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Chapter 2: Problem 26

Find an equation of variation for the given situation. \(y\) varies inversely as the square of \(x,\) and \(y=6\) when \(x=3\).

Short Answer

Expert verified
The equation is \( y = \frac{54}{x^2} \).

Step by step solution

01

Identify the Variation Type

Since it is given that y varies inversely as the square of x, we can write the relationship as \[ y = \frac{k}{x^2} \] where k is a constant.
02

Substitute Known Values

To find the value of the constant k, we substitute the given values of y and x into the equation. Here, y = 6 when x = 3.\[ 6 = \frac{k}{3^2} \]\[ 6 = \frac{k}{9} \]
03

Solve for the Constant

Multiply both sides of the equation by 9 to solve for k.\[ 6 \times 9 = k \]\[ k = 54 \]
04

Write the Final Equation

Using the value of k, substitute it back into the original equation of variation. The equation is\[ y = \frac{54}{x^2} \]

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Key Concepts

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

Understanding Variations in Mathematics
Variations in mathematics describe how one quantity changes in relation to another. There are two main types: direct and inverse variations.

In direct variation, as one variable increases, the other also increases. This is expressed as y = kx, where k is a constant. On the other hand, in inverse variation, as one variable increases, the other decreases. This can be written as y = k/x, where k is again a constant.

When dealing with variations, it鈥檚 essential to identify the type of relationship between the quantities. We often encounter more complex cases, like in our example, where y varies inversely as the square of x. The formula for an inverse square variation is y = k/x^2.
Inverse Proportionality
Inverse proportionality means that as one variable increases, the other variable decreases in such a way that their product remains constant. In simple terms, one quantity is inversely proportional to another if the product of the two is always the same number.

For our example, we see that y varies inversely with the square of x. This relationship is expressed with the equation y = k/x^2. Here, k is a constant that we need to find.

To find k, we substitute the given values into the equation. Given that y = 6 when x = 3, we have:
y = k/x^2
6 = k/3^2
6 = k/9

By multiplying both sides by 9, we find that k = 54. Therefore, the equation describing this relationship becomes y = 54/x^2.
Solving Equations
Solving equations involves finding the value of unknown variables that make the equation true. In our example, we need to determine the constant k to describe the inverse variation properly.

We start with the equation y = k/x^2 and substitute the given values. With y = 6 and x = 3, we solve for k as follows:

  • Step 1: Begin with the relationship y = k/x^2.
  • Step 2: Substitute the given values: 6 = k/3^2.
  • Step 3: Simplify the equation to get 6 = k/9.
  • Step 4: Multiply both sides by 9 to isolate k: 6 x 9 = k.
  • Step 5: Simplify to k = 54.


Now, we rewrite the original equation with the value of k. Therefore, the final equation is y = 54/x^2. This equation allows us to find y for any value of x using the established inverse variation between the variables.

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Most popular questions from this chapter

Fill in the blank with the correct term. Some of the given choices will not be used. $$\begin{array}{ll}\text { even function } & \text { relative maximum } \\\ \text { odd function } & \text { rlativie minimum } \\ \text { conssinnf_unction } & \text { solution } \\ \text { composite function } & \text { zerrodicular } \\ \text { direct variaion } & \text { perallel } \\\ \text { inverse variaition } & \text { parals }\end{array}$$ Nonvertical lines are ________ only if they have the same slope and different \(y\) -intercepts. [ 1.4]

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Determine whether the statement is true or false. The product of an even function and an odd function is odd.

Find an equation of variation for the given situation.The intensity \(I\) of light from a light bulb varies inversely as the square of the distance \(d\) from the bulb. Suppose that \(I\) is \(90 \mathrm{W} / \mathrm{m}^{2}\) (watts per square meter) when the distance is \(5 \mathrm{m}\). How much farther would it be to a point where the intensity is \(40 \mathrm{W} / \mathrm{m}^{2} ?\)

Find \((f \circ g)(x)\) and \((g \circ f)(x)\) and the domain of each. $$f(x)=\frac{6}{x}, g(x)=\frac{1}{2 x+1}$$
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