Mastering Direct Variation Problem Solving A Comprehensive Guide

Hey guys! Ever stumbled upon a math problem that looks simple but has a sneaky little twist? Well, buckle up because we're about to dissect one such problem involving direct variation. We'll not only pinpoint the mistake in the given solution but also embark on a comprehensive journey to truly understand the concept of direct variation and how to tackle these problems like seasoned math pros. So, let's dive in and unravel this mathematical mystery together!

The Problem: Direct Variation in Action

Before we jump into the solution, let's revisit the problem. It states that the variable s varies directly as the square of t. This is a crucial piece of information, as it sets the foundation for our entire approach. We are given that when s = 4, t = 12. The ultimate goal? To find the value of t when s = 48. A student, Nick, attempted to solve this, and his work is laid out for us to examine. Our mission is to not only identify where Nick went wrong but also to provide a crystal-clear, step-by-step solution that leaves no room for doubt.

Understanding Direct Variation: Direct variation, at its core, describes a relationship between two variables where one variable is a constant multiple of the other. In simpler terms, as one variable increases, the other increases proportionally, and vice versa. This relationship is mathematically expressed as y = kx, where y and x are the variables, and k is the constant of variation. This constant, k, is the key to unlocking the relationship between the variables. It tells us how much y changes for every unit change in x. In our specific problem, we have s varying directly as the square of t. This means the relationship is expressed as s = kt², where k is still the constant of variation, but now s is proportional to the square of t. This subtle difference is crucial and often a source of errors if overlooked. Remember, the square in t² significantly impacts how s changes with respect to t. For instance, if t doubles, s will quadruple (since 2² = 4). This non-linear relationship is what makes direct variation with squares a bit more intricate than simple direct variation. Now that we've refreshed our understanding of direct variation and its nuances, let's move on to dissecting Nick's solution and pinpointing the exact location of the error.

Dissecting Nick's Solution: Spotting the Mistake

Nick's work starts with the correct general equation for direct variation: s = kt². However, the critical error appears in the very next step. Nick substitutes s = 12 and t = 4 into the equation. This is where the mistake lies! The problem clearly states that s = 4 when t = 12, not the other way around. This incorrect substitution throws off the entire calculation. By plugging in the wrong values, Nick ends up calculating an incorrect constant of variation (k), which subsequently leads to a wrong answer for t when s = 48. It's like starting a journey with the wrong coordinates – you're bound to end up at the wrong destination. This seemingly small error highlights the importance of meticulous attention to detail when solving mathematical problems. Each value must be carefully placed in its corresponding variable to ensure the accuracy of the solution. The subsequent steps in Nick's solution, while mathematically correct in their execution (solving for k and then for t), are based on this flawed foundation, rendering the final answer incorrect. Let's break down why this initial error is so impactful. The constant of variation, k, is the bridge that connects s and t². It defines the specific relationship between these variables for this particular problem. By miscalculating k, we essentially distort this relationship, leading to inaccurate predictions of s for given values of t, and vice versa. It's like having a distorted map – it might look familiar, but it will lead you astray. Therefore, identifying and rectifying this initial substitution error is paramount to arriving at the correct solution. Now that we've pinpointed the mistake, let's move on to constructing the correct solution, step-by-step, ensuring clarity and accuracy at every stage.

The Correct Solution: A Step-by-Step Guide

Alright, guys, let's ditch the wrong turn and get back on the right track! To solve this problem accurately, we need to follow a precise, step-by-step approach. First, we start with the fundamental equation of direct variation, which, as we've established, is s = kt². This equation is our roadmap, guiding us towards the solution. Next, we carefully substitute the given values of s and t to determine the constant of variation, k. Remember, s = 4 when t = 12. Plugging these values into our equation, we get: 4 = k(12)². This is the crucial substitution, and it's imperative that we get it right. Now, let's simplify the equation and solve for k. We have 4 = k * 144. To isolate k, we divide both sides of the equation by 144, resulting in k = 4 / 144, which simplifies to k = 1/36. This value of k is the constant of variation for this specific problem, and it represents the specific relationship between s and t². With k in hand, we're halfway there! We now know the exact equation that governs the relationship between s and t: s = (1/36)t². The final step is to use this equation to find the value of t when s = 48. We substitute s = 48 into the equation: 48 = (1/36)t². To solve for t², we multiply both sides of the equation by 36, giving us t² = 48 * 36 = 1728. Finally, to find t, we take the square root of both sides: t = √1728. Now, let's simplify the square root. We can factor 1728 as 576 * 3, where 576 is a perfect square (24²). Therefore, t = √(576 * 3) = √576 * √3 = 24√3. So, the value of t when s = 48 is 24√3. There you have it! We've successfully navigated the problem, identified the mistake, and arrived at the correct solution. But remember, the journey doesn't end here. Understanding the underlying concepts is just as important as getting the right answer.

