0
$\begingroup$

I have a problem that arose in a kinematics context. Suppose, the Jacobian $J_{x}(q) = \frac{\partial{x}}{\partial{q}}$ of vector $x$ w.r.t. vector $q$ is known. I am interested in the Jacobian $J_{\dot{x}}(\dot{q}) = \frac{\partial{\dot{x}}}{\partial{\dot{q}}}$ of the vector $\dot{x}=\frac{\mathrm{d}{x}}{\mathrm{d}{t}}$ w.r.t. the vector $\dot{q}=\frac{\mathrm{d}{q}}{\mathrm{d}{t}}$, with $t$ being scalar.

(In the kinematics context, $x$ is a Cartesian position, $q$ is a set of joint angles, and $t$ is time, so $\dot{x}$ is Cartesian velocity and $\dot{q}$ is the vector of joint velocities.)

Is $J_{\dot{x}}(\dot{q}) = J_{{x}}({q})$?

$\endgroup$

1 Answer 1

0
$\begingroup$

Let's do it step by step $$\begin{align} \frac{\partial\dot x}{\partial\dot q}=\frac{\partial(\frac{\partial x}{\partial q}\dot q)}{\partial\dot q}=\frac{\partial^2x}{\partial\dot q\partial q}\dot q+\frac{\partial x}{\partial q} \end{align}$$ In kinematics, $q=q(x)$ is only a function of position, or $x=x(q)$ is irrelevant to $\dot q$. We can conclude that $\frac{\partial^2x}{\partial\dot q\partial q}=0$. So your conclusion is right.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.