About the nature of acceleration and velocity given a vector trajectory

Question

Ok, it's a simple question, i'm studying for a test and I am given a vector $\vec{r}(t)=<t^2-5t,2t+1,3t^2>$, and asked to determine the points where it's acceleration and velocity are orthogonal. Simple enough, but I ran into a problem. I thought I could compute it in two ways, either:$$\vec{v}=\vec{r'}\\\vec{a}=\vec{r''}$$ and then do $\vec{a}\cdot{\vec{v}}=0$, which apparently did not work, or use the formulas for the acceleration with $$\vec{v}=\vec{r'}\\\vec{a}=v'.\vec{T}+[v^2/\rho].\vec{N}$$and compute the same dot product. I do understand that $d\vec{v}/dt=d[v.\vec{T}]/dt$ and that requires you to use the chain rule to get to the second expression of the acceleration I proposed, but I don't quite see why doing my first approach should yield the wrong result. So, thats it: what make the first approach wrong, and the second right?

Answer 1

\begin{align} \mathbf{v} &= (2t-5,2,6t) \\ \mathbf{a} &= (2,0,6) \\ \dot{s} &= \sqrt{(2t-5)^2+2^2+(6t)^2} \\ &= \sqrt{40t^2-20t+29} \\ \ddot{s} &= \frac{10(4t-1)}{\sqrt{40t^2-20t+29}} \\ \end{align}

When $t=\dfrac{1}{4}$, $$\ddot{s}=0 \implies \mathbf{a} \perp \mathbf{v}$$

See also link 1 and link 2.