Problem 27 - Precalculus Mathematics for Calculus

Question

Express the statement as an equation. Use the given information to find the constant of proportionality. \(s\) is inversely proportional to the square root of \(t .\) If \(s=100\) then \(t=25\).

Step-by-Step Solution

Answer

The constant of proportionality \(k\) is 500.

Step 1: Understanding Inverse Proportionality

When a quantity \(s\) is inversely proportional to the square root of another quantity \(t\), it can be represented as \(s \propto \frac{1}{\sqrt{t}}\). This means that as \(t\) increases, \(s\) will decrease, and vice versa.

Step 2: Formulating the Equation

To convert the proportionality statement into an equation, introduce a constant of proportionality \(k\). The equation becomes \(s = \frac{k}{\sqrt{t}}\). Here, \(k\) is a constant value that relates the two variables.

Step 3: Substituting Known Values

We are given that when \(s = 100\), \(t = 25\). Substitute these values into the equation: \(100 = \frac{k}{\sqrt{25}}\). Calculate \(\sqrt{25}\), which equals 5.

Step 4: Solving for the Constant of Proportionality

With the values substituted and \(\sqrt{25} = 5\), the equation becomes \(100 = \frac{k}{5}\). Solve for \(k\) by multiplying both sides by 5, resulting in \(k = 100 \times 5\).

Step 5: Calculating \(k\)

Complete the calculation from the previous step to find \(k\). The equation \(k = 100 \times 5\) gives \(k = 500\).

Key Concepts

Constant of Proportionality

The concept of the constant of proportionality is essential when dealing with proportional relationships between variables. In our case, because the relationship between two variables, \( s \) and \( t \), is inverse, we can express this as \( s \propto \frac{1}{\sqrt{t}} \). This means that as one quantity increases, the other decreases. To make this relationship into an equation, we introduce a constant \( k \), which is called the constant of proportionality. So, the equation can be rewritten as \( s = \frac{k}{\sqrt{t}} \). This constant \( k \) helps us exactly quantify how much \( s \) changes with changes in \( t \). In simple terms:

• It's the factor that differentiates pure proportionality from an exact equation.
• Calculating \( k \) allows us to predict one variable if the other is known.

Learning to find \( k \) is vital as it makes the relationship between the involved quantities precisely measurable.

Square Root

The square root is a mathematical operation opposite of squaring. Taking the square root of a number \( t \) finds a value that, when multiplied by itself, results in \( t \). For example, the square root of 25 is 5 since \( 5 \times 5 = 25 \).In inverse proportionality statements, such as \( s \propto \frac{1}{\sqrt{t}} \), the square root is pivotal as it appears in the denominator of the expression. This affects how \( s \) responds to changes in \( t \). To simplify and solve equations involving square roots, it is helpful to:

• Know basic square roots, like \( \sqrt{4} = 2 \), \( \sqrt{9} = 3 \), etc.
• Understand properties like \( \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b} \).

Mastering these basics can ease solving equations and understanding the effects of changes in variables, which are crucial in problems like this one.

Mathematical Equation

A mathematical equation is a statement showing the equality between two expressions. It involves unknowns and constants and uses mathematical operations to establish a relationship among them. In our given problem, \( s = \frac{k}{\sqrt{t}} \) is the equation form.Translating proportionality into an exact equation involves converting statements of variance into precise equations using constants like \( k \).The process of solving involves:

• Recognizing the relationship (inverse in this case).
• Introducing a constant of proportionality, converting \( s \propto \frac{1}{\sqrt{t}} \) to \( s = \frac{k}{\sqrt{t}} \).
• Substituting known values to find unknowns like \( k \).

This process allows us to substitute numerical values into the equation, making it easier to solve for unknowns, establishing real-world relationships, and enhancing understanding of algebraic expressions.