A particle is moving in a velocity field $$V(x, y, z) = 5x\hat i + (x + 4y)\hat j + z^2\hat k$$
At time $t = 2$ the particle is located at the point $(1, 3, 2)$.
(a) What is the velocity v of the particle at $t = 2$?
(b) What is the approximate location of the particle at $t = 2.01$?
I'm not sure how to approach this. Any help would be appreciated. :)
On
Hints: For velocity at t=$2 sec$ simply put the coordinates to get $$\vec V=5i+13j+4k$$ $$|\vec V|=\sqrt{25+169+16}$$
$V_x=5x$ therefore for x coordinate at time t:
$$\int_1^x \frac{dx}{x}=5\int_2^t dt$$ $$\ x=e^{5(t-2)}\tag{1}$$
For $V_y=x+4y$ $$\int\frac{dy}{dt}=e^{5(t-2)}+4y\tag{Solve further via Bernoulli}$$ $$\text{Integrating factor: }e^{\int-4dt}$$ $$\int d(ye^{-4t})=\int e^{-4t}.e^{5t-10}.dt +c\tag{2}$$
For $V_z=z^2$ $$\int\frac{dz}{dt}=z^2$$ $$\int_2 ^z\frac{dz}{z^2}=\int_2 ^t dt$$ $$\frac{1}{z^2}-\frac{1}{4}=t-2\tag{3}$$
Now you have Equations for your position vector at any time $t$. So you are good to go! Cheers!
(a) can be obtained by simply plugging in your starting point $(1,3,2)$ into the velocity field formula.
As for (b), since the amount of time that has elapsed is small and the velocity field is not too large and doesn't change rapidly near $(1,3,2)$, we may approximate the field as constant near the point $(1,3,2)$. Specifically, we may approximate the velocity field near $(1,3,2)$ as being equal to whatever the answer to (a) is. Once you have that, you just need to calculate how far the particle moves (and in what direction) in $0.01$ time units, and add that to your starting point $(1,3,2)$. Can you take it from here?