Problem 24 - Algebra and Trigonometry

Question

Use the four-step procedure for solving variation problems given on page 445 to solve Exercises 21–36. The distance that a spring will stretch varies directly as the force applied to the spring. A force of 12 pounds is needed to stretch a spring 9 inches. What force is required to stretch the spring 15 inches?

Step-by-Step Solution

Answer

The force required to stretch the spring 15 inches is approximately 20 pounds.

Step 1: Identify the two quantities that are varying

In this problem, the two quantities that are varying are the force applied to the spring (F) and the distance that the spring stretches (d). They have a direct variation relationship: \( F = kd \), where k is the constant of variation.

Step 2: Plug in the known values to find the constant of variation

We know that a force of 12 pounds stretches the spring 9 inches. By substituting these values into the equation we get: \( 12 = k * 9 \). Solving for k gives \( k = 12 / 9 = 1.33 \) (rounded to two decimal places). So the constant of variation is about 1.33.

Step 3: Substitute the known distance and constant into the equation to find the force

We are asked to find the force required to stretch the spring 15 inches. We substitute d = 15 and k = 1.33 into the equation to get: \( F = 1.33 * 15 = 20 \) (rounded). So about 20 pounds of force are required to stretch the spring 15 inches.

Key Concepts

Variation in Algebra

One of the fundamental concepts in algebra is the idea of variation, which refers to how one quantity changes in relation to another.
Understanding variation is pivotal for solving real-world problems involving relationships between variables. In the context of algebra, there are two main types of variation: direct and inverse.

Direct variation occurs when two variables increase or decrease in the same ratio. It is represented by the equation \( y = kx \), where \( y \) varies directly with \( x \), and \( k \) is the constant of variation. This constant is what makes the relationship between the variables consistent. For example, in the original problem given, the distance a spring stretches varies directly with the force applied, demonstrating a direct variation.

Constant of Variation

The constant of variation, denoted as \( k \), is a crucial part of understanding direct variation. It is the consistent ratio between two variables that vary directly with each other.

• In a formula, it is expressed as \( y = kx \).
• It can be found by dividing the known value of one variable by the corresponding value of the other variable (\( k = \frac{y}{x} \)).

Once determined, this constant allows us to predict the behavior of the variables under different conditions.

Returning to our spring example from the exercise, once the constant of variation is known, it can be used to calculate either the required force for a given distance or the distance that a spring would stretch under a certain force. By calculating \( k = 1.33 \) from the provided values, we essentially unlock the ability to forecast the spring's behavior under different forces.

Importance of Accuracy

When dealing with physical applications like a stretched spring, the accuracy of the constant of variation is of paramount importance, as it ensures the precise prediction of how much force will produce a specific stretch.

Solving Algebraic Equations

Solving algebraic equations is a day-to-day task in algebra that involves finding the values of unknowns that make the equation true.
When solving equations dealing with direct variation:

• Firstly, identify the variables that are directly proportional to each other.
• Next, establish the constant of variation using known values.
• Then apply this constant to solve for unknown variables.

The process of equation-solving is not only about manipulation of numbers and letters; it requires logical thinking and understanding the nature of the relationships between variables.

Improving in this area happens through practice and by attempting different types of variation problems.
For instance, the exercise presented shows how algebra can solve practical questions such as determining the force needed to stretch a spring a certain distance.

By applying the methods step by step—identifying varying quantities, finding the constant of variation, and finally substitizing values to solve for the unknown—we demonstrate a structured approach to utilizing algebra to dissect and understand real-world scenarios.