At time t 1 a particle is located at position x y 1 5 If i

At time

t = 1,

a particle is located at position

(x, y) = (1, 5).

If it moves in the velocity field

F(x, y) =

xy 1, y² 11

find its approximate location at time

t = 1.04.

(x, y) =

Solution

at t=1 particle position is (1,5)

f(x,y) = (xy-1 , y^2 -11)

f(1,5) = (5 -1 , 25 -11)

f(1,5) = (4 , 14)

now velocity

(4,14) = (x-1 , y-5)/ 1.04 - 1

(4,14) = (x-1, y-5)/0.4

(x-1, y-5) = (1.6, 5.6)

x= 2.6 and y = 10.6

the location = (2.6 , 10.6)