Understanding Partial Derivatives in Position-Velocity Relationship

[mwspice]

Hi,

I'm a little confused about something. I have an object, and I want to take the partial derivative of its position wrt velocity and vice versa. I'm not sure how to begin solving this problem. Essentially, what I have is this:
$\frac{\partial x}{\partial \dot x}$
and
$\frac{\partial \dot x}{\partial x}$
where the position $x$ can be determined by its velocity $\dot x$ by:
$\int_0^t \! \dot x \, \mathrm{d}t$

Any help with this would be much appreciated.

 

[drjustian]

Hi mw, don't be distressed about not being sure how to begin to solve your problem, no one has figured it out yet. Most likely you'll be faced with a 'probability' of the objects position at any given velocity. I look forward to your reported efforts on the matter.

 

[davidmoore63@y]

I'm not sure partial derivatives make sense in this case. When you take partial derivatives of a function f of two variables, say f(x,t), you are essentially pretending the other variable is a constant. In your case, x(dot) is not given as a function of x and t, and x is not given as a function of x(dot) and t. On the other hand, the full differentials make sense, so (apologies no latex):

dx/dx(dot) =dx/dt / dx(dot)/dt = x(dot)/x(double dot)

and

dx(dot)/dx = reciprocal of the above = x(double dot)/x(dot).

 

[Mark44]

I'm not sure partial derivatives make sense in this case. When you take partial derivatives of a function f of two variables, say f(x,t), you are essentially pretending the other variable is a constant. In your case, x(dot) is not given as a function of x and t, and x is not given as a function of x(dot) and t. On the other hand, the full differentials make sense, so (apologies no latex):

dx/dx(dot) =dx/dt / dx(dot)/dt = x(dot)/x(double dot)

and

dx(dot)/dx = reciprocal of the above = x(double dot)/x(dot).

Some help with LaTeX...
Here is the first equation, above.
$$\frac{dx}{d\dot{x}} = \frac{\frac{dx}{dt}}{\frac{d\dot{x}}{dt}} = \frac{\dot{x}}{\ddot{x}}$$

In bare form (unrendered) it looks like this:
$ $\frac{dx}{d\dot{x}} = \frac{\frac{dx}{dt}}{\frac{d\dot{x}}{dt}} = \frac{\dot{x}}{\ddot{x}}$ $
The only difference between the rendered form and the unrendered form is the extra spaces between the $ pairs.

If these are supposed to be partial derivatives, use \partial x etc. instead of dx etc.

 

[davidmoore63@y]

ok so i copied your latex into the 'Write Latex code' box, and clicked 'Show Preview' and all it did was copy out the exact same code

 

[mwspice]

I'm not sure partial derivatives make sense in this case. When you take partial derivatives of a function f of two variables, say f(x,t), you are essentially pretending the other variable is a constant. In your case, x(dot) is not given as a function of x and t, and x is not given as a function of x(dot) and t. On the other hand, the full differentials make sense, so (apologies no latex):

dx/dx(dot) =dx/dt / dx(dot)/dt = x(dot)/x(double dot)

and

dx(dot)/dx = reciprocal of the above = x(double dot)/x(dot).

Thanks for the reply!
I think I understand where you're coming from; however, I don't follow why you say that $x$ is not given as a function of $\dot x$ and $t$ because the way I see it, the integral given above is a function of $\dot x$ and $t$

Also, the partial derivative in this case is coming from the Lagrangian formula, with the form I am using given below. I don't think I could change the partial derivatives to total derivatives without messing up the formula, is that correct?
$\frac{d}{dt} \left( \frac{\partial T}{\partial \dot q_{i}} \right) - \frac{\partial T}{\partial q_{i}} + \frac{\partial V}{\partial q_{i}} + \frac{\partial R}{\partial \dot q_{i}} = Q_i$

As far as latex, I've always used # instead of $, so I'm not sure if that makes a difference

 

[davidmoore63@y]

I would have thought in the Lagrangian framework that you have T(q, qdot, t) and so q, qdot are being treated as independent variables. Hence partial derivative with respect to each other are zero.

 

[mwspice]

I would have thought in the Lagrangian framework that you have T(q, qdot, t) and so q, qdot are being treated as independent variables. Hence partial derivative with respect to each other are zero.

Thanks, I will try that! I thought about saying that they were 0, but I wasn't sure if that was actually the case since they do have a dependence on each other.

 

[Mark44]

ok so i copied your latex into the 'Write Latex code' box, and clicked 'Show Preview' and all it did was copy out the exact same code

I'm not sure what you're doing. To get this -- $ $\frac{dx}{d\dot{x}} = \frac{\frac{dx}{dt}}{\frac{d\dot{x}}{dt}} = \frac{\dot{x}}{\ddot{x}}$ $ -- to render correctly, remove the space between each pair of $ characters.

With the extra spaces removed, this is what you get:
$$\frac{dx}{d\dot{x}} = \frac{\frac{dx}{dt}}{\frac{d\dot{x}}{dt}} = \frac{\dot{x}}{\ddot{x}}$$

 

[davidmoore63@y]

got it mark44 thanks