Chapter 4: Problem 28
Use the Mean Value Theorem to show that \(s=1 / t^{2}\) decreases on any interval to the right of the origin.
Short Answer
Expert verified
The function \( s(t) = \frac{1}{t^2} \) decreases for \( t > 0 \) since its derivative is negative.
Step by step solution
01
Understand the Mean Value Theorem
The Mean Value Theorem states that if a function \( f(x) \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists some \( c \) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). This theorem helps us relate the average rate of change of the function to its instantaneous rate of change at some point within that interval.
02
Verify Conditions of the Theorem
The function \( s(t) = \frac{1}{t^2} \) is continuous and differentiable on intervals \((a, b)\) where \(a > 0\) and \(b > 0\). This means we can apply the Mean Value Theorem for any interval \((a, b)\) to the right of the origin (where \( t = 0 \)).
03
Calculate the Derivative
Find the derivative of \( s(t) \). Using the power rule, we have \( s'(t) = \frac{d}{dt}(t^{-2}) = -2t^{-3} \). This derivative is defined for \( t > 0 \).
04
Analyze the Sign of the Derivative
The derivative \( s'(t) = -2t^{-3} \) simplifies to \( -\frac{2}{t^3} \). Since \( t > 0 \) in our chosen interval \((a, b)\), we observe that \( s'(t) \) will always be negative. A negative derivative indicates that the function \( s(t) \) is decreasing.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Continuous Function
In the world of calculus, a continuous function is one that is uninterrupted and smooth—much like a seamless path. Imagine drawing a graph of a function without lifting your pencil off the paper. That's what continuity looks like. For the function \( s(t) = \frac{1}{t^2} \), continuity is important because it ensures that there are no breaks or gaps within the interval we are examining.
For the formula \( s(t) = \frac{1}{t^2} \), continuity holds for all values of \( t > 0 \). Since we are interested in intervals to the right of the origin (i.e., \( t > 0 \)), the function fulfills the first requirement of the Mean Value Theorem. This is because, for a theorem that predicts something about the behavior of a function, we need to ensure that the function behaves predictably (without jumps or undefined points) within the interval we are studying.
Thus, continuity is essential for analyzing changes over a specific range of values.
For the formula \( s(t) = \frac{1}{t^2} \), continuity holds for all values of \( t > 0 \). Since we are interested in intervals to the right of the origin (i.e., \( t > 0 \)), the function fulfills the first requirement of the Mean Value Theorem. This is because, for a theorem that predicts something about the behavior of a function, we need to ensure that the function behaves predictably (without jumps or undefined points) within the interval we are studying.
Thus, continuity is essential for analyzing changes over a specific range of values.
Differentiable Function
Differentiation is all about the rates of change. When a function is differentiable, it means you can find its derivative—a formula that tells you how the function changes at every point. This derivative, much like a compass, gives you the direction of your function's graph at any specific point along its path.
The function \( s(t) = \frac{1}{t^2} \) is differentiable for all values of \( t > 0 \), allowing us to find \( s'(t) \). Differentiability is a key requirement for applying the Mean Value Theorem because it suggests the function’s behavior is predictable and smooth within the specific interval of interest.
If we're to apply this theorem successfully, we need the derivative to give us definite insights about whether the function is increasing or decreasing along the interval. It aids in linking the instantaneous change at a certain point to the overall behavior between two points.
The function \( s(t) = \frac{1}{t^2} \) is differentiable for all values of \( t > 0 \), allowing us to find \( s'(t) \). Differentiability is a key requirement for applying the Mean Value Theorem because it suggests the function’s behavior is predictable and smooth within the specific interval of interest.
If we're to apply this theorem successfully, we need the derivative to give us definite insights about whether the function is increasing or decreasing along the interval. It aids in linking the instantaneous change at a certain point to the overall behavior between two points.
Average Rate of Change
The average rate of change gives us a big-picture view of how a function behaves over an interval—much like finding the slope of a straight line connecting two points. It's calculated as the change in the function’s value over the change in its independent variable.
According to the Mean Value Theorem, for \( s(t) = \frac{1}{t^2} \) which is defined and differentiable on intervals to the right of the origin, there exists at least one point where the instantaneous rate of change equals the average rate of change across that interval.
This average rate forms the basis for understanding the Mean Value Theorem itself. It's the consistency within our interval that lets us say, "Somewhere along these points, the rate of change matches the average," thus connecting it to the instantaneous rate at a particular point within the interval.
According to the Mean Value Theorem, for \( s(t) = \frac{1}{t^2} \) which is defined and differentiable on intervals to the right of the origin, there exists at least one point where the instantaneous rate of change equals the average rate of change across that interval.
This average rate forms the basis for understanding the Mean Value Theorem itself. It's the consistency within our interval that lets us say, "Somewhere along these points, the rate of change matches the average," thus connecting it to the instantaneous rate at a particular point within the interval.
Instantaneous Rate of Change
This concept is essentially the heart of differentiation. The instantaneous rate of change, represented by the derivative, tells us what’s happening to the function at precisely one point. It's like looking through a microscope at the function's behavior.
For \( s(t) = \frac{1}{t^2} \), the instantaneous rate of change is expressed as its derivative \( s'(t) = -\frac{2}{t^3} \). For any \( t > 0 \), this derivative is negative, indicating a downward slope—in other words, the function is decreasing at each point.
This is important because the Mean Value Theorem links this instantaneous rate to the average rate over an interval. When every point on our interval shows a negative rate, it assures us that the function decreases consistently throughout. Using these insights, we can confidently assess how the function behaves within any interval, further substantiating its overall behavior.
For \( s(t) = \frac{1}{t^2} \), the instantaneous rate of change is expressed as its derivative \( s'(t) = -\frac{2}{t^3} \). For any \( t > 0 \), this derivative is negative, indicating a downward slope—in other words, the function is decreasing at each point.
This is important because the Mean Value Theorem links this instantaneous rate to the average rate over an interval. When every point on our interval shows a negative rate, it assures us that the function decreases consistently throughout. Using these insights, we can confidently assess how the function behaves within any interval, further substantiating its overall behavior.
