Problem 49 - Algebra 2

Question

Suppose that \(y\) varies inversely with the square of \(x,\) and \(y=50\) when \(x=4\) . Find \(y\) when \(x=5 .\)

Step-by-Step Solution

Answer

When \(x = 5\), \(y = 32\).

Step 1: Find the constant of variation

You begin with the general formula for inverse square variation \(y = k/x^2\). To find the constant of variation \(k\), substitute the given values of \(y = 50\) and \(x = 4\) into the formula. This gives you \(50 = k/(4^2)\), which simplifies to \(50 = k/16\). By mathematical principles, to find \(k\), you multiply both sides by 16, which gives \(k= 800\).

Step 2: Compute y when x=5

Substitute the found value of \(k = 800\) in the variation formula \(y = k/x^2\) to find \(y\) when \(x = 5\). This gives \(y = 800/5^2 = 800/25 = 32\).

Step 3: Interpret the result

The result value corresponds to the value of \(y\) when \(x = 5\) in the context of the problem. In this case, \(y = 32\). So, the value of \(y\) when \(x = 5\) is 32. This indicates that \(y\) decreases as \(x\) increases, in line with the properties of inverse square variation.

Key Concepts

Constant of Variation

The constant of variation, often denoted as

• k, is a crucial component in any form of mathematical variation.
• In this context, it helps maintain a consistent relationship between two variables.

For inverse variations, the constant establishes a bond between variables even as they change in values. In our example, the relationship is given by the equation \[ y = \frac{k}{x^2}\], which describes how one variable, y, varies inversely with the square of another, x.

To find k, you use the known values of the variables. Here, when x = 4 and y = 50, substituting these into the equation gives:\[50 = \frac{k}{4^2} \]Solving this equation provides k = 800.
This constant serves as a fixed point, allowing us to predict how y will change with different values of x. It's fundamental to consistently understanding and applying inverse variation formulas.

Inverse Square Variation

Inverse square variation describes how one quantity varies inversely with the square of another quantity.

• In the formula y = \frac{k}{x^2}, y decreases as the square of x increases.
• This means that when x becomes larger, y becomes smaller at a rate defined by the squaring of x.

This concept is especially relevant in physics with laws like gravity or light intensity, where effects diminish with the square of the distance.In this particular problem, the value of y when x = 5 is derived using the constant k. So, with k = 800, at x = 5: \[y = \frac{800}{5^2} = \frac{800}{25} = 32\]This illustrates how an increase in x results in a decrease in y, maintaining their inverse relationship.

Algebraic Manipulation

Algebraic manipulation allows us to rearrange equations to solve for a particular variable or constant. This skill is essential when dealing with any form of mathematical variation.
Here, algebraic manipulation was used twice:

• First, to find the constant of variation k by isolating it on one side of the given equation.
• Second, to compute a new value of y for different x using the established relationship.

For solving the exercise, the steps involved rearranging the equation \[50 = \frac{k}{16}\]to solve for k. This was done by multiplying both sides by 16:\[k = 50 \times 16 = 800\]Similarly, to find y when x = 5, k = 800 was substituted back into the original variation formula. The resulting equation\[y = \frac{800}{5^2}\]was simplified to find y = 32.Mastering algebraic manipulation in inverse variation problems enables solving complex relationships efficiently.