Problem 16 - Applied Calculus

Question

Values of \(f(t)\) are given in the following table. (a) Does this function appear to have a positive or negative first derivative? Second derivative? Explain. (b) Estimate \(f^{\prime}(2)\) and \(f^{\prime}(8)\) $$\begin{array}{c|c|c|c|c|c|c}\hline t & 0 & 2 & 4 & 6 & 8 & 10 \\\\\hline f(t) & 150 & 145 & 137 & 122 & 98 & 56 \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

(a) Negative first and negative second derivative; (b) \(f^{\prime}(2) \approx -3.25\), \(f^{\prime}(8) \approx -16.5\).

Step 1: Understanding the Function

The table lists values of the function \(f(t)\) at various points \(t\). Our task is to analyze this data to find information about the derivatives of \(f(t)\), which give us insights into the rate of change of the function at these points.

Step 2: Analyzing the First Derivative Sign

Observe the function values: \(f(t) = 150, 145, 137, 122, 98, 56\). They consistently decrease as \(t\) increases, indicating a decreasing trend. Therefore, the first derivative \(f'(t)\) appears to be negative, indicating a negative rate of change.

Step 3: Analyzing the Second Derivative Sign

Calculate the differences between successive values: \(f(2)-f(0)=145-150=-5\), \(f(4)-f(2)=137-145=-8\), \(f(6)-f(4)=122-137=-15\), \(f(8)-f(6)=98-122=-24\), \(f(10)-f(8)=56-98=-42\). Each difference is more negative than the last, suggesting that the rate of decrease is increasing. Thus, the second derivative \(f''(t)\) may be negative, indicating a concave down shape.

Step 4: Estimating \(f^{\prime}(2)\)

To estimate the derivative at \(t=2\), use values of \(f\) at adjacent points. Choose points around \(t=2\): Use \(f(4)-f(2)\) and \(f(2)-f(0)\):\[f^{\prime}(2) \approx \frac{f(4) - f(0)}{4 - 0} = \frac{137 - 150}{4} = \frac{-13}{4} = -3.25\]

Step 5: Estimating \(f^{\prime}(8)\)

To estimate the derivative at \(t=8\), use values at \(t = 6\) and \(t = 10\): \[f^{\prime}(8) \approx \frac{f(10) - f(6)}{10 - 6} = \frac{56 - 122}{4} = \frac{-66}{4} = -16.5\]

Key Concepts

First Derivative

The first derivative of a function provides important information about its behavior. It's the slope of the tangent line to the function at a specific point, indicating how the function's value changes as the input varies. In the context of our exercise, we observe that the function values decrease consistently from 150 at \(t = 0\) to 56 at \(t = 10\). This steady decrease suggests that the first derivative, also known as \( f'(t) \), is negative. A negative first derivative implies that as \( t \) increases, the function \( f(t) \) is decreasing overall. - This tells us the function is moving downward.- In real-world terms, if \( f(t) \) represented distance, the object would be slowing down or moving backwards.Therefore, analyzing the first derivative lets us see how the function behaves over its domain, specifically showing a negative rate of change.

Second Derivative

The second derivative provides deeper insights into a function's changing behavior. It is the derivative of the first derivative, meaning it reflects how the rate of change itself is changing. In our analysis, after computing the differences in function values, we saw increasingly negative differences: -5, -8, -15, -24, and -42. This pattern suggests that the rate at which \( f(t) \) is decreasing grows larger in magnitude. Calling on the concept of concavity:- A negative second derivative indicates the function is concave down.- This means the slope of \( f(t) \) is decreasing faster as \( t \) increases.Understanding the second derivative aids in predicting how a graph might curve and guides in identifying points of inflection where the graph's curvature changes.

Rate of Change

The rate of change in a function involves understanding how much the function's output changes for a change in input. This exercise shows a clear picture of a negative rate of change. By observing the first derivative, we identified a consistent downward trend. Key points about rate of change include:- It's calculated as the difference in function values divided by the difference in inputs: - Example: Between \( t = 0 \) and \( t = 2 \), - Rate = \( \frac{145-150}{2-0} = -2.5 \)By analyzing this rate, we gain comprehension of how quickly a function's value rises or falls, crucial in fields such as physics for velocity and economics for cost functions.

Function Values

Function values are the specific outputs of a function at given inputs. In this exercise, they were provided as fixed points like \( f(0) = 150 \), \( f(2) = 145 \), and so on. These values act as a foundation for analyzing trends in a function and determining derivatives. Recognizing function values means:- Knowing specific outputs helps estimate derivatives at those points.- They allow us to see overall trends in the function's graph, helping predict behavior.When we look at these given function values, trends such as increasing, decreasing, or fluctuating behavior can be easily spotted, paving the way for deeper derivative analysis.

Estimation of Derivatives

Estimating derivatives from table data is a practical approach when exact formulas are unavailable. This exercise demonstrated this through numerical methods by selecting adjacent points to approximate derivatives. For instance, to estimate \( f'(2) \), points at \( t=0 \) and \( t=4 \) were utilized:- Calculation: \[ f^{\prime}(2) \approx \frac{f(4) - f(0)}{4 - 0} = \frac{-13}{4} = -3.25 \]- This estimation technique is known as the difference quotient.Similar calculations help us estimate \( f'(8) \). By choosing nearby function values to compute the slope, we obtain a reasonable approximation of the derivative at any tabled point. This method is essential when dealing with empirical data lacking a straightforward analytical form.