## Section 3.5 | Lengths of Curves

As far as geometric uses of integrals, we have mainly used them to compute areas (for two-dimensional quantities) or volumes (for three-dimensional quantities). It turns out that there are several more geometric uses for integrals still waiting to be discovered. In this section, we will use integrals to compute the length of a curve, either defined as a function of \(x\) or as a function of \(y\).

Why do integrals come into play when computing the length of a curve? Let's first look at a Wolfram Demonstration to illustrate the idea.

Why do integrals come into play when computing the length of a curve? Let's first look at a Wolfram Demonstration to illustrate the idea.

Arc Length Approximation from the Wolfram Demonstrations Project by Chad Pierson

What you can see from this Demonstration is the basic concept behind using integrals to compute the length of a curve. We have some curve, say \(f(x)\), and we decide to cut this up into small

Suppose we are given a curve defined by the function \(f(x)\) and we want to find the arc length of the curve on the interval \([x_{i - 1},x_i]\). Define \(P_i\) to be the point \((x_i,f(x_i))\). Then we are asking for the

The fact that this curve length could depend on \(x\) or \(y\) and still holds the same form is reflected in the next definition, where I choose to use the independent variable \(\lambda\) to represent

*approximating*line segments \(|\overline{P_{i -1}P_i}|\). Once we feel satisfied with the number of slices we can get a decent approximation by adding these up: \(L \approx \sum_{i = 1}^n{|\overline{P_{i - 1}P_i}|}\). Finally, if we want to make this approximation exact, we let \(n\) approach \(\infty\). That is, \[L = \lim_{n \to \infty}{\sum_{i = 1}^n{|\overline{P_{i - 1}P_i}|}}.\] All we need now is to find a nice expression for \(|\overline{P_{i - 1}P_i}|\).Suppose we are given a curve defined by the function \(f(x)\) and we want to find the arc length of the curve on the interval \([x_{i - 1},x_i]\). Define \(P_i\) to be the point \((x_i,f(x_i))\). Then we are asking for the

*approximate*distance between \(P_{i - 1}\) and \(P_i\). By the Pythagorean Theorem, \[|\overline{P_{i - 1}P_i}| \approx \sqrt{\Delta x_i^2 + \Delta y_i^2}.\]Since \(\Delta y_i = f(x_i) - f(x_{i -1})\), we can use the Mean Value Theorem for Derivatives to state that \[\Delta y_i = f(x_i) - f(x_{i - 1}) = f'(x_i^*)(x_i - x_{i - 1}) = f'(x_i^*) \Delta x_i\] for some \(x_i^* \in [x_{i - 1},x_i]\). Hence, we can rewrite the length of this arc as \[| \overline{P_{i - 1}P_i} | \approx \sqrt{ \Delta x_i^2 + [f'(x_i^*) \Delta x_i]^2} = \sqrt{1 + f'(x_i^*)^2} \cdot \Delta x_i.\]Using a similar argument, it is easy to show that \[|\overline{P_{i - 1}P_i} | \approx \sqrt{g'(y_i^*)^2 + 1} \cdot \Delta y_i\] if \(g\) is a function of \(y\).The fact that this curve length could depend on \(x\) or \(y\) and still holds the same form is reflected in the next definition, where I choose to use the independent variable \(\lambda\) to represent

*either*\(x\) or \(y\).**DEFINITION | CURVE LENGTH FUNCTION**

If \(h(\lambda)\) is a function such that \(h'\) is continuous on \([\alpha,\lambda]\), then the

**curve length function**on this interval is defined to be \[s(\lambda) = \int_{t = \alpha}^{t = \lambda}{\sqrt{1 + [h'(t)]^2}dt}.\]

Taking the differential of \(s\) yields the following useful result.\[ds = s'(\lambda) d\lambda = \sqrt{1 + [h'(\lambda)]^2} d\lambda.\]

**THEOREM | DISTANCE ALONG A CURVE**

If \(h'\) is continuous on \([a,b]\), then the distance along the curve from \(\lambda = a\) to \(\lambda = b\) is \[L = \int_{\lambda = a}^{\lambda = b}{ds},\]where \[ds = \sqrt{1 + [h'(\lambda)]^2} d\lambda,\]and where \(\lambda\) is a "fill-in" variable for either \(x\) or \(y\).

**!WARNING!**

Because curve length always deals with radicals, integration can be difficult or impossible. Numerical integration is often used!

Using the \(ds\) notation will eventually prove to be very helpful. It turns out that we can always choose which variable to integrate with respect to.

Determine the length of \(y = \left(\frac{3x}{2}\right)^{2/3} + 1\) on \(0 \leq x \leq 2\sqrt{3}\) using \(dx\) and then again \(dy\).

**EXAMPLE 1 (Comprehension) | A basic arc length two ways**Determine the length of \(y = \left(\frac{3x}{2}\right)^{2/3} + 1\) on \(0 \leq x \leq 2\sqrt{3}\) using \(dx\) and then again \(dy\).

**EXAMPLE 2 (Application) | Linkin' logs**

Find the exact length of \(y=\ln{(\sec{(x)})}\) on \([0,\frac{\pi}{3}]\).

**EXAMPLE 3 (Application) | Approximating lengths**

Use Simpson's Method with \(n = 10\) to estimate the length of the curve \(y = x \sin{(x)}\) on \([0,2\pi]\). Compare this to the result that your calculator gives.

**EXAMPLE 4 (Synthesis) | Heavy metal**

Have you ever wondered how long a sheet of corrugated metal roofing would be if you could straighten it out? Me too! Suppose you have a sheet of corrugated metal that measures 56 inches wide and 2 inches from low point to high point. The ripple pattern looks identical to a sine wave and you notice that there four full periods of the sine with this piece of sheet metal you have.

- Write a sine function for this sheet metal.
- Find the "straightened" length of the sheet metal. [Hint: You must use numerical integration for this.]