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Chapter 5: Problem 38

For the following exercises, use the given information to find the unknown value. \(y\) varies jointly as \(x\) and \(z\) and inversely as \(w .\) When \(x=5, z=2,\) and \(w=20,\) then \(y=4 .\) Find \(y\) when \(x=3\) and \(z=8,\) and \(w=48\).

Short Answer

Expert verified
The value of \( y \) is 4.

Step by step solution

01

Set up the Joint and Inverse Variation Equation

Since the problem states that \( y \) varies jointly as \( x \) and \( z \), and inversely as \( w \), the relationship can be expressed as \( y = k \frac{xz}{w} \), where \( k \) is a constant of proportionality that we need to evaluate.
02

Insert Given Values to Find Constant k

Substitute \( x = 5 \), \( z = 2 \), \( w = 20 \), and \( y = 4 \) into the equation to find \( k \). This results in: \[ 4 = k \frac{5 \times 2}{20} \] This simplifies to: \[ 4 = k \frac{10}{20} = k \frac{1}{2} \] Then, solve for \( k \): \[ k = 4 \times 2 = 8 \]
03

Substitute New Values into the Equation

We now use the value of \( k \) we just found to solve for \( y \) using the new values. Substitute \( x = 3 \), \( z = 8 \), \( w = 48 \), and \( k = 8 \) into the equation: \[ y = 8 \frac{3 \times 8}{48} \] Simplifying the expression gives: \[ y = 8 \frac{24}{48} = 8 \times \frac{1}{2} \]
04

Calculate Final Value of y

Evaluate the expression from the previous step: \[ y = 8 \times \frac{1}{2} = 4 \] Thus, the value of \( y \) is 4.

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Key Concepts

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

Constant of Proportionality
In mathematics, when two variables are related in a way that one variable is a constant multiple of another, we describe the relationship using the term "constant of proportionality." This constant acts as a multiplier that scales one variable's effect on the other.
Consider a scenario with variables that exhibit joint and inverse variation. In this case, the constant of proportionality helps us express how these variables interact.
  • For joint variation, where a variable varies directly with more than one variable, the expression is of the form: \( y = k \cdot xz \), meaning \( k \) is the constant that scales the product of \( x \) and \( z \).
  • For inverse variation, the expression changes to: \( y = \frac{k}{w} \), indicating \( y \) reduces as \( w \) increases, modulated by \( k \).
When combining both relationships, we use an equation like \( y = k \frac{xz}{w} \), where \( k \) ensures the proportional interaction remains consistent across different values.
Solving Equations
Solving equations involves finding the unknown values that satisfy a given mathematical expression. Let's take a deeper look at how we can solve equations involving joint and inverse variations.
Suppose you have the initial equation \( y = k \frac{xz}{w} \), where you need to determine the unknown constant \( k \). Here鈥檚 a step-by-step method to solve such equations:
  • Plug in all known values into the equation. For initial conditions like \( x = 5 \), \( z = 2 \), \( w = 20 \), and \( y = 4 \), the equation becomes: \( 4 = k \frac{5 imes 2}{20} \).
  • Simplify the equation to isolate \( k \). Continue by simplifying fractions or solving the basic math, in this case: \( \frac{10}{20} = \frac{1}{2} \), leading to \( 4 = k \frac{1}{2} \).
  • Solve for \( k \) by clearing the isolation fraction or by performing arithmetic operations, here: \( k = 4 \times 2 = 8 \).
Now with \( k \) known, use this to rearrange and solve for other unknowns with new sets of values, highlighting the methodical approach of solving equations.
Algebraic Expressions
Algebraic expressions are combinations of variables, numbers, and operations that form a mathematical sentence. Understanding algebra involves recognizing how these expressions can model real-world situations like joint and inverse variations.
Let's break down the expression used in such variations: The expression \( y = k \frac{xz}{w} \) represents both a joint and inverse relationship among variables and is a compact way to express complex interactions.
  • **Joint Variation:** The product \( xz \) within the expression indicates how \( y \) scales with the direct interaction of both \( x \) and \( z \).
  • **Inverse Variation:** When \( w \) is in the denominator, it suggests how \( y \) will shrink or grow inversely with \( w \).
  • **Manipulation of Components:** With algebraic expressions, you can re-arrange components to find unknowns, by multiplying or dividing to isolate specific variables. Note how solving for \( k \) or \( y \) followed this approach.
Algebraic expressions are versatile, capable of handling unique situations by adapting this general formula to fit specific scenarios, making them essential in understanding and solving equations related to variations.

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Most popular questions from this chapter

For the following exercises, use the given information to answer the questions. The distance \(s\) that an object falls varies directly with the square of the time, \(t,\) of the fall. If an object falls 16 feet in one second, how long for it to fall 144 feet?

For the following exercises, find the dimensions of the right circular cylinder described. The height is one less than one half the radius. The volume is 72\(\pi\) cubic meters.

For the following exercises, determine the function described and then use it to answer the question. A container holds 100 \(\mathrm{ml}\) of a solution that is 25 \(\mathrm{ml}\) acid. If \(n \mathrm{ml}\) of a solution that is 60\(\%\) acid is added, the function \(C(n)=\frac{25+0.6 n}{100+n}\) gives the concentration, \(C,\) as a function of the number of ml added, \(n .\) Express \(n\) as a function of \(C\) and determine the number of \(m 1\) that need to be added to have a solution that is 50\(\%\) acid.

For the following exercises, use a graph to help determine the domain of the functions. $$f(x)=\sqrt{\frac{9-x^{2}}{x+4}}$$

For the following exercises, use the given information to find the unknown value. \(y\) varies inversely with the square of \(x .\) When \(x=4\) then \(y=3\). Find \(y\) when \(x=2\).
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