Continuously Differentiable Curve has Finite Arc Length

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Theorem

Let $y = \map f x$ be a real function which is continuous on the closed interval $\closedint a b$ and continuously differentiable on the open interval $\openint a b$.

The definite integral:

$s = \ds \int_{x \mathop = a}^{x \mathop = b} \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \rd x$

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:

$\sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$

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:

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

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

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 $\sqrt {\paren {\Delta x_i}^2 + \paren {\Delta y_i}^2}$, the sum of all these line segments is given by:

$\ds \sum_{i \mathop = 1}^k \ \sqrt {\paren {\Delta x_i}^2 + \paren {\Delta y_i}^2}$

\(\ds \)

\(=\)

\(\ds \sum_{i \mathop = 1}^k \ \sqrt {\paren {\Delta x_i}^2 + \paren {\frac {\Delta y_i} {\Delta x_i} {\Delta x_i} }^2}\)

multiply the second term in the radicand by $1 = \dfrac {\Delta x_i} {\Delta x_i}$

\(\ds \)

\(=\)

\(\ds \sum_{i \mathop = 1}^k \ \sqrt {1 + \paren {\frac {\Delta y_i} {\Delta x_i} }^2} \Delta x_i\)

factor $\Delta x_i$ out of the radicand

Thus the approximate arc length is given by the sum:

$\ds s \approx \sum_{i \mathop = 1}^k \sqrt {1 + \paren {\frac {\Delta y_i} {\Delta x_i} }^2}\Delta x_i$

Recall that by hypothesis:

$f$ is continuous on $\closedint a b$

$f$ is differentiable on $\openint a b$.

Thus the Mean Value Theorem can be applied.

In every open interval $\openint {x_i} {x_{i - 1} }$ there exists some $c_i$ such that:

$D_x \, \map f {c_i} = \dfrac {\Delta y_i} {\Delta x_i}$

Plugging this into the above sum we have:

$\ds s \approx \sum_{i \mathop = 1}^k \ \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \Delta x_i$

By hypothesis, $D_x f$ 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, $\sqrt {1 + \paren {\dfrac {\d y} {\d x} }^2}$ is continuous as well.

Because Continuous Real Function is Darboux 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:

$\ds s = \int_{x \mathop = a}^{x \mathop = b} \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \rd x$

$\blacksquare$

Sources

• 2005: Roland E. Larson, Robert P. Hostetler and Bruce H. Edwards: Calculus (8th ed.) $\S 7.4$