Aside: Without an introduction to integral calculus, this section may be too much. However, you should be able to understand the ideas as well as the numerical approximations.
Just as we have a formula for the length of a line, the distance formula, we can also develop a formula for the length of a curve. In this section, we will only be able to find the length of a curve between two points on the curve (in two dimensions) and when we have an equation for the curve. Also, we will stay in two dimensions.
Figure 1: Curve approximated by a few straight lines. By using the distance formula the curve length is approximately the sum of the line segment lengths. Obviously the more line segments that are used, the closer will be the approximation.
Figure 1 shows a curve approximated by line segments. The length of the curve is approximately the sum of the lengths of the segments and as more and more segments are added, the approximation will become more precise. The length of each segment is computed by the distance formula.
$$ \begin{aligned} & |S_{i-1} - S_{i}|= \\ & \sqrt{(x_{i}-x_{i-1})^{2}+(y_{i}-y_{i-1})^{2}} \end{aligned} \tag*{$\class{red}{(1)}$} \label{1}$$
Equation $\ref{1}$ can be rewritten as
$$\left|
S_{i-1} - S_{i}\right|=\sqrt{\Delta x^{2}+\Delta y^{2}} \tag{2} \label{2}$$
By the mean value theorem we can rewrite $\Delta y=f^{\prime}(x_{c})\cdot\Delta x$, where $c$ is a special value along the curve that causes the tangent at that point to be equal to the segment slope from $S_{i-1}$ to $S_{i}$. However, and very important, since this is an approximation anyway, as we choose more and more intervals, the value $x_{c},$ gets ever closer to each other for all possible choices. Consequently, we can choose the left or right endpoint of the interval, or the middle, or any consistent choice within the interval. At the limit, where $\Delta x\approx 0,$ there is but one choice.
$$\vert S_{i-1}-S_{i}\vert=\sqrt{\Delta x^{2}+\left(f^{\prime}(x_{c})\cdot\Delta x\right)^{2}} \tag{3} \label{3}$$
Factoring out $(\Delta x)^{2}$ and getting $\Delta x$ out from under the radical, we can express the length of each segment.
$$\vert S_{i-1}-S_{i}\vert=\Delta x\sqrt{1+\left(f^{\prime}(x_{c})\right)^{2}} \tag{4} \label{4}$$
It should be obvious that the length of the curve is the sum of the $n$ segment lengths. $$\sum_{i=1}^{n}\vert S_{i-1}-S_{i}\vert \tag{5} \label{5}$$
Without integral calculus, we might stop here and write computer code to get a quite good numerical answer. With calculus, we can go a step further.
Thus looking only at the right side of $\eqref{4}$ we can replace $f'(x_{c})$ with $dy/dx.$
$$=\Delta x\cdot\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}$$
Now, using the definite integral as the limit of sums
$$\text{Arc Length }=\lim\sum\vert S_{i-1}-S_{i}\vert_{i=1}^{\infty}=\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx \tag{6} \label{6}$$
where limits $a$ and $b$ represent the $x$ value (abscissa) endpoints of the function arc segment. This works well for $y=f(x)$ or if we want to just turn the graph sideways, then $x=g(y),$ but we have to be careful to only calculate arcs from true functions that do not have multiple values at a single abscissa or ordinate point. If we try to use this to compute the length of a spiral it will not go well. There is, however, a way to do that using parametric equations.
Exercise: Given the line $y=x+2$ and point $A=(1,3)\quad B=(4,6),$ obtain the distance between the points using the integral for standard arc length. Show that it is the same as would be found using the Euclidean norm.
Answer: We just need to form the expression
$$dist=\intop_{1}^{4}\left(\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\right)dx$$
The line equation is $y=x+2,$ so $\frac{dy}{dx}=1.$
$$\left.dist=x\sqrt{1+1^{2}}\right|_{1}^{4}=4\sqrt{2}-\sqrt{2}=3\sqrt{2}$$
Next we calculate the Euclidean norm.
$$\sqrt{(4-1)^{2}+(6-3)^{2}}=\sqrt{18}=3\sqrt{2}$$
They are the same.
Example 1: Given the function $$f(x)=\frac{1}{18}x^{9}+\frac{1}{14x^{7}}.$$ Find the arc length from $x=0.7$ to $x=1.3.$ Verify the answer with an estimate of summed pieces. (Although these equations are seldom integrable without numerical methods, this one is.)
Answer: Using equation $\eqref{6}$, we need the definite integral for $$\sqrt{1+\left(\frac{df(x)}{dx}\right)^{2}}dx.$$
Figure 2: Arc Length. Calculate the length of the curve between point $A(0.7,f(0.7))$ and point $B(1.3,f(1.3)).$ For estimation, 100 points have been placed on the curve using Geogebra and the command, $Pts=Sequence((x(A)+k,f(x(A)+k))$ $,k,0,1.3-0.7,(x(B)-x(A))/100).$ These were then summed with the command $Sum(sqrt((x(Pts(k)) - x(Pts(k + 1)))²$ $+ (y(Pts(k)) - y(Pts(k + 1)))²), k, 1, 100).$ The result was $1.44281.$
We can start with the derivative,
$$\frac{d\left(\frac{1}{18}x^{9}+\frac{1}{14x^{7}}\right)}{dx}=\frac{1}{2}x^{8}-\frac{1}{2x^{8}}$$
followed by squaring it and putting things back together.
