Continuously Differentiable Curve has Finite Arc Length

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Theorem

Let $y = f \left({x}\right)$ be a real function which is continuous on the closed interval $\left[{a \,.\,.\, b}\right]$ and continuously differentiable on the open interval $\left({a \,.\,.\, b}\right)$.

The definite integral:

$s = \displaystyle \int_{x \mathop = a}^{x \mathop = b} \sqrt{1 + \left({\frac {\mathrm d y}{\mathrm d x}}\right)^2}\ \mathrm d x$

exists, and is called the arc length of $f$ between $a$ and $b$.


Proof

It intuitively makes sense to define the length of a line segment to be the distance between the two end points, as given by the Distance Formula:

$\sqrt{\left({x_1 - x_2}\right)^2 + \left({y_1 - y_2}\right)^2}$

Similarly, it is reasonable to assume that the actual length of the curve would be approximately equal to the sum of the lengths of each of the line segments, as shown:

ArcLength1.png


To calculate the sum of the length of these line segments, divide $\left[{a \,.\,.\, b}\right]$ into any number of closed subintervals of the form $\left[{x_{i-1} \,.\,.\, x_i}\right]$ where:

$a = x_0 < x_1 < \cdots < x_{k-1} < x_k = b$

Define:

$\Delta x_i = x_i - x_{i-1}$
$\Delta y_i = y_i - y_{i-1}$

As the length of the $i$th line segment is $\sqrt{\left({\Delta x_i}\right)^2 + \left({\Delta y_i}\right)^2}$, the sum of all these line segments is given by:

$\displaystyle \sum_{i \mathop = 1}^k\ \sqrt{\left({\Delta x_i}\right)^2 + \left({\Delta y_i}\right)^2}$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^k\ \sqrt{\left({\Delta x_i}\right)^2 + \left(\frac{\Delta y_i}{\Delta x_i}{\Delta x_i}\right)^2}\) multiply the second term in the radicand by $1 = \dfrac {\Delta x_i}{\Delta x_i}$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^k\ \sqrt{1 + \left(\frac{\Delta y_i}{\Delta x_i}\right)^2}\Delta x_i\) factor $\Delta x_i$ out of the radicand

Thus the approximate arc length is given by the sum:

$\displaystyle s \approx \sum_{i \mathop = 1}^k \ \sqrt{1 + \left(\frac{\Delta y_i}{\Delta x_i}\right)^2}\Delta x_i$

Recall that by hypothesis:

$f$ is continuous on $\left[{a \,.\,.\, b}\right]$
$f$ is differentiable on $\left({a \,.\,.\, b}\right)$.

Thus the Mean Value Theorem can be applied.

In every open interval $\left({x_i \,.\,.\, x_{i-1}}\right)$ there exists some $c_i$ such that:

$D_x f \left({c_i}\right) = \dfrac {\Delta y_i}{\Delta x_i}$

Plugging this into the above sum we have:

$\displaystyle s \approx \sum_{i \mathop = 1}^k\ \sqrt{1 + \left({\frac{\mathrm dy}{\mathrm dx}}\right)^2}\Delta x_i$

By hypothesis, $D_xf$ is continuous.

As Square of Real Number is Non-Negative the radicand is always positive.

From Continuity of Root Function and Limit of Composite Function, $\sqrt{1 + \left({\dfrac{\mathrm d y} {\mathrm d x}}\right)^2}$ is continuous as well.

Because Continuous Function is Riemann Integrable, there exists a definite integral that confirms the intuitive notion that there is a value that represents the exact length of the curve as the limit of the above Riemann sum.

This definite integral is:

$\displaystyle s = \int_{x \mathop = a}^{x \mathop = b} \sqrt{1 + \left({\frac {\mathrm d y}{\mathrm d x}}\right)^2} \ \mathrm d x$

$\blacksquare$


Sources