Problem 65 - A Graphical Approach to Precalculus with Limits

Question

Rainbow trout are sensitive to zinc ions in the water. High concentrations are lethal. The average survival times \(x\) in minutes for trout in various concentrations of zinc ions \(y\) in milligrams per liter (mg/L) are listed in the table. $$\begin{array}{|c|c|c|c|c|} \hline x \text { (in minutes) } & 0.5 & 1 & 2 & 3 \\ \hline y \text { (in mg/L) } & 4500 & 1960 & 850 & 525 \end{array}$$ (a) The data can be modeled by $$ f(x)=a x^{b} $$ where \(a\) and \(b\) are constants. Determine \(a\). (Hint: Let \(f(1)=1960 .)\) (b) Estimate \(b\) (c) Evaluate \(f(4)\) and interpret the result.

Step-by-Step Solution

Answer

(a) \(a = 1960\). (b) \(b \approx -1.2\). (c) \(f(4) \approx 360\). It suggests zinc concentration is 360 mg/L.

Step 1: Understanding the Problem

We are given a mathematical model of the form \(f(x) = ax^b\) representing the average survival times \(x\) of trout for varying zinc ion concentrations \(y\). Our task is to determine the constants \(a\) and \(b\) using the data provided, and then evaluate the function for a new input \(x=4\).

Step 2: Solving for Constant \(a\)

We know \(f(1) = 1960\). Substituting this into the equation gives us \(1960 = a \cdot 1^b\). Since \(1^b = 1\), the equation simplifies to \(a = 1960\).

Step 3: Estimating Constant \(b\)

We use another data point to find \(b\). Let's use the point \((2, 850)\). Substituting into the model gives \(850 = 1960 \cdot 2^b\). Solving for \(b\), we get: \(2^b = \frac{850}{1960}\) or \(b = \log_2\left(\frac{850}{1960}\right)\). Use a calculator to find \(b\).

Step 4: Calculation of \(b\) using log

Calculate \(b\): \(b = \log_2(0.4337) \approx -1.2\). Now we have both \(a = 1960\) and \(b \approx -1.2\).

Step 5: Evaluate \(f(4)\)

Substitute \(x = 4\) into the model: \(f(4) = 1960 \cdot 4^{-1.2}\). Use a calculator to find \(f(4) \approx 360\).

Step 6: Interpretation of \(f(4)\)

The value \(f(4) \approx 360\) means that if the survival time of the trout is 4 minutes, the concentration of zinc in the water is approximately 360 mg/L.

Key Concepts

Understanding Exponential Functions in Mathematical Modeling

Exponential functions are essential in mathematical modeling because they can describe how one quantity changes in relation to another. In our exercise, the average survival time of trout, denoted as \( x \), varies with different concentrations of zinc ions, denoted as \( y \). This relationship is expressed by the exponential function \( f(x) = ax^b \). Here, \( a \) is a scaling constant, and \( b \) dictates how the rate of survival changes as zinc concentrations change.

• Exponential functions can grow rapidly, either increasing or decreasing.
• They are common in situations where one quantity affects another progressively, such as population growth or radioactive decay.
• The base \( x \) raised to the power of \( b \) determines the type of curve – steep or gradual.

In this scenario, the function helps us understand how sensitive the trout are to different zinc ions levels through their survival time.

Survival Analysis and Its Implications

Survival analysis is a statistical approach used to analyze the expected duration of time until one or more events happen, such as death in biological organisms. In the context of our exercise, it helps us understand the time frame over which trout can survive under different levels of zinc ion concentration.

Using the function \( f(x) = ax^b \), we can predict survival times by estimating the values of \( a \) and \( b \). This technique provides a way of modeling the leafing environment's harshness in terms of trout survival.

• We can distinguish differences in survival rates at various concentrations.
• Can help determine safe levels of zinc to ensure the health of aquatic organisms.
• Helps in planning conservation or remediation efforts accordingly.

By evaluating different scenarios using our model, we gain insights into likely outcomes, thus facilitating decision-making in environmental protection.

Applying Data Analysis in Problem Solving

Data analysis plays a pivotal role in translating raw data into functional insights. In this exercise, we have experimental data points relating survival time to zinc ion concentration.

By applying data analysis techniques, we developed estimates for the function's constants, \( a = 1960 \) and \( b \approx -1.2 \), enabling predictions of trout survival at any given concentration. Here's why data analysis is crucial:

• Data analysis enables identification of patterns and trends that are not immediately visible.
• Helps refine mathematical models by providing input for parameters.
• Offers evidence-based conclusions that support scientific inquiries and hypotheses.

Data analysis not only confirms the modeling's integrity but is also instrumental in accurately interpreting results, such as predicting a concentration of about 360 mg/L when survival is calculated for 4 minutes.