Arc Length Formula
Calculus How-To w/ Examples!

What is the calculus formula for arc length?

Excellent question, and it’s the we’re going to cover in today’s class.

Let’s go!

What do the following scenarios all have in common?

• The distance a soccer ball takes as it soars through the air on its way to the back of the net.
• The length of a rope bridge over a raging river.
• The exact perimeter of a vase of flowers.

They all describe the distance traveled between two points along a section of a curve.

Using Steps To Approximate Curve Length

But how do find the arc length of a curve that isn’t straight?

What if I said that, if you zoom in far enough on any curve, it will always appear straight.

This means that we can apply our knowledge of the Pythagorean Theorem to find the arc curve quickly!

How?

By dividing the curve into a small number of parts, we can approximate the length by calculating the change in y and the change in x and adding up each segment.

Isn’t it extraordinary how the method for finding the curve’s length is similar to finding the area between two curves and the volume of a solid of revolution?

We divide the curve into an infinite number of small distances, as Harvey Mudd College nicely states, and then sum their distances.

Using The Formula

Consequently, if f is a smooth curve and f’ is continuous on the closed interval [a,b], then the length of the curve is found by the following Arc Length Formula:

\begin{equation}
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
\end{equation}

Setup Only

Let’s see this formula in action by working on a few examples.

Suppose we are asked to set up an integral expression that will calculate the arc length of the portion of the graph between the given interval.

Example 1: \(y=5+x^{3}, 1 \leq x \leq 4\)

\begin{equation}
L=\int_{1}^{4} \sqrt{1+\underbrace{\left(3 x^{2}\right)^{2}}_{y^{\prime}(x)}} d x=\int_{1}^{4} \sqrt{1+9 x^{4}} d x
\end{equation}

Example 2: \(x=y^{5},-2 \leq y \leq 2\)

\begin{equation}
L=\int_{-2}^{2} \sqrt{1+\underbrace{\left(5 y^{4}\right)^{2}}_{x^{\prime}(y)}} d y=\int_{-2}^{2} \sqrt{1+25 y^{8}} d y
\end{equation}

See, it’s just that easy!

All we have to do is take the derivative of the function and toss it right into the formula!

Solve By Integration

Alright, so now that we know how to utilize the formula, let’s calculate the arc length using our integration skills!

\begin{equation}
\text { Find the length of the curve } y=\ln (\sec x) \text { from }\left[0, \frac{\pi}{3}\right]
\end{equation}

\begin{equation}
\text { First, we will find the derivative of the function: } \frac{d y}{d x}=\frac{\sec x \tan x}{\sec x}=\tan x
\end{equation}

Next, we substitute the derivative into our arc length formula, simplify, and integrate!

\begin{equation}
\begin{array}{l}
L=\int_{0}^{\pi / 3} \sqrt{1+(\tan x)^{2}} d x \\
L=\int_{0}^{\pi / 3} \sqrt{1+\tan ^{2} x} d x \quad \text { Pythagorean Identity } 1+\tan ^{2} x=\sec ^{2} x\\
L=\int_{0}^{\pi / 3} \sqrt{\sec ^{2} x} d x \\
L=\int_{0}^{\pi / 3}(\sec x) d x=\left.\ln (\sec x+\tan x)\right|_{0} ^{\pi / 3}=\ln \left(\sec \left(\frac{\pi}{3}\right)+\tan \left(\frac{\pi}{3}\right)\right)-\ln (\sec (0)+\tan (0)) \\
L=\ln (2+\sqrt{3})
\end{array}
\end{equation}

Not too terrible, right!

Arc Length And Surfaces Of Revolution

Okay, now let’s see how we can take our newfound knowledge of arc length and apply it to the rotation.

That means we can apply our knowledge of cylindrical shells and adapt it to create the area of a surface of revolution formula!

\begin{equation}
\begin{array}{l}
S A=2 \pi r l_{\Delta x}=2 \pi(\text { radius })(\text { length })(\text { thickness }) \\
S A=\int_{b}^{a} 2 \pi r(x) L(x) d x \\
S A=\int_{b}^{a} 2 \pi r(x) \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
\end{array}
\end{equation}

Example

Let’s suppose we have the curve.

\begin{equation}
f(x)=\sqrt{4-x^{2}},-1 \leq x \leq 1
\end{equation}

And we wish to find the area of the surface obtained by rotating this arc about the x-axis.

First, we will graph our curve and identify the derivative of the function and radius.

Next, we substitute everything into our surface area formula and integrate!

\begin{equation}
\begin{array}{l}
S A=\int_{-1}^{1} 2 \pi \underbrace{\left(\sqrt{4-x^{2}}\right)}_{\text {radius }} \underbrace{\sqrt{1+\left(\frac{-x}{\left.\sqrt{4-x^{2}}\right)^{2}}\right.}}_{\text {length }} d x \\
S A=\int_{-1}^{1} 2 \pi\left(\sqrt{4-x^{2}}\right) \sqrt{1+\left(\frac{x^{2}}{4-x^{2}}\right)} d x \\
S A=\int_{-1}^{1} 2 \pi\left(\sqrt{4-x^{2}}\right) \sqrt{\frac{\left(4-x^{2}\right)+x^{2}}{4-x^{2}}} d x \\
S A=\int_{-1}^{1} 2 \pi\left(\sqrt{4-x^{2}}\right) \frac{\sqrt{4}}{\sqrt{4-x^{2}}} d x \\
S A=\int_{-1}^{1} 4 \pi d x=\left.4 \pi x\right|_{-1} ^{1}=4 \pi(1)-4 \pi(-1) \\
S A=8 \pi
\end{array}
\end{equation}

Wow!

So we combined our knowledge of solids of revolution with arc length and calculated the surface area of an arc.

Summary

Together in this lesson, we will explore arc length and the area of a surface of revolution by developing their formulas and working through some rather challenging questions, all while applying our skills of integration.

Let’s get to it!

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