Problem 11
Question
Exer. 1-12: Express the statement as a formula that involves the given variables and a constant of proportionallty \(k,\) and then determine the value of \(k\) from the given conditions. \(y\) is directly proportional to the square root of \(x\) and inversely proportional to the cube of \(z\). If \(x=9\) and \(z=2\) then \(y=5\)
Step-by-Step Solution
Verified Answer
The formula is \( y = \frac{40}{3} \frac{\sqrt{x}}{z^3} \); \( k = \frac{40}{3} \).
1Step 1: Understanding Relationships
Since the problem states that \( y \) is directly proportional to the square root of \( x \) and inversely proportional to the cube of \( z \), we can express this relationship with the equation: \[ y = k \frac{\sqrt{x}}{z^3} \] where \( k \) is the constant of proportionality.
2Step 2: Substitute Given Values
We are given \( x = 9 \), \( z = 2 \), and \( y = 5 \). Substitute these values into the equation: \[ 5 = k \frac{\sqrt{9}}{2^3} \]
3Step 3: Simplify Equation
First, calculate \( \sqrt{9} = 3 \) and \( 2^3 = 8 \). Substitute these into the equation: \[ 5 = k \frac{3}{8} \]
4Step 4: Solve for Constant of Proportionality \(k\)
To find \( k \), multiply both sides of the equation by \( \frac{8}{3} \): \[ k = 5 \left( \frac{8}{3} \right) \] Simplifying, we get: \[ k = \frac{40}{3} \]
Key Concepts
Constant of ProportionalityProportional RelationshipsAlgebraic Equations
Constant of Proportionality
The constant of proportionality, often symbolized as \( k \), is a vital concept used when dealing with proportional relationships. In any equation or formula expressing a direct or inverse proportion, this constant represents the constant ratio or product that remains consistent within the relationship.
In our exercise example, the problem describes \( y \) as being directly proportional to the square root of \( x \) and inversely proportional to the cube of \( z \). This means that the product of \( y \) and \( z^3 \) divided by \( \sqrt{x} \) will always equal the constant \( k \).
To determine \( k \), we transformed the verbal relationship into the formula \( y = k \frac{\sqrt{x}}{z^3} \), then used the given values of \( x = 9 \), \( z = 2 \), and \( y = 5 \). By substituting these into the equation, we solved for \( k \) and found it to be \( \frac{40}{3} \). This constant is essential for maintaining the equation's balance across different conditions.
In our exercise example, the problem describes \( y \) as being directly proportional to the square root of \( x \) and inversely proportional to the cube of \( z \). This means that the product of \( y \) and \( z^3 \) divided by \( \sqrt{x} \) will always equal the constant \( k \).
To determine \( k \), we transformed the verbal relationship into the formula \( y = k \frac{\sqrt{x}}{z^3} \), then used the given values of \( x = 9 \), \( z = 2 \), and \( y = 5 \). By substituting these into the equation, we solved for \( k \) and found it to be \( \frac{40}{3} \). This constant is essential for maintaining the equation's balance across different conditions.
Proportional Relationships
Proportional relationships describe how two quantities change in relation to each other. These relationships can be either direct, where two variables increase or decrease together, or inverse, where one variable increases as the other decreases.
In our exercise, the relationship is a mix of both direct and inverse proportions. Here, \( y \) increases as the square root of \( x \) increases, portraying a direct relationship. On the other hand, \( y \) decreases as \( z^3 \) increases, illustrating an inverse relationship.
Understanding these two types of relationships allows us to build more complex expressions such as \( y = k \frac{\sqrt{x}}{z^3} \), which combines both elements.
In our exercise, the relationship is a mix of both direct and inverse proportions. Here, \( y \) increases as the square root of \( x \) increases, portraying a direct relationship. On the other hand, \( y \) decreases as \( z^3 \) increases, illustrating an inverse relationship.
Understanding these two types of relationships allows us to build more complex expressions such as \( y = k \frac{\sqrt{x}}{z^3} \), which combines both elements.
- Direct proportionality implies \( y \) is scaled by \( \sqrt{x} \).
- Inverse proportionality suggests \( y \) reduces by the cube of \( z \).
Algebraic Equations
Algebraic equations are fundamental tools for expressing mathematical relationships and finding unknown values. They allow us to translate verbal descriptions into mathematical statements that can be solved systematically.
In this particular problem, we derived the equation \( y = k \frac{\sqrt{x}}{z^3} \) from the description of \( y \)'s proportionality to \( x \) and \( z \). This equation is an expression where \( y \) depends on both \( \sqrt{x} \) and \( z^3 \), modulled by the constant \( k \).
To solve the equation, we used algebraic techniques such as substitution and simplification.
In this particular problem, we derived the equation \( y = k \frac{\sqrt{x}}{z^3} \) from the description of \( y \)'s proportionality to \( x \) and \( z \). This equation is an expression where \( y \) depends on both \( \sqrt{x} \) and \( z^3 \), modulled by the constant \( k \).
To solve the equation, we used algebraic techniques such as substitution and simplification.
- Substitute known variable values for \( x \), \( z \), and \( y \).
- Simplify the mathematical expressions (e.g., \( \sqrt{9} = 3 \) and \( 2^3 = 8 \)).
- Isolate \( k \) to solve for the constant of proportionality.
Other exercises in this chapter
Problem 10
Use the remainder theorem to find \(f(c)\). $$f(x)=2 x^{3}+4 x^{2}-3 x-1 ; \quad c=3$$
View solution Problem 11
Show that the equation has no rational root. $$x^{3}+3 x^{2}-4 x+6=0$$
View solution Problem 11
Sketch the graph of \(f\) $$f(x)=\frac{4 x-1}{2 x+3}$$
View solution Problem 11
Use the remainder theorem to find \(f(c)\). $$f(x)=x^{4}-6 x^{2}+4 x-8 ; \quad c=-3$$
View solution