When a curve is given by the graph of a function, the distance along it cannot be found as easily as the length of a line segment. However, the arc length of a curve over a given interval can be calculated using a definite integral. Let us see how the arc length formula is derived, what role the derivative plays in it, and how to apply it in practical examples.
Arc Length of a Curve: From a Polygonal Line to the Exact Length
Suppose the function
\[ y=f(x) \]
is defined on the interval \( [a,b] \).
We are interested in the part of its graph from the point with \( x \)-coordinate \( a \) to the point with \( x \)-coordinate \( b \). How can we determine the length of this curve if it is not a straight line?
Let us start with a simple idea. Divide the interval \( [a,b] \) into small parts using the points
\[ a=x_0<x_1<x_2<\ldots<x_n=b. \]
On the graph of the function, these values correspond to the points
\[ A_i=(x_i,f(x_i)). \]
Now we can connect each pair of neighboring points with a line segment. As a result, instead of a smooth curve, we obtain a polygonal line.
The length of each segment can be found using the distance formula between two points:
\[ \Delta l_i= \sqrt{(\Delta x_i)^2+(\Delta y_i)^2}, \]
where
\[ \Delta x_i=x_i-x_{i-1},\qquad \Delta y_i=f(x_i)-f(x_{i-1}). \]
Then the total length of the polygonal line is
\[ L_n= \sum_{i=1}^{n} \sqrt{(\Delta x_i)^2+(\Delta y_i)^2}. \]
What happens if we choose points on the curve that are closer and closer together? The segments of the polygonal line become shorter, while the polygonal line itself follows the shape of the curve more accurately. In the limit, its length approaches the length of the corresponding arc.
So, the problem comes down to studying the behavior of one small segment of the polygonal line and then moving from these segments to the curve itself.
Arc Length Element: From Increments to the Derivative
Now let us focus on one segment of the polygonal line and see what happens to its length as the partition points move closer together.
For a small segment, we have
\[ \Delta l= \sqrt{(\Delta x)^2+(\Delta y)^2}. \]
Factor \( (\Delta x)^2 \) out from under the square root:
\[ \Delta l= \sqrt{ (\Delta x)^2 \left( 1+\left(\frac{\Delta y}{\Delta x}\right)^2 \right) }. \]
If we move along the \( x \)-axis from \( a \) to \( b \), then \( \Delta x>0 \). Therefore,
\[ \Delta l= \sqrt{ 1+\left(\frac{\Delta y}{\Delta x}\right)^2 }\,\Delta x. \]
Now let us make the length of the interval smaller and smaller. The ratio of the change in the function to the change in \( x \) approaches the derivative:
\[ \frac{\Delta y}{\Delta x}\to f'(x). \]
Moving from increments to differentials, we obtain an expression for the arc length element:
\[ dl= \sqrt{1+\left(f'(x)\right)^2}\,dx. \]
Where does this expression come from? It is based on the Pythagorean theorem. A small change in the \( x \)-coordinate is represented by \( dx \), while the corresponding change in the \( y \)-coordinate is represented by \( dy \). At the same time, the differentials satisfy the relation
\[ dy=f'(x)\,dx. \]
Therefore, the arc length element depends not only on the change in the \( x \)-coordinate but also on the change in the value of the function. The derivative \( f'(x) \) shows how quickly the function changes.
The greater the absolute value of the derivative, the steeper the curve. As a result, the length of a small part of the curve is greater than the corresponding displacement along the \( x \)-axis.
We now have an expression for the length of an individual element of the curve. The next step is to combine these elements over the entire interval from \( a \) to \( b \).
Arc Length of a Curve: From the Formula to the Definite Integral
At each point of the curve, its length element is given by
\[ dl= \sqrt{1+\left(f'(x)\right)^2}\,dx. \]
To obtain the length of the entire arc, we need to add up these elements over the whole given interval. This is exactly what the definite integral does.
If the function \( y=f(x) \) has a continuous derivative on the interval \( [a,b] \), then its graph on this interval is a smooth curve whose length can be calculated using the formula
\[ L= \int_a^b \left( \sqrt{1+\left(f'(x)\right)^2} \right)dx. \]
Here, \( a \) and \( b \) determine the beginning and end of the part of the graph whose length we want to find. The derivative \( f'(x) \) accounts for how the function changes over this interval, while integration combines the lengths of all the small elements of the curve into a single value \( L \).
Thus, finding the arc length can be reduced to a clear sequence of steps. First, determine the endpoints of the required part of the curve and find the derivative \( f'(x) \). Then form the expression
\[ \sqrt{1+\left(f'(x)\right)^2} \]
and calculate its definite integral over the interval \( [a,b] \). The resulting value is the length of the given arc.
Arc Length of a Curve: Practical Calculations Using an Integral
The arc length formula becomes much easier to understand when we apply it step by step to different functions. The key is to find the derivative correctly, simplify the expression under the square root, and only then calculate the definite integral.
