# How to calculate the length of a line or curve

The length of a line or curve can be calculated for many functions, but not all of them (which is common for arc-length integrals) for example, the length of a sine wave has to be done numerically. Some integrals require trigonometric substitution.
There is a standard formula for the length of a curve. Here is where it came from.
If you take a line and approximate the length of a part of it, you would use pythagoras' theorem:

(Δs)2  ≈ (Δx)2 + (Δy)2
If I could get all the little line lengths and add them up, I can calculate the length of the curve between one point and another. I can achieve this first by finding an equation I can integrate. Integration is like adding tiny values together so what I want to do is make an equation that has a divide by Δx part. This is why.
You can read my article explaining the link between the summation symbol ∑ and integration so what I'm going to do once I've found an equation which has a divide by Δx part so when it is summed, it is equivalent of just adding the Δs parts all divided by Δx:
 Δs1 + Δs2 + Δs3 +  ... = Δs1 + Δs2 + Δs3 + ... Δx Δx Δx Δx

I can take the above, add all them up and multiply by Δx and I will get the length of the line.
 ∑ n Δsn Δx Δx = Δsn

## Finding an equation

I'm going to try to get the Δx on the left:

(Δs)2  ≈ (Δx)2 + (Δy)2 then divide by (Δx)2
 (Δs)2 ≈ 1 + (Δy)2 (Δx)2 (Δx)2 so now I can take the square root of both sides

 Δs ≈ √( 1 + (Δy)2 ) Δx (Δx)2
I can simplify it a little since (Δy)2 / (Δx)2 = (Δy / Δx)2
 Δs Δx ≈ √(1 + (Δy / Δx)2)
I can think of two ways to understand how you integrate this. First of all, you can take the √(1 + (Δy / Δx)2) equation and multiply it by Δx and add up each little line length, then change it to integration:
∑ √(1 + (Δy / Δx)2) Δx = the length of the line
So making Δx smaller (or as Δx → 0) then I have this equation:
 Line length between a and b = ∫ b √(1 + (dy / dx)2) dx a

You can do this because
 lim ∑ yi•Δx = ∫ y dx Δx → 0
Which means as the line lengths become unimaginally small, then it equals the integral equation.

Another way of thinking about it is getting this equation again:
 Δs Δx ≈ √(1 + (Δy / Δx)2)

and make Δx smaller Δx → 0 then I have this equation:
 ds dx ≈ √(1 + (dy / dx)2)

Now I can integrate the above:

 Line length between a and b = ∫ b √(1 + (dy / dx)2) dx a
Now for an example...

## Why is the circumference of a circle 2πr?

To illustrate what I have written above, I shall calculate the perimiter of a circle. I get an equation that draws half a circle, calculate its length, then I double it (the term 'r' is representing the radius):
y = √(r2 - x2)

The standard line length formula is:
 Line length = ∫ b √(1 + (dy / dx)2) dx a

so doubling the length of half a circle will get me the circumference if I draw it from -r to r:
y = √(r2 - x2)
 Circumference = 2 ∫ r √(1 + (dy / dx)2) dx -r

I want to calculate the gradient of 'y' first. I have written an article on differentiating functions which I will use here. I have to use the function of a function (you can look at it by clicking here):
if y(x) = b(a(x)) then y'(x) = a'(x) • b'(a(x))
so if a(x) = r2 - x2 then a'(x) = -2x
 and if b(x) = √a(x), then the gradient is b'(x) = -1 = -1 2√(a(x)) 2√(r2 - x2)

putting them together and simplifying, y'(x) = a'(x) • b'(a(x)):

 dy = -2x -1 = x dx 2√(r2 - x2) √(r2 - x2)

Now if you look at the equation, I need to square this gradient and add one:
 Circumference = 2 ∫ r √(1 + (dy / dx)2) dx -r
so squaring the gradient gives me:
 x2 r2 - x2

To add 1 to this, I actually add (r2 - x2) / (r2 - x2) which is the same as adding one (provided the denominator is not zero) so now I can say (after I have simplified the equation):
 1 + (dy / dx)2 = r2 r2 - x2
I need to take the square root of this. I am left with a constant 'r' which I will put outside the integral (to calculate this, I use trigonometric substitution where an example of this can be found on my home page under "Why does the area of a circle equal πr2?")
 Circumference = 2r ∫ r 1 dx = 2r(arcsin (1) - arcsin(-1) ) -r r2 - x2

= 2πr