Problem 10
Question
Express the statement as a formula that involves the given variables and a constant of proportionality \(k\), and then determine the value of \(k\) from the given conditions. \(y\) is directly proportional to \(x\) and inversely proportional to the sum of \(r\) and \(s\). If \(x=3, r=5\), and \(s=7\), then \(y=2\).
Step-by-Step Solution
Verified Answer
The value of \(k\) is 8. The formula is \(y = 8 \left( \frac{x}{r+s} \right)\).
1Step 1: Write the Proportionality Relationship
The problem states that \(y\) is directly proportional to \(x\), and inversely proportional to the sum of \(r\) and \(s\). This translates to the equation \(y = k \left( \frac{x}{r+s} \right)\) where \(k\) is a constant of proportionality.
2Step 2: Substitute the Given Values
We are given that when \(x = 3\), \(r = 5\), and \(s = 7\), then \(y = 2\). Substitute these values into the equation: \[2 = k \left( \frac{3}{5+7} \right).\] This simplifies to: \[2 = k \left( \frac{3}{12} \right).\]
3Step 3: Solve for the Constant \(k\)
To find the value of \(k\), solve the equation from Step 2:\[2 = k \left( \frac{1}{4} \right).\]Multiply both sides by 4 to solve for \(k\):\[k = 2 \times 4 = 8.\]
4Step 4: Rewrite the Formula with \(k\)
Now that you have determined the value of \(k\), rewrite the original relationship formula:\[y = 8 \left( \frac{x}{r+s} \right).\]
Key Concepts
Constant of ProportionalityProportionality RelationshipsSolving for Constants
Constant of Proportionality
In mathematics, the constant of proportionality plays a key role in understanding how variables relate to each other through direct or inverse relationships. Imagine you have a situation where two quantities change in relation to each other; the constant of proportionality, denoted typically by the letter \(k\), is the number that relates these variables in a proportion.
For instance, in the original exercise, the relationship stated is that \(y\) is directly proportional to \(x\) and inversely proportional to the sum of \(r\) and \(s\), creating the equation: \[ y = k \left( \frac{x}{r+s} \right) \] Here, \(k\) is the constant of proportionality that scales the ratio \( \frac{x}{r+s} \) to determine \(y\). In simple terms, \(k\) tells us how much \(y\) changes for given changes in \(x\) relative to \(r+s\). Understanding \(k\) is crucial because it helps us predict the behavior of the dependent variable \(y\) based on the independent variables \(x\), \(r\), and \(s\).
Finding \(k\) involves using known values to make the equation true, making it a step pursued after setting up the proportional relationship.
For instance, in the original exercise, the relationship stated is that \(y\) is directly proportional to \(x\) and inversely proportional to the sum of \(r\) and \(s\), creating the equation: \[ y = k \left( \frac{x}{r+s} \right) \] Here, \(k\) is the constant of proportionality that scales the ratio \( \frac{x}{r+s} \) to determine \(y\). In simple terms, \(k\) tells us how much \(y\) changes for given changes in \(x\) relative to \(r+s\). Understanding \(k\) is crucial because it helps us predict the behavior of the dependent variable \(y\) based on the independent variables \(x\), \(r\), and \(s\).
Finding \(k\) involves using known values to make the equation true, making it a step pursued after setting up the proportional relationship.
Proportionality Relationships
Proportionality relationships describe how one variable is affected when another variable increases or decreases. These relationships can either be direct or inverse.
**Direct Proportionality:** If one variable increases, the other increases in a constant ratio, given by \[ y = kx \] where \(k\) is the constant of proportionality. In our scenario, \(y\) increases as \(x\) does, if other factors remain constant.
**Inverse Proportionality:** On the other hand, if the increase in one variable leads to a decrease in another, then they are inversely proportional, described by \[ y = \frac{k}{x} \]. In our problem, \(y\) is inversely proportional to \(r+s\), which means as \(r+s\) increases, \(y\) would decrease given a constant \(x\).
Understanding the nature of these relationships allows us to form equations that model real-life scenarios effectively. For example, the joint statement in our exercise that \(y\) is directly proportional to \(x\) and inversely proportional to \(r+s\) translates into one cohesive equation that balances both relationships.
**Direct Proportionality:** If one variable increases, the other increases in a constant ratio, given by \[ y = kx \] where \(k\) is the constant of proportionality. In our scenario, \(y\) increases as \(x\) does, if other factors remain constant.
**Inverse Proportionality:** On the other hand, if the increase in one variable leads to a decrease in another, then they are inversely proportional, described by \[ y = \frac{k}{x} \]. In our problem, \(y\) is inversely proportional to \(r+s\), which means as \(r+s\) increases, \(y\) would decrease given a constant \(x\).
Understanding the nature of these relationships allows us to form equations that model real-life scenarios effectively. For example, the joint statement in our exercise that \(y\) is directly proportional to \(x\) and inversely proportional to \(r+s\) translates into one cohesive equation that balances both relationships.
Solving for Constants
Solving for constants involves determining the value of the constant of proportionality. In mathematical problems, this is achieved by applying given conditions to simplify the equation.
In the exercise provided, we found \(k\) using the formula: \[ y = k \left( \frac{x}{r+s} \right) \] Given specific values of \(x = 3\), \(r = 5\), \(s = 7\), and \(y = 2\), substituting them into the equation gives: \[ 2 = k \left( \frac{3}{12} \right) \] Breaking it down: The fraction \(\frac{3}{12}\) simplifies to \(\frac{1}{4}\), which gives us \[ 2 = k \left( \frac{1}{4} \right) \] To isolate \(k\), multiply both sides by 4 (the denominator of the fraction) yielding \(k = 8\). Thus, the constant \(k\) helps bridge the given values to derive \(y\) conclusively in other similar expressions.
Solving for constants is essential in making models reflective of real-life scenarios where values are not always given.
In the exercise provided, we found \(k\) using the formula: \[ y = k \left( \frac{x}{r+s} \right) \] Given specific values of \(x = 3\), \(r = 5\), \(s = 7\), and \(y = 2\), substituting them into the equation gives: \[ 2 = k \left( \frac{3}{12} \right) \] Breaking it down: The fraction \(\frac{3}{12}\) simplifies to \(\frac{1}{4}\), which gives us \[ 2 = k \left( \frac{1}{4} \right) \] To isolate \(k\), multiply both sides by 4 (the denominator of the fraction) yielding \(k = 8\). Thus, the constant \(k\) helps bridge the given values to derive \(y\) conclusively in other similar expressions.
Solving for constants is essential in making models reflective of real-life scenarios where values are not always given.
Other exercises in this chapter
Problem 9
Exer. 1-10: A polynomial \(f(x)\) with real coefficients and leading coefficient 1 has the given zero(s) and degree. Express \(f(x)\) as a product of linear and
View solution Problem 9
$$ \text { Use the remainder theorem to find } f(c) \text {. } $$ $$ f(x)=3 x^{3}-x^{2}+5 x-4 ; \quad c=2 $$
View solution Problem 10
Sketch the graph of \(f\). $$ f(x)=\frac{4 x}{2 x-5} $$
View solution Problem 10
Exer. 1-10: A polynomial \(f(x)\) with real coefficients and leading coefficient 1 has the given zero(s) and degree. Express \(f(x)\) as a product of linear and
View solution