Is $ d \mathbf v · d \mathbf v = d \mathit v^2 $? My teacher has proved the following:
$$ \mathit v^2 = \mathbf v·\mathbf v = \frac{d\mathbf r}{dt}·\frac{d\mathbf r}{dt} = \left(\frac {ds}{dt}\right)^2 \Rightarrow \mathit v = \frac{ds}{dt} $$
Because $\mathit v$ is always positive and:
$$\Vert d \mathbf r \Vert =ds$$
If I do the same with acceleration I get this:
$$ \mathit a^2 = \mathbf a·\mathbf a = \frac{d\mathbf v}{dt}·\frac{d\mathbf v}{dt} = \left(\frac {d\mathit v}{dt}\right)^2 \Rightarrow \mathit a = \frac{d\mathit v}{dt} $$
But that's not true, because:
$$\mathit a= \Vert \mathbf a \Vert = \left \Vert \frac{d\mathit v}{dt} \hat t + \frac{\mathit v^2}{\mathit R} \hat n \right \Vert $$
where $ \hat t $ is the unit tangent vector and $ \hat n $ is the normal vector. $ \frac{d\mathit v}{dt} $ is only the magnitude of the tangential acceleration, not the magnitude of $ \mathbf a $.
Where is the wrong step? Is it that $ d \mathbf v · d \mathbf v \not = d \mathit v^2 $?
 A: So you aren't doing the same thing as your teacher here.
Notice how in what your teacher did they have $\Vert d \mathbf r \Vert =ds$, so they did not say that
$$\frac{d\mathbf r}{dt}·\frac{d\mathbf r}{dt} = \left(\frac {dr}{dt}\right)^2 $$
this would not be true in say circular motion, where the position vector is changing but its magnitude is not.
However, you do this exact thing in your work
$$\frac{d\mathbf v}{dt}·\frac{d\mathbf v}{dt} = \left(\frac {d\mathit v}{dt}\right)^2 $$
so this is your invalid step. That $v$ in the above equation shouldn't be the magnitude of the velocity just how in your teacher's derivation they did not substitute in the magnitude of the position vector.
If you wanted to repeat what your teacher did you would need to similarly define a value $d\nu$ so that $\Vert d \mathbf v \Vert =d\nu$ and then
$$\mathit a^2 = \mathbf a·\mathbf a = \frac{d\mathbf v}{dt}·\frac{d\mathbf v}{dt} = \left(\frac {d\nu}{dt}\right)^2 \Rightarrow \mathit a = \frac{d\nu}{dt}$$
For circular motion this all works out then:
$$\mathbf v=R\dot\theta\,\hat\theta$$
$$d\mathbf v=R\ddot\theta\,\hat\theta\,dt-R\dot\theta^2\,\hat r\,dt$$
$$(d\nu)^2=R^2\ddot\theta^2(dt)^2+R^2\dot\theta^4(dt)^2=(a\,dt)^2$$
A: Yes, $\vec v=\frac{ds}{dt}\hat{t}$
So, $$ d\vec{v}\cdot d\vec {v}=(d^2s\hat t +ds\,d\hat t)\cdot(d^2s\hat t+ds\,d\hat t)=(d^2s)^2+(ds)^2$$
But ${dv}^2=(d^2s)^2$
A: I think that in the first case, a step is missing:
$$\frac{d\mathbf r}{dt}.\frac{d\mathbf r}{dt} = \frac{d\mathbf r}{ds}\frac{ds}{dt}.\frac{d\mathbf r}{ds}\frac{ds}{dt}$$
$\frac{d\mathbf r}{ds}$ is an unit vector in the direction of $\frac{d\mathbf r}{dt}$. So the dot product: $\frac{d\mathbf r}{ds}.\frac{d\mathbf r}{ds} = 1$
We can't do the same for $\frac{d\mathbf v}{dt}$, because $\frac{d\mathbf v}{dv} = \frac{d\mathbf v}{d|\mathbf v|}$ is not an unit vector. In the case of an uniform circular movement for example, it is not even defined, because the denominator is zero.
In the picture below, the ratio between $\Delta \mathbf v$ and $\Delta |\mathbf v|$. As the angle $A > 90^\circ$, $\frac{\Delta \mathbf v}{\Delta |\mathbf v|}$ has a modulus greater than one.

