A hanging bridge in the middle of the forest
Calculus has a wide range of applications, one of which is finding the properties of curves. Finding the length of a curve is a prime example of using both derivatives and integrals together. Let's see how derivatives and integrals pair together to find the length of a curve!
Finding the Arc Length of a Curve
Let's think for a moment about the length of a curve. If rather than a curve you had a straight line you could easily find its length in a given interval using the Pythagorean theorem.
Fig. 1. The Pythagorean Theorem can be used to find the length of a straight segment.
Just like you can approximate the area below a curve using rectangles, you can approximate the length of a curve using straight segments. Let's see an illustration on how this is done.
Fig. 2. Approximation of the length of the parabola using 4 segments.
If you use more segments you will get a better approximation.
Fig. 3. Approximation of the length of the parabola using 8 segments.
Sounds familiar? Just like in Riemann Sums, you start by making a partition of the interval, then you evaluate the function at each value of the partition. This time you do not have to deal with right or left-endpoints since both values are being used to find the segments. The length of each individual segment can be found using the Pythagorean theorem.
Fig. 4. The Pythagorean Theorem can be used to find the length of each segment.
Finally, all segments are added up, finding an approximation of the length of the curve. But what if we want the exact value of the curve's length? Then you need to integrate.
Formula for the Arc Length of a Curve
Suppose you need to find an approximation of the length of a curve in the interval \( [a,b] \). You can follow these steps:
Do a partition of the interval using \(N\) points.
Find the length of each segment that joins a pair of adjacent points of the partition.
Add the length of all the segments.
Let's name each individual segment \(s_{i}\) and the approximation will be \(S_N\). The length of the \(i\text{-}\)th segment is given by
$$s_{i}=\sqrt{(\Delta x)^2+(\Delta y_{i})^2}.$$
You can rewrite the above expression as
$$s_{i}=\Delta x\sqrt{1+\Big(\frac{\Delta y_{i}}{\Delta x}\Big)^2}$$
with the help of some algebra. By adding all the segments together you get an approximation for the length of the curve
$$S_N = \sum_{i=1}^{N}s_{i}.$$
For each segment \(s_{i}\), The Mean Value Theorem tells us that there is a point within each subinterval \(x_{i-1}\leq x_{i}^{*}\leq x_{i}\) such that \(f'(x_{i}^{*})=\frac{\Delta y_{i}}{\Delta x_i}\). This is where derivatives come into play! The length of each individual segment can then be rewritten as
$$s_{i}=\Delta x\sqrt{1+(f'(x_{i}^{*}))^2}.$$
By taking the limit as \(N\rightarrow\infty\), the sum becomes the integral
$$\begin{align}\text{Arc Length} &= \lim_{N\rightarrow\infty}\sum_{i=1}^{N}\Delta x\sqrt{1+(f'(x_{i}^{*})^2}\\ &=\int_{a}^{b}\sqrt{1+(f'(x))^2}\,\mathrm{d}x,\end{align}$$
giving you an expression for the length of the curve. This is the formula for the Arc Length.
Let \(f(x)\) be a function that is differentiable on the interval \( [a,b]\) whose derivative is continuous on the same interval. The Arc Length of the curve from the point \( (a,f(x))\) to the point \((b,f(b))\) is given by the following formula:
$$\text{Arc Length}=\int_{a}^{b}\sqrt{1+(f'(x))^2}\,\mathrm{d}x.$$
Please note that the expressions involved in finding arc lengths are sometimes hard to integrate. If you need a refresher be sure to check out our Integration Techniques article!
Arc Length of a Curve Examples
Let's see some examples of how to find the arc length of curves.
Find the length of \(f(x)=x^{\frac{3}{2}}\) on the interval \( [0,3]\).
