Problem 21 - College Algebra with Modeling and Visualization

Question

Heavier birds tend to have larger wings than smaller birds. For one species of bird, the table lists the area \(A\) of the bird's wing in square inches if the bird weighs \(w\) pounds. $$\begin{array}{rccccc}w(\mathrm{b}) & 2 & 6 & 10 & 14 & 18 \\\\\hline A(w)\left(\mathrm{in}^{2}\right) & 160 & 330 & 465 & 580 & 685\end{array}$$ (a) Find a function that models the data. (b) Graph \(A\) and the data. (c) What weight corresponds to a wing area of 500 square inches?

Step-by-Step Solution

Answer

The model function is \(A(w) = 32.8125w + 94.375\). A weight of approximately 12.36 pounds corresponds to a wing area of 500 square inches.

Step 1: Determine a relationship between weight and wing area

Examine the given data to find a possible mathematical relationship. The pattern suggests a linear relationship between weight \(w\) and wing area \(A(w)\) since the differences between the successive wing areas are roughly constant. This can be modeled by a linear function of the form \(A(w) = mw + b\).

Step 2: Calculate the slope (m)

Use any two points from the data table to calculate the slope \(m\) of the linear function. Let's use the points \((2, 160)\) and \((18, 685)\).\[m = \frac{685 - 160}{18 - 2} = \frac{525}{16} = 32.8125\]

Step 3: Identify the y-intercept (b)

Substitute the slope \(m = 32.8125\) and one point, say \((2, 160)\), into the linear equation \(A(w) = mw + b\) to solve for \(b\):\[160 = 32.8125 \times 2 + b\]\[160 = 65.625 + b\]\[b = 160 - 65.625 = 94.375\]. Thus, the equation is \(A(w) = 32.8125w + 94.375\).

Step 4: Confirm the model with other data points

Check if the derived linear equation fits the other points as a way to confirm the model. Calculate \(A(6)\) using the derived equation:\[A(6) = 32.8125 \times 6 + 94.375 = 330.25\] This is approximately 330, fitting the data point \((6, 330)\). Similar checks for other points validate the function.

Step 5: Graph the function and data points

Plot the data points \((2, 160), (6, 330), (10, 465), (14, 580), (18, 685)\) on a graph. Also, plot the line \(A(w) = 32.8125w + 94.375\) on the same graph to visually confirm that the model fits the data.

Step 6: Solve for weight corresponding to a wing area of 500 square inches

Set \(A(w) = 500\) and solve for \(w\) using the function \(A(w) = 32.8125w + 94.375\):\[500 = 32.8125w + 94.375\]\[405.625 = 32.8125w\] \[w = \frac{405.625}{32.8125} \approx 12.36\] pounds.

Key Concepts

Slope Calculation

When we engage with linear regression, determining the slope is a pivotal first step. The slope, often symbolized as \( m \), represents the rate of change between two variables. In this context, it indicates how much the wing area (\( A \)) changes as the bird's weight (\( w \)) increases. To find the slope, you use the formula:

• \( m = \frac{y_2 - y_1}{x_2 - x_1} \)

You select any two data points from the dataset. For the given exercise, the points \((2, 160)\) and \((18, 685)\) are chosen.
Substituting these into the formula gives:

• \( m = \frac{685 - 160}{18 - 2} = 32.8125 \)

This result shows that for each pound increase in weight, the wing area increases by approximately 32.8125 square inches. Understanding the slope helps us quantify the relationship between bird weight and wing area.

Linear Equation Modeling

Once we have the slope, the next task is to determine the linear equation that models the relationship. A linear equation takes the form:

• \( y = mx + b \)

Where \( m \) is the slope and \( b \) is the y-intercept. The y-intercept is where the line crosses the y-axis, i.e., the area when weight is zero. Using the calculated slope \( m = 32.8125 \) and one of the data points, say \((2, 160)\), we substitute them into the equation to find \( b \):

• \( 160 = 32.8125 \times 2 + b \)
• \( 160 = 65.625 + b \)
• \( b = 94.375 \)

Thus, the linear equation modeling the data is:

• \( A(w) = 32.8125w + 94.375 \)

This equation allows us to predict the wing area for any given weight within the context of our data.

Data Interpretation

Data interpretation is crucial to verify and make predictions using our derived model. We must check if our linear model holds for all given data. One way to do this is by plugging in other data points into the model equation to see if it outputs values that align closely with actual observations.

• For example, substituting \( w = 6 \) into \( A(w) = 32.8125w + 94.375 \) results in \( A(6) = 330.25 \).

This is close to the actual data point \((6, 330)\), demonstrating model accuracy. Data interpretation also involves using the model for predictions — such as determining weight for a known wing area. For a wing area of 500 square inches:

• Set \( A(w) = 500 \), solve: \( 500 = 32.8125w + 94.375 \).
• This gives \[ w \approx 12.36 \] pounds.

Such predictions are only as accurate as the original data's linearity. Thus, understanding the constraints and context of the data is essential.