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Chapter 3: Problem 62
Suppose that \(y\) varies inversely as the cube of \(x\). If the value of \(x\) isdecreased to \(\frac{1}{4}\) of its original value, what is the effect on \(y\) ?
Short Answer
Expert verified
y will be 64 times greater.
Step by step solution
01
Understand What Inverse Variation Means
When a variable y varies inversely as another variable x, it means that y is proportional to the reciprocal of x. Given that y varies inversely as the cube of x, we can express this as: \[ y = \frac{k}{x^3} \] where k is a constant.
02
Set Up the Initial and Final Relationships
Initially, let the original value of x be denoted as \( x_0 \). Thus, the initial relationship is: \[ y = \frac{k}{x_0^3} \] When x is decreased to \( \frac{1}{4} x_0 \), the new value of y can be represented as: \[ y_{new} = \frac{k}{\left(\frac{1}{4} x_0\right)^3} \]
03
Simplify the New Relationship
Simplify the expression for \( y_{new} \): \[ y_{new} = \frac{k}{\left(\frac{1}{4} \right)^3 \cdot x_0^3 } = \frac{k}{\frac{1}{64} x_0^3} = k \cdot \frac{64}{x_0^3} = 64 \cdot \frac{k}{x_0^3} \]
04
Compare With the Original Relationship
Notice that \( y = \frac{k}{x_0^3} \) represents the original value of y. Therefore, the new value of y is: \[ y_{new} = 64 \cdot y \]
05
State the Final Effect on y
The effect on y when x is decreased to \( \frac{1}{4} \) of its original value is that y will be multiplied by 64.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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