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I have been trying to differentiate the following function with respect to time

$$ t \mapsto \frac{1}{2}\dot{f}(t)^T A (f)\dot{f}(t) $$

but I am struggling with the chain rule for matrix calculus. Here, $f$ is a vector dependent on time and the matrix is a function of the vector $f$. I tried following these ideas but didn't come very far. Any help would be appreciated!

Rodrigo de Azevedo
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fishsticks
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2 Answers2

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$ \def\Diag{\operatorname{Diag}} \def\a{\alpha}\def\b{\beta}\def\l{\lambda} \def\h{\frac 12} \def\E{{\cal E}} \def\qiq{\quad\implies\quad} $For ease of typing define $$\eqalign{ v &= \dot f &\big({\rm velocity}\big) \\ a &= \dot v = \ddot f \qquad&\big({\rm acceleration}\big) \\ }$$ then it is straightforward to calculate the derivative of the given expression $$\eqalign{ \E &= \h \Big(v^TAv\Big) \\ \dot\E &= \h \Big(a^TAv+v^T\dot Av+v^TAa\Big) \\ &= v^TAa + \frac{v^T\dot Av}{2} \\ }$$ Now consider three plausible formulas for the symmetric $A$ matrix $$\eqalign{ A &= ff^T &\qiq \dot A = vf^T+fv^T \\ A &= \Diag(f) &\qiq \dot A = \Diag(v) \\ A &= fb^T + bf^T &\qiq \dot A = vb^T+bv^T \\ }$$ So the calculation of $\dot A$ is seldom a difficult problem.

greg
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Until someone else comes up with something better, here's how you might do this using Einstein index notation:

$\frac{d}{dt} \left( A^i_{~j}(f)\,\dot{f}^i\,\dot{f}_j \right) = A^i_{~j}\,\ddot{f}^i\,\dot{f}_j + A^i_{~j}\,\dot{f}^i\,\ddot{f}_j + \frac{\partial A^i_{~j}}{\partial f^k}\dot{f}^k\,\dot{f}^i\,\dot{f}_j\\ =\ddot{f}^T A \dot{f} + \dot{f}^T A \ddot{f} + (\text{not sure how to denote that last term in matrix notation}\overset{\otimes~\otimes}{\frown})$

Raad Shaikh
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