Question

section 16.2 line integrals 1131

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.
2. at time t = 1, a particle is located at position (1,3). if it moves in a velocity field f(x,y) = <xy - 2,y² - 10> find its approximate location at time t = 1.05.

Explanation:

Step1: Recall the approximation formula

The approximation formula for position is $\mathbf{r}(t + \Delta t)\approx\mathbf{r}(t)+\mathbf{v}(t)\Delta t$. Here, $\Delta t$ is the small - time increment.

Step2: For problem 37

First, find the velocity at the given point. Given $\mathbf{V}(x,y)=\langle x^{2},x + y^{2}
angle$ and the point $(x,y)=(2,1)$ at $t = 3$.
$x = 2,y = 1$, then $\mathbf{V}(2,1)=\langle2^{2},2 + 1^{2}
angle=\langle4,3
angle$.
$\Delta t=3.01 - 3=0.01$.
The position $\mathbf{r}(t)$ at $t = 3$ is $\langle2,1
angle$. Using the approximation formula $\mathbf{r}(3.01)\approx\langle2,1
angle+\langle4,3
angle\times0.01=\langle2 + 4\times0.01,1+3\times0.01
angle=\langle2.04,1.03
angle$.

Step3: For problem 38

First, find the velocity at the given point. Given $\mathbf{F}(x,y)=\langle xy - 2,y^{2}-10
angle$ and the point $(x,y)=(1,3)$ at $t = 1$.
$x = 1,y = 3$, then $\mathbf{F}(1,3)=\langle1\times3 - 2,3^{2}-10
angle=\langle1,-1
angle$.
$\Delta t=1.05 - 1 = 0.05$.
The position $\mathbf{r}(t)$ at $t = 1$ is $\langle1,3
angle$. Using the approximation formula $\mathbf{r}(1.05)\approx\langle1,3
angle+\langle1,-1
angle\times0.05=\langle1+1\times0.05,3+( - 1)\times0.05
angle=\langle1.05,2.95
angle$.

Answer:

For problem 37: $(2.04,1.03)$
For problem 38: $(1.05,2.95)$