In this lesson, we’ll use the concept of the definite integral to calculate the arc length of a curve. Let’s say that the curve \(P_1P_n\) in Figure 1 is any arbitrary curve. Let’s subdivide this curve into an \(n\) number of smaller arcs \(P_1P_2,P_2P_3,…,P_{n-1}P_n\) as illustrated in Figure 1. By drawing a straight line from \(P_i\) to \(P_{i+1}\) for each arc \(P_iP_{i+1}\) (where \(i=1,…,n\)), we can draw an \(n\) number of chords which will roughly be equal to the length of each arc \(P_iP_{i+1}\) if \(n\) is very large. If we use the notation \(L_i(P_iP_{i+1})\) to represent the length of each chord drawn in Figure 1, then by summing the lengths of the chords, \(P_1P_2,P_2P_3,…,P_{n-1}P_n\), we can obtain a rough estimate of the total arc length of the curve \(P_1P_n\). Thus,
$$\text{Arc length of }P_1P_n≈L_1(P_1P_2)+…+L_n(P_{n-1}P_n)$$
or, equivalently,
$$\text{Arc length of }P_1P_n≈\sum_{i=1}^nL_i(P_iP_{i+1}).\tag{1}$$
Figure 1: The curve \(P_1P_n\) split up into an \(n\) number of chords \(P_iP_{i+1}\) where \(i=1,...,n\). By taking the sum of the lengths of each chord (represented by \(\sum_{i=1}^nL(P_iP_{i+1})\)) and then taking the limit as \(n→∞\), we obtain the exact arc length \(s\) of the entire curve.
Let’s briefly review the concept of a limit. If I write the limit,
$$\lim_{z→k}\text{ (‘something’)}=?,$$
then the thing that the limit is equal to is whatever the “something” gets closer and closer to equaling as \(z\) approaches \(k\). Now, as \(n→∞\) (which is to say, as the number of subdivisions in the arc \(P_1P_n\) approaches infinity), what does the sum \(\sum_{i=1}^nL_i(P_iP_{i+1})\) get closer and closer to equaling? Answering this question is very important since whatever \(\sum_{i=1}^nL_i(P_iP_{i+1})\) gets closer and closer to equaling as \(n→∞\) must be the thing that the limit \(\lim_{n→∞}\sum_{i=1}^nL_i(P_iP_{i+1})\) is equal to. Well, as the number of subdivisions becomes greater and greater, the length of each chord, \(L_i(P_iP_{i+1})\), will get closer and closer to equaling the length of each arc \(P_iP_{i+1}\). Thus, as the number of subdivisions keeps increasing and as \(n→∞\), the sum \(\sum_{i=1}^nL_i(P_iP_{i+1})\) will get closer and closer to equaling the exact arc length of the curve \(P_1P_n\). Thus,
$$\lim_{n→∞}\sum_{i=1}^nL_i(P_iP_{i+1})=\text{Arc length of curve }P_1P_n.$$
Since it is typical to denote the arc length of a curve with the letter \(s\), we’ll rewrite the above equation as
$$s=\lim_{n→∞}\sum_{i=1}^nL_i(P_iP_{i+1}).\tag{2}$$
Equation (2) represents, conceptually, how the arc length can be obtained by taking the infinite sum of the lengths of infinitesimally small chords. But since a definite integral involves taking the limit of a sum of the form
$$S_n=\sum_{i=1}^ng(x_i)Δx,\tag{3}$$
we must find a way to represent Equation (1) of the same form as Equation (3). Then, after that, by taking the limit of the sum as \(n→∞\) as we did in Equation (2), we’ll obtain a definite integral of the form
$$\int_a^bg(x)dx.$$
Figure 2: A close up view of the \(i^{th}\) chord subdividing the curve \(P_1P_n\). Since the chord \(P_iP_{i+1}\) forms a right triangle, by using the Pythagorean theorem we can express \(L(P_iP_{i+1})\) as \(\sqrt{(Δx_i)^2+(Δy_i)^2}\).
Let’s try to represent the quantity \(L_i(P_iP_{i+1})\) in the same form as \(f(x_i)Δx_i\). In Figure 2, I have drawn a “zoomed in” image of the \(i^{th}\) chord \(P_iP_{i+1}\). As you can see from Figure 2, the typical chord \(P_iP_{i+1}\) can be expressed in terms of \(x\) and \(y\) as
$$L_i(P_iP_{i+1})=\sqrt{(Δx_i)^2+(Δy_i)^2}.\tag{4}$$
Equation (4) brought us one step closer to expressing \(L_i(P_iP_{i+1})\) in the form \(f(x_i)\); the only problem is that \(Δy_i\) term in Equation (4). Let’s try to represent the radical in Equation (4) entirely in terms of \(x\) by doing some algebraic manipulations. If we multiply the right-hand side of Equation (4) by \(\frac{\sqrt{(Δx_i)^2}}{\sqrt{(Δx_i)^2}}\) (\(=1\)), then we can simplify the radical in Equation (4) to
$$\frac{\sqrt{(Δx_i)^2}}{\sqrt{(Δx_i)^2}}\sqrt{(Δx_i)^2+(Δy_i)^2}=\sqrt{\frac{1}{(Δx_i)^2}\biggl((Δx_i)^2+(Δy_i)^2\biggr)}Δx_i$$
$$=\sqrt{1+\biggl(\frac{Δy_i(x)}{Δx_i }\biggr)^2}Δx_i $$
Thus,
$$L_i(P_iP_{i+1})=\sqrt{1+\biggl(\frac{Δy_i(x)}{Δx_i }\biggr)^2}Δx_i.\tag{5}$$
What is nice about Equation (5) is that it can be expressed entirely in terms of (x\) since the slope, \(Δy_i(x)/Δx_i\), of the chord \(P_iP_{i+1}\) can be represented entirely in terms of \(x\). Substituting Equation (5) into (2), we have
$$S=\lim_{n→∞}\sum_{i=1}^n=\sqrt{1+\biggl(\frac{Δy_i(x)}{Δx_i }\biggr)^2}Δx_i.\tag{6}$$
The sum in Equation (6) is precisely the same form as Equation (3). Thus, the limit in Equation (6) must be equal to the definite integral
$$\lim_{n→∞}\sum_{i=1}^n=\sqrt{1+\biggl(\frac{Δy_i(x)}{Δx_i }\biggr)^2}Δx_i=\int_a^b\sqrt{1+(f’(x))^2}dx.\tag{7}$$
(As the number \(n\) of subdivisions of the arc \(P_1P_n\) approaches infinity, the length of the typical chord, \(L_i(P_iP_{i+1})\), becomes infinitesimally small. This means that the other side lengths \(Δx_i\) and \(Δy_i\) in the right triangle in Figure 2 must also become infinitesimally small. In other words, as \(n→∞\), \(Δx_i→dx\) and \(Δy_i→dy\). Thus, \(Δy_i/Δx_i→dy/dx\) and the slope becomes the derivative \(f’(x)\). I thought I would take the time to explain this since it’s a common source of confusion as to where the \(f’(x)\) in Equation (7) comes from.)
The definite integral in Equation (7) is used to calculate the arc lenth \(s\) of any arbitrary curve. Provided that you know what the function \(f’(x)\) is, then you’ll be able to determine the arc length so long as the definite integral in Equation (7) is solvable. If the definite integral in Equation (7) is unsolvable (indeed, most integrals are), then you’ll have to resort to numerical methods (ideally using a computer) to estimate its value.