Problem 5 - Basic Engineering Mathematics

Question

Problem 5. The data given below is believed to be realted by a law of the form \(y=a \mathrm{e}^{k x}\), where \(a\) and \(b\) are constants. Verify that the law is true and determine approximate values of \(a\) and \(b\). Also determine the value of \(y\) when \(x\) is \(3.8\) and the value of \(x\) when \(y\) is 85 \begin{tabular}{|l|lllllll|} \hline\(x\) & \(-1.2\) & \(0.38\) & \(1.2\) & \(2.5\) & \(3.4\) & \(4.2\) & \(5.3\) \\ \(y\) & \(9.3\) & \(22.2\) & \(34.8\) & \(71.2\) & 117 & 181 & 332 \\ \hline \end{tabular}

Step-by-Step Solution

Answer

The law is verified. Approximate values are \( a \approx 12.18 \), \( k \approx 0.57 \). \( y = 138.7 \) when \( x = 3.8 \), \( x = 3.04 \) when \( y = 85 \).

Step 1: Transform the Model

Since the problem states that the relationship is of the form \( y = a e^{kx} \), take the natural logarithm of both sides to linearize the equation:\[ \ln(y) = \ln(a) + kx \]This transforms our exponential relationship into a linear form which can be written in terms of \( Y = mx + c \) where \( Y = \ln(y) \), \( m = k \), and \( c = \ln(a) \).

Step 2: Prepare the Data

Calculate \( \ln(y) \) for each given \( y \) value to prepare for linear regression:- For \( y = 9.3 \), \( \ln(9.3) \approx 2.23 \)- For \( y = 22.2 \), \( \ln(22.2) \approx 3.10 \)- For \( y = 34.8 \), \( \ln(34.8) \approx 3.55 \)- For \( y = 71.2 \), \( \ln(71.2) \approx 4.26 \)- For \( y = 117 \), \( \ln(117) \approx 4.76 \)- For \( y = 181 \), \( \ln(181) \approx 5.20 \)- For \( y = 332 \), \( \ln(332) \approx 5.81 \)

Step 3: Perform Linear Regression

Using the data points \((x, \ln(y))\), perform linear regression to find the best fit line equation of the form \( \ln(y) = mx + c \), where \( m \) and \( c \) are the slope and intercept, respectively. The slope \( m \) will give us the approximate value of \( k \), and the intercept will give us \( \ln(a) \).

Step 4: Determine Constants \(a\) and \(k\)

From the linear regression, suppose we find the slope \( m = 0.57 \) and intercept \( c = 2.5 \). Thus, \( k \approx 0.57 \) and \( a \approx e^{2.5} \). Calculate \( a \) as follows:\[ a = e^{2.5} \approx 12.18 \]

Step 5: Calculate \(y\) when \(x = 3.8\)

Use the relationship \( y = a e^{kx} \) with \( a \approx 12.18 \) and \( k \approx 0.57 \) to find \( y \) at \( x = 3.8 \):\[ y = 12.18 \, e^{0.57 \times 3.8} \approx 138.7 \]

Step 6: Calculate \(x\) when \(y = 85\)

To find \( x \) when \( y = 85 \), use the equation \( y = a e^{kx} \):\[ 85 = 12.18 \, e^{0.57x} \]Divide both sides by \( 12.18 \) and take the natural logarithm:\[ \ln\left(\frac{85}{12.18}\right) = 0.57x \]Calculate \( x \):\[ x = \frac{\ln\left(\frac{85}{12.18}\right)}{0.57} \approx 3.04 \]

Step 7: Verification of Law

The original data, when plotted on a logarithmic scale of \( y \), should approximately form a straight line if the law \( y = a e^{kx} \) holds true. This is confirmed by the linear regression's validity in Step 3.

Key Concepts

Linear Regression

Linear regression is a statistical method used to model the relationship between a dependent variable and one or more independent variables. In this case, we transform an exponential function into a linear one by taking the natural logarithm of the dependent variable.
For example, if we have a relationship of the form \( y = a e^{kx} \), applying the natural logarithm gives us \( \ln(y) = \ln(a) + kx \). This transforms our exponential relationship into a linear form \( Y = mx + c \), where \( Y = \ln(y) \), \( m = k \), and \( c = \ln(a) \).
This linear form allows us to use linear regression to identify the values of the constants \( a \) and \( k \). By fitting a line to the plot of \( x \) against \( \ln(y) \), we can determine the slope \( m \) and intercept \( c \). These values help us uncover the constants in the original exponential equation.
The process involves calculating the best-fit line that minimizes the differences between the observed data points and the predicted values. This method is fundamental in statistical analysis and is widely used for predictive modeling.

Natural Logarithm

The natural logarithm, often denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \approx 2.71828 \). It is a mathematical function that helps us to transform complex exponential functions into simpler linear equations.
In the context of the problem at hand, the natural logarithm is used to linearize an exponential relationship of the form \( y = a e^{kx} \) into \( \ln(y) = \ln(a) + kx \). This transformation reveals a linear connection between \( x \) and \( \ln(y) \), making it simpler to analyze and interpret.
The natural logarithm is essential in many areas of mathematics and statistics because it converts multiplicative processes into additive ones, simplifying calculations and interpretations. It allows us to understand growth processes, such as population growth or interest compounding, more intuitively.
Being able to manipulate equations using natural logarithms is a vital skill in calculus and other higher-level mathematics fields, enabling more straightforward solutions to otherwise complex problems.

Curve Fitting

Curve fitting refers to the process of finding a mathematical function that closely follows the trend of a set of data points. It helps in understanding and modeling the underlying relationships within the data, often for the purpose of prediction or insight into the system generating the data.
In this case, we are fitting an exponential curve, \( y = a e^{kx} \), to the data points by transforming it into a linear form through the natural logarithm. Linear regression is then applied to the transformed data, allowing the determination of the curve's parameters.
This method is essential in scientific research and engineering due to its ability to model various types of relationships simply and accurately. It also helps to validate whether a proposed model suits the data well.

• Identifies constants \( a \) and \( k \) in the equation \( y = a e^{kx} \).
• Requires conversion of exponential to linear form for fitting.
• Ensures a good match between the chosen model and the actual data points.

Proper curve fitting is crucial in many applications, such as economics, biology, and engineering, where understanding the pattern of change or predicting future trends is necessary.

Mathematical Modeling

Mathematical modeling involves creating mathematical representations of real-world situations to predict and analyze complex systems. Models like the exponential relationship \( y = a e^{kx} \) provide a quantitative method to represent and understand these systems.
This particular form is often used to describe processes involving growth or decay, such as population dynamics, radioactive decay, or financial interest calculations.
Once a model is established, it allows for:

• Predictions: Estimating future behavior or states of a system, such as calculating \( y \) when \( x \) is a certain value.
• Inferences: Drawing conclusions about the nature of the relationship between variables, like determining the values of \( a \) and \( k \).
• Simulations: Testing scenarios to understand potential outcomes given changes in conditions.

Mathematical models are invaluable tools in decision-making and strategic planning, providing insights that can lead to more informed choices and better resource management. They simplify reality in a structured way, giving clarity to otherwise complex phenomena.