Problem 12 - Precalculus: Functions and Graphs

Question

Exer. 1-12: Express the statement as a formula that involves the given variables and a constant of proportionallty \(k,\) and then determine the value of \(k\) from the given conditions. \(y\) is directly proportional to the square of \(x\) and inversely proportional to the square root of \(z\). If \(x=5\) and \(z=16\) then \(y=10\)

Step-by-Step Solution

Answer

The value of the constant of proportionality \(k\) is \(\frac{8}{5}\).

Step 1: Identify Relationships

Understand that when a variable is directly proportional to another, it is described by multiplying with a constant, while inversely proportional means dividing with a constant. So, since \(y\) is directly proportional to the square of \(x\), describe it as \(y = k \cdot x^2\). Furthermore, \(y\) is inversely proportional to the square root of \(z\), so adjust the formula to \(y = \frac{k \cdot x^2}{\sqrt{z}}\).

Step 2: Insert Given Values

We are given that \(x=5\), \(z=16\), and \(y=10\). Substitute these values into the equation we derived: \(10 = \frac{k \cdot 5^2}{\sqrt{16}}\).

Step 3: Simplify the Equation

First, calculate \(5^2 = 25\) and \(\sqrt{16} = 4\). Substitute these values into the equation to get \(10 = \frac{k \cdot 25}{4}\).

Step 4: Solve for the Constant of Proportionality (k)

To isolate \(k\), multiply both sides of the equation by 4 to obtain \(40 = k \cdot 25\). Then, divide both sides of the equation by 25 to solve for \(k\), resulting in \(k = \frac{40}{25}\).

Step 5: Simplify the Constant (k)

Divide the numerator and the denominator of \(\frac{40}{25}\) by their greatest common divisor, which is 5. This simplifies to \(k = \frac{8}{5}\).

Key Concepts

Direct Proportionality

When two variables are in direct proportionality, they increase or decrease together at a consistent rate. This means if one variable increases, the other does too, in a way that maintains a constant ratio. Mathematically, if \(y\) is directly proportional to \(x^2\), then we can express it with the formula \(y = k \cdot x^2\). Here, \(k\) is the constant of proportionality, demonstrating how much \(y\) will increase as \(x^2\) increases.
Understanding direct proportionality helps us predict one variable's change based on another. For example, in physics, force is directly proportional to acceleration. Similarly, in our problem, as the square of \(x\) changes, the value of \(y\) will change proportionally depending on \(k\).
This relationship is common in mathematics and science, making it a vital concept for understanding various natural patterns and occurrences.

Inverse Proportionality

Inverse proportionality describes a relationship where one variable increases while the other decreases. If a variable \(y\) is inversely proportional to \(z\), it can be expressed as \(y = \frac{k}{z}\). This means as \(z\) increases, \(y\) decreases, and vice versa.
In the context of our exercise, \(y\) is inversely proportional to the square root of \(z\). Thus, this relationship is expressed as \(y = \frac{k \cdot x^2}{\sqrt{z}}\). Here, unlike direct proportionality, one value increases while the other decreases, maintaining a constant product. Inverse relationships are useful in many scientific calculations, such as Boyle's Law that relates the pressure and volume of a gas.

Constant of Proportionality

The constant of proportionality, \(k\), is a key element in both direct and inverse proportionality. It serves as a scaling factor that quantifies how the variables relate. In our exercise, \(k\) bridges the square of \(x\) and the square root of \(z\) with \(y\).
To find \(k\), we inserted the known values of \(x\), \(z\), and \(y\) into the formula \(y = \frac{k \cdot x^2}{\sqrt{z}}\). After simplifying and solving, we determined \(k\) as \(\frac{8}{5}\).
Values of \(k\) help comprehend the direction and strength of relationships, instrumental in various mathematical and scientific models.

Square Root

The square root operation is used to find a number which, when multiplied by itself, gives the original number. For example, the square root of 16 is 4, because \(4 \times 4 = 16\).
In this exercise, \(y\) is inversely proportional to the square root of \(z\). Thus, we use \(\sqrt{z}\) in our equation as \(\frac{k \cdot x^2}{\sqrt{z}}\). Breaking down complex numbers through square roots aids in simpler calculations and solvability.

• Square root functions help in interpreting diverse phenomena such as areas in geometry.
• Square roots are crucial in real-world applications like physics and engineering.

Problem Solving

Problem-solving in mathematics involves a series of systematic steps to find solutions. Our exercise illustrates these steps with a clear, logical progression:
1. Begin by identifying relationships between variables and express them in mathematical terms, like direct and inverse proportionality.
2. Use the given values to substitute into your equation, simplifying the process.
3. Simplify and rearrange the equation to isolate the desired variable or constant, like \(k\).
4. Solve for your unknown, checking your work for accuracy.
By using stepwise problem solving, complex problems become tractable and logical, fostering deeper understanding and skill development in mathematics.