The staircase paradox, or why $\pi\ne4$ What is wrong with this proof?

Is $\pi=4?$

This problem illustrates the fact that two functions can be very close: $|f(x)-g(x)|<\epsilon$
for all $x\in [0,1]$, but their derivatives can still be far apart, $|f'(x)-g'(x)|>c$ for some
constant $c>0$. 
In our case, let $x=a(t),y=b(t),0\le t\le 1$ and $x=c(t),y=d(t), 0\le t\le 1$ be the 
parametrizations of the two curves. By smoothing the corners, we may assume that both
are smooth. $$ \|(a(t),b(t))\|\approx \|(c(t),d(t))\|$$ does not imply
$$  \|(a'(t),b'(t))\|\approx \|(c'(t),d'(t))\|$$
Therefore $\int_0^1 \|(a'(t),b'(t))\| dt$ need not be close to $\int_0^1 \|(c'(t),d'(t))\| dt.$
Here $\|(x,y)\|$ denotes $\sqrt{x^2+y^2}$.