Example 1. Find the arc length of the graph of the function
\[
y=2\cdot x+1
\]
on the interval \( [0,3] \).
To find the arc length, we use the formula
\[ L= \int_a^b \left( \sqrt{1+\left(f'(x)\right)^2} \right)dx. \]
First, find the derivative of the given function:
\[ f'(x)=2. \]
Substitute it into the formula:
\[ L= \int_0^3 \left( \sqrt{1+2^2} \right)dx. \]
Simplify the expression under the square root:
\[ L= \int_0^3 \left( \sqrt{5} \right)dx. \]
Since \( \sqrt{5} \) does not depend on \( x \), we can take it outside the integral:
\[ L= \sqrt{5} \int_0^3 dx. \]
Now calculate the definite integral:
\[ L= \sqrt{5}\cdot x\Big|_0^3. \]
Therefore,
\[ L= 3\cdot\sqrt{5}. \]
Thus, the length of the given part of the graph is
\[ L=3\cdot\sqrt{5}. \]
In this case, the graph is a straight line, so the resulting length is equal to the distance between its endpoints.
Example 2. Find the arc length of the graph of the function
\[
y=\frac{x^2}{2}
\]
on the interval \( [0,1] \).
Let us start by finding the derivative:
\[ f'(x)=x. \]
Then the arc length formula becomes
\[ L= \int_0^1 \left( \sqrt{1+x^2} \right)dx. \]
Unlike the previous case, the integrand now depends on \( x \). To calculate this integral, we use the known antiderivative formula:
\[ \int \left( \sqrt{1+x^2} \right)dx = \frac{x}{2}\cdot\sqrt{1+x^2} + \frac{1}{2}\ln\left(x+\sqrt{1+x^2}\right)+C. \]
Therefore,
\[ L= \left( \frac{x}{2}\cdot\sqrt{1+x^2} + \frac{1}{2}\ln\left(x+\sqrt{1+x^2}\right) \right)\Bigg|_0^1. \]
Substitute the upper limit:
\[ \frac{1}{2}\cdot\sqrt{2} + \frac{1}{2}\ln\left(1+\sqrt{2}\right). \]
For the lower limit, we have
\[ \frac{0}{2}\cdot\sqrt{1} + \frac{1}{2}\ln(1) = 0. \]
Therefore,
\[ L= \frac{\sqrt{2}}{2} + \frac{1}{2}\ln\left(1+\sqrt{2}\right). \]
The approximate value is
\[ L\approx1.148. \]
Thus, the arc length of the parabola over the given interval is
\[ L= \frac{\sqrt{2}}{2} + \frac{1}{2}\ln\left(1+\sqrt{2}\right) \approx1.148. \]
Example 3. Find the arc length of the graph of the function
\[
y=\sqrt{4-x^2}
\]
on the interval \( [0,1] \).
First, find the derivative:
\[ f'(x)= -\frac{x}{\sqrt{4-x^2}}. \]
Now substitute it into the arc length formula:
\[ L= \int_0^1 \left( \sqrt{ 1+ \left( -\frac{x}{\sqrt{4-x^2}} \right)^2 } \right)dx. \]
After squaring, the minus sign disappears:
\[ L= \int_0^1 \left( \sqrt{ 1+ \frac{x^2}{4-x^2} } \right)dx. \]
Write the expression under the square root with a common denominator:
\[ 1+\frac{x^2}{4-x^2} = \frac{4-x^2+x^2}{4-x^2} = \frac{4}{4-x^2}. \]
Therefore,
\[ L= \int_0^1 \left( \sqrt{\frac{4}{4-x^2}} \right)dx = \int_0^1 \left( \frac{2}{\sqrt{4-x^2}} \right)dx. \]
We use the antiderivative
\[ \int \left( \frac{1}{\sqrt{4-x^2}} \right)dx = \arcsin\left(\frac{x}{2}\right)+C. \]
Therefore,
\[ L= 2\cdot \arcsin\left(\frac{x}{2}\right)\Bigg|_0^1. \]
Substitute the limits:
\[ L= 2\cdot \left( \arcsin\left(\frac{1}{2}\right) – \arcsin(0) \right). \]
Since
\[ \arcsin\left(\frac{1}{2}\right)=\frac{\pi}{6}, \qquad \arcsin(0)=0, \]
we obtain
\[ L= 2\cdot\frac{\pi}{6} = \frac{\pi}{3}. \]
Thus,
\[ L=\frac{\pi}{3}\approx1.047. \]
This result agrees with the geometric meaning of the function: its graph is part of a circle with radius \( 2 \).
Example 4. Find the arc length of the graph of the function
\[
y=
\frac{x^3}{6}
+
\frac{1}{2\cdot x}
\]
on the interval \( [1,2] \).
Find the derivative of the function:
\[ f'(x) = \frac{x^2}{2} – \frac{1}{2\cdot x^2}. \]
Then
\[ L= \int_1^2 \left( \sqrt{ 1+ \left( \frac{x^2}{2} – \frac{1}{2\cdot x^2} \right)^2 } \right)dx. \]
At first glance, this integral looks quite complicated. However, it is worth simplifying the expression under the square root before trying to integrate it.
