Trying to show that $\vec{v}\cdot\frac{d||\vec{v}||}{dt}\hat{v} = 0\implies \frac{d||\vec{v}||}{dt} = 0$ This is (hopefully) the final step in a larger proof. I would like to show that:
$$\vec{v}\cdot\frac{d||\vec{v}||}{dt}\hat{v} = 0\implies \frac{d||\vec{v}||}{dt} = 0$$
Where: $\vec{v}$ is a velocity vector and $\hat{v}$ is the the unit velocity vector, i.e. has a magnitude of 1.
I don't know that $||\vec{v}||$ is a constant as this is what I'm trying to show so I can't simply take it out of the dot product. However, it seems apparent that it must be a constant as the only way (that I can see) to make the dot product equal to zero is for $\frac{d||\vec{v}||}{dt} = 0$.
I'm hoping to find a more mathematically rigorous way of showing the desired outcome.
Thank you for your time and assistance.
Best,
Eric
 A: Since $\|\vec{v}\| = \sqrt{\vec{v}\cdot \vec{v}}$, the chain rule gives $$\frac{d}{dt}\|\vec{v}\| = \frac{1}{2\|v\|} \frac{d}{dt}\left(\vec{v} \cdot \vec{v}\right) = \frac{1}{2\|\vec{v}\|}\left(2\vec{v}\cdot \frac{d\vec{v}}{dt}\right) = \hat{v}\cdot \frac{d\vec{v}}{dt}.$$ Therefore 
$$\vec{v} \cdot \frac{d\|\vec{v}\|}{dt}\hat{v} = \vec{v}\cdot \left(\hat{v}\cdot \frac{d\vec{v}}{dt}\right)\hat{v} = \left(\hat{v}\cdot \frac{d\vec{v}}{dt}\right)(\vec{v}\cdot \hat{v}) = \left(\hat{v}\cdot \frac{d\|\vec{v}\|}{dt}\right)\|\vec{v}\| = \vec{v}\cdot \frac{d\vec{v}}{dt}.$$
Hence, the condition $\vec{v} \cdot \frac{d\|\vec{v}\|}{dt}\hat{v} = 0$ implies $$\vec{v}\cdot \frac{d\vec{v}}{dt} = 0.$$ By the chain rule, this is same as $$\frac{1}{2}\frac{d}{dt}(\vec{v}\cdot \vec{v}) = 0.$$ That is, $$\frac{1}{2}\frac{d}{dt}\|\vec{v}\|^2 = 0.$$  Consequently, $\|\vec{v}\|^2$ is constant, so $\|\vec{v}\|$ is constant.
A: First
$$\frac{d}{dt}\hat{v}\cdot\hat{v}=0$$
which implies
$$\vec{v}\cdot\frac{d}{dt}\hat{v}=0$$
Then
$$\frac{d}{dt}\|v\|^2=\frac{d}{dt}\vec{v}\cdot\vec{v}=2\vec{v}\cdot\frac{d}{dt}\vec{v}=2\vec{v}\cdot\frac{d\|v\|}{dt}\hat{v}+2\vec{v}\cdot\|v\|\frac{d\hat{v}}{dt}=2\vec{v}\cdot\frac{d\|v\|}{dt}\hat{v}$$
Hence
$$\vec{v}\cdot\frac{d\|v\|}{dt}\hat{v}=0$$
implies
$$\frac{d}{dt}\|v\|=0$$
