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Topic: Second differentials and parametrized curves.
Replies: 4   Last Post: Oct 18, 2013 9:42 AM

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William Elliot

Posts: 1,703
Registered: 1/8/12
Re: Second differentials and parametrized curves.
Posted: Oct 16, 2013 3:20 AM
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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.




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