Problem 130 - College Algebra with Modeling and Visualization

Question

Predicting Wind Speed Wind speed typically varies in the first 20 meters above the ground. Close to the ground, wind speed is often less than it is at 20 meters above the ground. For this reason, the National Weather Service usually measures wind speeds at heights between 5 and 10 meters. For a particular day, let the formula \(f(x)=1.2 \ln x+2.3\) compute the wind speed in meters per second at a height \(x\) meters above the ground for \(x \geq 1 .\) (IMAGE CANNOT COPY) (a) Find the wind speed at a height of 5 meters. (b) Graph \(f\) in the window \([0,20,5]\) by \([0,7,1]\) Interpret the graph. (c) Estimate the height where the wind speed is 5 meters per second.

Step-by-Step Solution

Answer

(a) 4.23 m/s at 5 meters; (c) estimated height for 5 m/s is 9.49 meters.

Step 1: Finding the wind speed at 5 meters

To find the wind speed at a height of 5 meters, substitute \(x = 5\) into the formula \(f(x) = 1.2 \ln x + 2.3\). This gives us:\[ f(5) = 1.2 \ln 5 + 2.3 \]First find \( \ln 5 \approx 1.6094 \), then multiply it by 1.2 to get \(1.9313\). Now add 2.3:\[ f(5) = 1.9313 + 2.3 = 4.2313 \]Thus, the wind speed at 5 meters is approximately 4.23 meters per second.

Step 2: Graphing the function

The function \(f(x) = 1.2 \ln x + 2.3\) needs to be graphed over the range \([0, 20]\) for \(x\) and \([0, 7]\) for \(f(x)\) in steps of 5 for \(x\) and 1 for \(f(x)\).- Start by calculating values of \(f(x)\) at several key points, such as \(x = 1, 5, 10, 15, 20\).- Use a calculator or software to get these values: * \(f(1) = 1.2 \ln 1 + 2.3 = 2.3\) * \(f(10) = 1.2 \ln 10 + 2.3 \approx 5.06\) * \(f(15) = 1.2 \ln 15 + 2.3 \approx 5.93\) * \(f(20) = 1.2 \ln 20 + 2.3 \approx 6.40\)- Use these points to draw the curve on the graph.The graph will be logarithmic, starting at 2.3 when \(x=1\) and gradually increasing as \(x\) increases.

Step 3: Interpreting the graph

The graph of the function \(f(x) = 1.2 \ln x + 2.3\) shows that as the height \(x\) increases, the wind speed also increases. This is consistent with the logarithmic nature of the function, which grows slowly. We can see that wind speed starts at 2.3 m/s when \(x = 1\), increases more significantly between \(x = 5\) and \(x = 20\).

Step 4: Estimating the height for a wind speed of 5 m/s

Set \(f(x) = 5\) to find the height where the wind speed is 5 m/s.\[ 1.2 \ln x + 2.3 = 5 \]Subtract 2.3 from both sides:\[ 1.2 \ln x = 2.7 \]Divide by 1.2:\[ \ln x = 2.25 \]Exponentiate both sides to solve for \(x\):\[ x = e^{2.25} \approx 9.49 \]Thus, the wind speed is estimated to be 5 m/s at approximately 9.49 meters above the ground.

Key Concepts

Logarithmic Functions

Logarithmic functions are a cornerstone in understanding various natural phenomena, including wind speed prediction. A logarithm is essentially the inverse of exponentiation. In simpler terms, if you think about raising a number to an exponent to get another number, a logarithm asks the question "What power do I need to raise this base to, to get this other number?" When you see a logarithmic function, like \( f(x) = 1.2 \ln x + 2.3 \), it represents a relationship where the rate of increase slows as the independent variable increases. This is crucial for modeling things that grow quickly and then taper off, like wind speed increasing with height. With each additional meter, the increase in wind speed gets smaller compared to the previous meters climbed.

Graphing Techniques

Graphing a function like \( f(x) = 1.2 \ln x + 2.3 \) provides a visual representation of wind speed as height changes. When graphing logarithmic functions, it's useful to note these characteristics:

• The curve starts at a basic value (in our function, it begins at 2.3 when \( x = 1 \))
• It rises slowly, reflecting the natural logarithmic progression
• Key values along the \( x \)-axis should be selected to get a comprehensive shape of the graph (e.g., points at \( x = 5, 10, 15, \, \text{and} \, 20 \))

To plot the graph effectively:

• Calculate \( f(x) \) at each chosen \( x \)
• Mark these points on a coordinate plane
• Draw a smooth curve connecting the points

This visual aid makes it easier to predict wind speeds at other heights and illustrates the function's growth.

Mathematical Modeling

Mathematical modeling involves creating equations that represent real-world situations. Here, we're modeling how wind speed changes with height using \( f(x) = 1.2 \ln x + 2.3 \). This function simulates:

• Types of growth that are not constant but slow as the height increases
• Complex natural phenomena in a manageable way

Mathematical models let us predict outcomes, test hypotheses, and understand the underlying mechanics of a system. In this type of modeling:

• It's crucial to select appropriate parameters (like coefficients "1.2" and "2.3") to ensure the model aligns well with observed data
• We interpret the meaning of the parameters: Here, "1.2" influences how rapidly the speed changes, while "2.3" is the initial wind speed at 1 meter

These models can be adjusted as more data becomes available, enhancing their accuracy.

Algebraic Equations

Algebraic equations serve as the foundation for predicting and solving for unknown values. In our wind speed prediction scenario, setting up and solving the equation \( 1.2 \ln x + 2.3 = 5 \) helps to discover at what height the wind speed is 5 m/s. Follow these steps to solve:

• Isolate the logarithmic component: Subtract constants from both sides
• Divide to find \( \ln x \)
• Use the inverse operation (exponentiation) to solve for \( x \)

The result, approximately 9.49 meters, provides insight into the specific height where the desired wind speed is achieved. Solving algebraic equations often involves rearranging terms, performing arithmetic operations, and applying inverse functions—all vital skills for problem-solving in mathematics and physics.