$$\text{Arc Length }=\int_{0.7}^{1.3}\sqrt{1+\left(\frac{1}{2}x^{8}-\frac{1}{2x^{8}}\right)^{2}}dx=\int_{0.7}^{1.3}\sqrt{\frac{1}{2}+\frac{x^{16}}{4}+\frac{1}{4x^{16}}}dx$$
Pull out $1/2,$
$$\frac{1}{2}\int\sqrt{2+x^{16}+\frac{1}{x^{16}}}dx$$
Simplify,
$$\frac{1}{2}\int\left(\frac{x^{16}+1}{x^{8}}\right)dx=\frac{1}{2}\int\left(x^{8}+\frac{1}{x^{8}}\right)dx$$
$$=\frac{1}{2}\int x^{8}dx+\frac{1}{2}\int\frac{1}{x^{8}}dx$$
$$=\frac{1}{2}\cdot\frac{x^{9}}{9}-\frac{1}{2}\cdot\frac{1}{7x^{7}}$$
and implement the definite integral,
$$\begin{aligned}\left.\left(\frac{1}{2}\cdot\frac{x^{9}}{9}-\frac{1}{2}\cdot\frac{1}{7x^{7}}\right)\right|_{0.7}^{1.3}=\\
= & \left(\frac{1}{2}\cdot\frac{(1.3)^{9}}{9}-\frac{1}{2}\cdot\frac{1}{7(1.3)^{7}}\right)-\left(\frac{1}{2}\cdot\frac{(0.7)^{9}}{9}-\frac{1}{2}\cdot\frac{1}{7(0.7)^{7}}\right)\\
\approx & 1.4428
\end{aligned}
$$
For a numerical estimate of the value, see figure 2.
Example 2: Given $y(x)=2e^{-5x},$ find the length of the curve between $x=0$ and $x=2.$
Answer: Due to the integrals, these quickly get a little difficult and I will use a CAS to assist with the integrals.
We can plug into equation $\eqref{6}$ almost directly.
$$\frac{d\left(2e^{-5x}\right)}{dx}=-10e^{-5x}$$
$$(-10e^{-5x})^{2}=100e^{-10x}$$
$$\int_{a}^{b}\sqrt{1+\left(\frac{d\left(2e^{-5x}\right)}{dx}\right)^{2}}\,dx=3.468$$
Figure 3: Arc Length for this function is calculated by code using SageMath.
Rather than using pencil and paper and probably being unsuccessful at getting the integral, I will use SageMath to numerically get the answer.
from sage.symbolic.integration.integral import definite_integral
y(x)=2*e^(-5*x)
L1=plot(y,x,(0,2))
f(x)=sqrt(1+diff(y,x)^2)
Length=definite_integral(f(x),x,0,2).n()
print Length
show(L1,figsize=4)
Length=3.468
Example 3: Find the circumference for the ellipse given by $$\frac{x^{2}}{4}+\frac{y^{2}}{0.25}=1 \tag*{$\class{blue}{(7)}$} \label{7}$$
Figure 4: We only compute the length of the curve in quadrant 1 and then multiply by 4.
Answer: After earlier saying that this is a historically difficult problem we might wonder how we can solve it. The answer is that it will be done numerically. Although we are numerically following a different method than used in the earlier example, that method would work also. First, we want to graph the ellipse and get some approximation of the answer.
We see that the semi-major axis, $a$, has length $2$, and that the semi-minor axis has length $0.5.$ We could have gotten that just as easily from the equation, but there is a reason to see it. When graphed the ellipse consists of $4$ curves, one in each quadrant. Furthermore, the length of each curve must be at least $2$, since the $x$-axis distance is $\pm2.$ That implies that the minimum circumference for this ellipse is $8$ and we know it must really be a little more.
Figure 5: The curve is $$h=\sqrt{1+\left( -05\cdot \frac{x}{\sqrt{2^2-x^2}\cdot 2} \right)^2}.$$ The Riemann Sum was computed by Sagemath as the definite integral of $h$ between $0$ and $2-10^{-10}$ The integral was computed in Geogebra to get the same answer. Also, a simple addition of linear segments of the actual ellipse curve was performed with 500 divisions between $0$ and $2$ in quadrant 1. That sum was 2.1445.
Since we can see mirror symmetry across both axes, we will only get the length of the top part of the ellipse between $0$ and $2.$ Then we will multiply that by $4$ to get the whole circumference. Our length equation is
$$ArcLength=\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx$$
and it requires an equation of the form $y=f(x).$ We get that by solving equation $\ref{7}$ for $y.$ For the top half of the ellipse, the solved equation is
$$y=\sqrt{2^{2}-x^{2}}\cdot2\cdot\frac{0.5}{2^{2}} \tag*{$\class{blue}{(8)}$} \label{8}$$
Now according to $\eqref{6}$ we will need the derivative of $\ref{8}$ squared.
$$\left(\frac{dy}{dx}\right)^{2}=\left(-\frac{1}{2}\frac{x}{\sqrt{2^{2}-x^{2}}\cdot2}\right)^{2} \tag*{$\class{blue}{(9)}$} \label{9}$$
Finally, we can write the arc length of the top quarter.
$$\frac{1}{4}ArcLength=\intop_{0}^{2}\sqrt{1+\left(-\frac{1}{2}\frac{x}{\sqrt{2^{2}-x^{2}}\cdot2}\right)^{2}} \tag*{$\class{blue}{(10)}$} \label{10}$$
You might be wondering what we can do next, since despite all this work we still have a not evaluated integral. The answer is that we will need to graph $\ref{10}$ and use a method such as Riemann sums to approximate the integral. The graph is shown in figure 5.
