Problem 26 - College Algebra

Question

For the following exercises, use the given information to find the unknown value. \(y\) varies directly as the cube of \(x\). When \(x=3\), then \(y=5 .\) Find \(y\) when \(x=4\).

Step-by-Step Solution

Answer

When \(x=4\), \(y \approx 11.85\).

Step 1: Understand Direct Variation

When we say that \(y\) varies directly as the cube of \(x\), it means that \(y = kx^3\), where \(k\) is a constant. This is the formula for direct variation with the cube of \(x\).

Step 2: Substitute Known Values

We know that when \(x = 3\), \(y = 5\). Substitute these values into the equation to find \(k\): \[5 = k(3)^3\]which simplifies to:\[5 = k imes 27\].

Step 3: Solve for the Constant \(k\)

Rearrange the equation to solve for \(k\):\[k = \frac{5}{27}\].

Step 4: Substitute \(k\) and New \(x\) Value to Find \(y\)

Now that we know \(k = \frac{5}{27}\), substitute \(x = 4\) into the variation equation \(y = kx^3\):\[y = \frac{5}{27} imes 4^3\]\[y = \frac{5}{27} imes 64\].

Step 5: Perform the Calculation

Compute \(y\):\[y = \frac{5 imes 64}{27}\]\[y = \frac{320}{27}\]This simplifies to approximately \(11.85\).

Key Concepts

Cube of a Number

A cube of a number refers to multiplying a number by itself three times. It's represented as raising the number to the power of three, like this: \[x^3 = x \times x \times x\]This operation is very common in direct variation equations, especially when one quantity changes as the cube of another quantity. In our exercise, the term "cube of \(x\)" is crucial to understanding how \(y\) is influenced. Here, \(y\) changes depending on the cube of \(x\), or \(x^3\). To easily compute the cube of a number, simply multiply the number by itself, and then multiply the result by the number again. For instance, if \(x = 4\), then \(x^3 = 4 \times 4 \times 4 = 64\). This repetition symbolizes the profound way the variable \(x\) impacts \(y\) in scenarios involving cubes. Remember, larger numbers will yield exponentially larger results when cubed.

Constants in Equations

In equations involving direct variation, especially with more complex patterns like cubes, constants serve a critical role. A constant is a fixed value that does not change; it represents the relationship between the variables involved. In the equation \(y = kx^3\), \(k\) is our constant.To find the constant \(k\), we need known values of \(x\) and \(y\). From the problem, when \(x = 3\) and \(y = 5\), we can substitute these into the equation:\[5 = k(3)^3\]Solving for \(k\), we find that \(k = \frac{5}{27}\).

• Constants simplify the relationship between varying quantities.
• They allow easy calculation when new variables are introduced.

Thus, constants ensure that even as \(x\) changes, the proportional relationship remains consistent.

Solving Equations

Solving equations, particularly those involving cubes and direct variation, involves a series of simple steps that bring consistency and clarity. Let's break down the process using our example.The original equation is given as \(y = kx^3\). Once we determine the constant \(k\), we substitute it back into the equation with the new \(x\) value. For this problem, substituting \(k = \frac{5}{27}\) and \(x = 4\) gives:\[y = \frac{5}{27} \times 4^3\]Knowing \(4^3 = 64\), the equation becomes:\[y = \frac{5}{27} \times 64\]Carrying out this multiplication provides:\[y = \frac{320}{27}\]

• Solve step-by-step to find \(y\).
• Break complex operations into manageable parts.

This solution, rounding to two decimal places, is approximately \(11.85\). By understanding these steps, students can solve equations systematically, making a complex problem far more approachable.