Problem 11
Question
Use the values given for each function. (a) Does the derivative of the function appear to be positive or negative over the given interval? Explain. (b) Does the second derivative of the function appear to be positive or negative over the given interval? Explain. $$ \begin{array}{c|c|c|c|c|c|c} \hline t & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline s(t) & 12 & 14 & 17 & 20 & 31 & 55 \\ \hline \end{array} $$
Step-by-Step Solution
Verified Answer
(a) The derivative is positive. (b) The second derivative is positive.
1Step 1: Calculate Changes in Function Values
To determine whether the derivative \( s'(t) \) is positive or negative, examine the change in function values \( s(t) \) over the given intervals. Compute the differences \( \Delta s = s(t) - s(t-1) \). The values are as follows:- \( \Delta s(1) = 14 - 12 = 2 \)- \( \Delta s(2) = 17 - 14 = 3 \)- \( \Delta s(3) = 20 - 17 = 3 \)- \( \Delta s(4) = 31 - 20 = 11 \)- \( \Delta s(5) = 55 - 31 = 24 \)
2Step 2: Analyze the First Derivative
All the computed differences \( \Delta s \) are positive, indicating that the function consistently increases over each interval. Therefore, the first derivative \( s'(t) \) appears to be positive over the entire interval \( t = 0 \) to \( t = 5 \).
3Step 3: Calculate Changes in Differences
To analyze the second derivative \( s''(t) \), examine whether the rates of change \( \Delta s \) themselves are increasing or decreasing. Calculate the differences between consecutive \( \Delta s \):- \( \Delta^2 s(2) = 3 - 2 = 1 \)- \( \Delta^2 s(3) = 3 - 3 = 0 \)- \( \Delta^2 s(4) = 11 - 3 = 8 \)- \( \Delta^2 s(5) = 24 - 11 = 13 \)
4Step 4: Analyze the Second Derivative
All the differences \( \Delta^2 s \) are either positive or zero, indicating an increasing trend in the rates of change of \( s(t) \). This suggests that the second derivative \( s''(t) \) appears to be positive over the interval \( t = 0 \) to \( t = 5 \).
Key Concepts
First DerivativeSecond DerivativeFunction Analysis
First Derivative
The first derivative of a function represents the rate at which the function's value changes with respect to a change in the input variable, often denoted as \( s'(t) \). This rate of change helps us understand the behavior of the function over a given interval. If a derivative is positive over an interval, it means the function is increasing during that period. Conversely, a negative derivative indicates a falling behavior.
In the original exercise, the function values increase as \( t \) progresses from 0 to 5, with changes calculated as \( \Delta s(t) = s(t) - s(t-1) \). The differences are: 2, 3, 3, 11, and 24. Since all these differences are positive, this reveals that the first derivative \( s'(t) \) is positive, indicating that the function is continually increasing over this interval.
In practical terms, if you think of \( s(t) \) as a position function of an object in motion, the positive first derivative represents the object moving forward at an increasing rate. Understanding the first derivative offers insights into the basic trend of the function.
In the original exercise, the function values increase as \( t \) progresses from 0 to 5, with changes calculated as \( \Delta s(t) = s(t) - s(t-1) \). The differences are: 2, 3, 3, 11, and 24. Since all these differences are positive, this reveals that the first derivative \( s'(t) \) is positive, indicating that the function is continually increasing over this interval.
In practical terms, if you think of \( s(t) \) as a position function of an object in motion, the positive first derivative represents the object moving forward at an increasing rate. Understanding the first derivative offers insights into the basic trend of the function.
Second Derivative
The second derivative provides information about the curvature or the "concavity" of a function and its acceleration or deceleration. More formally, the second derivative \( s''(t) \) indicates how the rate of change of a function is itself changing. This is accomplished by examining changes in the differences, through \( \Delta^2 s \).
For example, in the original problem, these second differences are calculated as: 1, 0, 8, and 13, derived by subtracting subsequent \( \Delta s \) values from each other. Since all of these are positive or zero, the second derivative \( s''(t) \) can be considered positive. This indicates that the function's increase is accelerating, not just increasing at a constant rate.
Thus, understanding the second derivative is essential because it reveals whether a function's increases are becoming more dramatic or merely steady. It goes beyond the scope of the first derivative by considering the "slope of the slope," giving us more refined insights into the behavior of the function.
For example, in the original problem, these second differences are calculated as: 1, 0, 8, and 13, derived by subtracting subsequent \( \Delta s \) values from each other. Since all of these are positive or zero, the second derivative \( s''(t) \) can be considered positive. This indicates that the function's increase is accelerating, not just increasing at a constant rate.
Thus, understanding the second derivative is essential because it reveals whether a function's increases are becoming more dramatic or merely steady. It goes beyond the scope of the first derivative by considering the "slope of the slope," giving us more refined insights into the behavior of the function.
Function Analysis
Function analysis involves a comprehensive understanding of how a function behaves over a specific interval, using derivatives as primary tools. Through the first and second derivatives, one can infer several aspects of the function's behavior.
This specific analysis might examine whether a function is increasing or decreasing (via the first derivative) and whether it is accelerating or decelerating (via the second derivative). Combining information from both derivatives provides a clear picture of the function's growth pattern: Is it shooting up quickly? Is it flattening out? Is it dipping?
This specific analysis might examine whether a function is increasing or decreasing (via the first derivative) and whether it is accelerating or decelerating (via the second derivative). Combining information from both derivatives provides a clear picture of the function's growth pattern: Is it shooting up quickly? Is it flattening out? Is it dipping?
- The first derivative tells you about the function's direction of change.
- The second derivative informs you about the change in the rate of that direction.
Other exercises in this chapter
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