Total Derivative expression doubt

Question

I have a doubt about the expression of a derivative, I would like to know if I am doing something wrong.

I would like to compute the following:

$F=\frac{d}{dt}\left({h}u^TR_i^TR_jv\right)$

• $h$ is a scalar which is a function of a certain $p_i,p_j$

• $u,v$ are 2 constant column vectors

• $R_i,R_j$ are 2 rotation matrices $\in \mathbb{R}^{3\times3}$ which are functions of 2 angles respectively $\psi_i,\psi_j$

Can I write $F$, applying the chain rule, as:

$\left((\frac{\partial}{\partial p_i}h)\cdot\dot{p}_i+(\frac{\partial}{\partial p_j}h)\cdot\dot{p}_j\right)u^TR_i^TR_jv+hu^T\left((\frac{\partial}{\partial \psi_i}R_i^T)\cdot\dot{R}_i^T\right)R_jv+hu^TR_i^T\left((\frac{\partial}{\partial \psi_j}R_j^T)\cdot\dot{R}_j\right)v$

EDIT:

I think I did an error: $\dot{R_i}=\left((\frac{\partial}{\partial \psi_i}R_i)\cdot\dot{R}_i\right)$ is not true, because: $\dot{R_i}=\left((\frac{\partial}{\partial \psi_i}R_i)\cdot\dot{\psi}_i\right)$

This is valid if $R_i$ depends only on $\psi_i$.

If instead $R_i$ depends from both $\psi_i,\psi_j$ I will have

$\dot{R_i}=\left((\frac{\partial}{\partial \psi_i}R_i)\cdot\dot{\psi}_i\right) + \left((\frac{\partial}{\partial \psi_j}R_i)\cdot\dot{\psi}_j\right)$

Thanks a lot for your time

Answer 1

Let $\boldsymbol{A} := R_i$, $\boldsymbol{B} := R_j$, $\mathbf{a} := u$, and $\mathbf{b} := v$. Then, $$ F = \frac{d}{dt}\left( h\, \mathbf{a}^T \boldsymbol{A}^T \boldsymbol{B} \mathbf{b}\right) = \frac{d}{dt}\left[ h\, (\boldsymbol{A}\mathbf{a})^T (\boldsymbol{B} \mathbf{b})\right] = : \frac{d}{dt}\left[ h\, (\mathbf{m})^T \mathbf{n}\right] $$ For clarity, we can think of the above as tensor-type operations with a dot product defined as $$ \mathbf{a}^T\mathbf{b} \equiv \mathbf{a}\cdot\mathbf{b} = \sum_i a_i b_i \equiv a_i b_i \quad;\quad \boldsymbol{A} \mathbf{b} \equiv \boldsymbol{A} \cdot \mathbf{b} = \sum_j A_{ij} b_j \equiv A_{ij} b_j \,. $$ So we can write (with repeated indices assumed to be summed over) $$ F = \frac{d}{dt}\left[h\,(\boldsymbol{A}\cdot\mathbf{a})\cdot(\boldsymbol{B}\cdot\mathbf{b})\right] \equiv \frac{d}{dt}\left[h (A_{ij} a_j) (B_{ip} b_p)\right] = \frac{d}{dt}\left(h m_i n_i\right)\,. $$ Now we can easily apply the chain rule repeatedly. To start off, $$ F = m_i n_i\frac{dh}{dt} + h n_i \frac{dm_i}{dt} + h m_i \frac{dn_i}{dt} $$ Look at the first term above and let $\alpha := p_i$ and $\beta := p_j$ (just to avoid indices unless they are indices of a matrix or a vector). We have $$ \frac{dh}{dt} = \frac{dh}{d\alpha}\frac{d\alpha}{dt} + \frac{dh}{d\beta}\frac{d\beta}{dt} $$ Plug in all the definitions and you get the first term in your expression which appears to be correct.

Now look at the second term in the expression for $F$ and define $\theta := \psi_i$ and $\phi := \psi_j$ to get $$ \frac{dm_i}{dt} = \frac{dm_i}{d\theta}\frac{d\theta}{dt} + \frac{dm_i}{d\phi}\frac{d\phi}{dt} $$ Let's proceed further with the algebra,and recall that the vectors $\mathbf{a},\mathbf{b}$ are constant, to get $$ \frac{dm_i}{d\theta} = \frac{d}{d\theta}(A_{ij} a_j) = \frac{dA_{ij}}{d\theta} a_j + A_{ij} \frac{da_j}{d\theta} = \frac{dA_{ij}}{d\theta} a_j ~;~~ \frac{dm_i}{d\phi} = \frac{dA_{ip}}{d\phi} a_p $$ Therefore, $$ \frac{dm_i}{dt} = a_j \frac{dA_{ij}}{d\theta} \frac{d\theta}{dt} + a_p \frac{dA_{ip}}{d\phi}\frac{d\phi}{dt} $$ and $$ h n_i \frac{dm_i}{dt} = h n_i a_j \frac{dA_{ij}}{d\theta} \frac{d\theta}{dt} + h n_i a_p \frac{dA_{ip}}{d\phi}\frac{d\phi}{dt} $$ In index-free form, and using $\dot{a} := da/dt$, $$ h \,\mathbf{n}\cdot\dot{\mathbf{m}} = h\left[(\boldsymbol{B}\cdot\mathbf{b})\cdot\left(\frac{d\boldsymbol{A}}{d\theta}\cdot\mathbf{a}\right)\right]\dot{\theta} + h\left[(\boldsymbol{B}\cdot\mathbf{b})\cdot\left(\frac{d\boldsymbol{A}}{d\phi}\cdot\mathbf{a}\right)\right]\dot{\phi} $$ Similarly, for the third term in the expression for $F$, $$ h \,\mathbf{m}\cdot\dot{\mathbf{n}} = h\left[(\boldsymbol{A}\cdot\mathbf{a})\cdot\left(\frac{d\boldsymbol{B}}{d\theta}\cdot\mathbf{b}\right)\right]\dot{\theta} + h\left[(\boldsymbol{A}\cdot\mathbf{a})\cdot\left(\frac{d\boldsymbol{B}}{d\phi}\cdot\mathbf{b}\right)\right]\dot{\phi} $$ You can write these without $\dot{\theta}$ and $\dot{\phi}$ as $$ h \,\mathbf{n}\cdot\dot{\mathbf{m}} = h\left[(\boldsymbol{B}\cdot\mathbf{b})\cdot\left(\dot{\boldsymbol{A}}\cdot\mathbf{a}\right)\right] $$ and $$ h \,\mathbf{m}\cdot\dot{\mathbf{n}} = h\left[(\boldsymbol{A}\cdot\mathbf{a})\cdot\left(\dot{\boldsymbol{B}}\cdot\mathbf{b}\right)\right] $$ Plug in the definitions of $\boldsymbol{A}$ etc. into these and you will notice that you cannot have terms like $$ \frac{d}{d\psi_i}(R_i^T) \quad \text{and} \quad \dot{R_i}^T $$ simultaneously in your expressions. So the answer to your question is no.