Double differentials and some curious problems

  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. 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. tiny-tim

    tiny-tim 26,043
    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. 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. That is
    [tex]
    \frac{\Delta p }{\Delta t}}/\Delta p
    =\frac{\Delta \dot{\theta} }{\Delta t}}/\Delta \dot{\theta}
    [/tex]
     
  7. tiny-tim

    tiny-tim 26,043
    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. Mute

    Mute 1,391
    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. Mute

    Mute 1,391
    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. 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. 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. 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?
     

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