Dot product of a vector and a derivative of that vector

[Ashiataka]

I'm reading through Douglas Gregory's Classical Mechanics, and at the start of chapter 6 he says that $m \vec{v} \cdot \frac{d\vec{v}}{dt} = \frac{d}{dt}\left(\frac12 m \vec{v} \cdot \vec{v}\right)$, but I'm not sure how to get the right hand side from the left hand side.

If someone could point me in the direction of where to look for this I would be grateful.

Thank you.

 

[mathman]

I'm reading through Douglas Gregory's Classical Mechanics, and at the start of chapter 6 he says that $m \vec{v} \cdot \frac{d\vec{v}}{dt} = \frac{d}{dt}\left(\frac12 m \vec{v} \cdot \vec{v}\right)$, but I'm not sure how to get the right hand side from the left hand side.

If someone could point me in the direction of where to look for this I would be grateful.

Thank you.

$\vec{v} \cdot \vec{v}=v_x^2 + v_y^2 + v_z^2$
Just use elementary calculus to get derivative of the right hand side. Assumes m is constant.

 

[gopher_p]

I'm reading through Douglas Gregory's Classical Mechanics, and at the start of chapter 6 he says that $m \vec{v} \cdot \frac{d\vec{v}}{dt} = \frac{d}{dt}\left(\frac12 m \vec{v} \cdot \vec{v}\right)$, but I'm not sure how to get the right hand side from the left hand side.

If someone could point me in the direction of where to look for this I would be grateful.

Thank you.

It's the product rule, $\frac{d}{dt}[\mathbf{u}\cdot\mathbf{v}]=\frac{d\mathbf{u}}{dt}\cdot \mathbf{v}+\mathbf{u}\cdot\frac{d\mathbf{v}}{dt}$, that gives this result. It's probably easiest to see this if you work right to left.

 

[Ashiataka]

Yes, of course it is. Thank you.