Intuition behind equation for finding arc length in polar coordinate I know how to derive the equation for finding arc length in polar coordinates but I don't understand this:
Given a parametric equation let L be the length of the arc from point t = a to to t = b we have:
$L = \int_{a}^{b} \sqrt{dx^2 + dy^2}\frac{dt}{dt} = \int_{a}^{b} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}dt$. 
To turn this equation into the polar coordinate version we can assume the parametric equation is $x = rcos\theta$ and $y = rsin\theta$ (so t = $\theta$) and after substitution we should get the equation $L = \int_{\theta_{1}}^{\theta_{2}}\sqrt{r^2 +r'^2} d\theta$. 
So what I don't' understand is the "intuition" behind why $dx^2 + dy^2 \neq dr^2$. Clearly, if  $dx^2 + dy^2 = dr^2$ L would equal $\int_{\theta_{1}}^{\theta_{2}}|r'| d\theta$ which "intuitively" can't be correct (length cannot be just a function of slope). 
I'm not entirely sure how to explain what I mean by "intuition" but I what I'm hoping to figure out is what I'm not understanding that makes me think $dx^2 + dy^2 = dr^2$ should be correct/ is not making me see why this should be incorrect.
 A: Whether Cartesian or Polar, we use Pythagoras thm hypotenuse to find differential arc length and thereafter integrate it.
Just think of differentials temporarily as if they are finite side lengths to start with and use Pythagoras:

Cartesian
$$ PQ^2 = ds^2= dx^2+dy^2 = (1+y^{'2})\; dx^2 $$
Polar
$$ PQ^2= ds^2= dr^2 + (r d\theta)^2 = (r^{'2}+ r^2)\; d\theta^2$$
Do not try Cartesian to Polar conversion at the raw stage. Think of differential lengths independently.
Next integrate the differential arc length $ds$ with respect to the independent variable $x$ or $ \theta$.
Cartesian
$$ s= \int \sqrt{ (1+y^{'2})}\;dx\;$$
Polar
$$ s= \int \sqrt{ r^{'2}+ r^2}\;  d \theta $$
I got rid of my doubts about arc length and individual differential lengths in this trigonometrical way much earlier once and for all.
A: $(dx,dy)$ describes a displacement in an arbitrary direction. It can be decomposed in a radial displacement ($dr$) and a trangential one ($r\,d\theta$) and we have the identity
$$dx^2+dy^2=dr^2+r^2d\theta^2.$$

We can check this by direct computation by expanding
$$dx^2+dy^2=(dr\cos\theta-r\sin\theta\,d\theta)^2+(dr\sin\theta-r\cos\theta\,d\theta)^2.$$
In other words,
$$\frac{ds}{dt}=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=\sqrt{\left(\frac{dr}{dt}\right)^2+r^2\left(\frac{d\theta}{dt}\right)^2}$$
and $$\frac{ds}{d\theta}=\sqrt{\left(\frac{dr}{d\theta}\right)^2+r^2}.$$
