MHB How to find the derivative of the absolute value of a vector?

[A330NEO]

Why is \frac{d}{dt}\left | r(t) \right | =\frac{1}{\left | r(t) \right |} r(t)\cdot r'(t) ?
A hint is given to me, saying \left | r(t)^2 \right | = r(t)\cdot r(t) . I think it's something to do with differentiating both side of the equation given as 'hint', but I have no idea how to execute it.
Just a little bit of hint would be appreciated. Thank you

 

[mathmari]

Why is \frac{d}{dt}\left | r(t) \right | =\frac{1}{\left | r(t) \right |} r(t)\cdot r'(t) ?
A hint is given to me, saying \left | r(t)^2 \right | = r(t)\cdot r(t) . I think it's something to do with differentiating both side of the equation given as 'hint', but I have no idea how to execute it.
Just a little bit of hint would be appreciated. Thank you

$$\frac{d}{dt} \left (r(t) \cdot r(t)\right ) = \frac{d}{dt} |r(t)|^2 = 2|r(t)| \frac{d}{dt} |r(t)|
\\ \Rightarrow \frac{d}{dt} |r(t)| = \frac{1}{2|r(t)|} \frac{d}{dt} \left ( r(t) \cdot r(t) \right )= \frac{1}{2|r(t)|} \left ( r'(t) \cdot r(t) + r(t) \cdot r'(t) \right )= \frac{1}{2|r(t)|} 2r(t) \cdot r'(t) \\ \Rightarrow \frac{d}{dt} |r(t)|= \frac{1}{|r(t)|} r(t) \cdot r'(t)$$

 

[Euge]

Why is \frac{d}{dt}\left | r(t) \right | =\frac{1}{\left | r(t) \right |} r(t)\cdot r'(t) ?
A hint is given to me, saying \left | r(t)^2 \right | = r(t)\cdot r(t) . I think it's something to do with differentiating both side of the equation given as 'hint', but I have no idea how to execute it.
Just a little bit of hint would be appreciated. Thank you

Let $u = |r(t)|^2$. Then $|r(t)| = \sqrt{u}$; by the chain rule,

$$\frac{d}{dt}|r(t)| = \frac{1}{2\sqrt{u}} \frac{du}{dt}$$.

Again by the chain rule,

$$ \frac{du}{dt} = 2r(t) \cdot r'(t)$$.

In case you don't see this, let $x(t)$, $y(t)$, and $z(t)$ be the $x$-, $y$-, and $z$-components of $r(t)$, respectively (assuming $r(t)$ is a space vector for all $t$). Then $u = x^2 + y^2 + z^2$, and thus

$$\frac{du}{dt} = 2xx' + 2yy' + 2zz' = 2\langle x,y,z \rangle \cdot \langle x', y', z' \rangle = 2r \cdot r'.$$

Therefore

$$\frac{d}{dt}|r(t)| = \frac{1}{\cancel{2}|r(t)|} (\cancel{2}r(t)\cdot r'(t)) = \frac{1}{|r(t)|} r(t)\cdot r'(t).$$