Arc Length Confusion: What is the Idea Behind it?

[Char. Limit]

Why is arc length of a function $$f(x)$$ from a to b defined as $$\int_a^b \sqrt{1+(f'(x))^2} dx$$?

Where they get the idea of squaring the derivative, adding 1, taking the square root, and then integrating it is beyond me.

 

[tiny-tim]

They got the idea from Pythagoras …

√(1 + (dy/dx)²) dx = √((dx)² + (dy)²) = ds

 

[Char. Limit]

But aren't dx, dy, and ds all infinitesimals with no real meaning here?

 

[tiny-tim]

They're all infinitesimals, but they are the limit of very small increments ∆x ∆y and ∆s with ∆s² = ∆x² + ∆y²

how would you define arc-length? ​

 

[HallsofIvy]

If you don't want to use differentials, which you say are "infinitesimals with no real meaning here" then you will have to use the standard "Riemann sum" definition. Given a curve y= f(x) with x from a to b, divide the x-axis from a to b into n intervals, each of length $\Delta x$. We can approximate the curve from $(x_i, f(x_i))$ to $(x_i+ \Delta x, f(x_i+ \Delta x))= (x_i+ \Delta x, f(x_i)+ \Delta y)$ where I have taken $\Delta y= f(x_i+ \Delta y)- f(x_i)$, by the straight line between those points. It's length, by the Pythagorean theorem, is $\sqrt{(\Delta x)^2+ (\Delta y)^2}$. Now, factor $\Delta x$ out of that:
$$\sqrt{(\Delta x)^2+ (\Delta y)^2}= \sqrt{1+ \left(\frac{\Delta y}{\Delta x}\right)^2}\Delta x$$

So the Riemann sum is
$$\sum\sqrt{1+ \left(\frac{\Delta y}{\Delta x}\right)^2}\Delta x$$
which, in the limit, becomes
$$\int_a^b \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx$$

I would be surprised if your textbook didn't give all of that.

 

[tiny-tim]

awww, you gave him the answer!

 

[Char. Limit]

Well, since this isn't a homework problem, I think giving the answer is all right...

I honestly have never seen the derivation before.

Thanks for the help, both of you.

 

[evagelos]

If you don't want to use differentials, which you say are "infinitesimals with no real meaning here" then you will have to use the standard "Riemann sum" definition. Given a curve y= f(x) with x from a to b, divide the x-axis from a to b into n intervals, each of length $\Delta x$. We can approximate the curve from $(x_i, f(x_i))$ to $(x_i+ \Delta x, f(x_i+ \Delta x))= (x_i+ \Delta x, f(x_i)+ \Delta y)$ where I have taken $\Delta y= f(x_i+ \Delta y)- f(x_i)$, by the straight line between those points. It's length, by the Pythagorean theorem, is $\sqrt{(\Delta x)^2+ (\Delta y)^2}$. Now, factor $\Delta x$ out of that:
$$\sqrt{(\Delta x)^2+ (\Delta y)^2}= \sqrt{1+ \left(\frac{\Delta y}{\Delta x}\right)^2}\Delta x$$

So the Riemann sum is
$$\sum\sqrt{1+ \left(\frac{\Delta y}{\Delta x}\right)^2}\Delta x$$
which, in the limit, becomes
$$\int_a^b \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx$$

I would be surprised if your textbook didn't give all of that.

Certainly in the limit $$\sum\sqrt{1+ \left(\frac{\Delta y}{\Delta x}\right)^2}\Delta x$$ does not become $$\int_a^b \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx$$ according to the definition of the Rieman integral.

BUT ,by the mean value theorem we have that:

Δy = f'(ξ)Δx ,where ξ is $$x\leq\xi\leq x+\Delta x$$ and now ,

$$\sum\sqrt{1+ \left(\frac{\Delta y}{\Delta x}\right)^2}\Delta x$$ becomes :$$\sum\sqrt{1+ \left(f'(\xi)\right)^2}\Delta x$$ .This a Rieman sum which in the limit it becomes:


$$\int_a^b \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx$$