Problem 20 - Algebra and Trigonometry

Question

Express the statement as an equation. Use the given information to find the constant of proportionality. \(P\) is directly proportional to \(T .\) If \(T=300,\) then \(P=20\)

Step-by-Step Solution

Answer

The equation is \(P = \frac{1}{15}T\).

Step 1: Understand Direct Proportionality

When a variable \(P\) is directly proportional to another variable \(T\), it means that \(P\) can be expressed as \(P = kT\), where \(k\) is the constant of proportionality. This relationship implies that as \(T\) increases, \(P\) will also increase by the same factor determined by \(k\), and vice versa.

Step 2: Set Up the Proportionality Equation

Using the relationship \(P = kT\), we need to find the value of \(k\) using the given conditions. We have the equation in the form of \(P = kT\).

Step 3: Substitute Given Values into the Equation

We are given that \(T = 300\) and \(P = 20\). Substitute these values into the equation \(P = kT\): \(20 = k \times 300\).

Step 4: Solve for the Constant of Proportionality \(k\)

Simplify the equation \(20 = k \times 300\) to find \(k\). Divide both sides of the equation by 300: \(k = \frac{20}{300}\).

Step 5: Calculate the Value of \(k\)

Perform the division \(\frac{20}{300}\) to find \(k = \frac{1}{15}\). Thus, the constant of proportionality is \(\frac{1}{15}\).

Key Concepts

Constant of Proportionality

In mathematics, when two variables, such as \(P\) and \(T\), are directly proportional, they change in relation to each other by a constant factor. This constant is what we call the "constant of proportionality," represented by \(k\). Imagine two gears turning together - as one turns, the other turns at a speed determined by the constant ratio between them.

The notation \(P = kT\) expresses this relationship, indicating that \(P\) increases or decreases consistently with \(T\). Every time \(T\) grows, so does \(P\) by a multiplication of \(k\), making \(k\) a crucial element in understanding the linkage between the variables. To find \(k\), you measure the change in \(P\) in relation to \(T\). In our example, using the values given, \(k\) is computed to be \(\frac{1}{15}\).

Recognizing the constant of proportionality is essential because it

• Defines how strongly connected two variables are.
• Helps predict the behavior of one variable if the other changes.

This concept is a cornerstone in algebra and helps with modeling equations across sciences.

Proportionality Equation

A proportionality equation helps frame the relationship between variables that change together. In the form \(P = kT\), this equation tells us that \(P\) depends directly on \(T\), scaled by \(k\). It is like setting up a rulebook where \(k\) defines the rules of how \(P\) and \(T\) interact.

Using our example, we start with \(P = kT\). Then, we input our known values, such as \(T = 300\) and \(P = 20\), to form a complete equation: \(20 = k \times 300\).

A proportionality equation is useful because:

• It provides a clear formula for predicting outcomes.
• It maintains consistency across different scenarios involving similar variables.

By solving this equation, you affirm the direct relationship between two quantities and find \(k\), which guides how to adjust one variable based on changes in another. Understanding and setting up a correct proportionality equation is key to solving many algebraic problems.

Solving Equations

Solving equations, like \(20 = k \times 300\), involves finding the value of an unknown variable, in this case, \(k\). The goal is to isolate the variable to determine its value. In our example, to solve for \(k\), we rearrange the equation by dividing both sides by 300. This cancellation simplifies the equation to \(k = \frac{20}{300}\).

Solving equations often includes stages:

• Simplifying the equation by performing operations, such as addition, subtraction, multiplication, or division.
• Ensuring all calculations maintain equality across both sides of the equation.
• Performing arithmetic to compute the numerical value, as seen with \(k = \frac{1}{15}\).

By solving, we turn a math statement into a solution, translating our calculated result into a practical understanding of how variables interact. Each time we solve such an equation, we build skills essential for tackling more complex algebraic challenges!