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Chapter 17: Problem 9

If \(y\) is inversely proportional to the square root of \(x,\) by what percentage will \(y\) change when \(x\) is decreased by \(50.0 \% ?\)

Short Answer

Expert verified
Y will increase by approximately 41.0%.

Step by step solution

01

Understand the Inverse Proportionality

Given that y is inversely proportional to the square root of x, we can write the relationship as y = k / 鈭歺, where k is the constant of proportionality.
02

Relate the Change in x to Change in y

If x is decreased by 50%, the new value of x is x' = 0.5x. We can then express the new value of y (let's call it y') in terms of the new x using the same proportionality, y' = k / 鈭歺'.
03

Calculate the Percentage Change in y

To find the percentage change in y, we need to calculate (y' - y) / y * 100%. Using our proportionality relationships, we get (k / 鈭歺' - k / 鈭歺) / (k / 鈭歺) * 100% = (鈭歺 / 鈭歺' - 1) * 100%.
04

Substitute Values and Simplify

Substituting x' with 0.5x, we get (鈭歺 / 鈭�(0.5x) - 1) * 100% = (鈭歺 / (鈭�0.5 * 鈭歺) - 1) * 100% = (1 / 鈭�0.5 - 1) * 100% 鈮� (1 / 0.707 - 1) * 100% 鈮� (1.41 - 1) * 100% 鈮� 0.41 * 100% = 41.0%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proportionality Constants
When studying inverse proportionality in mathematics, the proportionality constant is a key player. This constant, often denoted as 'k', represents the consistent value in the relationship between two variables that are inversely proportional to each other. Let's break it down:
  • For the given relationship, if one variable increases, the other decreases in such a way that the product of the two remains constant 鈥� that's where our constant 'k' comes into play.
  • When presented with an equation like y = k / \(\sqrt{x}\), we identify 'k' as the proportionality constant linking the variable 'y' and the square root of 'x'.
  • Even when 'x' varies, 'k' keeps the inverse relationship in check, ensuring that 'y' adjusts accordingly to maintain the equation's balance.
Understanding the role of 'k' is essential, as it helps us manipulate the equation to find new values or effects of changes on 'y' when 'x' is altered.
Percentage Change Calculation
The concept of percentage change is crucial in understanding fluctuations within different contexts, such as economics, population studies, and as seen here, inverse proportionality in math. Here's how it's applied in our exercise:
  • Percentage change essentially measures the extent of variation in a quantity over time, relative to its original value.
  • To calculate it, we take the difference between the new value and the original value, divide by the original value, and then multiply the result by 100 to express it as a percentage.
  • In formulaic terms, this is represented as ((new value - original value) / original value) * 100%.
Applying this to the problem at hand, when 'x' decreases by 50%, we calculate the corresponding change in 'y' to understand the impact of that 50% reduction in 'x'.
Square Root Functions
Square root functions provide a clear example of non-linear relationships in mathematics. These functions are critical when dealing with quantities that exhibit squared relationships.
  • A square root function appears as f(x) = \sqrt{x}, where 'f(x)' usually corresponds to the output value as 'x' changes.
  • Within inverse proportionality, instead of increasing proportionally to the square of 'x', the response variable, in this case 'y', decreases when 'x' increases, due to the presence of \(\sqrt{x}\) in the denominator.
  • This kind of non-linear relationship can sometimes be more challenging to grasp, but understanding how to work with square root functions in equations is crucial to solving a diverse array of mathematical problems.
In our exercise, \(\sqrt{x}\) dictates how 'y' changes when 'x' is altered, making it essential to comprehend the properties of square root functions to predict and understand these changes.
Quantitative Relationships
The examination of quantitative relationships involves understanding how one quantity changes in response to another. These relationships can be direct, inverse, or even more complex in nature.
  • In a direct relationship, as one quantity increases, so does the other, while in an inverse relationship, as one quantity increases, the other decreases.
  • The problem provided examines an inverse quantitative relationship between 'y' and the square root of 'x'. To grasp this relationship, it's beneficial to visualize or tabulate how changes in 'x' influence 'y' and vice versa.
  • Understanding these dynamics is not only foundational for solving textbook problems but is also key in real-world applications where variables are often dependent on one another.
Illustrating these relationships, especially through graphs or proportionality constants, can offer tangible insights into how variables coexist and affect each other within mathematical models.

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Most popular questions from this chapter

Boyle's law states that for a confined gas at a constant temperature, the product of the pressure and the volume is a constant. Another way of stating this law is that the pressure is inversely proportional to the volume, or that the volume is inversely proportional to the pressure. Assume a constant temperature in the following problems. A certain quantity of gas, when compressed to a volume of \(2.50 \mathrm{m}^{3},\) has a pressure of 184 Pa. The pascal (Pa) is the SI unit of pressure. It equals 1 newton per square meter. Find the pressure resulting when that gas is further compressed to \(1.60 \mathrm{m}^{3}\).

If \(y\) is directly proportional to the cube of \(x,\) and \(y\) is 638 when \(x\) is \(145,\) find \(y\) when \(x\) is 68.3

If \(y\) varies jointly as \(w\) and \(x,\) and \(y\) is 483 when \(x\) is 742 and \(w\) is \(383,\) find \(y\) when \(x\) is 274 and \(w\) is 756.

The inverse square law states that for a surface illuminated by a light source, the intensity of illumination on the surface is inversely proportional to the square of the distance between the source and the surface. A light source located \(7.50 \mathrm{m}\) from a surface produces an illumination of 426 lux on that surface. At what distance must that light source be placed to give an illumination of 850 lux?

A certain solar house stores heat in 155 metric tons of stone which are in a chamber beneath the house. Another solar house is to have a chamber of similar shape but with all dimensions increased by \(15 \% .\) How many metric tons of stone will it hold?
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