# Definition of Arc Length Function

#### wubie

Hello,

I am having trouble remembering some of the material required for my current calculus course so I am reviewing some of the previous material that I have forgotten.

I am having trouble following the definition of The Arc Length Function as presented in James Stewart's "Calculus: Fourth Ed." page 579.

I already follow how to derive the formula for arc length. But I am having problems with the concept of the arc length function. I am given the following:

We will find it useful to have a function that measures the arcl ength of a curve from a paraticular starting point to any other point on the curve. This, if a smooth curve c has the equation y = f(x), a =< x =< b, let s(x) be the distance along C from the intial point Psub0(a,f(a)) to the point Q(x,f(x)). The s is a function, called the arc length function, and, by the formula

L = integral of (1+[f'(x)]^2)^1/2 dx

then

(1) s(x) = integral of (1+[f'(t)]^2)^1/2 dt

with the limits of [a,b].

(We have replaced the variable of integration by t so that x does not have two meanings.) We can use Part 1 of the Fundamental Theorem of Calculus to differentiate (1) (since the integrand is continuous):

(2) ds/dx = (1+[f'(x)]^2)^1/2 = (1+ (dy/dx)^2)^1/2.

Equation (2) shows that the rate of change of x with respect to x is always at least 1 and is equal to 1 when f'(x), the slope of the curve, is 0. The differential of arc length is

(3) ds = (1+ (dy/dx)^2)^1/2 dx

and this equation is somtime written in the symmetric form

(4) (ds)^2 = (dx)^2 + (dy)^2.

The geometric interpretation of equation (4) can be used as a mnemonic device for remembering (1). If we write

(5) L = integral ds,

then from equation (4) we can solve to get (3).
Now I am not following the purpose of this section on arc length function. I can already figure out (5) with ds being equal to (3). So what is the purpose of this section? And I am also not following what the text is doing at the beginning by replacing the variable of integration with t and then differentiating and re-replacing t with x again. What the hey? I would really like to know what they are doing in this step.

Any help is appreciated. Thankyou.

#### Zurtex

Homework Helper
Well the purpose of the function is simply to work out what the length of a curve is between two points. Imagine the curve $y = x^2$ from the point 0 to 1. There is no mathematical way to accurately work out the length of the curve that you have probably come across, you could draw it very accurately and use a string and a ruler but that is hardly mathematically sound.

Imagine a very small section of the curve, the horizontal distance being a very small section of x, $\delta x$ the vertical distance being a very small section of y, $\delta y$ and the length of this very small section of curve being $\delta s$/

Now because you have such a small section of the curve it is almost a right angle triangle. So we can say:

$$(\delta s)^2 \approx (\delta x)^2 + (\delta y)^2$$

Dividing both sides by $(\delta x)^2$ we get:

$${\left( \frac{\delta s}{\delta x} \right)}^2 \approx 1 + {\left( \frac{\delta y}{\delta x} \right)}^2$$

If we let $\delta x$ approach 0 we can say that:

$${\left( \frac{ds}{dx} \right)}^2 = 1 + {\left( \frac{dy}{dx} \right)}^2$$

And the rest is manipulation, hope that helps.

#### wubie

Hmmmm....

Perhaps my question is not a good question - too trivial?

See, I completely understand what you are saying. I have no problem with that. In fact I have no problem deriving L. What I don't follow is the way that they are presenting the first half of the above passage.

What is going on with the replacement of x with t and then back to x again?

Also, am I to see that

L = s(x) = integral ds = integral (1+ (dy/dx)^2)^1/2 dx?

If so, I can see it. But like I said again I am not following the replacement of integrating variable x with t then back again. What are they doing here in the above passage?

(We have replaced the variable of integration by t so that x does not have two meanings.)
I know this is extremely trivial but it is confusing me. Perhaps it is inconsequential since I can follow the derviation of arc length. But I would like to know how they go from (1) to (2).

#### Hurkyl

Staff Emeritus
Gold Member
(We have replaced the variable of integration by t so that x does not have two meanings.)
You might be having trouble because (1) is wrong! The formula for s is:

$$s(x) := \int_a^x \sqrt{1 + f'(t)^2} \, dt$$

#### wubie

Nope. That is not my trouble. Simply a typo.

8)

What are they doing with the exchange of variables back and forth?

#### Hurkyl

Staff Emeritus
Gold Member
Do you see the trouble inherent in this expression:

$$g(x) = \int_a^x f(x) \, dx$$

?

#### wubie

No I don't. I don't see the problem. What am I missing here? It's been a while since my last calculus course. And a while before that one as well, so I have gaps in my foundation. And it is causing me trouble.

g(x) is dependent on the limits [a,x].

But it is not the same as f(x). Let y = f(x). Then y changes as x changes. x in this case will change over the interval [a,x]. And there lies the problem. x is changing while x in the interval does not.

Correct?

I am sort of confused.

I would have been more comfortable to give the limit a different variable say t then the other way around. That is

$$g(t) = \int_a^t f(x) \, dx$$

What the hey. I am still confused.

I think you caught what was wrong.

Hurkyl's expression for g(x) has x in both the boundaries of integration and the integrand, which is absurd. The expression you made for g(t) is better.

#### wubie

Yes. This is a problem for me. I can see it. But I can't explain it properly. It's causing me problems in my current course. There are small parts of concepts that I am not entirely clear on but I still seem to get the whole of concept.

You probably can't explain the meaning of a definite integral with the same variable in both the boundaries and the integrand because it doesn't make any sense anyway. =]

The variable in the integrand for indefinite integrals is just a dummy variable. It doesn't matter what it is. You can make it whatever you want, like calling it "cookiemonster" for all anybody cares. The true variable is the upper boundary of the integral.