Problem 21
Question
Find the constant of variation for each of the stated conditions. \(y\) varies directly as \(x\) and inversely as \(z\), and \(y=45\) when \(x=18\) and \(z=2\).
Step-by-Step Solution
Verified Answer
The constant of variation is 5.
1Step 1: Understand the Variation Relationship
The problem states that \(y\) varies directly with \(x\) and inversely with \(z\). This can be expressed with the formula \(y = k \frac{x}{z}\), where \(k\) is the constant of variation.
2Step 2: Substitute the Given Values
We know \(y=45\), \(x=18\), and \(z=2\). Substitute these values into the variation equation:\[45 = k \frac{18}{2}\].
3Step 3: Simplify the Equation
Simplify the fraction on the right side of the equation: \(\frac{18}{2} = 9\). The equation becomes:\[45 = 9k\].
4Step 4: Solve for the Constant of Variation, \(k\)
To find \(k\), divide both sides of the equation by 9:\[k = \frac{45}{9}\].Simplify the fraction: \(k = 5\).
Key Concepts
Direct VariationInverse VariationVariation Equations
Direct Variation
Direct variation occurs when two variables change in the same proportion. Specifically, if one variable increases, the other variable increases at the same rate. The relationship between these variables can be expressed as \( y = kx \), where \( k \) is the constant of variation. This constant represents the ratio of \( y \) to \( x \).
You can think of direct variation like the relationship between distance and speed. If you travel faster, the distance you cover also increases proportionally. For example, if you double your speed, the distance covered in the same time doubles as well.
You can think of direct variation like the relationship between distance and speed. If you travel faster, the distance you cover also increases proportionally. For example, if you double your speed, the distance covered in the same time doubles as well.
- Straight-line relationships: The graph of a direct variation is a straight line that passes through the origin.
- Proportional change: As one variable doubles, the other doubles too.
Inverse Variation
Inverse variation is a relationship between two variables where one variable increases as the other decreases. The product of the two variables always remains constant. It can be expressed using the formula \( y = \frac{k}{z} \), where \( k \) is the constant of variation.
Imagine a seesaw: if one side goes up, the other must come down to keep balance. This is similar to how inverse variation works—the product of certain changing variables remains constant, maintaining balance.
Imagine a seesaw: if one side goes up, the other must come down to keep balance. This is similar to how inverse variation works—the product of certain changing variables remains constant, maintaining balance.
- Inverse proportion: As one variable doubles, the other must halve to maintain equality.
- Graph interpretation: Inverse variation graphs are hyperbolas. Curves that decrease as they approach the axes.
Variation Equations
Variation equations combine elements of both direct and inverse variations to show how changes in one or more variables affect another variable. These equations help clarify these relationships by using a blend of both variation types.
In our exercise, the variation equation is given as \( y = k \frac{x}{z} \). This illustrates that \( y \) directly varies with \( x \) (meaning \( y \) increases as \( x \) increases) and inversely varies with \( z \) (meaning \( y \) decreases as \( z \) increases).
In our exercise, the variation equation is given as \( y = k \frac{x}{z} \). This illustrates that \( y \) directly varies with \( x \) (meaning \( y \) increases as \( x \) increases) and inversely varies with \( z \) (meaning \( y \) decreases as \( z \) increases).
- Application of multiple concepts: Variation equations often combine expectations from direct and inverse relationships.
- Solving Variation Equations: Identify known values and solve for the unknown variable, typically the constant \( k \).
Other exercises in this chapter
Problem 20
Graph each of the following linear and quadratic functions. $$f(x)=-4 x^{2}-8 x$$
View solution Problem 20
Specify the domain for each of the functions. $$f(x)=\frac{7}{x^{2}-8 x-20}$$
View solution Problem 21
Find the inverse of the given function by using the "undoing process," and then verify that \(\left(f \circ f^{-1}\right)(x)=x\) and \(\left(f^{-1} \circ f\righ
View solution Problem 21
Show that \((f \circ g)(x)=x\) and \((g \circ f)\) \((x)=x\) for each pair of functions. \(f(x)=4 x+2\) and \(g(x)=\frac{x-2}{4}\)
View solution