# Arc Length

**Using Calculus to find the length of a curve**.*(Please read about Derivativesand Integrals first) *

Imagine we want to find the length of a curve between two points. And the curve is smooth (the derivative is continuous).

First we break the curve into small lengths and use the Distance Between 2 Points formula on each length to come up with an approximate answer:

The distance from **x _{0}** to

**x**is:

_{1}S_{1} = √ (x_{1} − x_{0})^{2} + (y_{1} − y_{0})^{2}

And let's use ** Δ** (delta) to mean the difference between values, so it becomes:

S_{1} = √ (Δx_{1})^{2} + (Δy_{1})^{2}

Now we just need lots more:

S_{2} = √(Δx_{2})^{2} + (Δy_{2})^{2}

S_{3} = √(Δx_{3})^{2} + (Δy_{3})^{2}

...

...

S_{n} = √(Δx_{n})^{2} + (Δy_{n})^{2}

We can write all those many lines in just **one line** using a Sum:

_{i})

^{2}+ (Δy

_{i})

^{2}

But we are still doomed to a large number of calculations!

Maybe we can make a big spreadsheet, or write a program to do the calculations ... but lets try something else.

We have a cunning plan:

- have all the
**Δx**be_{i}**the same**so we can extract them from inside the square root - and then turn the sum into an integral.

Let's go:

First, divide *and* multiply **Δy _{i}** by

**Δx**:

_{i}_{i})

^{2}+ (Δx

_{i})

^{2}(Δy

_{i}/Δx

_{i})

^{2}

Now factor out **(Δx _{i})^{2}**:

_{i})

^{2}(1 + (Δy

_{i}/Δx

_{i})

^{2})

Take **(Δx _{i})^{2}** out of the square root:

_{i}/Δx

_{i})

^{2}Δx

_{i}

Now, as **n approaches infinity** (as we head towards an infinite number of slices, and each slice gets smaller) we get:

_{i}/Δx

_{i})

^{2}Δx

_{i}

We now have an integral and we write **dx** to mean the **Δx** slices are approaching zero in width (likewise for **dy)**:

^{2}dx

And **dy/dx** is the derivative of the function f(x), which can also be written** f’(x)**:

^{2}dx

**The Arc Length Formula**

And now suddenly we are in a much better place, we don't need to add up lots of slices, we can calculate an exact answer (if we can solve the differential and integral).

*Note: the integral also works with respect to y, useful if we happen to know x=g(y):*

^{2}dy

So our steps are:

- Find the derivative of
**f’(x)** - Solve the integral of
**√1 + (f’(x))**^{2}dx

Some simple examples to begin with:

### Example: Find the length of f(x) = 2 between x=2 and x=3

f(x) is just a horizontal line, so its derivative is **f’(x) = 0**

Start with:

^{2}dx

Put in **f’(x) = 0**:

^{2}dx

Simplify:

Calculate the Integral:

So the arc length between 2 and 3 is 1. Well of course it is, but it's nice that we came up with the right answer!

Interesting point: the "(1 + ...)" part of the Arc Length Formula guarantees we get **at least** the distance between x values, such as this case where ** f’(x)** is zero.

### Example: Find the length of f(x) = x between x=2 and x=3

The derivative **f’(x) = 1**

Start with:

^{2}dx

Put in **f’(x) = 1**:

^{2}dx

Simplify:

Calculate the Integral:

And the diagonal across a unit square really is the square root of 2, right?

OK, now for the harder stuff. A real world example.

### Example: Metal posts have been installed **6m apart** across a gorge.

Find the length for the hanging bridge that follows the curve:

f(x) = 5 cosh(x/5)

Here is the actual curve:

Let us solve the general case first!

A hanging cable forms a curve called a **catenary**:

f(x) = a cosh(x/a)

Larger values of **a** have less sag in the middle

And "cosh" is the hyperbolic cosine function.

The derivative is **f’(x) = sinh(x/a)**

The curve is symmetrical, so it is easier to work on just half of the catenary, from the center to an end at "b":

Start with:

^{2}dx

Put in **f’(x) = sinh(x/a)**:

^{2}(x/a) dx

Use the identity **1 + sinh ^{2}(x/a) = cosh^{2}(x/a):**

^{2}(x/a) dx

Simplify:

Calculate the Integral:

S = a sinh(b/a)

Now, remembering the symmetry, let's go from −b to +b:

S = 2a sinh(b/a)

In our **specific case** a=5 and the 6m span goes from −3 to +3

S = 2×5 sinh(3/5)

= **6.367 m** (to nearest mm)

This is important to know! If we build it exactly 6m in length there is **no way** we could pull it hard enough for it to meet the posts. But at 6.367m it will work nicely.

### Example: Find the length of y = x^{(3/2)} from x = 0 to x = 4.

The derivative is **y’ = (3/2)x ^{(1/2)}**

Start with:

^{2}dx

Put in **(3/2)x ^{(1/2)}**:

^{(1/2)})

^{2}dx

Simplify:

We can use integration by substitution:

- u = 1 + (9/4)x
- du = (9/4)dx
- (4/9)du = dx
- Bounds: u(0)=1 and u(4)=10

And we get:

Integrate:

S = (8/27) u^{(3/2)} from 1 to 10

Calculate:

S = (8/27) (10^{(3/2)} − 1^{(3/2)}) = **9.073...**

## Conclusion

The Arc Length Formula for a function f(x) is:

^{2}dx

Steps:

- Take derivative of f(x)
- Write Arc Length Formula
- Simplify and solve integral