Chapter 4: Problem 178
Construct a mathematical model given the following: \(y\) is directly proportional to the square of \(x\) and inversely proportional to \(Z\), where \(y=-6\) when \(x=2\) and \(z=-8\).
Short Answer
Expert verified
The model is \( y = 12 \frac{x^2}{z} \).
Step by step solution
01
Understand Direct and Inverse Proportions
The statement 'y is directly proportional to the square of x' means we have the relationship \( y \propto x^2 \), and 'inversely proportional to Z' gives \( y \propto \frac{1}{Z} \). These relationships combined give the proportional equation \( y \propto \frac{x^2}{Z} \).
02
Introduce a Constant of Proportionality
To convert the proportionality into an equation, introduce a constant \( k \). Thus, the equation becomes \( y = k \frac{x^2}{Z} \).
03
Substitute Known Values
Use the given values \( y = -6 \), \( x = 2 \), and \( z = -8 \) to find \( k \). Substitute these into the equation: \( -6 = k \frac{2^2}{-8} \). Simplifying, \( -6 = k \frac{4}{-8} \) which gives \( -6 = k \times (-0.5) \).
04
Solve for the Constant k
Solve \( -6 = -0.5k \) to find \( k \). Divide both sides by \(-0.5\) to get \( k = 12 \).
05
Write the Final Equation
Substitute \( k = 12 \) back into the equation \( y = k \frac{x^2}{z} \) to get \( y = 12 \frac{x^2}{z} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Direct Proportionality
Direct proportionality describes a specific relationship where one quantity increases at a consistent rate as another quantity increases. This is expressed in the equation format as \( y \propto x \), meaning "\( y \) is proportional to \( x \)."
In our exercise, it is slightly modified to state that \( y \) is proportional to the square of \( x \). This changes the expression to \( y \propto x^2 \). Simply put, if \( x \) doubles, \( y \) increases by four times since squaring doubles the effect.
Direct proportionality is all about maintaining a constant ratio between two quantities.
In our exercise, it is slightly modified to state that \( y \) is proportional to the square of \( x \). This changes the expression to \( y \propto x^2 \). Simply put, if \( x \) doubles, \( y \) increases by four times since squaring doubles the effect.
Direct proportionality is all about maintaining a constant ratio between two quantities.
- If \( x \) increases, \( y \) increases.
- If \( x \) decreases, \( y \) decreases.
Inverse Proportionality
Inverse proportionality is an interesting concept where one quantity increases while the other decreases. Mathematically, it's shown as \( y \propto \frac{1}{z} \). This means if one quantity increases, the related one decreases such that their product remains constant.
Imagine inflating a balloon: as you add more air inside (volume increases), the pressure of the air within decreases, assuming the temperature is constant. This is akin to inverse proportionality.
In our exercise, \( y \) is inversely proportional to \( z \), implying that if \( z \) becomes larger, \( y \) becomes smaller as a reaction. This helps in modeling relationships where there is an inverse cause-and-effect impact between involved variables.
Imagine inflating a balloon: as you add more air inside (volume increases), the pressure of the air within decreases, assuming the temperature is constant. This is akin to inverse proportionality.
In our exercise, \( y \) is inversely proportional to \( z \), implying that if \( z \) becomes larger, \( y \) becomes smaller as a reaction. This helps in modeling relationships where there is an inverse cause-and-effect impact between involved variables.
- If \( z \) increases, \( y \) decreases.
- If \( z \) decreases, \( y \) increases.
Mathematical Modeling
Mathematical modeling allows us to describe a real-world situation or a theoretical problem using mathematical language. This eliminates ambiguity and focuses the solution on numerical representation.
In our provided problem, the given conditions describe a combination of relationships that are woven together in a single equation: \( y = k \frac{x^2}{z} \).
This model encapsulates both direct and inverse proportionality, utilizing mathematical symbols to represent these dynamics effectively. The beauty of mathematical modeling lies in:
In our provided problem, the given conditions describe a combination of relationships that are woven together in a single equation: \( y = k \frac{x^2}{z} \).
This model encapsulates both direct and inverse proportionality, utilizing mathematical symbols to represent these dynamics effectively. The beauty of mathematical modeling lies in:
- Simplifying complex relationships
- Predicting outcomes based on known variables
- Providing a clear, quantifiable way to study real-world phenomena
Constant of Proportionality
The constant of proportionality is a crucial element when dealing with proportional relationships. It provides the specific multiplier needed to transition from a proportional relationship to an equation.
In our exercise, by substituting the known values of \( y = -6 \), \( x = 2 \), and \( z = -8 \), we solve for this constant \( k \), yielding \( k = 12 \).
This constant unifies the relationship described, making the equation \( y = 12 \frac{x^2}{z} \) complete and usable in calculating \( y \) for any given \( x \) and \( z \).
The constant ensures that the modeled equation can precisely predict or describe how \( y \) changes as \( x \) and \( z \) vary.
In our exercise, by substituting the known values of \( y = -6 \), \( x = 2 \), and \( z = -8 \), we solve for this constant \( k \), yielding \( k = 12 \).
This constant unifies the relationship described, making the equation \( y = 12 \frac{x^2}{z} \) complete and usable in calculating \( y \) for any given \( x \) and \( z \).
The constant ensures that the modeled equation can precisely predict or describe how \( y \) changes as \( x \) and \( z \) vary.
- Provides accuracy in the mathematical model
- Reflects the true proportion between the quantities
- Acts as a bridge from theoretical relationships to practical applications