Solving Direct Variation Problems A Step By Step Guide
The problem at hand involves direct variation, a fundamental concept in mathematics. Understanding direct variation is crucial for solving a variety of problems in algebra and beyond. In this case, the variable s varies directly as the square of t. This means that there's a constant k such that s = kt². We are given that when s = 4, t = 12. Nick attempts to find the value of t when s = 48. Let's analyze Nick's work step-by-step and identify any potential errors, and then delve into the correct methodology for solving this type of problem. This exploration will not only help us find the right answer but also solidify our understanding of direct variation and its applications. We will cover the initial setup, the calculation of the constant of variation, and the final step of solving for the unknown variable t. By the end of this discussion, you will have a clear grasp of how to approach direct variation problems with confidence.
Analyzing Nick's Work
Nick's work starts with the correct general equation for direct variation: s = kt². This is the foundational relationship that we need to work with. However, the subsequent steps reveal a critical error in substituting the given values. Nick substitutes s = 12 and t = 4 into the equation, which is incorrect. The problem states that s = 4 when t = 12. This initial mistake propagates through the rest of the solution, leading to an incorrect value for k and ultimately an incorrect value for t. To emphasize the importance of accurate substitution, let's reiterate the given information: when s = 4, t = 12. Using these values correctly is paramount to finding the accurate constant of variation and the final answer. Before we proceed with the correct solution, it's essential to understand why this initial error is so significant. In direct variation problems, the constant of variation, k, defines the specific relationship between the variables. An incorrect k value will distort this relationship, leading to a flawed solution. Therefore, meticulous attention to detail during substitution is absolutely necessary.
The Correct Approach to Solving Direct Variation Problems
To correctly solve this problem, we need to follow a systematic approach. The first step, as Nick correctly identified, is to write the general equation for direct variation: s = kt². This equation represents the fundamental relationship between s and t. The next crucial step is to substitute the given values (s = 4 and t = 12) into the equation. This will allow us to solve for the constant of variation, k. Substituting these values, we get: 4 = k(12)². Now, we need to simplify the equation and isolate k. 12² is equal to 144, so the equation becomes 4 = 144k. To solve for k, we divide both sides of the equation by 144: k = 4/144. Simplifying this fraction, we find that k = 1/36. This is the correct constant of variation for this particular problem. Now that we have the value of k, we can move on to the final step: finding the value of t when s = 48. We will use the same direct variation equation, s = kt², but this time we will substitute s = 48 and k = 1/36. This will give us an equation that we can solve for t. The methodical approach ensures accuracy and clarity in solving direct variation problems.
Solving for t when s = 48
Now that we have the correct constant of variation, k = 1/36, we can proceed to find the value of t when s = 48. We substitute these values into the direct variation equation, s = kt², resulting in: 48 = (1/36)t². To solve for t², we multiply both sides of the equation by 36: 48 * 36 = t². This gives us t² = 1728. To find t, we take the square root of both sides of the equation: t = ±√1728. Now, we need to simplify the square root. We can factor 1728 as 576 * 3, and the square root of 576 is 24. Therefore, t = ±24√3. This means there are two possible values for t: 24√3 and -24√3. In many real-world applications, we would consider the context of the problem to determine if both solutions are valid or if only the positive solution makes sense. For example, if t represents time, a negative value might not be physically meaningful. However, in this purely mathematical context, both solutions are valid. This final step highlights the importance of not only algebraic manipulation but also the interpretation of results in the context of the problem.
Common Mistakes and How to Avoid Them
Direct variation problems, while conceptually straightforward, can be tricky if certain common mistakes are made. One of the most frequent errors, as seen in Nick's work, is the incorrect substitution of given values. It's crucial to carefully read the problem statement and ensure that the correct values are assigned to the corresponding variables. A simple way to avoid this is to double-check the substitution before proceeding with the calculations. Another common mistake is in the algebraic manipulation of the equation. For instance, when solving for k or t, it's essential to perform the same operations on both sides of the equation to maintain equality. This includes correctly applying the order of operations (PEMDAS/BODMAS) and paying attention to signs (positive and negative). Additionally, when taking the square root to solve for a variable, remember to consider both positive and negative solutions, as we saw in this problem. Overlooking the negative solution can lead to an incomplete answer. To avoid these pitfalls, practice is key. Working through a variety of direct variation problems will help you develop a strong understanding of the concepts and the necessary problem-solving skills. Furthermore, it's always a good idea to check your work, either by plugging your solution back into the original equation or by using a different method to solve the problem.
The Importance of Understanding Direct Variation
Understanding direct variation is not just about solving textbook problems; it's a fundamental concept that has wide-ranging applications in various fields. In mathematics and science, direct variation is used to model relationships between quantities that change proportionally. For example, the distance traveled at a constant speed varies directly with time, and the force required to stretch a spring varies directly with the amount of stretch (Hooke's Law). In economics, direct variation can be used to model the relationship between supply and demand. In everyday life, we encounter direct variation in situations such as calculating the cost of buying multiple items at the same price or determining the amount of ingredients needed when scaling a recipe. The ability to recognize and work with direct variation is therefore a valuable skill. It allows us to make predictions, solve practical problems, and gain a deeper understanding of the world around us. Mastering direct variation lays the groundwork for more advanced mathematical concepts, such as inverse variation and other types of proportional relationships. By developing a solid foundation in direct variation, you'll be well-equipped to tackle a wide range of mathematical and real-world challenges. The key takeaways are careful substitution, accurate algebraic manipulation, and the consideration of all possible solutions.
Conclusion
In summary, the problem we analyzed highlights the importance of a thorough understanding of direct variation and the need for careful attention to detail when solving mathematical problems. Nick's initial error in substituting the values for s and t demonstrates how a small mistake early on can lead to an incorrect solution. By correctly applying the principles of direct variation, we were able to determine the constant of variation, k, and subsequently solve for the value of t when s = 48. The correct solution involves recognizing the direct variation relationship, setting up the equation s = kt², substituting the given values to find k, and then using k to solve for the unknown variable. We also discussed common mistakes, such as incorrect substitution and overlooking negative solutions, and how to avoid them. Furthermore, we emphasized the broad applicability of direct variation in various fields, underscoring its importance as a fundamental mathematical concept. By mastering direct variation, you'll not only improve your problem-solving skills but also gain a valuable tool for understanding and modeling relationships in the real world. Remember to practice consistently, double-check your work, and always consider the context of the problem when interpreting your results.
