Answer
$(2.04, 1.03)$
Work Step by Step
$x = 2$
$y = 1$
$V(x,y) = (x^2, x + y^2)$
$= ((2)^2, 2 + (1)^2)$
$= (4, 3)$
$t = 3.01 - 3$
$ = 0.01$
$t \times V(2, 1) = 0.01(4, 3)$
$ = (0.04, 0.03)$
Location at time $t$ is calculated as follows:
$V(x, y) = (x, y) + (t_x, t_y)$
$= (2, 1) + (0.04, 0.03)$
$= (2.04, 1.03)$
