Fundamental Theorem of Calculus 1: differentiating $\int_a^b \sqrt{(1+(dy/dx)^2)} dx$

Question

I got given the equation

$$\int_a^b \sqrt{(1+(dy/dx)^2)} dx$$ (length of a curve/arc length)

How do I use that to show that

$$ds/dx=(1+(dy/dx)^2)^{(1/2)}$$

Answer 1

You should start by relating arc length $s$ to the integral. Never in calculus should you write any math without a meaningful equals sign.

$$s = \int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$$

In instances like this, it helps to think of $ds$ as a super tiny increment of $s$, of $dx$ as a super tiny increment of $x$, etc. This way, you can see that $ds^2 = dx^2+dy^2$ is really just an infinitely small application of Pythagorean theorem. Integrating sums up all the tiny $ds$ and gives you the total $s$.

Of course, this way of thinking is not mathematically rigorous, although it will help your comprehension.

In a formal proof, you should take your relationship and differentiate both sides of the equation with respect to $x$:

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

At this point, you should apply the tautology that if $F(x)=\int f(x)\,dx$ then $F^\prime(x)=f(x)$. This is such rudimentary calculus that you would not need to invoke it by explicitly stating it or naming it. In fact, it doesn’t really have a name, and it’s not considered one of the fundamental theorems of calculus. So, the the derivative with respect to $x$ and the integral with respect to $x$ “cancel” each other out, leaving you with

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

Sure enough, you can algebraically manipulate this (if necessary) to see that it simply comes form the Pythagorean theorem.

Your second error was trying to include limits of integration. Doing so turns the integral definite—that is, a constant—and you can’t manipulate a constant in the way we want to (namely, the derivative of a constant is $0$). Keeping the integral indefinite gives us a nice malleable “function” of sorts.

Answer 2

I think there is some circular reasoning going on here. We should begin with

$$s=\int ds$$

Then, if we look at small section of curve, say $ds$ and complete a right triangle parallel to the axes we would see that

$$ds^2=dx^2+dy^2$$

or

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