# Double differentials and some curious problems

by Ultimâ
Tags: curious, differentials, double
 P: 35 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. $$\frac{d}{d\theta}(\dot{\theta}) =\frac{d \dot{\theta}}{dt} \times \frac{dt}{d\theta} =\ddot{\theta} \times \dot{\theta}^{-1} =\ddot{\theta} / \dot{\theta} =\frac{\Delta p }{\Delta t}}/\Delta p =\Delta t \approx 0$$ 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. $$\frac{d}{d\theta}(\dot{\theta}cos\theta) =\frac{dF}{d\theta} =\frac{\partial F}{\partial \dot{\theta}} \times \frac{d\theta}{dt} + \frac{\partial F}{\partial \theta} \times \ddot{\theta} =\dot{\theta}cos\theta - \ddot{\theta}\dot{\theta}sin\theta$$ Last one has me flummaxed... 3. $$\frac{d}{d\theta}(\theta+\dot{\theta}dt)=?$$ And finally 4. $$\frac{d}{d\dot{\theta}}(q sin\phi tan\theta + r cos\phi tan\theta) =\frac{1}{\ddot{\theta}}\times \frac{d}{dt}(q(t) sin\phi (t) tan\theta (t)+ r (t) cos\phi (t) tan\theta (t)) =?$$ Number 4 I arrive at from the chain rule (an example below): $$\frac{dy}{d\dot{\theta}}=\frac{dy}{dt} \times \frac{dt}{d\dot{\theta}} =\frac{dy}{dt} \times \left(\frac{d\dot{\theta}}{dt}\right)^{-1} =\frac{\dot{y}}{\ddot{\theta}}$$ Could anyone confirm what I've done so far (or point out any mistakes)? Cheers.
 P: 35 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) $$\frac{d}{dz}(xy)= x\frac{dy}{dz}+y\frac{dx}{dz}$$
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P: 25,474
 Quote by Ultimâ 1.$$\frac{d}{d\theta}(\dot{\theta}) =\frac{d \dot{\theta}}{dt} \times \frac{dt}{d\theta} =\ddot{\theta} \times \dot{\theta}^{-1} =\ddot{\theta} / \dot{\theta} =\frac{\Delta p }{\Delta t}}/\Delta p =\Delta t \approx 0$$
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.

P: 35

## Double differentials and some curious problems

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

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:

$$\mathbf{x}_{k|k-1}=\mathbf{A}_k\mathbf{x}_{k-1|k-1}$$

and I need to find

$$\mathbf{J}_k=\frac{d(\mathbf{x}_{k|k-1})}{d(\mathbf{A}_k\mathbf{x}_{k-1|k-1})}$$

Thinking about 3. a bit more I believe $$1$$ would be a fairly accurate approximate.
 P: 35 That is $$\frac{\Delta p }{\Delta t}}/\Delta p =\frac{\Delta \dot{\theta} }{\Delta t}}/\Delta \dot{\theta}$$
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P: 25,474
Hi Ultimâ!
 Quote by 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/...00000000000000)
 HW Helper P: 1,391 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$\ddot{\theta}(t) = \ddot{\theta}(t(\theta)) = \ddot{\theta}(\theta)$.
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P: 1,391
 Quote by 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 $\theta = a + bt$.
 P: 35 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....
 P: 341 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...
P: 35
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
 appEJacobianTemp.pdf (88.0 KB, 4 views)

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