Answer
$S = \lt{2.04},{1.03}\gt$
Work Step by Step
$V(x,y) =\lt{x^2}, {x+y^2}\gt$
The particle's initial position is $\lt2,1\gt$, therefore:
$x=2$
$y=1$
$V(2,1) =\lt{4}, {3}\gt$
We can consider the velocity as being constant, taking into account how small $\triangle{t}$ is.
$\triangle{S} =0,01\cdot\lt{4}, {3}\gt$
$\triangle{S} =\lt{0.04}, {0.03}\gt$
The particle's final position can then be estimated at:
$\lt{2.04},{1.03}\gt$