Problem 45 - Algebra and Trigonometry

Question

Describe in words the variation shown by the given equation. $$z=\frac{k \sqrt{x}}{y^{2}}$$

Step-by-Step Solution

Answer

The variable \(z\) is directly proportional to the square root of \(x\) and to the constant \(k\), and inversely proportional to the square of \(y\). Thus as \(x\) or \(k\) increases, \(z\) also increases albeit \(x\) at a slower rate due to the square root, and as \(y\) increases, \(z\) decreases.

Step 1: Identify the Variables

The equation has three variables. \(z\) is the dependent variable which is a function of independent variables \(x\) and \(y\). Additionally, we have a constant \(k\). The form implies that \(z\) is determined by the square root of \(x\) divided by \(y^{2}\), with the whole expression multiplied by \(k\).

Step 2: Discuss \(x\) variable variance

Since \(x\) is under a square root in the numerator, as \(x\) increases, \(z\) also increases. \(z\) is directly proportional to the square root of \(x\), and this is valid for \(x > 0\). Therefore, for any positive change in \(x\), there will be a corresponding increase in \(z\). Additionally, the rate of increase in \(z\) will be slower than the rate of increase of \(x\) due to the presence of the square root.

Step 3: Discuss \(y\) variable variance

The variable \(y\) is in the denominator and is squared, so as \(y\) increases, \(z\) decreases. Thus, \(z\) is inversely proportional to the square of \(y\). For \(y > 0\), if \(y\) doubles, \(z\) would decrease to a quarter of its previous value, assuming \(x\) and \(k\) remain the same.

Step 4: Discuss the role of \(k\)

The constant \(k\) is a proportionality constant in this equation. It determines the rate at which \(z\) changes with respect to \(x\) and \(y\). If \(k\) is increased, \(z\) would increase for fixed \(x\) and \(y\), and vice versa. Therefore, \(z\) is directly proportional to \(k\).