Proof that $ \boldsymbol{v} \cdot \boldsymbol{a} = vv' $

Question

Assume $\alpha$ is a path in $\mathbb{R^3}$ with velocity $\boldsymbol{v}(t)$, speed $v(t)$, and acceleration $ \boldsymbol{a}(t)$. Assume $\boldsymbol{v}(t) \neq 0$ and $\boldsymbol{T}'(t) \neq 0$ for all $t$. Prove that $\boldsymbol{v} \cdot \boldsymbol{a} = vv'$

By the way, right off the bat, I'm thrown by one thing: Speed $\left (\text{i.e.}, ~ v(t) = || \boldsymbol{v}|| \right)$ is a scalar. So shouldn't $\frac{d}{dt}v(t) = 0$?

Anyway, $ \text{velocity} = \boldsymbol{v}(t) = \alpha'(t) = (x'(t), y'(t), z'(t))$ and $ \text{acceleration} = \boldsymbol{a}(t) = \alpha''(t) = (x''(t), y''(t), z''(t))$

But I don't see how the dot product of the two $(x'(t)x''(t) + y'(t)y''(t) + z'(t)z''(t))$ would equal speed times its own derivative $\sqrt{x'(t) + y'(t) + z'(t)} * \frac{(x'(t) + y'(t) + z'(t))^{-1/2}}{2}$.

What is the correct approach to this proof?

Answer 1

Let $v=v(t)=||\boldsymbol{v}||$, we have

$$\boldsymbol{v}\cdot\boldsymbol{a}=\boldsymbol{v}\cdot \frac{d\boldsymbol{v}}{dt}=\frac12\frac{d}{dt}(\boldsymbol{v}\cdot\boldsymbol{v})=\frac12\frac{dv^2}{dt}=\frac12\cdot2vv'=vv'$$

Answer 2

Regarding the statement that threw you off ($\frac{d}{dt}v=0$), it is useful to point out that you wrote "$v(t)=\lVert \vec{v} \rVert$" in your question. But it might make more sense if we rewrite this as $v(t)=\lVert \vec{v}(t) \rVert$ (note the addition of the $t$), which lets us see that while yes, it is true that $v(t)$ is a scalar-valued function, the value of $v$ depends on time, $t$, and as such is not necessarily constant. If it so happens that for some real valued constant $c$, $v(t)=c$ for all values of time $t$, then you get $\frac{d}{dt}v(t)=0$. Hope this helps.