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Double differentials and some curious problems 
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#1
May1508, 03:30 PM

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. [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 anglerate 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
May1608, 04:07 AM

P: 35

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] 


#3
May1608, 05:02 AM

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P: 26,157

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. 


#4
May1608, 05:43 AM

P: 35

Double differentials and some curious problems
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}_{kk1}=\mathbf{A}_k\mathbf{x}_{k1k1} [/tex] and I need to find [tex] \mathbf{J}_k=\frac{d(\mathbf{x}_{kk1})}{d(\mathbf{A}_k\mathbf{x}_{k1k1})} [/tex] Thinking about 3. a bit more I believe [tex]1[/tex] would be a fairly accurate approximate. 


#5
May1608, 10:56 AM

P: 35

That is
[tex] \frac{\Delta p }{\Delta t}}/\Delta p =\frac{\Delta \dot{\theta} }{\Delta t}}/\Delta \dot{\theta} [/tex] 


#6
May1608, 12:49 PM

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Hi Ultimâ!
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) 


#7
May1608, 01:39 PM

HW Helper
P: 1,391

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]. 


#8
May1708, 01:12 PM

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P: 1,391

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]. 


#9
May1808, 01:57 PM

P: 35

tinytim  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 doable, though I don't actually have theta_doubledot available and would need to use [tex]\frac{\dot{\theta}_t  \dot{\theta}_{(t1)}}{dt}[/tex] to estimate this.... 


#10
May1908, 04:20 AM

P: 341

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... 


#11
Jun2408, 10:50 AM

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?



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