Taking a derivative of a magnitude of a vector

Question

I have found several questions including this as a step in the explanation, but have not been able to find an explicit explanation of how to take a derivative of a magnitude of a vector.

I am trying to take the derivative d/dt ||r'(t)|| but don't know how to address the magnitude signs.

Answer 1

I will assume that you are working with the euclidian norm and the dot product. Let $v(t)$ be a vector: $$v(t) \cdot v(t)=|v(t)|^2$$ And $$\frac{\mathrm{d}|v(t)|^2}{\mathrm{d}t}=2|v(t)|\frac{\mathrm{d}|v(t)|}{\mathrm{d}t}$$ Which implies that $$\begin{align} \frac{\mathrm{d}|v(t)|}{\mathrm{d}t}&=\frac{1}{2|v(t)|}\frac{\mathrm{d}|v(t)|^2}{\mathrm{d}t}\\ &=\frac{1}{2|v(t)|}\frac{\mathrm{d}(v(t) \cdot v(t))}{\mathrm{d}t}\\ &=\frac{1}{2|v(t)|}\left(v(t) \cdot v'(t)+v'(t) \cdot v(t)\right)\\ &=\frac{1}{2|v(t)|}\left(2v(t) \cdot v'(t)\right)\\ &=\frac{v(t) \cdot v'(t)}{|v(t)|} \end{align}$$

Answer 2

If the norm is derived from the dot product $\langle \cdot , \cdot \rangle$, then

$$\frac{d\Vert r^\prime(t) \Vert}{dt} = \frac{\langle r^\prime(t),r^{\prime \prime}(t)\rangle}{\Vert r^\prime(t)\Vert}.$$

Answer 3

$$ \eqalign{ & \left| {r(t)} \right|^2 = r_{\,x} (t)^2 + r_{\,y} (t)^2 + \cdots \cr & 2\left| {r(t)} \right|{d \over {dt}}\left| {r(t)} \right| = 2\left( {r_{\,x} (t)r_{\,x} '(t) + r_{\,y} (t)r_{\,y} '(t) + \cdots } \right) \cr & {d \over {dt}}\left| {r(t)} \right| = {{r_{\,x} (t)r_{\,x} '(t) + r_{\,y} (t)r_{\,y} '(t) + \cdots } \over {\left| {r(t)} \right|}} \cr} $$