Answer:
To find the arc length of the given function you will need first to find its derivative, which can be found using The Power Rule, that is
$$f'(x)=\frac{3}{2}x^{\frac{1}{2}}.$$
Since the derivative resulted in a continuous function you can freely use the formula for finding the Arc Length
$$\text{Arc Length}=\int_a^b \sqrt{1+(f'(x))^2}\,\mathrm{d}x,$$
and then substitute \(a=0\), \(b=3\), and \(f'(x)=\frac{3}{2}x^{\frac{1}{2}}\) into the formula, giving you
$$\begin{align} \text{Arc Length} &= \int_0^3 \sqrt{1+\Big(\frac{3}{2}x^{\frac{1}{2}}\Big)^2}\,\mathrm{d}x \\[0.5em] &=\int_0^3 \sqrt{1+\frac{9}{4}x}\,\mathrm{d}x. \end{align}$$
You can find the antiderivative using Integration by Substitution. Start by letting
$$u=1+\frac{9}{4}x,$$
use The Power Rule to find its derivative
$$\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{9}{4},$$
and use it to find \( \mathrm{d}x \)$$\mathrm{d}x=\frac{4}{9}\mathrm{d}u.$$
This way you can write the integral in terms of \(u\) and \(\mathrm{d}u\)
$$\int\sqrt{1+\frac{9}{4}x}\,\mathrm{d}x=\frac{4}{9}\int\sqrt{u}\,\mathrm{d}u,$$
so you can integrate it using the power rule
$$\int\sqrt{1+\frac{9}{4}x}\,\mathrm{d}x=\frac{4}{9}\cdot\frac{2}{3}u^{\frac{3}{2}},$$
and substitute back \(u=1+\frac{9}{4}x\) while simplifying
$$\int\sqrt{1+\frac{9}{4}x}\,\mathrm{d}x=\frac{8}{27}(1+\frac{9}{4}x)^{\frac{3}{2}}.$$
You can now go back to the arc length formula and evaluate the definite integral using The Fundamental Theorem of Calculus
$$\text{Arc Length}=\frac{8}{27}\left(1+\frac{9}{4}(3)\right)^{\frac{3}{2}}-\frac{8}{27}\left(1+\frac{9}{4}(0)\right)^{\frac{3}{2}}.$$
The above expression can be evaluated using a calculator. Here we will round down to 2 decimal places for illustrative purposes, so
$$\text{Arc Length}\approx 6.1$$
Most of the integrals we need to evaluate in order to find the arc length of a curve are hard to do. We can use a Computer Algebra System to evaluate the resulting definite integrals!
Find the arc length of \(f(x)=\frac{1}{2}x^2\) on the interval \( [1,2]\). Evaluate the resulting definite integral using a Computer Algebra System or a graphing calculator.
Answer:
Begin by using The Power Rule to find the derivative of the function
$$f'(x)=x,$$
and use the arc length formula
$$\text{Arc Length}=\int_a^b \sqrt{1+(f'(x))^2}\,\mathrm{d}x.$$
Now you can substitute \(a=1\), \(b=2\) and \(f'(x)=x\) into the arc length formula to get
$$\text{Arc Length}=\int_1^2 \sqrt{1+x^2}\,\mathrm{d}x,$$
which can be done with Trigonometric Substitution. Unfortunately, it is rather complicated, so you can use a Computer Algebra System instead to evaluate the definite integral:
$$\text{Arc Length}\approx 1.8101.$$
Arc Length of a Curve described by an equation
So far, you have been studying the Arc Length of curves that can be described using functions. However, it is also possible to find the arc length of curves that are described using equations, like the equation of a circumference
$$x^2+y^2=r^2.$$
The above equation, despite not being a function, can also be graphed on a coordinate system. You can also find its Arc Length! The approach is quite similar, but you need to consider different factors. Take a look at our Arc Length in Polar Coordinates article for a review on the subject!
Arc Length of a Plane Curve
A plane curve is a curve that you can draw on a plane. All the above examples are curves on a plane.
It is important to stress this out because it is also possible to have curves in three-dimensional space, which is unfortunately out of the scope of this article.
Arc Length of a Parametric Curve
When studying about the arc length of a curve you might come upon the Arc Length of a Parametric Curve. This refers to another subject and is out of the scope of this article. For more information take a look at our Calculus of Parametric Curves and Length of Parametric Curves articles.
Summary
Arc Length of a Curve - Key takeaways
- The length of a curve can be approximated by splitting the curve into straight segments.
- For a function \(f(x)\) that is differentiable, and whose derivative is continuous, the exact Arc Length of the curve in the interval \( [a,b] \) is given by $$\text{Arc Length}=\int_a^b \sqrt{1+(f'(x))^2}\,\mathrm{d}x.$$
- The definite integrals involved in the calculation of Arc Length are rather complex. The use of Computer Algebra Systems can be extremely helpful when evaluating such integrals.
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