Mastering Direct Variation: Key Takeaways

Okay, guys, we've cracked this specific problem, but the real victory lies in understanding the broader concepts. So, let's recap the key takeaways to truly master direct variation problems. First and foremost, always start with the correct general equation. Identifying the relationship between the variables is the cornerstone of solving direct variation problems. Whether it's y = kx, s = kt², or any other variation, understanding the underlying equation is crucial. Next, pay meticulous attention to detail when substituting values. As we saw in Nick's solution, a simple mix-up in substitution can lead to a completely incorrect answer. Double-check and triple-check that you're plugging the values into the correct variables. This might seem like a small thing, but it can make all the difference. Another crucial aspect is understanding the meaning of the constant of variation, k. This constant is the bridge that connects the variables, defining their specific relationship in the problem. Calculating k accurately is essential for solving for unknown values. Furthermore, practice, practice, practice! The more you work through different types of direct variation problems, the more comfortable and confident you'll become. Try variations with squares, cubes, and even inverse variations. Challenge yourself to apply the concepts in different contexts. Don't just memorize the steps; understand the reasoning behind them. This will allow you to adapt your approach to tackle any direct variation problem that comes your way. Finally, don't be afraid to break down complex problems into smaller, manageable steps. As we demonstrated in the solution, systematically working through each step, from identifying the equation to simplifying the final answer, can make even the most challenging problems seem less daunting. Remember, mathematics is a journey, not a sprint. Embrace the challenges, learn from your mistakes, and celebrate your successes. With a solid understanding of the fundamental concepts and a meticulous approach, you'll be well-equipped to conquer any direct variation problem that crosses your path. So, keep practicing, keep exploring, and keep unlocking the beauty of mathematics!

Practice Problems: Put Your Skills to the Test

Alright, guys, now that we've dissected the problem and laid out the key takeaways, it's time to put your newfound knowledge to the test! Practice is the key to mastery, so let's dive into a couple of practice problems to solidify your understanding of direct variation. These problems will challenge you to apply the concepts we've discussed and hone your problem-solving skills. Remember, the goal isn't just to get the right answer; it's to understand the process and the underlying principles. So, grab a pencil and paper, and let's get started!

Practice Problem 1:

The variable y varies directly as the cube of x. When y = 24, x = 2. Find the value of y when x = 3.

Practice Problem 2:

The variable p varies directly as the square of q. When p = 5, q = 10. Find the value of q when p = 20.

These problems are designed to reinforce your understanding of the concepts we've covered. Remember to start by identifying the correct general equation, then carefully substitute the given values to find the constant of variation, k. Once you have k, you can use the equation to solve for the unknown variable. Don't be afraid to break down the problems into smaller steps and work through them systematically. And most importantly, don't be discouraged if you encounter challenges along the way. Every mistake is an opportunity to learn and grow. After you've given these problems a shot, check your answers and review your work. If you're unsure about any part of the solution, revisit the key takeaways and the step-by-step guide we discussed earlier. You can also try working through similar problems from textbooks or online resources. The more you practice, the more confident you'll become in your ability to tackle direct variation problems. So, keep at it, and remember to embrace the challenge! Now, let's move on to discussing the solutions to these practice problems, so you can check your work and ensure you're on the right track.

Solutions to Practice Problems: Check Your Understanding

Alright, guys, let's see how you did with those practice problems! It's time to unveil the solutions and check your understanding. Remember, the goal is not just to have the right answer, but also to understand the process and reasoning behind it. So, let's dive in and break down the solutions step-by-step.

Solution to Practice Problem 1:

Problem: The variable y varies directly as the cube of x. When y = 24, x = 2. Find the value of y when x = 3.

Solution: First, we identify the general equation: y = kx³. Next, we substitute the given values to find k: 24 = k(2)³. Simplifying, we get 24 = 8k. Dividing both sides by 8, we find k = 3. Now we have the specific equation: y = 3x³. Finally, we substitute x = 3 to find y: y = 3*(3)³ = 3 * 27 = 81. Therefore, the value of y when x = 3 is 81.

Solution to Practice Problem 2:

Problem: The variable p varies directly as the square of q. When p = 5, q = 10. Find the value of q when p = 20.

Solution: First, we identify the general equation: p = kq². Next, we substitute the given values to find k: 5 = k(10)². Simplifying, we get 5 = 100k. Dividing both sides by 100, we find k = 1/20. Now we have the specific equation: p = (1/20)q². Finally, we substitute p = 20 to find q: 20 = (1/20)q². Multiplying both sides by 20, we get q² = 400. Taking the square root of both sides, we find q = ±20. Therefore, the values of q when p = 20 are 20 and -20.

How did you do? If you got the correct answers, congratulations! You've demonstrated a solid understanding of direct variation. If you encountered any difficulties, don't worry! Review the steps we took to solve the problems and identify any areas where you might have made a mistake. Perhaps you need to revisit the concept of substituting values, or maybe you need to brush up on your algebra skills. Whatever the case, remember that learning is a process, and every challenge is an opportunity to grow. If you're still feeling unsure, try working through similar problems from textbooks or online resources. You can also seek help from a teacher, tutor, or classmate. Don't be afraid to ask questions and clarify any concepts that are unclear. The more you practice and the more you seek out understanding, the more confident you'll become in your ability to tackle direct variation problems. So, keep up the great work, and remember to celebrate your progress! You're well on your way to mastering direct variation and other mathematical concepts. Now that we've wrapped up the practice problems and their solutions, let's move on to a final recap of the key takeaways from this deep dive into direct variation.

Fix the input keyword: What was Nick's mistake and what is the correct value of t when s=48, given that s varies directly as the square of t, s=4 when t=12?