Why is there only one right length of an arc? This is the way I was taught to (non-rigorously) derive the length of an arc given in polar coordinates, i.e. $r(\theta)$. Take a small slice $d\theta$, and approximate the curve in this slice by a straight line segment, so: 
$$
ds^2 = dx^2 + dy^2
$$
Divide both sides by $d\theta^2$, plug in $x = r\cos\theta$ and $y=r\sin\theta$, and simplify and rearrange to obtain:
$$
ds = \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta \tag{1}
$$
However, once we have  a small slice $d\theta$, it would seem to me that, instead of a straight line segment, an equally sensible alternative approximation to the true curve is an arc, in which case we have simply:
$$ds = r d\theta \tag{2}
$$
Obviously, (1) and (2) give different results in general. I can inuitively understand why this happens - it's because the ratio of the length or the arc to that of the straight line segment does not go to zero as $d\theta$ goes to zero.
But what I can't really understand is why one is right and the other is wrong. Is it simply a matter of definition? Do we just agree to define the length of a curve in terms of straight-line segments? I don't think so, because I strongly suspect that if we had to do a physical experiment using string, it would agree with (1) not with (2).
 A: When you approximate the length of the curve from $(r_1,\theta_1)$ to a close point $(r_2,\theta_2),$ you could correctly use an arc between those two points instead of a straight line. But your arc goes from $(r_1,\theta_1)$ to $(r_{\color{red}{\,1}},\theta_2),$ not to $(r_{\color{red}{\,2}},\theta_2).$
What you're doing is similar to saying that one leg of a right triangle is a good approximation to the hypotenuse, which isn't true in general.
A: As noted in the comments the formula (1) is correct for any curve of equation $r=r(\theta)$. The formula (2) is valid  only for a circle with center at the origin and radius $r$ , it is a special case of (1), because for a circle of equation $r=k$ we have $\frac{dr}{d\theta}=0$
A: Of course one could give a full proof of the length formula for a parametric curve
$$\gamma:\quad\theta\mapsto r(\theta)(\cos\theta,\sin\theta)\qquad(\alpha\leq\theta\leq\beta)\ ,$$
but this is not what you want.
Note that your problem already comes up when studying the length of graphs $y=f(x)$. Here $ds=\sqrt{1+f'^2(x)}\>dx$, and not $ds=dx$. The reason is the following: When the tangent at $\bigl(x,f(x)\bigr)$  makes an angle $\alpha$ with the $x$-axis then the length $\Delta s$ of the tiny arc over $[x,x+\Delta x]$ satisfies
$$\Delta s\approx {\Delta x\over \cos\alpha}=\sqrt{1+\tan^2\alpha}\ \Delta x\  .$$
Here the factor $\sqrt{1+\tan^2\alpha}=\sqrt{1+f'^2(x)}\ $ does not become smaller when we split $\Delta x$ in two halves, resp., let $\Delta x\to0$.
