- Transform the power model \(y = a \cdot x^b\) into a linear form by taking the natural log: \(\ln(y) = \ln(a) + b \cdot \ln(x)\).
- Map the data points \((x, y)\) to their logarithmic counterparts \((X, Y)\).
- Perform a linear regression on these transformed points to determine the values of the intercept \(K = \ln(a)\) and the slope \(b\).
- Exponentiate the intercept to find the coefficient \(a\), resulting in the final model \(\boxed{y = 42.2326 \cdot x^{-0.4494}}\).
Explanation
- Linearizing the Power Model
To find the power regression model of the form \(y = a \cdot x^b\), we first linearize the equation. By taking the natural logarithm of both sides, we get:
\[\ln(y) = \ln(a) + b \cdot \ln(x)\]
Let \(Y = \ln(y)\), \(X = \ln(x)\), and \(K = \ln(a)\). This transforms the relationship into a linear equation:
\[Y = K + b \cdot X\]
where \(b\) is the slope and \(K\) is the y-intercept. We will use the given data points \((x, y)\) to find these constants.
The data provided is: \((15, 11.8), (30, 9.9), (45, 8.2), (60, 6.3), (75, 5.9)\).
- Performing Linear Regression
Transform the data points \((x, y)\) to \((X, Y)\) using the natural logarithm:
- \(\ln(15) \approx 2.7081, \ln(11.8) \approx 2.4681\)
- \(\ln(30) \approx 3.4012, \ln(9.9) \approx 2.2925\)
- \(\ln(45) \approx 3.8067, \ln(8.2) \approx 2.1041\)
- \(\ln(60) \approx 4.0943, \ln(6.3) \approx 1.8406\)
- \(\ln(75) \approx 4.3175, \ln(5.9) \approx 1.7750\)
Using these transformed values, we calculate the slope \(b\) and the intercept \(K\) for the line of best fit \(Y = K + bX\). Performing the least-squares linear regression calculations, we determine the slope \(b \approx -0.4494\) and the intercept \(K \approx 3.7432\).
- Determining the Regression Parameters
Now that we have \(K = \ln(a) \approx 3.7432\), we can solve for \(a\) by exponentiating:
\[a = e^K = e^{3.7432} \approx 42.2326\]
Using our values for \(a\) and \(b\) rounded to four decimal places, the power regression model is:
\[y = 42.2326 \cdot x^{-0.4494}\]
We substitute these values into the model form \(y = a \cdot x^b\).
- Final Power Regression Model
The power regression model for the given data, with constants rounded to four decimal places, is:
\[y = 42.2326 \cdot x^{-0.4494}\]
The final expression is \(\boxed{y = 42.2326 \cdot x^{-0.4494}}\).
Examples
Power regression is widely used in physics and engineering to model relationships that do not grow or decay linearly. For instance, if you are measuring the cooling rate of a liquid or the depletion of a resource over time, a power law might better describe the process than a linear trend. By identifying the parameters of such a model, you can make more accurate predictions about how the system will behave in the future, helping to optimize resource management or physical processes.