# How do you find the length of a curve in calculus?

May 6, 2018

In Cartesian coordinates for y = f(x) defined on interval $\left[a , b\right]$ the length of the curve is

$\implies L = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

In general, we could just write:

$\implies L = {\int}_{a}^{b} \mathrm{ds}$

#### Explanation:

Let's use Cartesian coordinates for this explanation.

If we consider an arbitrary curve defined as $y = f \left(x\right)$ and are interested in the interval $x \in \left[a , b\right]$, we can approximate the length of the curve using very tiny line segments.

Consider a point on the curve ${P}_{i}$. We can compute the distance of a line segment by finding the difference between two consecutive points on the line $| {P}_{i} - {P}_{i - 1} |$ for $i \in \left[1 , n\right]$ where $n$ is the number of points we've defined on the curve.

This means that the approximate total length of curve is simply a sum of all of these line segments:

$L \approx {\sum}_{i = 1}^{n} | {P}_{i} - {P}_{i - 1} |$

If we want the exact length of the curve, then we can make the assumption that all of the points are infinitesimally separated. We now take the limit of our sum as $n \to \infty$.

$L = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} | {P}_{i} - {P}_{i - 1} |$

Since we are working in the $x y$-plane, we can redefine our distance between points to take on the typical definition of Euclidean distance.

$| {P}_{i} - {P}_{i - 1} | = \sqrt{{\left({y}_{i} - {y}_{i - 1}\right)}^{2} + {\left({x}_{i} - {x}_{i - 1}\right)}^{2}} = \sqrt{\delta {y}^{2} + \delta {x}^{2}}$

We can now apply the Mean Value Theorem, which states there exists a point ${x}_{i}^{'}$ lying in the interval $\left[{x}_{i - 1} , {x}_{i}\right]$ such that

$\implies f \left({x}_{i}\right) - f \left({x}_{i - 1}\right) = f ' \left({x}_{i}^{'}\right) \left({x}_{i} - {x}_{i - 1}\right)$

which we could also write (using the notation we are using) as

$\implies \delta y = f ' \left({x}_{i}^{'}\right) \delta x$

Applying this means we now have

$| {P}_{i} - {P}_{i - 1} | = \sqrt{{\left[f ' \left({x}_{i}^{'}\right) \delta x\right]}^{2} + \delta {x}^{2}}$

Simplifying this expression a bit gives us

$| {P}_{i} - {P}_{i - 1} | = \sqrt{{\left[f ' \left({x}_{i}^{'}\right)\right]}^{2} \delta {x}^{2} + \delta {x}^{2}}$

$| {P}_{i} - {P}_{i - 1} | = \sqrt{\left({\left[f ' \left({x}_{i}^{'}\right)\right]}^{2} + 1\right) \delta {x}^{2}}$

$| {P}_{i} - {P}_{i - 1} | = \sqrt{\left(1 + {\left[f ' \left({x}_{i}^{'}\right)\right]}^{2}\right)} \delta x$

We can now use this new distance definition for our points in our summation.

$L = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \sqrt{\left(1 + {\left[f ' \left({x}_{i}^{'}\right)\right]}^{2}\right)} \delta x$

Sums are nice, but integrals are nicer for continuous circumstances! It's easy to just write this as a definite integral since both integrals and sums are "summation" tools. In the integral, we can drop our sum index as well.

$L = {\int}_{a}^{b} \sqrt{\left(1 + {\left[f ' \left(x\right)\right]}^{2}\right)} \delta x$

Writing this a little bit more typically yields

$\textcolor{b l u e}{L = {\int}_{a}^{b} \sqrt{\left(1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}\right)} \mathrm{dx}}$

We have arrived at our result! In general, the length is usually defined for a differential of arclength $\mathrm{ds}$

$L = {\int}_{a}^{b} \mathrm{ds}$

where $\mathrm{ds}$ is defined accordingly for whatever type of coordinate system you are working in. However, I wanted the explanation to be clearer, so I just chose Cartesian ones for simplicity. You could use polar coordinates or spherical coordinates as well by simply making the necessary substitutions.

In general, you need to take the derivative of the function defining your curve to substitute into the integral. Then the trick is to find a way (usually) to try and get a perfect square inside the square root to simplify the integral and find your solution. It varies for every type of curve.

Let me know if you have any further questions in the comments!