I have a question on the process for which to solving this question. It is a homework question, and I already have the answer, but I am not sure on the correct process to attaining that answer.
The question in full:
A particle has position given by $r(t)=\left<3t, 5t^2, 3t^3 \right>$ at time $t$ for $0 \le t \le 1$. At time $t = 1$ the particle departs the curve and flies off along the line tangent to the curve at $r(1)$. If the particle maintains a constant speed given by $|v(1)|$, what is the trajectory for the particle for $t\ge 1$.
My process:
So first I thought I'd get the position at $t=1$, and find $r'(t)$ in order to find $|v(1)|$ ($r'(1)$).
$r(t)=\left<3t, 5t^2, 3t^3 \right>$ $\space$ $r(1)= \left< 3, 5, 3 \right>$
$r'(t)=\left<3, 10t, 9t^2 \right>$ $\space$ $r'(1)= \left< 3, 10, 9 \right>$
So having these, I thought I'd just set up parametric equations for the new trajectory and got $x=3+3t, y=5+10t, z=3+9t$ and in turn got the new position vector function to be $r(t)=\left<3+3t, 10t+5, 3+9t \right>$. But the answer is $r(t)=\left<3t, 10t-5, 9t-6 \right>$ and that is wrong, so I'm not sure on how to go about getting those values. Thanks for all the help in advance.
You did everything right except for one thing. If you'd like to find the position at $t \geq 1$, you start with $r(1)$ as you correctly did, and add $t-1$ times $r'(1)$. Why $t-1$ and not $t$? Because $t-1$ is exactly how much time passes since $t=1$. So the parametrization is
\begin{align} r(t) &= r(1)+(t-1)r'(1)\\\\ &=\left<3,5,3\right>+(t-1)\left<3,10,9\right>\\\\ &=\left<3+3t-3,5+10t-10,3+9t-9\right>\\\\ &=\boxed{\left<3t,10t-5,9t-6\right>}\\ \end{align}
What you calculated was $r(1)+tr'(1)$.