Differentiating vector dot product

[dyn]

Hi.
If I have a vector v , say for velocity for example then v.v = v² and I differentiate wrt t v.v I get 2v.dv/dt but if I differentiate v² I get 2v dv/dt but v.dv.dt is not the same as v dv/dt so what am I doing wrong ?
Thanks

 

[kuruman]

Why do you say they are not the same? Remember that $v =\left( v_x^2+v_y^2+v_z^2 \right )^{1/2}$.

 

[mfb]

$\frac{dv}{dt}= \frac{d}{dt} \sqrt{\vec v \vec v} = \frac{1}{2\sqrt{\vec v \vec v}} \left(\vec v \frac{d\vec v}{dt} + \frac{d\vec v}{dt} v\right) = \frac{1}{v} \vec v \frac{d \vec v}{dt}$.
The two expressions are the same.

 

[dyn]

Why do you say they are not the same? Remember that $v =\left( v_x^2+v_y^2+v_z^2 \right )^{1/2}$.

Because for vdv/dt to be zero needs v or dv/dt to be zero but v.dv/dt could be zero if v and dv/dt are orthogonal

 

[kuruman]

Because for vdv/dt to be zero needs v or dv/dt to be zero but v.dv/dt could be zero if v and dv/dt are orthogonal

To paraphrase @mfb,
$$v\frac{dv}{dt} =\left( v_x^2+v_y^2+v_z^2 \right )^{1/2} \frac{d}{dt}\left( v_x^2+v_y^2+v_z^2 \right )^{1/2} \\=
\frac{1}{2} \frac{\left( v_x^2+v_y^2+v_z^2 \right )^{1/2}}{\left( v_x^2+v_y^2+v_z^2 \right )^{1/2}}\left(2v_x\frac{dv_x}{dt} +2v_y\frac{dv_y}{dt}+2v_z\frac{dv_z}{dt} \right) =\left(v_x\frac{dv_x}{dt} +v_y\frac{dv_y}{dt}+v_z\frac{dv_z}{dt} \right)\\=\vec v \cdot \frac{d \vec v}{dt}$$.
They are the same. Furthermore, the above proof shows that, contrary to your assertion, $v\frac{dv}{dt} =0$ when $\vec v$ and $\frac{d\vec v}{dt}$ are orthogonal.

 

[mfb]

If $\vec v$ and $\frac{d \vec v}{dt}$ are orthogonal, then $\frac{dv}{dt} = 0$. This can be derived from what I posted, but it can also be understood geometrically: An acceleration orthogonal to the velocity does not change the speed.

 

[dyn]

Thanks everyone. Much appreciated