/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Express the meaning of the given... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Express the meaning of the given equation in a verbal statement, using the language of variation. ( \(k\) and \(\pi\) are constants.) $$f=\frac{k L}{\sqrt{m}}$$

Short Answer

Expert verified
The variable \( f \) varies directly with \( L \) and inversely with \( \sqrt{m} \), scaled by \( k \).

Step by step solution

01

Identify the Components of the Equation

The given equation is \( f = \frac{kL}{\sqrt{m}} \). Here, \( f \) is the dependent variable, \( k \) is a constant, \( L \) is a variable, and \( m \) is another variable. The constant \( k \) represents a proportionality factor, \( L \) can be any factor that affects \( f \) directly, and \( m \) affects \( f \) inversely.
02

Describe the Relationship between the Variables

The equation demonstrates that \( f \) is directly proportional to \( L \) and inversely proportional to the square root of \( m \). This means as \( L \) increases (while \( m \) remains constant), \( f \) increases. Conversely, as \( m \) increases (while \( L \) remains constant), \( f \) decreases, but the rate of decrease is slower due to the square root.
03

Formulate the Verbal Statement

The verbal statement of the equation \( f = \frac{kL}{\sqrt{m}} \) is: 'The variable \( f \) varies directly with \( L \) and inversely with the square root of \( m \), and it is scaled by the constant \( k \)'.

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Key Concepts

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

Direct variation
In mathematics, direct variation describes a relationship between two variables where one variable is a constant multiple of the other. When one variable increases, the other also increases proportionally. This can be expressed in the formula \( y = kx \), where \( y \) and \( x \) are the two variables and \( k \) is the constant of proportionality.
In the context of the equation \( f = \frac{kL}{\sqrt{m}} \), the direct variation is between \( f \) and \( L \). This means that as the variable \( L \) increases, the value of \( f \) also increases when all other factors remain constant.
**Key Points:**
  • Direct variation implies a linear relationship between the two variables.
  • As one variable increases, so does the other, provided multiplication by the constant \( k \).
  • The proportionality constant \( k \) indicates how steeply the increase happens.
Inverse variation
Inverse variation is a mathematical relationship between two variables whereby one variable is a constant divided by the other variable. This signifies that as one variable increases, the other decreases, and vice versa. In a formula, this is represented as \( y = \frac{k}{x} \), where \( y \) and \( x \) are the variables and \( k \) is a constant of proportionality.
In the equation \( f = \frac{kL}{\sqrt{m}} \), \( f \) varies inversely with the square root of \( m \) (i.e., \( \sqrt{m} \)). Therefore, when \( m \) increases, \( \sqrt{m} \) increases, causing \( f \) to decrease, as \( \sqrt{m} \) is in the denominator.
The impact of \( m \) on \( f \) is moderated by its square root, meaning changes in \( m \) have a less pronounced effect.
**Key Points:**
  • Inverse variation implies one variable decreases as the other increases.
  • The effect is less drastic when a square root is involved.
  • The constant \( k \) still scales this relationship, adjusting the degree of change.
Proportionality in equations
Proportionality in equations revolves around the idea that one variable changes in correspondence with another, either directly or inversely. This principle helps in understanding how different factors affect each other in mathematical models.
Understanding proportionality is key to interpreting equations like \( f = \frac{kL}{\sqrt{m}} \). This equation showcases both direct and inverse proportionality:
  • \( f \) is directly proportional to \( L \).
  • \( f \) is inversely proportional to \( \sqrt{m} \).

The constant \( k \) integrates the two forms of proportionality, scaling the impact that \( L \) and \( m \) have on \( f \).
Using this knowledge, you can predict how changes in one variable will influence the variable of interest, \( f \). Mastering these relationships means you can often solve complex problems and apply these principles in real-world situations.
**Key Points:**
  • Proportionality concerns the relational change between variables.
  • Understanding the constants and their role is crucial in interpreting these relationships.
  • This type of analysis can be applied to various fields, from physics to economics.

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Most popular questions from this chapter

Answer the given questions by setting up and solving the appropriate proportions. When a bullet is fired from a loosely held rifle, the ratio of the mass of the bullet to that of the rifle equals the negative of the reciprocal of the ratio of the velocity of the bullet to that of the rifle. If a \(3.0 \mathrm{kg}\) rifle fires a \(5.0 \mathrm{g}\) bullet and the velocity of the bullet is \(300 \mathrm{m} / \mathrm{s}\) what is the recoil velocity of the rifle?

Set up the general equations from the given statements. The average electric power \(P\) entering a load varies jointly as the resistance \(R\) of the load and the square of the effective voltage \(V\) and inversely as the square of the impedance \(Z\)

Solve the given applied problems involving variation. In a physics experiment, a given force was applied to three objects. The mass \(m\) and the resulting acceleration \(a\) were recorded as follows: $$\begin{array}{l|c|c|c} m(\mathrm{g}) & 2.0 & 3.0 & 4.0 \\ \hline a\left(\mathrm{cm} / \mathrm{s}^{2}\right) & 30 & 20 & 15 \end{array}$$ (a) Is the relationship \(a=f(m)\) one of direct or inverse variation? Explain. (b) Find \(a=f(m)\)

Find the required quantities from the given proportions. In an electric instrument called a "Wheatstone bridge," electric resistances are related by $$\begin{aligned} &\frac{R_{1}}{R_{2}}=\frac{R_{3}}{R_{4}}\\\ &\text { Find } R_{2} \text { if } R_{1}=6.00 \Omega\\\ &R_{3}=62.5 \Omega, \text { and }\\\ &R_{4}=15.0 \Omega . \text { See Fig. } 18.4 \end{aligned}$$

Solve the given applied problems involving variation. To cook a certain vegetable mix in a microwave oven, the instructions are to cook 4.0 oz for 2.5 min or 8.0 oz for 3.5 min. Assuming the cooking time \(t\) is proportional to some power (not necessarily integral) of the weight \(w\), use logarithms to find \(t\) as a function of \(w\)

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