Solving Direct And Inverse Variation Problems Finding Z

In mathematics, direct and inverse variations describe how quantities relate to each other. Understanding these relationships is crucial for solving various problems in algebra, physics, and other fields. This article delves into a problem involving both direct and inverse variation, providing a step-by-step solution and explaining the underlying concepts. Let's begin by defining direct and inverse variation.

Direct Variation: Two variables are said to be in direct variation if one variable increases as the other variable increases, or one variable decreases as the other decreases. Mathematically, if $z$ varies directly as $x$, we can write this relationship as $z = kx$, where $k$ is the constant of variation. This means that $z$ is directly proportional to $x$. For instance, the distance traveled at a constant speed varies directly with time; the longer you travel, the greater the distance covered.

Inverse Variation: Two variables are said to be in inverse variation if one variable increases as the other variable decreases, and vice versa. Mathematically, if $z$ varies inversely as $y$, we can write this relationship as $z = \frac{k}{y}$, where $k$ is the constant of variation. This means that $z$ is inversely proportional to $y$. An example of inverse variation is the time it takes to complete a job and the number of workers; the more workers you have, the less time it takes to finish the job, assuming everyone works at the same rate.

When dealing with problems involving both direct and inverse variation, we combine these concepts into a single equation. For example, if $z$ varies directly as $x$ and inversely as $y$, the relationship can be expressed as $z = \frac{kx}{y}$, where $k$ is the constant of variation. This constant is essential for understanding the specific relationship between the variables in a given problem. The first step in solving these types of problems typically involves finding the value of this constant using initial conditions.

The problem we're addressing states that $z$ varies directly as the square of $x$ and inversely as the cube of $y$. Mathematically, this can be written as:

$z = k \frac{x^2}{y^3}$

where:

• $z$ is the dependent variable
• $x$ is one independent variable
• $y$ is another independent variable
• $k$ is the constant of variation

We are given that $z = 171$ when $x = 8$ and $y = 9$. Our goal is to find the value of $z$ when $x = 9$ and $y = 6$. This type of problem requires a systematic approach involving identifying the relationships, finding the constant of variation, and applying the equation to new conditions.

To find the constant of variation $k$, we use the initial conditions provided: $z = 171$, $x = 8$, and $y = 9$. Substitute these values into the equation:

$171 = k \frac{8^2}{9^3}$

Now, we solve for $k$:

$171 = k \frac{64}{729}$

To isolate $k$, multiply both sides of the equation by $\frac{729}{64}$:

$k = 171 \times \frac{729}{64}$

$k = \frac{171 \times 729}{64}$

$k = \frac{124659}{64}$

So, the constant of variation $k$ is $\frac{124659}{64}$. This constant represents the specific relationship between $z$, $x$, and $y$ in this particular problem. Understanding and correctly calculating $k$ is vital, as it forms the basis for solving the rest of the problem.

Now that we have the value of $k$, we can rewrite the variation equation with the constant of variation included:

$z = \frac{124659}{64} \times \frac{x^2}{y^3}$

This equation represents the specific relationship between $z$, $x$, and $y$ for this problem. It's crucial to have this equation accurately stated before proceeding to the next step, where we'll use it to find the value of $z$ under new conditions. The correct equation allows us to confidently substitute the new values of $x$ and $y$ and calculate the corresponding value of $z$.

We are asked to find the value of $z$ when $x = 9$ and $y = 6$. We now substitute these values into the equation we derived in the previous step:

$z = \frac{124659}{64} \times \frac{9^2}{6^3}$

First, calculate the squares and cubes:

$z = \frac{124659}{64} \times \frac{81}{216}$

Now, we perform the multiplication:

$z = \frac{124659 \times 81}{64 \times 216}$

$z = \frac{10097379}{13824}$

Now, divide to find the value of $z$:

$z ≈ 730.421$

Finally, round the answer to two decimal places as requested:

$z ≈ 730.42$

Therefore, when $x = 9$ and $y = 6$, the value of $z$ is approximately 730.42. This result demonstrates how changes in $x$ and $y$ affect $z$ based on the established direct and inverse relationships.

To verify our answer, we can plug the values back into the original equation and see if they hold true. Although we won't perform the full verification here, it's a good practice to ensure the result makes sense within the context of the problem.

In conclusion, when $x = 9$ and $y = 6$, the value of $z$ is approximately 730.42. This problem demonstrates the application of direct and inverse variation principles. By systematically finding the constant of variation and substituting the given values, we can solve for unknown variables in these relationships. Mastering these concepts is essential for further studies in mathematics and related fields.

This step-by-step solution provides a clear understanding of how to tackle problems involving direct and inverse variation. It highlights the importance of each step, from setting up the initial equation to finding the constant of variation and finally solving for the desired variable. Such problems are common in introductory algebra and physics courses, and a firm grasp of these concepts is crucial for success in these areas.

Let's summarize the steps we took to solve this problem:

1. Identify the relationship: Express the given variation in the form of an equation. In this case, $z = k \frac{x^2}{y^3}$.
2. Find the constant of variation (k): Substitute the initial values of $z$, $x$, and $y$ into the equation and solve for $k$.
3. Substitute k into the equation: Rewrite the equation with the calculated value of $k$.
4. Find z with new values of x and y: Substitute the new values of $x$ and $y$ into the equation and solve for $z$.
5. Round the answer: If necessary, round the final answer to the specified number of decimal places.
6. Verification: Although not explicitly shown, verify the answer by plugging the values back into the original equation.

To further enhance your understanding of direct and inverse variation, consider trying these practice problems:

1. $a$ varies directly as $b$ and inversely as the square of $c$. If $a = 5$ when $b = 10$ and $c = 2$, find $a$ when $b = 15$ and $c = 3$.
2. The volume $V$ of a gas varies directly as the temperature $T$ and inversely as the pressure $P$. If $V = 200$ cubic centimeters when $T = 300$ Kelvin and $P = 1.5$ atmospheres, find the volume when $T = 350$ Kelvin and $P = 2$ atmospheres.
3. The intensity of illumination $I$ from a light source varies inversely as the square of the distance $d$ from the source. If the intensity is 25 lux at a distance of 2 meters, find the intensity at a distance of 5 meters.

Working through these problems will solidify your understanding and improve your problem-solving skills in this area. Remember to follow the steps outlined above and pay close attention to the details of each problem. Understanding and applying these concepts effectively is invaluable in mathematics and various scientific fields.

Solving problems involving direct and inverse variation requires a clear understanding of the relationships between variables and a systematic approach. By following the steps outlined in this article, you can confidently tackle these types of problems. Remember to identify the type of variation, set up the equation, find the constant of variation, and substitute the given values to solve for the unknown variable. Practice is key to mastering these concepts, so be sure to work through additional problems to solidify your understanding. Direct and inverse variation are fundamental concepts that have wide applications in various fields, making their mastery crucial for academic and professional success.