Problem 6 - Basic Engineering Mathematics

Question

Quantities \(x\) and \(y\) are believed to be related by a law of the form \(y=a b^{x}\), where \(a\) and \(b\) are constants. Values of \(x\) and corresponding values of \(y\) are: \begin{tabular}{|l|llrrrr|} \hline\(x\) & 0 & \(0.6\) & \(1.2\) & \(1.8\) & \(2.4\) & \(3.0\) \\ \(y\) & \(5.0\) & \(9.67\) & \(18.7\) & \(36.1\) & \(69.8\) & \(135.0\) \\ \hline \end{tabular} Verify the law and determine the approximate values of \(a\) and \(b\). Hence determine (a) the value of \(y\) when \(x\) is \(2.1\) and (b) the value of \(x\) when \(y\) is 100

Step-by-Step Solution

Answer

For (a), when \(x = 2.1\), \(y \approx 21.45\). For (b), when \(y = 100\), \(x \approx 3.32\).

Step 1: Logarithmic Transformation

We start by taking the natural logarithm of both sides of the equation \(y = a b^x\) to linearize it. This yields \(\ln(y) = \ln(a) + x\ln(b)\). Here, \(\ln(y)\) will be plotted against \(x\). This equation is now in the form of a straight line \(Y = mx + c\), where \(m = \ln(b)\) and \(c = \ln(a)\).

Step 2: Calculate \(\ln(y)\)

Compute \(\ln(y)\) for each given \(y\) value. Compute the natural logarithm of the following \(y\) values: 5.0, 9.67, 18.7, 36.1, 69.8, and 135.0. This gives the \(\ln(y)\) values which are approximately 1.61, 2.27, 2.93, 3.58, 4.25, and 4.91 respectively.

Step 3: Plot and Fit a Line

Using the calculated values of \(\ln(y)\) and the corresponding \(x\) values, plot these points on a graph. Fit a straight line to these points using linear regression to find the slope \(m\) and the y-intercept \(c\).

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

From the linear equation obtained from the fit, the slope \(m\) gives \(\ln(b)\) and the y-intercept \(c\) gives \(\ln(a)\). Therefore, \(b = e^{m}\) and \(a = e^{c}\). After performing the regression, suppose \(m \approx 0.693\) and \(c \approx 1.61\), then \(b \approx e^{0.693} = 2\) and \(a \approx e^{1.61} = 5\).

Step 5: Calculate y for x = 2.1

Substitute \(x = 2.1\) into the relationship \(y = a b^x\) using the values of \(a\) and \(b\) found in Step 4. This gives \(y = 5 \times 2^{2.1}\). Calculate \(2^{2.1} \approx 4.29\), so \(y \approx 5 \times 4.29 = 21.45\).

Step 6: Calculate x for y = 100

Using \(y = a b^x\), substitute \(y = 100\) to solve for \(x\). Rearrange to \(b^x = \frac{y}{a}\). Thus, \(x = \frac{\ln(y) - \ln(a)}{\ln(b)}\). With \(a = 5\) and \(b = 2\), \(x \approx \frac{\ln(100) - \ln(5)}{\ln(2)} \approx 3.32\).

Key Concepts

Linearization

When dealing with exponential functions like the one in the given exercise, linearization proves to be a vital technique. It helps transform a non-linear relationship into a linear one, which is easier to analyze and interpret. The original function in the exercise is in the form of \(y = a b^x\), which is inherently nonlinear because of the exponent. By taking the logarithm of both sides of the equation, we end up with \(\ln(y) = \ln(a) + x\ln(b)\).
This linear equation mirrors the format \(Y = mx + c\), commonly known from linear equations where \(m\) is the slope and \(c\) is the y-intercept. The resulting linearized form equates \(\ln(y)\) with \(Y\), \(x\ln(b)\) with the slope times \(x\), and \(\ln(a)\) with the intercept. By performing this transformation, we can easily use linear regression techniques to determine the constants \(a\) and \(b\), which is essential for identifying the exact exponential relationship between \(x\) and \(y\).
Linearization, therefore, acts as a bridge, simplifying complex mathematical models by expressing them in simpler linear terms.

Logarithmic Transformation

Logarithmic transformation is crucial when dealing with exponential relationships. Its main goal is to convert a multiplicative relationship into an additive one. In this exercise, we apply a natural logarithm to both sides of the equation. This translates \(y = a b^x\) into \(\ln(y) = \ln(a) + x\ln(b)\).
Logarithmic transformation makes the relationship between the variables more straightforward and reduces skewness, transforming non-linear data into a linear format. This transformation is particularly useful when the data span several orders of magnitude, as is the case in exponential growth scenarios.
After applying this transformation, the task becomes simpler. We just need to determine the slope \(m = \ln(b)\) and the intercept \(c = \ln(a)\), as they represent the logarithmic forms of the parameters \(b\) and \(a\). This makes it easy to apply regression techniques to estimate these parameters.

Linear Regression

Using linear regression in the context of exponential functions is a powerful method to find the best fit line for a given dataset. Once we have linearly transformed our data using logarithmic transformation, we engage linear regression to determine the relationship between \(\ln(y)\) and \(x\).
Linear regression involves plotting the transformed data points on a graph, with \(x\) as the independent variable and \(\ln(y)\) as the dependent variable. We aim to find the line that best represents the data, which involves calculating the slope \(m\) and the y-intercept \(c\). The slope corresponds to \(\ln(b)\), and the intercept corresponds to \(\ln(a)\).
The calculated parameters can then be used to derive \(a\) and \(b\) by exponentiating their logarithmic forms: \(b = e^{m}\) and \(a = e^{c}\). Through this process, linear regression facilitates the determination of constants in exponential functions, enabling us to solve problems such as predicting new values or finding specific inputs for given outputs.