Double differentials and some curious problems

[Ultimâ]

Hello, I'm toying around with a Jacobian that has raised some interesting problems. It's a case of differentiating rates of some variable x, with respect to itself.

First one I suspect the answer is zero, though perhaps my reasoning is a bit flawed.

1.
<br /> \frac{d}{d\theta}(\dot{\theta})<br /> =\frac{d \dot{\theta}}{dt} \times \frac{dt}{d\theta}<br /> =\ddot{\theta} \times \dot{\theta}^{-1}<br /> =\ddot{\theta} / \dot{\theta}<br /> =\frac{\Delta p }{\Delta t}}/\Delta p<br /> =\Delta t \approx 0<br />

The second I think you apply the total derivative rule to, but maybe not, should the angle and angle-rate be considered as two separate variables?

2.
<br /> \frac{d}{d\theta}(\dot{\theta}cos\theta)<br /> =\frac{dF}{d\theta}<br /> =\frac{\partial F}{\partial \dot{\theta}} \times \frac{d\theta}{dt} +<br /> \frac{\partial F}{\partial \theta} \times \ddot{\theta}<br /> =\dot{\theta}cos\theta - \ddot{\theta}\dot{\theta}sin\theta<br />

Last one has me flummaxed...
3.
<br /> \frac{d}{d\theta}(\theta+\dot{\theta}dt)=?<br />

And finally
4.
<br /> \frac{d}{d\dot{\theta}}(q sin\phi tan\theta + r cos\phi tan\theta)<br /> =\frac{1}{\ddot{\theta}}\times \frac{d}{dt}(q(t) sin\phi (t) tan\theta (t)+ r (t) cos\phi (t) tan\theta (t))<br /> =?<br />

Number 4 I arrive at from the chain rule (an example below):

<br /> \frac{dy}{d\dot{\theta}}=\frac{dy}{dt} \times \frac{dt}{d\dot{\theta}}<br /> =\frac{dy}{dt} \times \left(\frac{d\dot{\theta}}{dt}\right)^{-1}<br /> =\frac{\dot{y}}{\ddot{\theta}}<br />


Could anyone confirm what I've done so far (or point out any mistakes)? Cheers.

 

[Ultimâ]

Just to clarify, 3. slightly, the dt is actually the sample period, so \dot{\theta}dt\approx \Delta\theta, but I'm unsure how this affects the derivative...

Also, in 2. maybe I should use the product rule, but I think the term you are using for differentiating needs to be different...

(product rule)
<br /> \frac{d}{dz}(xy)= x\frac{dy}{dz}+y\frac{dx}{dz}<br />

 

[tiny-tim]

1.\frac{d}{d\theta}(\dot{\theta})<br /> =\frac{d \dot{\theta}}{dt} \times \frac{dt}{d\theta}<br /> =\ddot{\theta} \times \dot{\theta}^{-1}<br /> =\ddot{\theta} / \dot{\theta}<br /> =\frac{\Delta p }{\Delta t}}/\Delta p<br /> =\Delta t \approx 0

Hi Ultimâ!

I'm confused … what are ∆p and ∆t?

And what is the context that this is a part of? Is it something like a Lagrangian, where θ and θ' are treated as independent variables, so that ∂θ'/∂θ = ∂θ/∂θ' = 0 anyway?

If not, I don't understand how you get from θ''/θ' to (∆p/∆t)/∆p.

 

[Ultimâ]

Sorry! I just jumped into a shorthand replace with the following:
<br /> \frac{\Delta p }{\Delta t}}/\Delta p<br /> =\frac{\Delta \dot{\theta} }{\Delta t}}/\Delta \ddot{\theta}<br />

The context is trying to create a Jacobian matrix to estimate the covariance for angular rates. I don't really want to delve into to much detail as the matrix is rather large, but in a simplified form:

<br /> \mathbf{x}_{k|k-1}=\mathbf{A}_k\mathbf{x}_{k-1|k-1}<br />

and I need to find

<br /> \mathbf{J}_k=\frac{d(\mathbf{x}_{k|k-1})}{d(\mathbf{A}_k\mathbf{x}_{k-1|k-1})}<br />


Thinking about 3. a bit more I believe 1 would be a fairly accurate approximate.

