**Measuring Arc Length**

In this section we will look two related approaches using integration to measure the length of a curve in the plane. In particular the curves in our discussion are the graphs of functions (either explicit or implicit) with continuous first derivatives.

In our first approach the key tool for finding the arc length will be
the approximation of this length as a **sum of line segment lengths**
determined by the given curve. These estimating sums will have a limiting
value which by definition is a definite integral.

Our second approach to the arc length question uses an estimate of**
the rate at which the arc length is changing** as a function of one variable.
In other words, imagine an object moving along the curve with its first
coordinate changing at a unit speed. We estimate **the speed of the object
moving on the curve**. This estimate also uses a line segment length
to approximate the change in position along the arc.

**Arc Length as the Limit of Sums**

Perhaps the easiest way to estimate the distance between two points
along the curve `Y = f (X)` is to ignore the curve and find the distance
between the two points along a line segment between them . One might say
we use the distance between them ``as the crow flies.'' If the points have
coordinates `(x,y)` and

`(x+Delta x ,y + Delta y)` then the distance between them is

When `Delta x` is close to 0 we know that `{Delta y}
/{Delta x}` approaches `f '(x)`, and therefore `sqrt{ 1 + ({Delta y}/ {Delta x})^2}`
approaches

`sqrt {1 + (f'(x))^2}`.
Now we only need to consider how we accumulate these estimates to approximate
the length of the arc.

Keep in mind the definition of the definite integral, namely:

So we continue to estimate the length of the arc of the graph of the function between `(a, f (a))` and `(b, f (b))`. Partition the interval `[a,b]`, make the same type of estimate for each subinterval as above using`sqrt{Delta x_k ^2 + Delta y_k ^2 } = Delta x_k sqrt{ 1 + ({Delta y_k}/ {Delta x_k})^2}` so the arc length is approximated by

`int _a^b sqrt{1+ (f'(x))^2} dx ` .

Thus the arc length of the graph of the function between `(a,f (a))` and `(b,f (b))` is `int _a^b sqrt{1+ (f'(x))^2} dx `.

Here's an example to make this discussion more concrete.

**Generalization Using Riemann Sums. **The work in the last
example can be generalized
and made more rigorous in the estimation without much difficulty by
using Riemann sums to find the definite integral. In a general
situation the arc will be the graph
of a function f between the points `(a,f(a))` and `(b,f(b))` where `f`
has a
continous derivative function `f'` on `[a,b]`. We partition the
interval `[a,b]`
into `n` pieces of length `Delta x = (b-a)/n` with `x_k = a + k Delta
x` , `Delta y_k = f (x_{k+1} ) - f (x_k)`. We consider
the line segment length `Delta s_k` for each subinterval `[x_k,x_{k+1}
]`. `Delta s_k = sqrt{ (Delta x)^2 + (Delta y_k)^2} =
sqrt{ 1 + ({Delta y_k}/ {Delta x })^2} Delta x`.

Using
the Mean Value Theorem for `f` on each of these subintervals gives
points `z_k in [x_k ,x_{k+1}]` and we have

Imagine in the example we considered for arc length that we are moving a light along the X-axis which is shining on the point vertically above it on the graph of `Y=X^{3/2}`. The light is moving at a constant unit rate along the X-axis . How fast is the light image moving along the curve? This question can be addressed in two different ways, one using the arc length formula we just discovered and the other leading to a different way to find the arc length formula.

From our previous work the length of arc that the light has travelled from when it

was at `(a, a ^{3/2} )` to when it is at `(t, t^{3/2} )`
is s(t) = `int _1^t sqrt{1 +9/4 x} dx`.

Now a simple application of the derivative form of the Fundamental Theorem of Calculus shows
that `s'(t) = sqrt{1 +9/4 t}`.

In general the same result is true: The speed of an object moving along
a curve which is the graph of `Y = f(X)` and with its position determined
by `s(t)` where

`s(t) = int _a^t sqrt{1+ (f'(x))^2}` dx
is given by `s'(t) = sqrt{1+ (f'(t))^2}`.

