Understanding Inverse Variation: A Step-by-Step Guide

Inverse variation is a fundamental concept in mathematics that describes a relationship between two variables where an increase in one leads to a decrease in the other, and vice-versa, in a specific, predictable way. When we say that variable $p$ varies inversely as the square of $q$, we're establishing a direct link between how these two quantities interact. This means that as $q$ gets larger, $p$ gets smaller, and as $q$ gets smaller, $p$ gets larger. The "square of $q$" part is crucial; it tells us that the relationship isn't just inversely proportional to $q$, but to $q^2$. This mathematical relationship can be expressed as an equation: $p = \frac{k}{q^2}$, where $k$ is a constant of variation. This constant, $k$, is the key to solving problems involving inverse variation. It represents the specific scaling factor for the given relationship. To find $k$, we use a pair of known values for $p$ and $q$. In this problem, we are given that when $p=36$, $q=25$. We can plug these values into our equation to solve for $k$: $36 = \frac{k}{(25)^2}$. Calculating $(25)^2$ gives us $625$. So, the equation becomes $36 = \frac{k}{625}$. To isolate $k$, we multiply both sides of the equation by $625$: $k = 36 \times 625$. Performing this multiplication, we find that $k = 22500$. Now that we have the value of our constant $k$, we have a complete equation that governs the relationship between $p$ and $q$: $p = \frac{22500}{q^2}$. This equation will allow us to find the value of one variable when the other is known. Understanding this initial setup is the bedrock of solving any inverse variation problem. It requires careful attention to the wording, particularly terms like "inversely as the square," and a systematic approach to solving for the constant of variation using the provided data. Once the constant is determined, the relationship is fully defined, paving the way for further calculations and predictions within the scope of this specific variation.

Finding $q$ When $p$ is Known

Now that we've established the governing equation for our inverse variation, $p = \frac{22500}{q^2}$, we can confidently tackle the first question: "When $p=4$, what is $q$?". Our goal here is to isolate $q$. We start by substituting the given value of $p$ into our equation: $4 = \frac{22500}{q^2}$. To solve for $q$, we need to get $q^2$ out of the denominator and by itself. A common strategy is to multiply both sides of the equation by $q^2$: $4q^2 = 22500$. Now, we want to isolate $q^2$. We do this by dividing both sides by $4$: $q^2 = \frac{22500}{4}$. Performing the division, we find that $q^2 = 5625$. The final step to find $q$ is to take the square root of both sides: $q = \sqrt{5625}$. Calculating the square root of $5625$, we find that $q = 75$. It's important to remember that when taking the square root, there are technically two possible answers (a positive and a negative one). However, in many practical contexts, we consider the positive root, especially if $q$ represents a physical quantity. So, when $p=4$, $q=75$. This process illustrates how the inverse square variation works in practice. A significant decrease in $p$ (from $36$ to $4$) corresponds to a change in $q$ that is not linear but rather related to the square root of the inverse change in $p$. The initial value of $p=36$ and $q=25$ allowed us to find the constant $k=22500$. This constant is the heart of the relationship. With $p=4$, we found that $q^2$ had to be $5625$ to satisfy the equation, leading us to $q=75$. This shows that a decrease in $p$ requires an increase in $q$, and the specific way $q$ increases is related to the square root of the decrease in $p$ relative to the constant.

Calculating $p$ When $q$ is Known

Let's now address the second part of the problem: "When $q=10$, what is $p$?". With our established equation $p = \frac{22500}{q^2}$, this calculation is more straightforward. We simply substitute the given value of $q$ into the equation: $p = \frac{22500}{(10)^2}$. First, we calculate the square of $q$: $(10)^2 = 100$. Now, substitute this back into the equation: $p = \frac{22500}{100}$. Performing the division, we find that $p = 225$. This result demonstrates the inverse relationship clearly. When $q$ decreases from $25$ to $10$, $p$ increases from $36$ to $225$. The change in $p$ is not a simple multiplication or division; it's directly influenced by the square of $q$. In this case, $q$ was reduced by a factor of $2.5$ (from $25$ to $10$), but since $p$ varies inversely with the square of $q$, $p$ increases by a factor of $(2.5)^2 = 6.25$. To verify this: $36 \times 6.25 = 225$. This confirms our calculation and reinforces the understanding of the inverse square relationship. The options provided for this part of the question are A. 75, B. 225, C. 5,625. Our calculated value for $p$ is $225$, which corresponds to option B. The value $75$ is what we found for $q$ in the previous step, and $5625$ is the value of $q^2$ when $p=4$. Therefore, when $q=10$, $p$ is $225$. This detailed breakdown shows how to use the constant of variation to predict values of $p$ or $q$ given the other, always keeping in mind the inverse square relationship.

Conclusion: Mastering Inverse Variation

In summary, understanding inverse variation, especially when involving squares, is a powerful tool in mathematics. We began by defining the relationship $p = \frac{k}{q^2}$ and used the initial conditions ($p=36, q=25$) to find the constant of variation, $k=22500$. This led us to the specific equation $p = \frac{22500}{q^2}$. We then applied this equation to solve for unknown variables. First, when $p=4$, we found $q=75$. This involved rearranging the equation to solve for $q^2$ and then taking the square root. Second, when $q=10$, we calculated $p=225$. This was a more direct substitution into the equation. The exercise highlighted how changes in one variable necessitate specific, squared-inverse changes in the other, governed by the constant $k$. The relationship $p \propto \frac{1}{q^2}$ means that if $q$ is multiplied by a factor of $x$, $p$ is divided by a factor of $x^2$. Conversely, if $q$ is divided by a factor of $x$, $p$ is multiplied by a factor of $x^2$. This is precisely what we observed. The ability to solve these types of problems is crucial in many scientific and engineering fields where relationships between quantities are often not linear. For further exploration into variations and proportional reasoning, you can visit resources like Khan Academy.