The Quantity Y Is Partly Constant And Partly Varies Inversely As The Square Of X.
In the realm of mathematics, particularly in algebra and calculus, understanding the relationships between variables is crucial for solving a wide range of problems. One such relationship is inverse variation, where one quantity decreases as another increases, and vice versa. This article delves into the intricacies of inverse variation, using a specific problem as a case study to illustrate the concepts and techniques involved. We will explore how to express inverse variation mathematically, how to solve for unknown constants, and how to apply these principles to real-world scenarios.
Deciphering the Problem: Quantity y, Constant, and Inverse Variation
The given problem presents a scenario where a quantity, denoted as y, exhibits a dual nature. It is partly constant, meaning a portion of its value remains unchanged regardless of the value of another variable, x. Simultaneously, y partly varies inversely as the square of x, indicating that another portion of y changes in inverse proportion to the square of x. To unravel this complex relationship, we need to break it down into its fundamental components and express them mathematically.
Expressing the Relationship Between y and x Mathematically
The first step in solving this problem is to translate the verbal description into a mathematical equation. Since y is partly constant, we can represent this constant portion as k, where k is a constant value. The phrase "partly varies inversely as the square of x" implies that there is a term that is inversely proportional to x2. We can represent this term as c/ x2, where c is another constant of proportionality. Combining these two components, we arrive at the following equation:
y = k + c/ x2
This equation encapsulates the relationship between y and x, where y is the sum of a constant term (k) and a term that varies inversely as the square of x (c/ x2). The constants k and c are crucial parameters that determine the specific behavior of this relationship. To fully understand this relationship, we need to determine the values of these constants.
Unveiling the Constants: Utilizing Given Conditions
The problem provides us with two sets of conditions that will allow us to solve for the constants k and c. These conditions act as data points that we can plug into our equation to create a system of equations. The first condition states that when x = b, y = f1. Substituting these values into our equation, we get:
f1 = k + c/ b2
The second condition states that when x = 2b, y = f2. Similarly, substituting these values into our equation, we get:
f2 = k + c/ (2b)2
Simplifying the second equation, we have:
f2 = k + c/ 4b2
Now we have a system of two equations with two unknowns (k and c):f1 = k + c/ b2
f2 = k + c/ 4b2
Solving this system of equations will give us the values of k and c, thereby fully defining the relationship between y and x.
Solving the System of Equations: A Step-by-Step Approach
To solve this system of equations, we can use various methods, such as substitution or elimination. Here, we will use the elimination method. Subtracting the second equation from the first equation, we eliminate k:
f1 - f2 = (k + c/ b2) - (k + c/ 4b2)
Simplifying, we get:
f1 - f2 = 3c/ 4b2
Solving for c, we have:
c = 4b2 (f1 - f2) / 3
Now that we have found c, we can substitute it back into either of the original equations to solve for k. Substituting c into the first equation:
f1 = k + (4b2 (f1 - f2) / 3) / b2
Simplifying, we get:
f1 = k + 4(f1 - f2) / 3
Solving for k, we have:
k = f1 - 4(f1 - f2) / 3
k = (3f1 - 4f1 + 4f2) / 3
k = (4f2 - f1) / 3
Thus, we have found the values of both constants, k and c. Substituting these values back into our original equation, we obtain the complete relationship between y and x:
y = ((4f2 - f1) / 3) + (4b2 (f1 - f2) / 3) / x2
Putting it all Together: The Complete Solution
Now that we have solved for the constants k and c, we can express the relationship between y and x in its entirety. The equation we derived, y = ((4f2 - f1) / 3) + (4b2 (f1 - f2) / 3) / x2, provides a complete mathematical description of how y varies with x. This equation can be used to predict the value of y for any given value of x, and vice versa.
Exploring the Implications: Real-World Applications of Inverse Variation
Inverse variation is not just a mathematical concept; it has numerous applications in the real world. From physics to economics, inverse relationships are prevalent in various phenomena. For instance, the relationship between pressure and volume of a gas at constant temperature is an example of inverse variation. As the volume of the gas decreases, the pressure increases proportionally. Similarly, in economics, the relationship between price and demand often exhibits inverse variation. As the price of a product increases, the demand for that product tends to decrease.
Conclusion: Mastering Inverse Variation for Problem-Solving
In conclusion, the problem of quantity y being partly constant and partly varying inversely as the square of x provides a valuable case study for understanding inverse variation. By breaking down the problem into its components, expressing the relationship mathematically, solving for unknown constants, and exploring real-world applications, we have gained a comprehensive understanding of this important concept. Mastering inverse variation is essential for problem-solving in various fields, and this article serves as a solid foundation for further exploration and application.
Let's clarify the given mathematical problem concerning quantity y and its relationship with x. Here's a breakdown of the original problem statement and a slightly rephrased version for better understanding:
Original Problem:
The quantity y is partly constant and partly varies inversely as the square of x.
(i) Write down the relationship between y and x.
(ii) When x = b, y = f1, and when x = 2b, y = f2.
Rephrased Problem:
The value of y is determined by two components: a constant value and a value that varies inversely with the square of x.
(i) Express the relationship between y and x in an equation.
(ii) Given that y = f1 when x = b, and y = f2 when x = 2b, use these conditions to further define the relationship.
The rephrased version emphasizes the two components contributing to y's value and uses clearer language like "express in an equation" instead of "write down the relationship." This makes the problem statement more accessible while retaining the original meaning.
Key Improvements in the Rephrased Version
- Clearer Language: Replacing "write down the relationship" with "express the relationship in an equation" is more direct and specific.
- Emphasis on Components: The rephrased version explicitly states that y is determined by two components, making the problem's structure clearer.
- Concise Wording: Minor adjustments like replacing "When x = b, y = f1, and when x = 2b, y = f2" with "Given that y = f1 when x = b, and y = f2 when x = 2b" make the statement more concise.
By making these subtle changes, the rephrased problem statement is easier to understand and sets the stage for solving the problem effectively.