At what time is the speed minimum? The position function of a particle is given by $r(t) = \langle-5t^2, -1t, t^2 + 1t\rangle$. At what time is the speed minimum?

$$v(t)=r'(t)=<-10t,-1, 2t+1>$$
$$|v(t)|=\sqrt{100t^2+1+4t^2+4t+1}=\sqrt{104t^2+4t+1}$$
Now you want to minimize what's underneath the square root which we will call $f(t)$.
$$f'(t)=208t+4=0$$
Which gives $$t=\frac{-4}{208}=\frac{-1}{52}$$
