I am trying to show that $d/dt$ $|r(t)|^2 = r(t)*r'(t)$, where $r(t)= <x(t), y(t), z(t)>$ and $r(t) \neq 0$.
I first tried using the fact that $|r(t)|^2 = (x(t))^2+(y(t))^2+(z(t))^2$ and then deriving which is equivalent to $2*x(t)+2*y(t)+2*z(t)$. However, $r(t)*r'(t) = x(t)*x'(t)+y(t)*y'(t)+z(t)*z'(t)$
I then tried using the chain rule on $d/dt |r(t)|^2$ , but I was unsure how to derivate $\sqrt{x(t)^2+y(t)^2+z(t)^2}$. Is this second method the correct method, or were both approaches wrong? Any hints on how to get on the right track will be appreciated.
As you've written, $|r(t)|^2=x(t)^2 + y(t)^2 + z(t)^2$. However your next step was incorrect; deriving both sides leads to $$ \frac{d}{dt} |r(t)|^2 = 2x(t) x'(t) + 2y(t) y'(t) + 2z(t)z'(t), $$ by the chain rule. Therefore we have shown that $\frac{d}{dt} |r(t)|^2=2r(t)\cdot r'(t)$.