Why does dv/dt = v dv/dx = .5 d^2v/dx?
Update:
oh well... dv/dt = (dv/dx)(dx/dt). You can either cancel out the dx leaving it as dv/dt, or turn the dx/dt part in to V for velocity. Making it into v(dv/dx) or dv/dt. I'm more concerned how to get to the third term in my question from v(dv/dx)
Comments
The derivative of v^2 with respect to x is 2v dv/dx by the chain rule.
So the derivative of 0.5v^2 with respect to x is v dv/dx.
I'm not sure what the rest of your question is. Why is that equal to dv/dt? It's not in general. Perhaps you were told that for your particular physical situation.
This is something you either "see" you or you do not, solutions in physics/math can be furthered by a sharp eye that can recognize one form is equivalent to perhaps a simpler form. If you do not see it, it comes with practice, your eyes will sharpen quickly with some practice. Either way, given the result you can prove it is true.
the "claim" if we do not believe it, is that dv/dt = v dv/dx = (1/2) d(v^2) / dx
that is we need to show that (1/2) d(v^2)/dx = v dv / dx
compute the derivative on the left-hand side:
(1/2) { 2 v dv/dx} = v dv/dx
so it really is the same.
If that derivative is a little tricky to see, we can relabel it to make the chain rule more transparent:
write y = y(v(x)), where y = v^2, and v = v(x)
then chain rule requires
dy/dx is what we need to compute
dy/dx = dy/dv dv/dx
dy/dv = d/dv (v^2) = 2 v
so we have
dy/dx = 2 v dv/dx
and we see that if we divide by 2 (i.e. multiply by 0.5) we get v dv/dx as claimed.