Second derivative calculation

Question

Consider $\boldsymbol{\psi}(\boldsymbol{q})\in R^2$ is a function of vector $\boldsymbol{q}(t)\in R^3$ which is a function of time $t\in R^+$. The first time derivative of function $\boldsymbol{\psi}$ can be defined as following equation using chain rule: $\dot{\boldsymbol{\psi}}=\frac{\partial{\boldsymbol{\psi}}}{\partial{\boldsymbol{q}}}\dot{q}$. In this equation $\frac{\partial{\boldsymbol{\psi}}}{\partial{\boldsymbol{q}}}\in R^{2\times3}$ is the Jacobian matrix. How it is possible to define the second time derivative of $\ddot{\boldsymbol{\psi}}$ using chain rule? For example if the function $\psi$ and $q$ were scalars we could define its second derivative easily by the following equation $\ddot{\psi}=\frac{\partial^2{\psi}}{\partial{q^2}} \dot{q}^2+\frac{\partial{\psi}}{\partial{q}}\ddot{q}$. But what about this case? Is there any similar equation for vector forms of $\boldsymbol{\psi} \in R^2$ and $\boldsymbol{q}(t) \in R^3$ ?

Answer 1

You can derive needed identity for second derivative if you express matrix multiplication as summation. Let $q:(0, \infty) \to \mathbb R^3, \psi: \mathbb R^3 \to \mathbb R^2$. We have \begin{align} \frac {\operatorname d^2}{\operatorname d\! t^2} \psi(q(t)) &= \frac {\operatorname d}{\operatorname d\! t}\left(\frac {\partial \psi}{\partial q}\bigg|_{q(t)}\dot q(t)\right) \\ &= \frac {\operatorname d}{\operatorname d\! t}\left( \sum_{i=1}^2\sum_{j=1}^3 \frac {\partial \psi^i}{\partial x^j}\bigg|_{q(t)}\dot q^j(t){\bf e}_i\right) \\ &= \sum_{i=1}^2\sum_{j=1}^3 \left( \frac {\partial \psi^i}{\partial x^j}\bigg|_{q(t)}\ddot q^j(t)+\sum_{k=1}^3\frac{\partial^2 \psi^i}{\partial x^k\partial x^j}\bigg|_{q(t)}\dot q^k(t)\dot q^j(t)\right){\bf e}_i \\ &= \frac{\partial \psi}{\partial q}\bigg|_{q(t)}\ddot q(t) + \sum_{i=1}^2 \dot q(t)^TH(\psi^i)|_{q(t)}\dot q(t) {\bf e}_i, \end{align} where $H(\psi^i)$ is the Hessian matrix of $\psi^i$ component of $\psi$ function.

If you know what you are doing, you could shorten this notation and write

$$ \frac {\operatorname d^2}{\operatorname d\! t^2} \psi(q(t)) = \frac{\partial \psi}{\partial q}\ddot q(t) + \dot q(t)^TH(\psi)\dot q(t).$$