Direct And Inverse Variation Equation For S, T, And W

In the realm of mathematics, variation plays a crucial role in describing how quantities relate to each other. Understanding the concepts of direct and inverse variation is essential for solving a wide range of problems in algebra, physics, and other fields. This article delves into a specific scenario involving direct and inverse variation, providing a comprehensive explanation and step-by-step solution to determine the equation that best represents the given relationship.

Deciphering Direct and Inverse Variation

Before we tackle the specific problem at hand, let's refresh our understanding of direct and inverse variation. These concepts are fundamental to grasping the equation that governs the relationship between the variables s, t, and w.

Direct Variation: When One Quantity Increases, the Other Follows Suit

Direct variation describes a relationship where two quantities increase or decrease together at a constant rate. In simpler terms, if one quantity doubles, the other quantity also doubles, and so on. Mathematically, we represent direct variation using the equation:

y = kx

where:

• y is directly proportional to x
• x is the independent variable
• k is the constant of variation, representing the constant ratio between y and x

For example, the distance traveled by a car at a constant speed varies directly with the time spent traveling. The faster the car goes (higher speed), the further it travels in the same amount of time.

Inverse Variation: As One Quantity Increases, the Other Decreases

Inverse variation, on the other hand, describes a relationship where an increase in one quantity leads to a decrease in another, and vice versa. Think of it as a seesaw – as one side goes up, the other side goes down. The mathematical representation of inverse variation is:

y = k/x

where:

• y is inversely proportional to x
• x is the independent variable
• k is the constant of variation, representing the constant product of y and x

An example of inverse variation is the time it takes to complete a task and the number of people working on it. The more people working on the task, the less time it takes to finish.

Problem Statement: Unraveling the Relationship Between s, t, and w

Now that we've established the groundwork for direct and inverse variation, let's address the specific problem we aim to solve. The problem states:

The variable s varies directly as t and inversely as w. If s = 30 when t = 10, and w = 1, what equation best represents this variation?

A. s = rac{3t}{w} B. s = rac{3w}{t} C. $st = 3$ D. s = rac{t}{3w}

This problem combines both direct and inverse variation, adding a layer of complexity. Our goal is to determine the equation that accurately captures the relationship between s, t, and w based on the given information. To solve this, we'll break down the problem into smaller steps, applying our knowledge of direct and inverse variation.

Step 1: Expressing the Combined Variation

The problem tells us that s varies directly as t and inversely as w. This means that s is proportional to t and inversely proportional to w. We can express this combined variation using the following equation:

s = k * (t/w)

where:

• s is the dependent variable
• t is a variable that s varies directly with
• w is a variable that s varies inversely with
• k is the constant of variation, which we need to determine

This equation captures the essence of the combined variation. As t increases, s also increases (direct variation), and as w increases, s decreases (inverse variation). The constant of variation, k, determines the specific strength of this relationship.

Step 2: Determining the Constant of Variation (k)

To find the equation that best represents the variation, we need to calculate the constant of variation, k. We are given the values of s, t, and w for a specific scenario: s = 30 when t = 10 and w = 1. We can substitute these values into our equation:

30 = k * (10/1)

Simplifying the equation, we get:

30 = 10k

Now, we can solve for k by dividing both sides of the equation by 10:

k = 30 / 10

k = 3

Therefore, the constant of variation, k, is 3. This value is crucial for defining the specific relationship between s, t, and w.

Step 3: Substituting k into the Equation

Now that we've determined the constant of variation, k, we can substitute it back into our combined variation equation:

s = 3 * (t/w)

This equation can be rewritten as:

s = (3t)/w

This equation represents the specific relationship between s, t, and w, given the information provided in the problem. It states that s is equal to 3 times t divided by w. This equation perfectly captures the direct variation of s with t and the inverse variation of s with w.

Identifying the Correct Answer Choice

Now, let's compare our derived equation with the answer choices provided in the problem:

A. s = rac{3t}{w} B. s = rac{3w}{t} C. $st = 3$ D. s = rac{t}{3w}

By comparing our equation, s = (3t)/w, with the answer choices, we can clearly see that option A, s = rac{3t}{w}, matches our derived equation. Therefore, option A is the correct answer.

The other options do not represent the given variation:

• Option B, s = rac{3w}{t}, suggests that s varies directly with w and inversely with t, which is the opposite of what the problem states.
• Option C, $st = 3$, implies a relationship where the product of s and t is constant, which doesn't align with the combined direct and inverse variation described in the problem.
• Option D, s = rac{t}{3w}, suggests that s varies directly with t but inversely with 3w, which means that the constant of variation would be 1/3 and does not match the calculated k = 3.

Conclusion: The Equation that Binds s, t, and w

In conclusion, the equation that best represents the variation where s varies directly as t and inversely as w, given the specific conditions (s = 30 when t = 10 and w = 1), is:

A. s = rac{3t}{w}

This equation accurately captures the combined direct and inverse variation relationship between the variables. By understanding the fundamental concepts of variation and applying them systematically, we can solve complex problems and gain valuable insights into the relationships between different quantities. This approach is crucial in various fields, from mathematics and physics to economics and engineering.

This problem demonstrates the power of mathematical reasoning and the importance of understanding fundamental concepts. By breaking down the problem into smaller, manageable steps, we were able to identify the correct equation and gain a deeper understanding of the relationship between direct and inverse variation. The ability to analyze and solve such problems is a valuable skill in many areas of life.