Why is dx/dt*(\delta x*dx/dt) equal to 1/2\delta(dx/dt)^2?

[Gloyn]

Hello.

I don't understand one transformation that is made on page 25 of this paper:

http://www.atm.ox.ac.uk/user/read/mechanics/LA-notes.pdf

It is the second equation from the top, ont the one marked as '2', but just the second one.

dx/dt*(\delta x*dx/dt)=1/2\delta(dx/dt)^2

Why is that?

 

[haruspex]

Hello.

I don't understand one transformation that is made on page 25 of this paper:

http://www.atm.ox.ac.uk/user/read/mechanics/LA-notes.pdf

It is the second equation from the top, ont the one marked as '2', but just the second one.

dx/dt*(\delta x*dx/dt)=1/2\delta(dx/dt)^2

Why is that?

You've slightly misrepresented what the paper says. It's dx/dt*(δ dx/dt), not dx/dt*(δx dx/dt). To make it easier to read, let's write v for dx/dt. The transformation is then from v δv to δ(v²)/2. Is that clearer?

 

[Gloyn]

Right, I've made a mistake during writing the equation. Unfortunately, the transformation I don't get is the one you wrote. It's probably sth. very basic and I can't figure it out :s.

 

[haruspex]

Well, what do you get if you differentiate v²?

 

[Gloyn]

Oh. My. God. I'm dumb, thanks.