Particle position starting from a point other than the origin

Question

I'm working with a particle problem that I mostly understand but am questioning the starting point of a particle in motion.

I am given the velocity vector of a particle,

$v(t) = \langle 1, 6t, 9t^2 \rangle$

and its starting point of $(0 , 1, 2).$

I know that the particle's position vector can be found by the antiderivative of the velocity vector, I calculated this to be: $r(t) = \langle t, 3t^2, 3t^3 \rangle$ and at time = $1$ I've found the particle's position to be $(1, 3, 3).$

My question is, how does the starting point of this particle relate to the position at a given time? Is my assumed answer of $(1, 3, 3)$ correct, or does that assume starting from the origin and I should be adding the initial position of $(0, 1, 2)$ to my calculated point?

Answer 1

Well to answer your question. You did the right thing by integrating the velocity function $\mathbf{v}(t)$ however remember that when integrating you need to add a constant of integration which in your case is a constant vector $\mathbf{c}$ so

$$ \int \mathbf{v}(t) dt = \int (1,6t,9t^2)dt=(t,3t^2,3t^3)+(c_1,c_2,c_3)$$

Given that at $t=0$ we have

$$ (0,1,2) = (0,0,0)+(c_1,c_2,c_3)\Rightarrow \mathbf{c} = (0,1,2)$$

Hence,

$$ \mathbf{r}(t) = (t,3t^2+1,3t^3+2) , t\geq 0$$

position relation to your starting point. At $t=0$ this is the start of your trajectory. These position vectors are written with respect to the origin. Does that make sense. I hope I explained myself clearly. Any Questions get to me!!...Ciao!!

Answer 2

The path you’ve computed has the particle starting at the origin. You appear to have forgotten that there’s a family of antiderivatives of $\mathbf v$ that differ by an arbitrary constant of integration. That’s where the particle’s initial position comes in: use it to determine the value of this constant. Essentially, you need to offset that origin-initial path by the actual initial position.