 

[Ultimâ]

That is
<br /> \frac{\Delta p }{\Delta t}}/\Delta p<br /> =\frac{\Delta \dot{\theta} }{\Delta t}}/\Delta \dot{\theta}<br />

 

[tiny-tim]

Hi Ultimâ!

The context is trying to create a Jacobian matrix to estimate the covariance for angular rates. I don't really want to delve into to much detail …

I think you'd better delve a little, as I've really no idea what you're doing.

You seem to be trying to differentiate one element of a matrix with respect to another.

And does the J in Jk mean Jacobian, or angular momentum?

(btw, the LaTeX for ∂ is \partial … see http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000 )

 

[Mute]

For your first one:

\frac{d}{d\theta}(\dot{\theta})=\frac{d \dot{\theta}}{dt} \times \frac{dt}{d\theta}=\ddot{\theta} \times \dot{\theta}^{-1}=\ddot{\theta} / \dot{\theta}= \frac{d}{dt}\ln \dot{\theta}

If that total derivative is equal to zero, then it means you must have \ln \dot{\theta} = \mbox{const}, which means \dot{\theta} = \mbox{const}, which means \theta(t) = a + bt, which won't be the case in general. The reason it's not zero in general is that for general cases you can in principle invert \theta(t) to get t(\theta), and so one could then write\ddot{\theta}(t) = \ddot{\theta}(t(\theta)) = \ddot{\theta}(\theta).

 

[Mute]

For your first one:

\frac{d}{d\theta}(\dot{\theta})=\frac{d \dot{\theta}}{dt} \times \frac{dt}{d\theta}=\ddot{\theta} \times \dot{\theta}^{-1}=\ddot{\theta} / \dot{\theta}= \frac{d}{dt}\ln \dot{\theta}

If that total derivative is equal to zero, then it means you must have \ln \dot{\theta} = \mbox{const}, which means \dot{\theta} = \mbox{const}, which means \theta(t) = a + bt, which won't be the case in general. The reason it's not zero in general is that for general cases you can in principle invert \theta(t) to get t(\theta), and so one could then write\ddot{\theta}(t) = \ddot{\theta}(t(\theta)) = \ddot{\theta}(\theta).

Actually, I didn't even need to do the bit with the ln. From the very first term, if d\dot{\theta}/d\theta is zero, then \dot{\theta} = \mbox{const} and \theta = a + bt.

 

[Ultimâ]

tiny-tim - Sorry if I didn't make things very clear, but I was just hoping people could check what I had done seemed reasonable - that is simplifying the derivatives in 1.->4. These happen to be four of the elements of J (yes it is a Jacobian) that I'm inputting as a matrix for an Extended Kalman filter (EKF) I'm working with. This simplification means having things in terms of p q r \phi \ \theta \ \psi or their rates (values of which the EKF has available for making calculations).

I think Mute is suggesting I shouldn't simplify after the fourth part of No. 1, which do-able, though I don't actually have theta_doubledot available and would need to use \frac{\dot{\theta}_t - \dot{\theta}_{(t-1)}}{dt} to estimate this...

 

[trambolin]

This is a little bit guessing, but if you are playing around with jacobian then it is most probably
\frac{\partial \dot{\theta}}{\partial \theta} = 0

and the jacobian is evaluated at some point... Also you cannot always invert the function \theta(t) and you don't check if \frac{dt}{d\theta} is invertible...

 

[Ultimâ]

Well since I was asked for it, here's the full problem I have (see pdf),...Anything wrong with my reasoning here for the elements I have calculated for the Jacobian?