We can determine the arc of any curve using definite integrals. In this post, let's explore how it expands the approximate solution of solving arc lengths.
One critical application of integration is finding the arc length of a curve. This post explains the theory behind it:
Let's find the length of a curve between two points. It's best to illustrate it with real examples; say we have a function \(f(x)=x^2\):
We must solve the length from endpoint \(x_1=1\) to \(x_2=3\). How are we to accomplish this using integration?
Approximation
Arc of a Curve
Before discussing integration, say we approximate it first. Say that we connect the endpoints of the function \(f(x)=x^2\) between \(x_1=1\) to \(x_2=3\) with a straight line. We first solve for \(y_1\) and \(y_2\) by evaluating it, then use the distance formula to solve for the length \(L\).
The answer we got is only an approximation. For a better result, let's put intermediate points along the curve between the endpoints; Consequently, this divides the function curve into multiple-length sections. We can compute the distance between these points \(\Delta d_1, \Delta d_2, \Delta d_3, \ldots, \Delta d_{n}\) using the distance formula. Afterward, we add these distances to get an arc length \(s\).
The answer we will get is still inexact; however, it will be nearer to the actual value. Based on this, our approximation will be more accurate as we introduce more points between endpoints.
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Building from our approximate solution, we can solve the actual length of a curve. We will only need to modify two things in the approximate solution:
We need to set up a minuscule distance that is so small that we can express its length as the differential \(dL\)
We need an operation that will sum up all of these tiny divisions
We can combine all \(dL\) through integration; hence, the general equation to solve for the arc length \(s\) between two points along the function is:
\(s=\int_{a}^{b}{dL}\)
\(s\) is the arc length
\(dL\) is the differential length
\(a\) is the lower limit position (endpoint with the lowest value)
\(b\) is the upper limit position (endpoint with the highest value)
Differential Length
The differential length \(dL\) is the division we use to sum all points along the curve. We can express it in two ways - along the x- or y-axis.
We can express this length in terms of the distance formula. If \(dL\) is the hypotenuse, then we have differential \(dx\) and differential \(dy\) as the legs along the x and y-axes, respectively. Therefore, \(dL\) is equal to:
\(dL^2=dx^2+dy^2\)
We can express \(dL\) with respect to \(dx\) or \(dy\). From this, the arc equation \(s=\int_{x_1}^{x_2}{dL}\) can have two forms:
Differential Width (Along x)
Let's divide \(dL^2=dx^2+dy^2\) by \(dx^2\) and isolate \(dL\), we will have:
\(d L=\sqrt{1+\left(\frac{d x}{d y}\right)^2} d y\)
From here, the arc length equation becomes:
\(s=\int_{y_1}^{y_2}\sqrt{1+\left(\frac{d x}{d y}\right)^2} d y\)
\(s\) is the arc length
\(y_1\) is the lower limit along the y-axis
\(y_2\) is the upper limit along the y-axis
\(\frac{dx}{dy}\) is the first derivative of the curve with respect to \(y\)
\(dy\) is differential along y
Summary
Solving the arc length by integration expands on the approximation solution.
The logic is to subdivide the length needed with multiple points, solve for the length of each part, and add all of these distances.
It is necessary to keep the length small to increase the accuracy of the solution; hence, we must represent the length as differential \(dL\).
We can combine all \(dL\) through integration; hence, the general equation to solve for the arc length between two points: \(s=\int_{a}^{b}{dL}\). \(s\) is the arc length, \(dL\) is the differential length, \(a\) is the lower limit position (endpoint with the lowest value), \(b\) is the upper limit position (endpoint with the highest value)
The line with length \(dL\) is the differential length, which consists of differential \(dx\) and \(dy\).
There are two ways to express the arc length equation.
In terms along the x-axis, \(s\) is equal to: \(s=\int_{x_1}^{x_2}\sqrt{1+\left(\frac{d y}{d x}\right)^2} d x\)
In terms along the y-axis, \(s\) is equal to: \(s=\int_{y_1}^{y_2}\sqrt{1+\left(\frac{d x}{d y}\right)^2} d y\)
