Problem 30 - Calculus On Manifolds: A Modern Approach To Classical Theorems Of Advanced Calculus

Question

5-30. (a) If \(f:[0,1] \rightarrow \mathbf{R}\) is differentiable and \(c:[0,1] \rightarrow \mathbf{R}^{2}\) is defined by \(c(x)=(x, f(x))\), show that \(c([0,1])\) has length \(\int_{0}^{1} \sqrt{1+\left(f^{\prime}\right)^{2}} .\) (b) Show that this length is the least upper bound of lengths of inscribed broken lines. Hint: If \(0=t_{0} \leq t_{1} \leq \cdots \leq t_{n}=1\), then $$ \begin{aligned} \left|c\left(t_{i}\right)-c\left(t_{i-1}\right)\right| &=\sqrt{\left(t_{i}-t_{i-1}\right)^{2}+\left(f\left(t_{i}\right)-f\left(t_{i-1}\right)\right)^{2}} \\\ &=\sqrt{\left(t_{i}-t_{i-1}\right)^{2}+f^{\prime}\left(s_{i}\right)^{2}\left(t_{i}-t_{i-1}\right)^{2}} \end{aligned} $$ for some \(s_{i} \in\left[t_{i-1}, t_{i}\right]\).

Step-by-Step Solution

Answer

Length of\(c([0,1])\) is \(\int_{0}^{1} \sqrt{1 + (f')^2}.\) and it is the least upper bound of inscribed broken line lengths.

Step 1: Understand the Problem Statement (Part a)

Given a differentiable function \(f: [0, 1] \rightarrow \mathbb{R}\) and a vector function \(c(x) = (x, f(x))\), we need to show that the length of the curve \(c([0, 1])\) is \(\int_{0}^{1} \sqrt{1 + (f')^2} \, dx\).

Step 1: Parametric Curve and Arc Length Formula

Recall the formula for the arc length of a curve \(c(x) = (x, f(x))\) from \(x = 0\) to \(x = 1\): \[L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt\].

Step 2: Apply to the Given Curve

For the curve \(c(x) = (x, f(x))\), we have \( \frac{dx}{dx} = 1\) and \( \frac{dy}{dx} = f'(x)\). Substitute these expressions into the arc length formula:

Step 3: Simplify the Integral

\[L = \int_{0}^{1} \sqrt{(1)^2 + (f'(x))^2} \, dx = \int_{0}^{1} \sqrt{1 + (f'(x))^2} \, dx\]. This proves the required length for part (a).

Step 5: Understand the Problem Statement (Part b)

We need to show that the length \(\int_{0}^{1} \sqrt{1 + (f')^2} \, dx\) is the least upper bound of the lengths of inscribed broken lines.

Step 4: Length of Broken Lines

Consider a partition of \([0, 1]\) given by \(0 = t_0 \leq t_1 \leq \cdots \leq t_n = 1\). The length of the inscribed broken lines is the sum of distances between consecutive points:

Step 5: Distance Formula

\( \left|c(t_i) - c(t_{i-1})\right| = \sqrt{(t_i - t_{i-1})^2 + (f(t_i) - f(t_{i-1}))^2} \).

Step 6: Mean Value Theorem Application

Using the Mean Value Theorem, we have \(f(t_i) - f(t_{i-1}) = f'(s_i) \, (t_i - t_{i-1})\) for some \(s_i \in [t_{i-1}, t_i]\). Therefore, the distance can be written as \( \sqrt{(t_i - t_{i-1})^2 + (f'(s_i) \, (t_i - t_{i-1}))^2} = (t_i - t_{i-1}) \sqrt{1 + (f'(s_i))^2}\).

Step 7: Sum of Distances

Sum these distances for all intervals to get the total length of the inscribed broken line: \[ \sum_{i=1}^{n} (t_i - t_{i-1}) \sqrt{1 + (f'(s_i))^2} \].

Step 8: Integral Approximation

As the partition becomes finer, the sum approximates the integral \(\int_{0}^{1} \sqrt{1 + (f'(x))^2} \, dx\). Therefore, \( \int_{0}^{1} \sqrt{1 + (f')^2} \, dx\) is the least upper bound of these sums, which are the lengths of the inscribed broken lines.

Key Concepts

Differentiable Functions

In calculus, a function is called differentiable if it has a derivative at every point in its domain. This implies the function is smooth and has no sharp corners or discontinuities in the interval we are considering.
When we say a function is differentiable on \[ [0,1] \], it means we can compute the derivative at any point within this range. For example, given a function \( f(x) \), if we can find \( f'(x) \) for all \( x \) in \[ [0,1] \], then \( f \) is differentiable on \[ [0,1] \].
In the exercise's context, we used this property of \( f \) to evaluate the arc length formula. Differentiability ensures that \( f \) behaves nicely, allowing us to apply calculus techniques such as the Mean Value Theorem efficiently. This theorem helps us express differences \( f(t_i) - f(t_{i-1}) \) in terms of derivatives, simplifying the calculation of lengths for inscribed broken lines.

Parametric Curves

Parametric curves define points in terms of a parameter. Rather than expressing a curve solely with \( y \) as a function of \( x \) (e.g., \( y = f(x) \)), parametric curves use a third variable, often \( t \), to represent both \( x \) and \( y \). For instance, \( c(t) = (x(t), y(t)) \).
In the given exercise, the parametric curve is defined as \( c(x) = (x, f(x)) \). Here, \( x \) serves as the parameter, and both components of the curve ( \( x \) and \( f(x) \)) are expressed in terms of this parameter. This allows us to compute the arc length using the formula: \[ L = \begin{1}{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt \].
By plugging in the derivatives of \( x \) and \( y = f(x) \) with respect to \( x \), we simplify the integral form to get the required length.

Least Upper Bound

The concept of a least upper bound is fundamental in calculus. For a set of numbers, an upper bound is a number that is greater than or equal to every number in the set. The least upper bound (also called the supremum) is the smallest of these upper bounds.
In our exercise, we consider the lengths of inscribed broken lines of a curve. Each of these broken lines approximates the curve's true length from \[ [0,1] \]. As we make the partitions finer, our broken lines become a closer approximation of the curve. The least upper bound of these lengths, as partitions get finer and finer, is the arc length integral \[ \int_{0}^{1} \sqrt{1 + (f'(x))^2} \, dx \].
This least upper bound represents the true arc length of the curve, as no longer inscribed broken lines can surpass it. Essentially, it is the convergence point where the sums of the lengths of finer partitions lead to the curve's exact length.