Given the positional vector $ r(t)=[ln(t), t^2 - 4t]$,
find the potential maximum/minimum points using the velocity vector.
$v(t)=r´(t) = [1/t, 2t-4]$
Plotted into my graphing software of choice, we can see that the x-point where v(t) has its y-value equal to 0 (marked green), is not equal to the x-point where r(t) has its minimum point (marked red).
Why is this?
