. The formula to calculate FENa is given by: \( F = \frac{100 S_{u} C_{p}}{S_{p} C_{u}} \).
The numerator \(S_{u} C_{p}\) represents the product of the sodium concentration in the urine and the plasma concentration of creatinine. The denominator \(S_{p} C_{u}\) represents the product of the sodium concentration in the plasma and the urine concentration of creatinine.
In the formula, \(C_{p}\), \(S_{p}\), and \(C_{u}\) are constants. By evaluating how these constants interact with the variable \(S_{u}\) (sodium concentration in the urine), we can determine the efficiency of sodium excretion.

algebraic formulas

Algebraic formulas are equations that utilize variables to represent relationships between different quantities. In the given exercise, the formula \( F = \frac{100 S_{u} C_{p}}{S_{p} C_{u}} \) is an algebraic formula representing the fractional excretion of sodium.
Using algebra, we can manipulate the formula to isolate variables and identify relationships among them. By expressing the formula as \( F = k S_{u} \) where \( k = \frac{100 C_{p}}{S_{p} C_{u}} \), we simplify the relationship between \(F\) and \(S_{u}\) and identify the type of variation.
Understanding how to work with algebraic formulas is essential because they help us solve real-world problems by translating them into mathematical expressions.

proportional relationships

Proportional relationships describe how two quantities increase or decrease together in a specific ratio. In this exercise, the relationship between \(F\) (fractional excretion of sodium) and \(S_{u}\) (sodium concentration in the urine) is directly proportional.
To determine this, we observe that as \(S_{u}\) increases, \(F\) increases proportionally, following the equation \( F = k S_{u} \). Here, \( k = \frac{100 C_{p}}{S_{p} C_{u}} \) is the constant of proportionality.
In practical terms, if the sodium concentration in the urine doubles, the fractional excretion of sodium also doubles, as long as \(C_{p}\), \(S_{p}\), and \(C_{u}\) remain constant.
Recognizing proportional relationships is key in fields like science and engineering, where understanding how changes in one variable affect another is crucial.