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

Explanation:

Step1: Evaluate velocity at given point

We first find $V(2,1)$. Given $V(x,y)=\langle x^{2},x + y^{2}\rangle$, substitute $x = 2$ and $y=1$. $V(2,1)=\langle2^{2},2 + 1^{2}\rangle=\langle4,3\rangle$.

Step2: Use linear approximation for displacement

The displacement $\Delta\vec{r}$ of the particle in a short - time interval $\Delta t$ is approximately $\Delta\vec{r}\approx\vec{V}(x_0,y_0)\Delta t$. Here, $\Delta t=3.01 - 3=0.01$, and $\vec{V}(2,1)=\langle4,3\rangle$. So $\Delta\vec{r}\approx\langle4,3\rangle\times0.01=\langle4\times0.01,3\times0.01\rangle=\langle0.04,0.03\rangle$.

Step3: Find new position

If the initial position is $\vec{r}_0=\langle2,1\rangle$ and the displacement is $\Delta\vec{r}=\langle0.04,0.03\rangle$, the new position $\vec{r}$ is $\vec{r}=\vec{r}_0+\Delta\vec{r}$. $\vec{r}=\langle2 + 0.04,1+0.03\rangle=\langle2.04,1.03\rangle$.

Answer:

$(2.04,1.03)$