Problem 74 - Calculus for Biology and Medicine

Question

The following table is based on a functional relationship be tween \(x\) and \(y\) that is either an exponential or a power function: \begin{tabular}{lc} \hline \(\boldsymbol{x}\) & \(\boldsymbol{y}\) \\ \hline \(0.5\) & \(1.21\) \\ 1 & \(0.74\) \\ \(1.5\) & \(0.45\) \\ 2 & \(0.27\) \\ \(2.5\) & \(0.16\) \\ \hline \end{tabular} Use an appropriate logarithmic transformation and a graph to decide whether the table comes from a power function or an exponential function, and find the functional relationship between \(x\) and \(y\).

Step-by-Step Solution

Answer

The table comes from an exponential function, approximately \( y = 2.06 \times 0.36^x \).

Step 1: Recognize the Functional Forms

There are two possible functional forms: an exponential function, which can be expressed as \( y = ab^x \), or a power function, which can be expressed as \( y = ax^b \). To determine which form the data fits, we'll use logarithmic transformations.

Step 2: Perform Logarithmic Transformation for Exponential

For an exponential function \(y = ab^x\), taking the natural logarithm of both sides gives \( \ \ln y = \ln a + x \cdot \ln b \). Plot \( \ln y \) versus \( x \) and look for a linear relationship, which would indicate an exponential function.

Step 3: Transform Data and Plot for Exponential

Calculate \( \ln y \) for each given \( y \) value, then plot these values against the corresponding \( x \) values.- For \( x = 0.5, y = 1.21 \), \( \ln 1.21 \approx 0.19 \).- For \( x = 1, y = 0.74 \), \( \ln 0.74 \approx -0.30 \).- For \( x = 1.5, y = 0.45 \), \( \ln 0.45 \approx -0.80 \).- For \( x = 2, y = 0.27 \), \( \ln 0.27 \approx -1.31 \).- For \( x = 2.5, y = 0.16 \), \( \ln 0.16 \approx -1.83 \).Plot the points \((0.5, 0.19), (1, -0.30), (1.5, -0.80), (2, -1.31), (2.5, -1.83)\). Check the fit to a line.

Step 4: Logarithmic Transformation for Power Function

For a power function \(y = ax^b\), taking logarithms gives \( \ \ln y = \ln a + b \ln x \). Plot \( \ln y \) versus \( \ln x \) to see if there is a linear relationship, which indicates a power function.

Step 5: Transform Data and Plot for Power Function

Calculate \( \ln x \) and \( \ln y \):- For \( x = 0.5, y = 1.21 \), \( \ln x = -0.69 \), \( \ln y = 0.19 \).- For \( x = 1, y = 0.74 \), \( \ln x = 0 \), \( \ln y = -0.30 \).- For \( x = 1.5, y = 0.45 \), \( \ln x = 0.41 \), \( \ln y = -0.80 \).- For \( x = 2, y = 0.27 \), \( \ln x = 0.69 \), \( \ln y = -1.31 \).- For \( x = 2.5, y = 0.16 \), \( \ln x = 0.92 \), \( \ln y = -1.83 \).Plot these transformed values and determine if they form a linear relationship.

Step 6: Analyze Graphs and Determine Function Type

After plotting the transformed values from both transformations, compare the linearity of the graphs. The graph that exhibits a clearer straight-line relationship will indicate the form of the function (either exponential or power).

Step 7: Identify and Solve for Parameters

Assuming the exponential form best fits our data (which is apparent because it shows a straighter line), solve the fitted line \( \ln y = \ln a + x \cdot \ln b \) for \( b \) using two points, for example, (1, -0.30) and (2, -1.31).- Slope \( m = \frac{-1.31 - (-0.30)}{2 - 1} = -1.01 \), so \( \ln b = -1.01 \rightarrow b = e^{-1.01} \approx 0.36 \).- Substitute a point to find \( a \). Using \( x = 1 \), \( y = 0.74 \), we find \( 0.74 = a \cdot 0.36^1 \rightarrow a \approx 2.06 \).- Function is \( y \approx 2.06 \times 0.36^x \).

Key Concepts

Logarithmic Transformation

Logarithmic transformations are a key technique in analyzing data that might adhere to exponential or power functions. By transforming data, we can simplify complex relationships and lines become easier to identify. For exponential functions, the form is typically \(y = ab^x\). Applying a logarithmic transformation involves taking the natural log of both sides, resulting in \(\ln y = \ln a + x \cdot \ln b\). This transformation converts the exponential relationship into a linear one, where the slope and intercept can be easily determined.

A similar transformation applies to power functions, given by \(y = ax^b\). Taking the natural log gives \(\ln y = \ln a + b \ln x\). Here too, the power relationship becomes linear with a slope equivalent to the exponent \(b\). The logs make multiplicative relationships additive.

These transformations are crucial as they allow you to choose the best functional fit by comparing linearly transformed data.

Power Functions

Power functions describe relationships where the variable \(y\) depends on the variable \(x\) raised to some power. The general form is \(y = ax^b\), where:\

\
• \(a\) is a constant multiplier, affecting the vertical stretch of the function.
\
• \(b\) is the exponent, determining the slope's steepness and whether the function increases or decreases.
\

The beauty of power functions is their simplicity and ubiquity in natural phenomena, from population growth to physical laws like those governing gravity.

When dealing with datasets, applying a logarithmic transformation can reveal possible power relationships. Plotting \(\ln y\) against \(\ln x\) helps in identifying this function type. If the graph result suggests a straight line, the likelihood of an underlying power function is high. Understanding these transformations can simplify your analysis by allowing complex multiplicative relationships to be perceived in a straightforward manner.

Graphical Analysis

Graphical analysis is a visual method used to identify relationships within data and assess functions, such as exponential or power forms. Plotting data after logarithmic transformations, we utilize graphs to look for linear patterns. In practical applications, this helps distinguish which function best fits the data.

In graphical analysis, different plots are made for exponential and power functions to see which graph is most linear. The linearity of a transformed dataset visually confirms which functional form the original dataset follows:\

\
• For exponential transformations, plotting \(\ln y\) versus \(x\) is tested for linearity.
\
• For power functions, the line resulting from \(\ln y\) versus \(\ln x\) is analyzed.
\

This visual approach makes it easier to spot errors and inconsistencies as well as provide intuitive understanding through the manipulation and viewing of data in different lights. By doing so, the most suitable model for the data can be confidently selected, ensuring a more accurate depiction of the underlying relationship.