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Re: Second differentials and parametrized curves.
Posted:
Oct 16, 2013 3:20 AM


On Tue, 15 Oct 2013, Hetware wrote:
> Thomas briefly introduces the concept of a second differential, but, so far, > makes little use of it. I write ddy for d^2y because it seems more clear in > ASCII. > It's not very important.
> I decided to explore Thomas's assertion that finding ddy/dx^2 is difficult if > done directly. > > x = f(t) > y = g(t) > > dx = (dx/dt)dt > dy = (dy/dt)dt > > dy/dx = (dy/dt)/(dx/dt) = (dy/dt)(dt/dx) > > ddy/dx^2 = d(dy/dt)(dt/dx)/dx > = (dy/dt) d(dt/dx)/dx + d(dy/dt)/dx (dt/dx) > = (dy/dt) (ddt/dx^2) + (ddy/dt^2) (dt/dx)^2 [use dx^2 = ((dx/dt)dt)^2] > = (dy/dt) (ddt/((dx/dt)dt)^2) + (ddy/dt^2) (dt/dx)^2 > = (dy/dt) (ddt/dt^2) (dt/dx)^2 + (ddy/dt^2) (dt/dx)^2 > = (ddy/dt^2)(dt/dx)^2 [because ddt/dt^2 = 0.] > = ddy/dx^2 [because dt and dx are real values.] > > I believe that series of transformations is correct. > > I find it somewhat disconcerting that ddt/dx^2 = 0. I certainly don't expect > ddx/dt^2 = 0, in general. I know I should have a question about all of this, > but I'm not sure what that question should be.
Where did you come up with that and no, usually d^2 t/dx^2 /= 0. For example: y = t = x^3; dt/dx = 3x^2; d^2 t/dx^2 = 6x.



