Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Double differentials and some curious problems

  1. May 15, 2008 #1
    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.
    [tex]
    \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{\Delta p }{\Delta t}}/\Delta p
    =\Delta t \approx 0
    [/tex]

    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.
    [tex]
    \frac{d}{d\theta}(\dot{\theta}cos\theta)
    =\frac{dF}{d\theta}
    =\frac{\partial F}{\partial \dot{\theta}} \times \frac{d\theta}{dt} +
    \frac{\partial F}{\partial \theta} \times \ddot{\theta}
    =\dot{\theta}cos\theta - \ddot{\theta}\dot{\theta}sin\theta
    [/tex]

    Last one has me flummaxed...
    3.
    [tex]
    \frac{d}{d\theta}(\theta+\dot{\theta}dt)=?
    [/tex]

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

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

    [tex]
    \frac{dy}{d\dot{\theta}}=\frac{dy}{dt} \times \frac{dt}{d\dot{\theta}}
    =\frac{dy}{dt} \times \left(\frac{d\dot{\theta}}{dt}\right)^{-1}
    =\frac{\dot{y}}{\ddot{\theta}}
    [/tex]


    Could anyone confirm what I've done so far (or point out any mistakes)? Cheers.
     
  2. jcsd
  3. May 16, 2008 #2
    Just to clarify, 3. slightly, the [tex]dt[/tex] is actually the sample period, so [tex]\dot{\theta}dt\approx \Delta\theta[/tex], 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)
    [tex]
    \frac{d}{dz}(xy)= x\frac{dy}{dz}+y\frac{dx}{dz}
    [/tex]
     
    Last edited: May 16, 2008
  4. May 16, 2008 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Ultimâ! :smile:

    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. :confused:
     
  5. May 16, 2008 #4
    Sorry! I just jumped into a shorthand replace with the following:
    [tex]
    \frac{\Delta p }{\Delta t}}/\Delta p
    =\frac{\Delta \dot{\theta} }{\Delta t}}/\Delta \ddot{\theta}
    [/tex]

    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:

    [tex]
    \mathbf{x}_{k|k-1}=\mathbf{A}_k\mathbf{x}_{k-1|k-1}
    [/tex]

    and I need to find

    [tex]
    \mathbf{J}_k=\frac{d(\mathbf{x}_{k|k-1})}{d(\mathbf{A}_k\mathbf{x}_{k-1|k-1})}
    [/tex]


    Thinking about 3. a bit more I believe [tex]1[/tex] would be a fairly accurate approximate.
     
  6. May 16, 2008 #5
    That is
    [tex]
    \frac{\Delta p }{\Delta t}}/\Delta p
    =\frac{\Delta \dot{\theta} }{\Delta t}}/\Delta \dot{\theta}
    [/tex]
     
  7. May 16, 2008 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Ultimâ! :smile:
    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. :confused:

    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)
     
  8. May 16, 2008 #7

    Mute

    User Avatar
    Homework Helper

    For your first one:

    [tex]\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}[/tex]

    If that total derivative is equal to zero, then it means you must have [itex] \ln \dot{\theta} = \mbox{const}[/itex], which means [itex] \dot{\theta} = \mbox{const}[/itex], which means [itex]\theta(t) = a + bt[/itex], 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 [itex]\theta(t)[/itex] to get [itex]t(\theta)[/itex], and so one could then write[itex]\ddot{\theta}(t) = \ddot{\theta}(t(\theta)) = \ddot{\theta}(\theta)[/itex].
     
    Last edited: May 16, 2008
  9. May 17, 2008 #8

    Mute

    User Avatar
    Homework Helper


    Actually, I didn't even need to do the bit with the ln. From the very first term, if [itex]d\dot{\theta}/d\theta[/itex] is zero, then [itex] \dot{\theta} = \mbox{const}[/itex] and [itex]\theta = a + bt[/itex].
     
  10. May 18, 2008 #9
    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 [tex]\phi \ \theta \ \psi [/tex] 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 [tex]\frac{\dot{\theta}_t - \dot{\theta}_{(t-1)}}{dt}[/tex] to estimate this....
     
    Last edited: May 18, 2008
  11. May 19, 2008 #10
    This is a little bit guessing, but if you are playing around with jacobian then it is most probably
    [tex]\frac{\partial \dot{\theta}}{\partial \theta} = 0[/tex]

    and the jacobian is evaluated at some point... Also you cannot always invert the function [itex]\theta(t)[/itex] and you don't check if [itex]\frac{dt}{d\theta}[/itex] is invertible...
     
  12. Jun 24, 2008 #11
    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?
     

    Attached Files:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?