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Magazine Chapter 8: Problem 101
Are the statements true or false? If a statement is true, explain how you know. If a statement is false, give a counterexample. If \(f\) is differentiable on the interval \([0,10],\) then the arc length of the graph of \(f\) on the interval [0,1] is less than the arc length of the graph of \(f\) on the interval [1,10].
Short Answer
Expert verified
False. The intervals' numeric lengths don't determine arc length; the derivative's behavior is crucial.
Step by step solution
01
Understanding the Problem
We need to determine whether the arc length of the graph of a differentiable function on the interval \([0, 1]\) is always less than the arc length on the interval \([1, 10]\). This requires evaluating the lengths of these two segments of the arc based on the properties of differentiable functions.
02
Recall Arc Length Formula
The arc length \( L \) of a function \( f \) over an interval \([a, b]\) is given by the formula \( L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx \), where \( f'(x) \) is the derivative of \( f \). We apply this formula separately to the intervals \([0, 1]\) and \([1, 10]\).
03
Analyzing the Formula for Both Intervals
We calculate the arc length for each segment: \( L_{0,1} = \int_0^1 \sqrt{1 + (f'(x))^2} \, dx \) and \( L_{1,10} = \int_1^{10} \sqrt{1 + (f'(x))^2} \, dx \). The integrals depend on \(f'(x)\), making it possible for the shorter interval \([0,1]\) to have a larger or smaller arc length than the longer interval \([1,10]\) based on the behavior of \( f'(x) \).
04
Construct a Counterexample
Consider \( f(x) = x^2 \). Compute \(f'(x) = 2x\). For the interval \([0, 1]\), the arc length is: \( \int_0^1 \sqrt{1 + (2x)^2} \, dx \approx 1.147\). For the interval \([1,10]\), it is: \( \int_1^{10} \sqrt{1 + (2x)^2} \, dx \approx 101.347\). So, in this case, the arc length of \([0, 1]\) indeed is much less than that of \([1, 10]\), however due to the dependence on \(f'(x)\), we need an opposite example to show falsity.
05
True or False Evaluation
Evaluate a function where the concise arc drastically changes, such as \( f(x) = \sin(100x)\) in \([0, 1]\), generating lengthy oscillation versus some less curved path over \([1, 10]\) like \( f(x) = 0\). This example produces more arc length in \([0, 1]\), proving the original statement false.
06
Conclusion
The statement is false. Counterexamples like the oscillating function \( f(x) = \sin(100x) \) in \([0, 1]\) can generate a larger arc length than the simpler function on \([1, 10]\). Thus, reliance on the intervals' numerical lengths alone is inadequate.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiable Function
A differentiable function is a central concept in calculus. It means that at every point in its domain, the function has a defined derivative. This derivative is essentially the slope of the tangent line to the function's graph at any given point. Here are a few key aspects of differentiable functions:
- They must be continuous. If a function is not continuous, it cannot be differentiable. There's no point in even checking for a derivative at spots where the function jumps or is undefined.
- They typically have smooth graphs without any corners or cusps. This smoothness means the function doesn鈥檛 just have tangents (derivatives) everywhere, but these tangents change gradually.
- Examples include polynomial functions like \( f(x) = x^2 \) and trigonometric functions like \( g(x) = \sin(x) \) over their regular intervals. Functions like \( f(x) = |x| \) have places where they aren't differentiable (in this case, \( x = 0 \)).
Arc Length Formula
The arc length formula is a pivotal tool in calculus for determining the length of a curve over a certain interval. This formula is particularly useful for differentiable functions. For a function \( f(x) \), the arc length \( L \) over the interval \([a, b]\) is calculated using:\[ L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx \]Here's a breakdown of the formula:
- The term \( f'(x) \) represents the derivative of the function \( f(x) \). It measures how quickly or slowly \( f(x) \) changes as \( x \) changes.
- The expression \( \sqrt{1 + (f'(x))^2} \) adjusts our standard linear arc length formula to account for the function's rate of change. This ensures that the curve's length is accurately measured, including all its bends and wobbles.
- Integrating this expression over the interval \([a, b]\) adds up the infinitely small arc segments, resulting in the total arc length.
Integral Calculus
Integral calculus is the branch of calculus concerned with the accumulation of quantities and the spaces under and between curves. It's the counterpart to differential calculus, which deals with rates of change. In the context of calculating arc lengths:
- Integral calculus allows the summation of infinitely small data points. This is key in finding the total arc length, where we integrate the infinitesimally small linear arc lengths \( \sqrt{1 + (f'(x))^2} \, dx \).
- Through integration, we incorporate the entire behavior of the function鈥檚 derivative across an interval, providing a holistic view of the curve's length.
- Applications go beyond arc length, helping solve real-world problems involving areas, volumes, and other concepts where summation is crucial.
Counterexample
A counterexample in mathematics is used to show that a particular statement is false. By presenting a specific example that refutes a general claim, its validity is negated.Consider the statement from the original exercise, which suggests that arc lengths of differentiable functions on different intervals can be straightforwardly compared based only on those intervals' lengths. The counterexample used was the function \( f(x) = \sin(100x) \) on the interval \([0, 1]\).
- This function oscillates rapidly within the small interval, leading to a much longer arc length than one might expect given the interval's small numerical distance.
- Comparatively, a function that remains constant or doesn't change much within \([1,10]\), like \( f(x)=0 \), proves the original claim false due to its minimal arc length in a longer numerical interval.
- By illustrating these properties, counterexamples prove to be extremely useful tools in mathematical problem-solving, allowing us to better understand the limitations of broad statements.
