let $\Delta s_i$ be a piece of arc length hence:
$$\Delta s_i = \int_{i-1}^i \sqrt{[x'(t)]^2+[y'(t)]^2+[z'(t)]^2}dt$$ Why is that the length of $\Delta s_i$
in 2d I know that the as $\Delta x$ -> dx then the arc length is $$ds =\sqrt{dx^2+dy^2}$$ because of the Pythagorean theorem
First of all, ask yourself what "length" of a curve means. We surely have an intuitive picture; if you could somehow straighten the curve out,the length of the curve would be the length of that line (which we know how to calculate by geometric means). So imagine doing that to the line, and suppose that that length is given by an interval $$\Delta s=\int_{\text{curve}}ds$$ (the meaning of this integral is not really clear, so just take this formally, as an analogy).
Now, straighten it up. This action is effectively mapping the curve into some interval, in a smooth(differentiable) way. The curve is parameterized by some function $\gamma$ (assume this function is invertible); then it makes sense to call the "straightening out" function $\gamma ^{-1}$ and say it maps the curve to the interval $[a,b]$. Then, the change of variables formula gives you $$\Delta s=\int_a^b |\gamma'|dt$$ where ' indicates the derivative and the $||$ indicate the norm of the derivative. Then, writing this out in the components $\gamma(t)=(x(t),y(t),z(t))$, you get $$\Delta s = \int_a^b \sqrt{[x'(t)]^2+[y'(t)]^2+[z'(t)]^2}dt$$ and the intuition for this is now clear. If you treat $t$ as some kind of time, then all this formula is telling you is "the length of a curve is the length of the line one obtains by straightening it, and that length is of course the velocity of the curve multiplied by the time it took the curve to complete".
Of course, this explanation is merely illustrative. The first integral I wrote, as is, doesn't really mean anything, and I only used it to appeal to your intuition and sense of "form". The right way to arrive at this result involves defining the notion of line integral, in which one defines the integral of a vector function over a curve like the Riemann integral of the dot product between the function and the velocity of the curve (making use of the parameterization, but effectively then being independent of that choice). Then the length of a curve is just the line integral of the normalized velocity, $$\Delta s=\int_{\text{curve}} |\gamma'|^{-1}\gamma ' \cdot ds=\int_a^b |\gamma'|^{-1}\gamma ' \cdot \gamma' dt=\int_a^b |\gamma'|dt$$ This of course also makes sense, it's the same prescription, but it's significantly more opaque, hence why I chose to lead with the other explanation.