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An exponential model with unit adjustment: Show that the following data are exponential and find a formula for an exponential model. (Note: It will be necessary to make a unit adjustment. For this problem, round your answers to three decimal places.) $$ \begin{array}{|c|c|c|c|} \hline t & g(t) & t & g(t) \\ \hline 0 & 38.30 & 12 & 16.04 \\ \hline 4 & 28.65 & 16 & 11.99 \\ \hline 8 & 21.43 & 20 & 8.97 \\ \hline \end{array} $$

Short Answer

Expert verified
The data follows an exponential decay with formula \( g(t) = 38.30 \, e^{-0.102 \, t} \).

Step by step solution

01

Analyze the Data

Given the pairs \( (t, g(t)) \), check if the data can be fitted to an exponential model. Commonly, an exponential model takes the form \( g(t) = a \, e^{bt} \), where \( a \) and \( b \) are constants. Plotting the data, we see a rapid initial decrease in \( g(t) \), indicative of exponential decay.
02

Unit Adjustment for Better Fit

To enable a simpler model fitting, make a unit adjustment. Divide all \( t \) values by 4, so the new time unit is 4 times smaller. Now, calculate new \( t' = t/4 \), which results in the pairs (0, 38.30), (1, 28.65), (2, 21.43), (3, 16.04), and (4, 11.99), (5, 8.97). This adjustment helps in linearizing the data with respect to the model.
03

Linearize the Model

To verify the exponential nature, take the natural logarithm of \( g(t) \), which gives \( \, \ln(g(t)) = \ln(a) + b \cdot t \). We linearize by plotting \( t' \) against \( \ln(g(t)) \). This should yield a straight line if the data is indeed exponential.
04

Find the Line Equation from Logged Values

Using the pairs (0, \( \ln(38.30) \)), (1, \( \ln(28.65) \)), (2, \( \ln(21.43) \)), (3, \( \ln(16.04) \)), and so on, estimate the slope \( b \) and y-intercept \( \ln(a) \) using linear regression on the logged values. This involves finding the best-fit line to these points.
05

Calculate Exponential Model Parameters

The slope \( b \) obtained from regression and y-intercept \( \ln(a) \) give the exponential model in the form \( g(t') = e^{y-intercept} \cdot e^{(slope) \cdot t'} \). Calculate \( a \) by exponentiating the y-intercept and \( b \) directly from the slope.
06

Convert Back to Original Time Units

Since \( t' = t/4 \), translate the model back to the original \( t \) scale to find \( g(t) = a \, e^{(b/4) \, t} \). This adjustment accounts for the original time scale in the problem set.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Adjustment
When dealing with exponential models, unit adjustment plays a crucial role even though it may seem like a minor detail. In this exercise, we begin with the time values, given as \( t = 0, 4, 8, 12, 16, \) and \( 20 \). However, to help fit an exponential function more effectively, it makes sense to adjust the time unit. By dividing each \( t \) by 4, we simplify the calculations for constructing the model.

This adjustment doesn't alter the relationship between the variables, but it helps in examining trends more clearly. Now, our new time units \( t' \) become \( 0, 1, 2, 3, 4, \) and \( 5 \). These adjusted units are smaller by a factor of 4, allowing us to better see and interpret the exponential decay pattern. Unit adjustment is not just a mathematical trick; it’s a strategy to enhance clarity and improve data fitting.
Linearization
Linearization is a powerful technique used especially when dealing with curved datasets, like the exponential ones. By transforming the data, we can apply linear regression to find the best fit line.

In our exponential model, we transform the function using the natural logarithm so that an exponential decay equation \( g(t) = a \, e^{bt} \) becomes a linear one, \( \ln(g(t)) = \ln(a) + b \cdot t' \). This transformation transforms non-linear relationships into linear ones, which are much simpler to analyze and use statistical methods on.

When you plot the converted pairs \( (t', \ln(g(t'))) \), a straight line indicates a valid exponential relationship. This technique is crucial in validating the assumption of exponential behavior before proceeding with further analysis.
Exponential Decay
Exponential decay describes processes where quantities reduce at a constant multiplicative rate over equal time intervals. Commonly, exponential decay is used to model natural processes such as radioactive decay or cooling.

In our given data set, the rapid decrease in \( g(t) \) values initially suggests an exponential decay model. We observe decreasing values from 38.30 to 8.97 as time progresses, characteristic of exponential behavior. Remember, when modeled mathematically, exponential decay has the form \( g(t) = a \, e^{bt} \), with the constant \( b \) being negative, reflecting the decay.

Identifying such patterns in raw data is crucial; this understanding sets the stage for creating predictive models that capture the underlying processes of the phenomena.
Linear Regression
Linear regression is a method used to find the best-fitting straight line through a set of data points. It minimizes the differences between the observed and predicted values. In this exercise, after transforming the data using a natural logarithm, we apply linear regression to estimate the parameters of the exponential model.

By plotting \( t' \) against \( \ln(g(t')) \), we apply regression to find the slope \( b \) and y-intercept \( \ln(a) \). The slope represents the rate of exponential decay in the transformed model, while the y-intercept exponentiates back to give the scaling constant \( a \) in the original model.

Linear regression is one of the most foundational techniques in statistics and forms the basis for more complex analyses and forecasting models, offering insights into underlying data trends.

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Most popular questions from this chapter

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