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I have a 2D motion problem. Suppose $(x,y) = (x(t), y(t))$ are the coordinates of a particle of a 2D plane at time $t$. How can we express $d^2 t/ d x^2$, $d^2 t/ d x d y$ and $d^2 t/ d y^2$ in terms of $$x'(t) := \frac{dx}{dt}, x''(t) := \frac{d^2 x}{d t^2}, y'(t) := \frac{dy}{dt}, y''(t) := \frac{d^2 y}{d t^2}~~?$$

My answer is the following. For example, for $d^2 t/ d x^2$, I first find the 1st derivative as follows.

$$\frac{dt}{dx} = [x'(t)]^{-1}$$

and the 2nd derivative with respect to $x$ only is

\begin{align} \frac{dt}{dx} = \frac{d}{dx} (\frac{dt}{dx}) &=\frac{d}{dx} [x'(t)]^{-1} \\ &= \frac{dt}{dx} \frac{d}{dt} [x'(t)]^{-1} \\ &~~ \vdots \\ &= - [x'(t)]^{-3} [x''(t)] \end{align}

where $...$ means simplifying the required algebra by using the 1st derivative. Am I going through the right way? If not, are there any other ways to do so? Thank you very much for your answer.

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(The follow-up part)

Suppose the (differential) equations of motion are the following.

$3 H^2 = \frac{1}{2}([x'(t)]^2 + [y'(t)]^2) + V$

$x''(t) + 3 H x'(t) + \frac{\partial V}{\partial x} = 0$

$y''(t) + 3 H y'(t) + \frac{\partial V}{\partial y} = 0$

where $H = H(x, y, x' ,y')$, $x, y, x', y'$ are defined above, $V = V(x, y)$ and one prime means the derivative with respect to time $t$. So, we can find $\frac{dt}{dx} = [x'(t)]^{-1}$ and $\frac{dt}{dy} = [y'(t)]^{-1}$ and their 2nd derivatives in terms of $x', y', x'', y''$, right? But, what about $\frac{d^2 t}{dx dy}$? Does $\frac{d^2 t}{dx dy}$ exist based on this system of differential equations? If yes, how can it be expressed in terms of $x', y', x'', y''$? If no, can we take it as $\frac{d^2 t}{dx dy} = \frac{d}{dx} \left( \frac{dt}{dy} \right)= 0$ as $\frac{dt}{dy}$ is independent of $x$?

(I asked such a thing is that I am going to express $dt/dx$, $dt/dy$, $d^2 t/dx^2$, $d^2 t/dy^2$ and $d^2 t/dx dy$ in terms of $x', y', x'', y''$ and then solve them numerically after defining $V(x, y)$. Thank you very much for your answer. )

Ellgan
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  • You cant do the partial of t w.r.t. x and y as t cannot be expressed as a function of x and y, its entirely separate. For x and y, you have different functions. But the answer should be 0 for both the cases since x(t) = x and y(t) = y. Hence, if you take 1st order derivative, it will give you 1 for both cases and 0 when you take the 2nd order derivative. Hence, both will be 0. Hope this helps. – Math_Is_Fun Feb 16 '20 at 06:17

1 Answers1

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For partial differential $\frac{dt^2}{dxdy}$ does not make sense, i.e. $t$ can not be regarded as a function of $x,~y$.

DLIN
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  • So, do you mean that d2t/dxdy and d2t/dydx do not exist? Or can I say d2t/dxdy = d2t/dydx = 0? – Ellgan Feb 16 '20 at 23:40
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    If $t,x,y$ satisfy some relation $F(t,x,y)=0$, where $F$ is an equation of three variables. And, If $\partial F/\partial t$ neat a point, then locally $t$ can be regarded as a function of $x,y$. And your notation is nor correct, $\partial t/\partial x$ – DLIN Feb 17 '20 at 04:47