This establishes that the arc length is the solution to the differential
equation

It is possible to arrive at this differential equation directly from
the knowledge of the function `f`. Consider `s(t)` at
the distance travelled on the arc at time `t`. Then `s(t+Delta
t)-s(t)` is the distance travelled along the arc between time `t` and
`t + Delta t`. When `Delta t` is close to zero, this distance
can be estimated by the distance between the points `(t,f(t))` and
`(t+Delta t, f(t+Delta t) )`. It should not be hard to see at
this stage that this distance is

`sqrt{(Delta t)^2 + (f(t+ Delta t) - f(t))^2 }=sqrt{(Delta t)^2 + (f'(z) Delta t)^2 }`
where `z ` is between `t` and `t + Delta t`.
As `Delta t -> 0` we see
that

`{s(t + Delta t) - s(t)}/{Delta t } = sqrt{
(Delta t)^2 + (f(t+ Delta t) - f(t))^2 }/{Delta t} = {Delta t sqrt{1 +(f'(z ))^2 }}/{Delta t}->sqrt{1+ (f'(t))^2}`.

Therefore `s'(t) = sqrt{1+ (f'(t))^2}`.

Optional: Arc Length for Curves Determined by Motion in the Plane

Suppose the position of an object moving in the plane determines a curve so that at time `t` the position of the object on the curve has coordinates `(x,y) =(x(t),y(t))` where `x` and `y` are continuously differentiable functions of the variable `t`. To find the length of the arc of this curve between time `t = a` and `t = b` we can use an analysis by estimation analogous with that done for the graph of a function. For a small interval of time we can estimate the length with the distance between the end points of the curve:

`Delta x = x(b) - x(a)` and `Delta y = y(b)-y(a)` so the length is estimated by `sqrt{(Delta x) ^2 + (Delta y) ^2}` `=sqrt{({Delta x}/{Delta t}) ^2 + ({Delta y}/{Delta t})^2}Delta t`.

Partition the time interval `[a,b]` with resulting cuts on the
curve at points `(x(t_k), y(t_k))`. We estimate the length of each
section of the curve as we did for the single interval of time:`Deltax_k = x(t_{k+1}) - x(t_k)` and
`Delta y_k = y(t_{k+1})-y(t_k)` so the
length is estimated by
`sqrt{(Delta x_k)^2+(Delta y_k)^2}=sqrt{({Delta x_k}/{Delta t})^2+({Delta y_k}/{Delta t})^2}Delta t`
and the total arc length is estimated by the sum: `Sigma _0^{n-1}
sqrt{({Delta x_k}/{Delta t}) ^2 + ({Delta y_k}/{Delta t})^2}Delta t`.

Now as `n->oo` these sums estimate the arc length for the curve and an integral as well- namely: `int_{t=a}^{t=b} sqrt{({dx}/{dt}) ^2 + ({dy}/{dt}) ^2 }dt.`

So the arc length for this curve is
`int_{t=a}^{t=b} sqrt{({dx}/{dt})^2 + ({dy}/{dt}) ^2 } dt = int_{t=a}^{t=b} sqrt{(x'(t)) ^2 + (y'(t)) ^2} dt.`

**Note**: The previous formula for arclength for the graph of a function `f` can
be found in this new formula by using `x(t) = t` and `y(t)= f(t)`.

**Example VIII.B.2** Suppose `x(t) = 3cos(t)` and `y(t) = 3sin(t)`
with `0=< t =< 2pi`. Then the curve is a circle of radius 3 and the
arclength is the circumference.

Using the formula just derived- the arclength is given by
`int_0^{2pi} sqrt { 9sin^2(t) + 9cos^2(t)} dt = int_0^{2pi}3 dt = 6pi.`

**Summary**

If `f` is a function with a continuous derivative on the interval `[a,b]` then the length of the arc of the graph of `f` over the interval `[a,b]` is given by `int _a^b sqrt{1+ (f'(x))^2} dx `.

If `s(t)` is the length of the arc of the graph of `f` over the interval `[a,t]`, then `s'(t) = sqrt{1+ (f'(t))^2}` , and `s(t) = int _a^t sqrt{1+ (f'(x))^2} dx `.

Exercises.

In each of the following problems

a) sketch a graph of the curve described,

b) use four chords to estimate the arc length of the curve,

c) express the arc length of the curve using a definite integral,

d) when possible find the exact arc length of the curve described.

1. The graph of Y^{ 2} = X^{3} between (0,0) and (1,1).

2. The graph of Y^{ 2} = X^{3} between (0,0) and (4,8).

3. The graph of Y = X^{ 2} between (0,0) and (1,1).

4. The graph of Y = X^{ 2} between (0,0) and (2,4).

5. The graph of Y = X^{ 2} between (-1,1) and (1,1).

6. The graph of Y ^{2} = X between (0,0) and (1,1).

7. The graph of Y ^{2} = X between (0,0) and (4,2).

8. The graph of Y = 1/X between (1,1) and (2,1/2).

9. The graph of Y = 1/X between (1,1) and (3,1/3).