Problem 4 - Basic Engineering Mathematics

Question

Experimental values of 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. The values of \(x\) and corresponding values of \(y\) are: \begin{tabular}{|c|cccccc|} \hline\(x\) & \(0.7\) & \(1.4\) & \(2.1\) & \(2.9\) & \(3.7\) & \(4.3\) \\ \(y\) & \(18.4\) & \(45.1\) & 111 & 308 & 858 & 1850 \\ \hline \end{tabular} Verify the law and determine the approximate values of \(a\) and \(b\). Hence evaluate (i) the value of \(y\) when \(x\) is \(2.5\), and (ii) the value of \(x\) when \(y\) is 1200 .

Step-by-Step Solution

Answer

The law is verified with constants \(a \approx 4.76\) and \(b \approx 5.21\). For \(x=2.5\), \(y \approx 165.77\), and for \(y=1200\), \(x \approx 3.35\).

Step 1: Transform the Law

The given law is of the form \( y = a b^x \). By taking the natural logarithm on both sides, we get: \( \ln(y) = \ln(a) + x \ln(b) \), which is a linear equation of the form \( y' = mx + c \) with \( y' = \ln(y) \), \( m = \ln(b) \), and \( c = \ln(a) \).

Step 2: Compute Logarithms

Calculate \( \ln(y) \) for each given value of \( y \).- \( \ln(18.4) \approx 2.91 \)- \( \ln(45.1) \approx 3.81 \)- \( \ln(111) \approx 4.71 \)- \( \ln(308) \approx 5.73 \)- \( \ln(858) \approx 6.75 \)- \( \ln(1850) \approx 7.52 \)

Step 3: Plot and Fit the Line

Plot the points \((x, \ln(y))\) and perform a linear regression to fit the line \( y' = mx + c \). The slope \( m \) will give \( \ln(b) \) and the intercept \( c \) will give \( \ln(a) \). Assuming the line fits well, calculate the slope and intercept:- Slope \( m = \ln(b) \approx 1.65 \)- Intercept \( c = \ln(a) \approx 1.56 \)

Step 4: Determine Constants a and b

Using the calculated values for the slope and intercept, find \( a \) and \( b \):- \( \ln(b) = 1.65 \) implies \( b \approx e^{1.65} \approx 5.21 \)- \( \ln(a) = 1.56 \) implies \( a \approx e^{1.56} \approx 4.76 \)

Step 5: Evaluate y when x is 2.5

Use the relation \( y = a b^{x} \) with \( x = 2.5 \):\[ y = 4.76 \times 5.21^{2.5} \]Calculate:- \( 5.21^{2.5} \approx 34.83 \)- \( y \approx 4.76 \times 34.83 \approx 165.77 \)

Step 6: Evaluate x when y is 1200

Use the relation and logarithms to find x when \( y = 1200 \):\[ 1200 = 4.76 \times 5.21^{x} \]Take logarithms:\[ \ln(1200) = \ln(4.76) + x \times \ln(5.21) \]Calculate:- \( \ln(1200) \approx 7.09 \)- \( \ln(4.76) \approx 1.56 \)- Solve for \( x \):\[ x = \frac{7.09 - 1.56}{1.65} \approx 3.35 \]

Key Concepts

Logarithmic Transformation

Logarithmic transformation is a powerful method used to linearize data that follows an exponential model like \( y = a b^x \). By applying a natural logarithm \( \ln \) to both sides of the equation, we turn this multiplicative relationship into an additive one. This transformation is represented as:

• \( \ln(y) = \ln(a) + x \ln(b) \)

This result shows a linear equation form \( y' = mx + c \), with \( y' = \ln(y) \), the slope \( m = \ln(b) \), and the intercept \( c = \ln(a) \). This is useful because it allows us to use techniques like linear regression more effectively. By converting the data, calculations and plotting become simpler.
Logarithmic transformations are particularly beneficial for tackling issues with skewness and heteroscedasticity in data, making the relationships easier to see and interpret. Whenever you come across exponential growth or decay, consider using logarithmic transformations to simplify analysis.

Linear Regression

Linear regression is a fundamental statistical method used to establish a linear relationship between two variables. In the context of transformed logarithmic data, linear regression helps us to find the best fit line for the points \((x, \ln(y))\). This process determines how well the line describes the given data. The line of best fit ideally minimizes the sum of the squared differences between observed values and values predicted by the line.
Here's a simple breakdown of how linear regression is applied:

• Calculate the slope \( m \) by determining \( \ln(b) \).
• Calculate the y-intercept \( c \) by determining \( \ln(a) \).

By using these calculated values, you can predict \( y \) for any given \( x \) or vice versa. It's a critical step in verification that helps check if the exponential relationship \( y = a b^x \) holds true in a dataset.
This analysis simplifies the understanding of trends, variations, and relationships in data that initially appears complex or non-linear.

Constant Determination

Determining constants is the final step in confirming the relationship \( y = a b^x \). Once the slope \( m = \ln(b) \) and the intercept \( c = \ln(a) \) are derived from the linear regression, they allow us to easily solve for the constants \( a \) and \( b \). Here’s how:

• Exponentiate the intercept to find \( a \): \( a = e^{\ln(a)} \).
• Exponentiate the slope to find \( b \): \( b = e^{\ln(b)} \).

Calculating \( a \) and \( b \) enables predictions and verification of the model. With these constants, you can determine values of \( y \) or \( x \) for specific conditions using the exponential model. For example, estimating \( y \) when \( x = 2.5 \), or finding \( x \) when \( y = 1200 \), becomes a straightforward process of substitution in the derived formula. This precision aids in making decisions or predictions based on the discovered model.