why do we need to multiply derivative by tiny change in the parameter surface to get the difference

Question

I cannot understand why we need to multiply by $ds$ or $dt$ to get this vector of change between two vectors that are on surface $S$

What do we get if we do not multiply by $ds$ or $dt$ ???

Answer 1

This is informal.

Locally, you have $v(x+h) \approx v(x)+ {\partial v(x) \over \partial x} h$ (here $x = (s,t) \in \mathbb{R}^2$, and $h \in \mathbb{R}^2$).

In particular, the point $x+(\delta_1,0)$ is mapped to $v(x)+ \delta_1 {\partial v(x) \over \partial x_1}$ and the point $x+ (0,\delta_2)$ is mapped to $v(x)+ \delta_2 {\partial v(x) \over \partial x_2}$.

Note that the (signed) area of the paralleogram (rectangle) $x,x+(\delta_1,0), x+(0, \delta_2), x+ (\delta_1,\delta_2)$ is $\delta_1 \delta_2$, and the area of the mapped paralleogram is given by $\delta_1\delta_2 \det {\partial v(x) \over \partial x}$. That is, locally $v$ scaled area by $\det {\partial v(x) \over \partial x}$. This is why $\det$ appears in change of variables formulae.