Question

1. a particle moves in a velocity field v(x, y) = <x²,x + y²>. if it is at position (2, 1) at time t = 3, estimate its location at time t = 3.01.

Explanation:

Step1: Find velocity at given point

We substitute $x = 2$ and $y = 1$ into $\mathbf{V}(x,y)=\langle x^{2},x + y^{2}
angle$.
$\mathbf{V}(2,1)=\langle2^{2},2 + 1^{2}
angle=\langle4,3
angle$

Step2: Use linear approximation for displacement

The displacement $\Delta\mathbf{r}$ over a small time - interval $\Delta t$ is approximately $\Delta\mathbf{r}\approx\mathbf{V}(x,y)\Delta t$. Here, $\Delta t=3.01 - 3=0.01$.
$\Delta\mathbf{r}\approx\langle4,3
angle\times0.01=\langle4\times0.01,3\times0.01
angle=\langle0.04,0.03
angle$

Step3: Estimate new position

If the initial position is $\mathbf{r}_{0}=\langle2,1
angle$, the new position $\mathbf{r}$ is $\mathbf{r}=\mathbf{r}_{0}+\Delta\mathbf{r}$.
$\mathbf{r}=\langle2 + 0.04,1+0.03
angle=\langle2.04,1.03
angle$

Answer:

$(2.04,1.03)$