# Arc Length formula

1. Mar 10, 2004

### Townsend

Find a function F(x) whose arc length L(x) from (1,1/2) to (x,F(x)), x>1 is (1/2)x^2 + (1/4)Ln(x).

First some short hand notation.

Int[f(x),dx], means the indefinite integral of the function f(x).

Int[f(x),dx,a,b], means the definite integral where a is the lower bound and b is the upper bound.

d/(dx)[f(x)], is the derivate of f(x).

Sqrt[f(x)], is the square root of f(x).

Basically I am trying to follow Mathematica conventions and adding a touch of Leibniz to it.

Ok, now here is what I have done and where I am at thus far.

Arc length formula is Int[Sqrt[1+(d/dt)[f(t)]^2],dt,a,x]

So I have Int[Sqrt[1+((d/dt)[f(t)])^2],dt,1,x] = (1/2)x^2 + (1/4)Ln(x)

So by the FTC I can say that

d/(dx)[(1/2)x^2 + (1/4)Ln(x)] = Sqrt[1+(d/dt)[f(t)]^2]

Which is

x + 1/(4x) = Sqrt[1+(d/dt)[f(t)]^2]

Squaring both sides I have

(x + 1/(4x))^2 = abs[1+(d/dt)[f(t)]^2]

I drop the absolute value here so I have.

(x + 1/(4x))^2 = 1+(d/dt)[f(t)]^2

x^2 + 1/2 + (1/(4x))^2 = 1+(d/dt)[f(t)]^2

Subtracting 1 from both sides and factoring the RHS

(x - 1/(4x))^2 = ((d/dt)[f(t)])^2

Taking square root of both sides and discarding the absolute value I have.

(d/dt)[f(t)] = x - 1/(4x)

Integrating both sides I get.

f(x) =(1/2)x^2 - (1/4)Ln(x) +c

I use the point (1,(1/2)) that is given to find c. Thus c = 0.

Seems ok, right?

Well that means that if I put f(x) back into my arc length formula I should get back to the given function.

Ok.... Did that and now I have an extra -(1/2) on my function? What am I missing here? One thing to notice is

Int[Sqrt[1+((d/dt)[f(t)])^2],dt,1,x] = (1/2)x^2 + (1/4)Ln(x)

When x=1 you have 0 = 1/2

Of course the restriction was x>1

Any help would be great, thanks.

cheers

2. Mar 10, 2004

Interesting. I'm getting the same thing.

You'll notice that the method fails right after applying the FTCs. You get an expression for Sqrt[1+f'[t]^2] that, if you integrate from 1 to x with respect to t, has the extra -1/2. Additionally, maybe I just can't use my calculator, but whenever I asked it to find the arclength numerically between two finite points for x^2/2 + 1/(4x), the answer did not agree with x + Log[x]/4 - 1/2.

Sorry I'm not more help, but mathematical rigor is most definitely not my strong point.

3. Mar 11, 2004

### phoenixthoth

i'm tempted to think there is no solution because if x=1, the arc length formula would give 1/2. the restriction x>1 may be the resolution to that little dilemma. seems odd though that the arc length function L(x) is discontinuous at x=1, jumping from 0 to 1/2. L(1) should be 0.

4. Mar 11, 2004

### Gunni

I don't see the problem, I don't get an extra 1/2 at the end. I get the same result as you do finding the function but I don't have any problem inputing back into the equation.

$$f(x) = x^2 - \frac{ln|x|}{4}$$

$$\frac{df(x)}{dx} = x - \frac{1}{4x}$$

$$\frac{df(x)}{dx}^2 = x^2 - \frac{1}{2} + \frac{1}{16x^2}$$

$$\frac{df(x)}{dx}^2 + 1 = x^2 + \frac{1}{2} + \frac{1}{16x^2} = (x + \frac{1}{4x})^2$$

$$L = \int_{1}^{x} \sqrt(x + \frac{1}{4x})^2 dx = \int_{1}^{x} (x + \frac{1}{4x}) dx$$

$$L = [ \frac{x^2}{2} + \frac{ln|x|}{4} ]_{1}^{x}$$

Like I said, unless I missed something in your post, I don't see any problem.

5. Mar 11, 2004

### phoenixthoth

in $$L = [ \frac{x^2}{2} + \frac{ln|x|}{4} ]_{1}^{x}$$, when you subtract off the result of plugging in the 1, you get -1/2.

6. Mar 11, 2004

### Gunni

Oh, I see. That's odd. Are you sure that in the text they didn't mean that the result from the arclength formula should be x^2/2 + lnx/4 before you put in the limits?

7. Mar 11, 2004

### phoenixthoth

or that the initial point might be (1,0)? if it were (1,0), then x^2/2+(logx)/4 would work.

kinda neat, actually. if A(f)(x) gives the arc length of f as a function of x and the initial point was (1,0), then this particular function x^2/2+(logx)/4 is a fixed "point" of the operator A.

8. Mar 12, 2004

### arildno

Your arclength formula is wrong. Starting with the derivative of the arclength, you have:
dL/dx=sqrt(1+f'(x)^(2)).
Integrating from 1 to x yields:
L(x)=L(1)+int^(x)_(1)sqrt(1+f'(t)^(2))dt
When L(1) is non-zero, as in your case, it simply means that the graph (t,f(t)) (on 1<t<x) represents a segment of a curve starting at some point (x0,y0) such that the arclength from (x0,y0) to (1,1/2) is L(1)..