Question

If \(y\) is directly proportional to the square of \(x\) and inversely proportional to the cube of \(w,\) and \(y\) is 11.6 when \(x\) is 84.2 and \(w\) is \(28.4,\) find \(y\) when \(x\) is 5.38 and \(w\) is 2.28.

Short answer

First find the constant of proportionality (k) using the given y, x, and w values, then use this constant along with the new x and w values to calculate the new y.

Step-by-step solution

Understanding direct and inverse proportionality

When we say that 'y' is directly proportional to the square of 'x', it means that y = k * x^2 for some constant k. Similarly, 'y' being inversely proportional to the cube of 'w' means that y = k / w^3 for the same constant k. Combining these, we have y = k * x^2 / w^3.

Find the proportionality constant k

Using the provided values (y = 11.6, x = 84.2, w = 28.4), we can substitute them into the equation from Step 1 to find k: 11.6 = k * 84.2^2 / 28.4^3. We solve this equation for k.

Calculate the value of k

We solve the equation 11.6 = k * 84.2^2 / 28.4^3 by isolating k: k = 11.6 * 28.4^3 / 84.2^2. We calculate the value of k using a calculator.

Use k to find y for the new values of x and w

With the value of k found in Step 3, we plug in the new values of x (5.38) and w (2.28) into the equation y = k * x^2 / w^3 to find the new value of y.

Compute the final value of y

Perform the calculations using k from Step 3, and the new x and w values. Finally, we get the value of y for x = 5.38 and w = 2.28.

Key concepts

Proportionality Constant

The proportionality constant is the 'k' in equations of proportionality. It connects two variables in a way that their ratio remains constant. For example, if we have a directly proportional relationship like \( y = kx \), 'k' represents how much 'y' changes for a unit change in 'x'.

Similarly, in inversely proportional relationships such as \( y = \frac{k}{x} \), 'k' also serves as the constant of proportionality. However, in this case, an increase in 'x' results in a decrease in 'y' and vice versa, maintaining the product of the two variables constant. To find the value of 'k', you need at least one set of values for 'x' and 'y'.

In our exercise, we determined 'k' using the given values of 'x', 'y', and 'w'. Once 'k' is found, it can be used to predict other values of 'y' for different 'x' and 'w', keeping the relationship intact.

Solving Proportional Equations

Solving proportional equations involves manipulating the equation to find the unknown variable while keeping the proportionality constant fixed. This process often requires algebraic skills such as cross-multiplication and isolation of variables.

To solve these equations, you typically have to clear fractions or decimals first and then isolate the variable of interest. In the given exercise, after finding the proportionality constant 'k', we used it to solve for 'y' when 'x' and 'w' were given new values. This step is crucial because it allows us to apply the proportionality constant to different scenarios, predicting outcomes based on our initial observations.

Directly Proportional Relationships

Directly proportional relationships occur when two variables change at the same rate. If one variable increases, the other does too, and the ratio between them stays the same. In other words, if variable 'y' is directly proportional to variable 'x', then as 'x' increases or decreases, 'y' does so as well by the same factor.

This is expressed in the formula \( y = kx \), where 'k' is the proportionality constant. In our exercise, 'y' was directly proportional to the square of 'x' (\( x^2 \)), which means that 'y' incrases as the square of 'x' incrases, illustrating a parabolic growth rather than a linear one.

Inversely Proportional Relationships

Inversely proportional relationships describe a situation where one variable decreases as the other increases such that their product remains constant. The mathematical representation is \( y = \frac{k}{x} \), where 'k' is the proportionality constant.

In these relationships, doubling one variable results in halving the other. This concept was illustrated in our exercise where 'y' was inversely proportional to the cube of 'w' (\( w^3 \)). Thus, when 'w' value is increased, the value of 'y' decreases at a rate that is cubic, not linear, showing a steeper reaction to changes in 'w'.