Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the Mean Value Theorem to show that \(s=1 / t^{2}\) decreases on any interval to the right of the origin.

Step-by-Step Solution

Answer

The function \( s = \frac{1}{t^2} \) decreases for \( t > 0 \) by MVT.

Step 1: Understand the Problem Statement

We are given the function \( s(t) = \frac{1}{t^2} \) and asked to show it's decreasing on any interval to the right of the origin using the Mean Value Theorem.

Step 2: State the Mean Value Theorem

The Mean Value Theorem states that if a function \( f(x) \) is continuous on a closed interval \([a,b]\) and differentiable on the open interval \((a,b)\), then there exists at least one point \( c \in (a,b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).

Step 3: Verify Conditions for MVT

To apply the Mean Value Theorem, the function \( s(t) = \frac{1}{t^2} \) must be continuous and differentiable on any interval \((a,b)\), where \( a, b > 0 \). The function \( s(t) \) is continuous and differentiable for \( t > 0 \).

Step 4: Find the Derivative of the Function

The derivative of the function \( s(t) = \frac{1}{t^2} \) is \( s'(t) = -\frac{2}{t^3} \).

Step 5: Analyze the Sign of the Derivative

Examine \( s'(t) = -\frac{2}{t^3} \). For \( t > 0 \), \( t^3 > 0 \) and hence \( -\frac{2}{t^3} < 0 \). So, \( s'(t) < 0 \) for \( t > 0 \).

Step 6: Conclusion Using the MVT

Since \( s'(t) < 0 \) for all \( t > 0 \), the function \( s(t) = \frac{1}{t^2} \) is decreasing on any interval to the right of the origin. This satisfies the Mean Value Theorem condition as \( f'(c) \) is negative, indicating a decreasing function.

Key Concepts

Calculus

Calculus is a branch of mathematics that studies how things change. It’s like a toolkit for finding out things such as how fast something is moving or how a curve behaves. Calculus is fundamentally about manipulation and understanding of change and it mainly deals with two concepts, derivatives and integrals.
If you imagine a car moving along a road, calculus helps us calculate the car’s speed (using derivatives) or the total distance it has traveled (using integrals). Many functions describing real-world phenomena are best understood through calculus, just like in this exercise, where we used calculus to prove a function decreases in a certain interval.

• Derivatives help us find the slope of functions.
• Integrals help us find areas under curves.

Understanding how these concepts work is essential for analyzing functions like the one in this problem, as they describe the behavior of functions.

Derivative

A derivative represents the rate at which a function is changing at any given point. Think of it as the slope of a function at a specific point. If a function is like a curve on a graph, the derivative tells us how steep that curve is.
In this exercise, we used derivatives to show that a function is decreasing. We found the derivative of the function \( s(t) = \frac{1}{t^2} \), which is \( s'(t) = -\frac{2}{t^3} \).
This derivative tells us:

• For \( t > 0 \), the derivative \( s'(t) \) is negative.
• A negative derivative indicates a decreasing function.

So for any positive \( t \), the function \( s(t) \) gets smaller as \( t \) gets larger. Understanding derivatives can tell you if a function is increasing, decreasing, or constant at a point.

Continuity

Continuity is an important condition for many theorems in calculus, including the Mean Value Theorem. A function is continuous if there are no breaks, jumps, or holes in its graph. It is like drawing a curve on a paper without lifting the pen.
In this exercise, the function \( s(t) = \frac{1}{t^2} \) is continuous for all \( t > 0 \). This means on any interval over these values, there are no jumps or gaps in the function.
Continuity ensures that each point in the interval can be connected with a smooth curve.

• Continuous functions do not jump abruptly.
• They can be graphed without lifting your pen.

Checking for continuity is a crucial step when applying the Mean Value Theorem as it confirms that the function behaves nicely on the given interval.

Function Analysis

Function analysis involves examining various attributes of a function, such as its rate of change, continuity, and limits. To thoroughly analyze a function, we need to consider how it behaves across intervals and at specific points.
In this exercise, function analysis involved checking if the function \( s(t) = \frac{1}{t^2} \) is decreasing. We achieved this by:

• Using derivatives to study the rate of change.
• Confirming continuity to apply the Mean Value Theorem.

Understanding the behavior of functions helps us see the bigger picture and predict how they’ll behave in different situations.
This kind of analysis is quite common in calculus, especially when dealing with real-world applications where we want to understand trends and behaviors over time or intervals.