Square the expression in parentheses:
\[ \left( \frac{x^2}{2} – \frac{1}{2\cdot x^2} \right)^2 = \frac{x^4}{4} – \frac{1}{2} + \frac{1}{4\cdot x^4}. \]
Add one:
\[ 1+ \left( \frac{x^2}{2} – \frac{1}{2\cdot x^2} \right)^2 = \frac{x^4}{4} + \frac{1}{2} + \frac{1}{4\cdot x^4}. \]
The resulting expression is a perfect square:
\[ \frac{x^4}{4} + \frac{1}{2} + \frac{1}{4\cdot x^4} = \left( \frac{x^2}{2} + \frac{1}{2\cdot x^2} \right)^2. \]
On the interval \( [1,2] \), the expression in parentheses is positive. Therefore,
\[ \sqrt{ 1+ \left( f'(x) \right)^2 } = \frac{x^2}{2} + \frac{1}{2\cdot x^2}. \]
Thus,
\[ L= \int_1^2 \left( \frac{x^2}{2} + \frac{1}{2\cdot x^2} \right)dx. \]
Find the antiderivative:
\[ L= \left( \frac{x^3}{6} – \frac{1}{2\cdot x} \right)\Bigg|_1^2. \]
Substitute the limits:
\[ L= \left( \frac{8}{6} – \frac{1}{4} \right) – \left( \frac{1}{6} – \frac{1}{2} \right). \]
Now perform the calculations:
\[ L= \frac{8}{6} – \frac{1}{4} – \frac{1}{6} + \frac{1}{2} = \frac{17}{12}. \]
Thus,
\[ L=\frac{17}{12}\approx1.417. \]
This example shows why it is important not to rush into integration. Sometimes an expression under the square root that looks complicated at first becomes much simpler after some algebraic transformations.
Example 5. Find the arc length of the graph of the function
\[
y=\ln(\cos(x))
\]
on the interval \( \left[0,\frac{\pi}{4}\right] \).
First, find the derivative:
\[ f'(x) = -\tan(x). \]
Substitute it into the formula:
\[ L= \int_0^{\frac{\pi}{4}} \left( \sqrt{1+\tan^2(x)} \right)dx. \]
Use the trigonometric identity
\[ 1+\tan^2(x)=\sec^2(x). \]
Therefore,
\[ L= \int_0^{\frac{\pi}{4}} \left( \sqrt{\sec^2(x)} \right)dx. \]
In general,
\[ \sqrt{\sec^2(x)}=|\sec(x)|. \]
On the interval
\[ \left[0,\frac{\pi}{4}\right], \]
we have \( \cos(x)>0 \), and therefore \( \sec(x)>0 \). So we can remove the absolute value:
\[ L= \int_0^{\frac{\pi}{4}} \left( \sec(x) \right)dx. \]
The antiderivative of this function is
\[ \int \left( \sec(x) \right)dx = \ln|\sec(x)+\tan(x)|+C. \]
Therefore,
\[ L= \ln|\sec(x)+\tan(x)| \Bigg|_0^{\frac{\pi}{4}}. \]
For the upper limit, we have
\[ \sec\left(\frac{\pi}{4}\right)=\sqrt{2}, \qquad \tan\left(\frac{\pi}{4}\right)=1. \]
For the lower limit,
\[ \sec(0)=1, \qquad \tan(0)=0. \]
Therefore,
\[ L= \ln(\sqrt{2}+1)-\ln(1). \]
Since
\[ \ln(1)=0, \]
we obtain
\[ L= \ln(1+\sqrt{2}). \]
Its approximate value is
\[ L\approx0.881. \]
Thus, the length of the given arc is
\[ L=\ln(1+\sqrt{2})\approx0.881. \]
From Arc Length to New Problems: What to Study Next
Arc length is just one of the important applications of the definite integral. Next, it is worth exploring other geometric problems and then moving on to the integration of functions of two variables.
- Area of a Region Using an Integral: Formulas and Examples — You will learn how the definite integral can be used to find the areas of regions bounded by graphs of functions.
- Volume of a Solid of Revolution: Applications of the Definite Integral — You will see how the definite integral can be used to find the volumes of solids formed by rotating plane regions.
- Double Integrals: Meaning, Region of Integration, and Examples — You will become familiar with integrating functions of two variables, defining the region of integration, and the main methods of evaluating double integrals.
Arc Length of a Curve: From a Mathematical Formula to Your Own Algorithm
If you enjoy programming, try putting what you have learned into practice. The flowchart shows an algorithm for approximately calculating the arc length of the function
\[ y=\frac{x^2}{2} \]
on the interval \( [0,1] \) using the midpoint rule. Follow the sequence of steps and then implement this algorithm in your favorite programming language. Compare the value obtained by the program with the result found using the definite integral.
