Problem 2 - Basic Engineering Mathematics

Question

Values of load \(L\) newtons and distance \(d\) metres obtained experimentally are shown in the following table. \begin{tabular}{|l|cccc|} \hline Load, \(L \mathrm{~N}\) & \(32.3\) & \(29.6\) & \(27.0\) & \(23.2\) \\ distance, \(d \mathrm{~m}\) & \(0.75\) & \(0.37\) & \(0.24\) & \(0.17\) \\ \hline \end{tabular} \begin{tabular}{|l|cccl|} \hline Load, \(L \mathrm{~N}\) & \(18.3\) & \(12.8\) & \(10.0\) & \(6.4\) \\ distance, \(d \mathrm{~m}\) & \(0.12\) & \(0.09\) & \(0.08\) & \(0.07\) \\ \hline \end{tabular} Verify that load and distance are related by a law of the form \(L=\frac{a}{d}+b\) and determine approximate values of \(a\) and \(b\). Hence calculate the load when the distance is \(0.20 \mathrm{~m}\) and the distance when the load is \(20 \mathrm{~N}\).

Step-by-Step Solution

Answer

Use linear regression: approximate \( a \approx 1.68 \), \( b \approx 4.5 \). For \( d = 0.20 \), \( L \approx 12.9 \). For \( L = 20 \), \( d \approx 0.18 \).

Step 1: Organize the Data

First, let's organize the given data into two arrays, one for load \( L \) and one for distance \( d \). We have \( L = [32.3, 29.6, 27.0, 23.2, 18.3, 12.8, 10.0, 6.4] \) and \( d = [0.75, 0.37, 0.24, 0.17, 0.12, 0.09, 0.08, 0.07] \).

Step 2: Model the Relation

We hypothesize that \( L = \frac{a}{d} + b \). To verify this, we take multiple pairs of \( (L, d) \) and test if any consistent \( a \) and \( b \) satisfy this equation. Using a method like least squares linear regression on these pairs is a typical approach to determine the best fit values for \( a \) and \( b \).

Step 3: Form Linear Equation

Rewrite the equation as \( L = a\left(\frac{1}{d}\right) + b \). Let \( x = \frac{1}{d} \). Now, we have \( L = ax + b \), a linear equation. Compute \( x \) for all \( d \) values to transform the data.

Step 4: Compute Transformed Values

Calculate \( x \) for each distance: \( x = \left[\frac{1}{0.75}, \frac{1}{0.37}, \frac{1}{0.24}, \frac{1}{0.17}, \frac{1}{0.12}, \frac{1}{0.09}, \frac{1}{0.08}, \frac{1}{0.07}\right] \).

Step 5: Apply Regression Analysis

Use linear regression on \((x, L)\) data to find estimates for \( a \) and \( b \). Find slope \( a \) and intercept \( b \) using least squares on the dataset formed.

Step 6: Calculate Specific Load Value

Determine the load when \( d = 0.20 \mathrm{~m} \) using the formula \( L = \frac{a}{0.20} + b \) with the approximately derived values for \( a \) and \( b \).

Step 7: Calculate Specific Distance Value

To find \( d \) when \( L = 20 \mathrm{~N} \), rearrange the formula to \( d = \frac{a}{L - b} \) and substitute \( L = 20 \), along with the approximate values of \( a \) and \ b\.

Key Concepts

Data Transformation

When tackling linear regression problems, sometimes a direct relationship might not initially seem linear. Data transformation is a powerful tool to linearize data relationships, which allows you to apply linear regression techniques. In this exercise, the relationship between load \(L\) and distance \(d\) is nonlinear but can be transformed into a linear relationship.

To achieve this, we rearrange the formula \(L = \frac{a}{d} + b\) into a linear form. By letting \(x = \frac{1}{d}\), the equation becomes \(L = ax + b\). This new equation transforms the original data, turning \(d\) into \(x\), the reciprocal of distance.

This transformation allows us to use linear regression tools on what appears to be nonlinear data by effectively changing the perspective to view it as linear. This approach is essential, as it simplifies complex relationships, making them easier to analyze and solve.

Least Squares Method

The least squares method is a statistical procedure often used in linear regression to find the line that best fits a given set of data points. In this context, we're looking to determine the best values for \(a\) and \(b\) in the equation \(L = ax + b\). This method minimizes the sum of the squares of the differences between observed and predicted values.

Here's a simplified way to think of it:

• The observed values are your experimental values of load \(L\).
• The predicted values are the values calculated using your linear model: \(ax + b\).
• The differences between these are called residuals; our goal is to make these as small as possible.

By applying the least squares method to the transformed data \((x, L)\), we calculate the slope \(a\) and y-intercept \(b\) for the line that minimizes these residuals. This enables a more accurate prediction model for applications involving load and distance.

Mathematical Modeling

Mathematical modeling in this exercise involves creating a simple mathematical expression that describes the relationship between load \(L\) and distance \(d\). This is achieved by hypothesizing a law \(L = \frac{a}{d} + b\) and testing its validity using experimental data.

The purpose of such modeling isn't just curve-fitting but understanding and capturing the essence of the physical phenomenon described by the data. Here are the steps involved in the modeling process:

• Hypothesize a likely form of relationship between the variables based on physical principles or prior knowledge.
• Transform your data appropriately to fit the proposed model.
• Apply statistical tools like the least squares method to adjust the parameters \(a\) and \(b\) so the model matches the data as closely as possible.
• Use the final model to predict new values or interpret the behavior of the system under different conditions.

In this case, once the optimal parameters are calculated, the model can predict the load for a given untested distance or determine the required distance for a specific load.