A Nice Relationship

I came upon this today while solving another problem. It's kind of a cool property. @Karan Chatrath also has this in his solution.

$v = \sqrt{v_x^2 + v_y^2 + v_z^2} \\ \frac{dv}{dt} = \frac{1}{2} (v_x^2 + v_y^2 + v_z^2)^{-1/2} (2 v_x a_x + 2 v_y a_y + 2 v_z a_z) \\ = \frac{\vec{v} \cdot \vec{a}}{v}$

Consider the following:

$\vec{V} = v_x \ \hat{i} + v_y \ \hat{j} + v_z \ \hat{k}$ $\vec{a} = a_x \ \hat{i} + a_y \ \hat{j} + a_z \ \hat{k}$ $S = \lvert \vec{V} \rvert = \sqrt{v_x^2 + v_y^2 + v_z^2}$

Now, the velocity vector:

$\vec{V} = S \left(\frac{v_x}{\sqrt{v_x^2 + v_y^2 + v_z^2}} \ \hat{i} + \frac{v_y}{\sqrt{v_x^2 + v_y^2 + v_z^2}} \ \hat{j}+\frac{v_z}{\sqrt{v_x^2 + v_y^2 + v_z^2}} \ \hat{k}\right)$

$\implies \ \vec{V}=S \ \hat{V}$ $\implies \ \frac{d\vec{V}}{dt} = \vec{a} = \frac{dS}{dt} \ \hat{V} + S \ \frac{d\hat{V}}{dt} \ \dots \ (1)$

Consider:

$\frac{d\hat{V}}{dt} = \frac{d}{dt}\left(\frac{v_x}{\sqrt{v_x^2 + v_y^2 + v_z^2}}\right) \ \hat{i} + \frac{d}{dt}\left(\frac{v_y}{\sqrt{v_x^2 + v_y^2 + v_z^2}}\right) \ \hat{j}+ \frac{d}{dt}\left(\frac{v_x}{\sqrt{v_x^2 + v_y^2 + v_z^2}}\right) \ \hat{k}$

Crunching out the derivatives and computing the following relation results in:

$\vec{a}-S\frac{d\hat{V}}{dt} = \frac{{v_x}\,\left({a_x}\,{v_x}+{a_y}\,{v_y}+{a_z}\,{v_z}\right)}{{{v_x}}^2+{{v_y}}^2+{{v_z}}^2} \ \hat{i} + \frac{{v_y}\,\left({a_x}\,{v_x}+{a_y}\,{v_y}+{a_z}\,{v_z}\right)}{{{v_x}}^2+{{v_y}}^2+{{v_z}}^2} \ \hat{j} + \frac{{v_z}\,\left({a_x}\,{v_x}+{a_y}\,{v_y}+{a_z}\,{v_z}\right)}{{{v_x}}^2+{{v_y}}^2+{{v_z}}^2} \ \hat{k}$ $\vec{a}-S\frac{d\hat{V}}{dt} = \frac{a_x v_x+a_y v_y+ a_z v_z}{v_x^2 + v_y^2 + v_z^2} \left(v_x \ \hat{i} + v_y \ \hat{j} + v_z \ \hat{k}\right)$ $\vec{a}-S\frac{d\hat{V}}{dt} = \frac{a_x v_x+a_y v_y+ a_z v_z}{v_x^2 + v_y^2 + v_z^2} \vec{V}$ $\vec{a}-S\frac{d\hat{V}}{dt} = \frac{a_x v_x+a_y v_y+ a_z v_z}{\sqrt{v_x^2 + v_y^2 + v_z^2}} \hat{V}$

The above expression, when compared to (1), shows that:

$\frac{dS}{dt} \ \hat{V} = \frac{a_x v_x+a_y v_y+ a_z v_z}{\sqrt{v_x^2 + v_y^2 + v_z^2}} \hat{V}$ $\implies \frac{dS}{dt} = \frac{a_x v_x+a_y v_y+ a_z v_z}{\sqrt{v_x^2 + v_y^2 + v_z^2}}$

Plugging in numbers gives:

$v_x = 1 \ ; \ v_y = 2 \ ; \ v_z = 3 \ ; \ a_x =-1 \ ; \ a_y = -1 \ ; \ a_z = 2$

$\boxed{\frac{dS}{dt} = \frac{3}{\sqrt{14}}}$

Note:

$\frac{dS}{dt} = \frac{a_x v_x+a_y v_y+ a_z v_z}{\sqrt{v_x^2 + v_y^2 + v_z^2}}$ $\frac{dS}{dt} = \vec{a} \cdot \hat{V}$

I do not see the intuition behind this yet.