Vector functions and differentiation

Question

I am confused with the solution to the following question. I'm not sure how to follow the steps from the 2nd line onwards and how it relates to the chain rule. Can someone please break this down for me? Why has ${\bf|{a}|}^2 = \bf{a}\cdot{a}$ been used?

Exercise:

Let a be a vector function. Calculate $\frac{d(f({\bf{|a|}})}{dt}$.

Solution:

\begin{align} \frac{d(f({|\bf{a}|})}{dt} &= \frac{d}{dt}({|\bf{a}|}) f'({|\bf{a}|}) \\ {|\bf{a}|}^2 &= \bf{a}\cdot{a} \quad \\ 2{|\bf{a}|} \frac{d}{dt} ({|\bf{a}|}) &= 2{\bf{a}\cdot{\dot{a}}} \\ \frac{d}{dt}(f({|\bf{a}|}) &= \frac{{\bf{a} \cdot{a}}}{{|\bf{a}|}} f'({|\bf{a}|}) \end{align}

Answer 1

Maybe if you expand in components it will help

$$ |{\bf a}|^2 = {\bf a}\cdot {\bf a} = \sum_k a_k a_k $$

So that

$$ \frac{{\rm d}}{{\rm d}t}|{\bf a}|^2 = \frac{{\rm d}}{{\rm d}t}\sum_k a_k a_k = \sum_k a_k \frac{{\rm d}a_k}{{\rm d}t} + \frac{{\rm d}a_k}{{\rm d}t}a_k = 2\sum_k a_k \frac{{\rm d}a_k}{{\rm d}t} = 2\sum_k a_k \dot{a}_k = 2 {\bf a}\cdot \dot{\bf a} $$

Answer 2

$|\mathbf a|^2=\mathbf a\cdot\mathbf a$ has been used because you want to express the final answer in terms of $\displaystyle\mathbf{\dot a}=\frac{d\mathbf a}{dt}$.

Chain rule gives$$\frac{df(|\mathbf a|)}{dt}=\frac{df(|\mathbf a|)}{d|\mathbf a|}\frac{d|\mathbf a|}{dt}=f'(|\mathbf a|)\frac{d|\mathbf a|}{dt}$$

$|\mathbf a|$ is also a function of $\mathbf a$. Use the chain rule to differentiate $|\mathbf a|^2=\mathbf a\cdot\mathbf a$ with respect to $t$ to get,$$2|\mathbf a|\frac{d|\mathbf a|}{dt}=\mathbf a\cdot\frac{d\mathbf a}{dt}+\frac{d\mathbf a}{dt}\cdot\mathbf a=2\mathbf a\cdot\dot{\mathbf a}\implies \frac{d|\mathbf a|}{dt}=\frac{\mathbf a\cdot\dot{\mathbf a}}{|\mathbf a|}$$Alternatively, you can also differentiate $|\mathbf a|=\sqrt{\mathbf a\cdot\mathbf a}$ with respect to $t$ to get$$\frac{d|\mathbf a|}{dt}=\frac12\cdot\frac{\mathbf a\cdot\frac{d\mathbf a}{dt}+\frac{d\mathbf a}{dt}\cdot\mathbf a}{\sqrt{\mathbf a\cdot\mathbf a}}=\frac{\mathbf a\cdot\dot{\mathbf a}}{|\mathbf a|}$$

This gives the answer as$$\frac{df(|\mathbf a|)}{dt}=f'(|\mathbf a|)\frac{\mathbf a\cdot\dot{\mathbf a}}{|\mathbf a|}$$

Answer 3

The reason why the identity is used is to avoid having to deal directly with the norm function. Instead the product rule is used and the calculation is simpler, as we are taking the derivative of known functions.

We could apply the same idea to compute the derivative of $\sqrt{x}$ from scratch by only knowing the derivative of $x$: Using that (for positive $x$): $$(\sqrt{x})^2=x$$ we apply the derivative on both sides and use the product rule $$2\sqrt{x}\cdot \frac{d}{dx}(\sqrt{x})^=\frac{d}{dx}((\sqrt{x})^2)=\frac{d}{dx}x=1$$ and thus $$\frac{d}{dx}\sqrt{x}=\frac1{2\sqrt{x}}$$