Question

A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its location at time t = 3.01.

Answer 1

Answer:

New location at time 3.01 is given by: (7.49, 2.11)

Explanation:

Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:

$V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11$

With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:

$distance=v\,*\, t$

$distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11$

Therefore, adding these displacements in component form to the original particle's position, we get:

New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)