Problem 62 Suppose that \(y\) varies invers... [FREE SOLUTION] (2024)

Short Answer

inverse variation

Inverse variation is a type of mathematical relationship where an increase in one quantity leads to a proportional decrease in another quantity. In simpler terms, if one value goes up, the other must go down to maintain a constant product. This concept is represented by the formula: y = \( \frac{k}{x} \), where k is a constant.
In this case, since the variable y varies inversely with the cube of x, we use y = \( \frac{k}{x^3} \).
Think of k as a fixed number that ensures y and x^3 maintain the inverse relationship. So, the product of y and x3 always equals k, adjusting automatically to changes in x or y.

proportional relationships

Proportional relationships are fundamental in understanding inverse variations. In an inverse variation, two quantities are proportionally related through a multiplicative constant. These quantities move in opposite directions: as one increases, the other decreases at a consistent rate.
For example, in our problem, y is inversely proportional to the cube of x (y = \( \frac{k}{x^3} \)). This shows a proportional relationship, but the proportion involves the cube of x. This type of relationship ensures that even though the variables change, the product you get by multiplying y and x^3 stays constant.

algebraic manipulation

To solve problems involving inverse variations, algebraic manipulation is often needed. This involves rearranging equations to isolate the variable of interest or to compare different states of a variable.
In the exercise provided, we started with the initial equation y = \( \frac{k}{x^3} \) and then found the new value of y after changing x to \(\frac{1}{4}x_0 \).
By substituting \(\frac{1}{4}x_0 \) into the equation, we performed algebraic manipulation and simplified the equation: \[ y_{new} = \frac{k}{\left(\frac{1}{4} x_0\right)^3} = 64 \cdot \frac{k}{x_0^3} = 64 \cdot y \]
These steps helped transform the equation to compare the effect on y easily.

effect of changes in variables

Understanding the effect of changing variables in inverse variations is crucial for problem-solving. If the variable x is decreased by a certain factor, say from x to \(\frac{1}{4}x\), it’s essential to understand how this affects y.
When x is reduced to \(\frac{1}{4}{x_0} \), its cube \( \left(\frac{1}{4} x_0\right)^3 \) will dramatically reduce to \( \frac{1}{64}{x_0^3} \).
This manipulation indicates that the new y will be multiplied by 64 because the cube term becomes extremely small, boosting y significantly:
y_{new} = 64 \cdot y.
Hence, when x decreases to \(\frac{1}{4} \), y is greatly increased by multiplying it by 64.