Time derivative of a vector of fixed magnitude
This post is about finding the derivative of a vector whose magnitude is fixed but the direction is changing. For simplicity, we consider unit vectors of a frame that is rotating with respect to an inertial frame. The result is applicable to any arbitrary vector.
Let \(XYZ\) be an inertial frame. Let \(\mathbf{\hat{I}}\), \(\mathbf{\hat{J}}\), and \(\mathbf{\hat{K}}\) be the unit vectors along \(X\), \(Y\), and \(Z\) axes respectively. Let \(xyz\) be another frame that is rotating at a constant angular velocity with respect to the \(Z-\) axis of the inertial frame. Let \(\mathbf{\hat{i}}\), \(\mathbf{\hat{j}}\), and \(\mathbf{\hat{k}}\) be the unit vectors along \(x\), \(y\), and \(z\) directions. Let the origins and axes of both the frames conincide at \(t=0\) and let the \(xyz\) frame rotate at an angular velocity \(\mathbf{\Omega}\). After time \(t\), the \(x\) and \(y-\) axes make an angle \(\mathbf{\Omega}\, t = \theta(t)\) with \(X\) and \(Y\) axes respectively. The orientation of the \(xyz\) frame relative to the inertial frame at time \(t\) is shown in figure below.
We want to find the time derivative of \(\mathbf{\hat{i}}\).
At the instant shown in figure, let \((x_{i}, y_{i}, z_{i})\) be the coordinates of \(\mathbf{\hat{i}}\). From firgure, we can write,
\[\begin{equation} \cos \theta = \frac{x}{\mathbf{\hat{||i||}}} \qquad \sin \theta = \frac{y}{\mathbf{\hat{||j||}}} \end{equation}\]Since \(\mathbf{\hat{i}}\) and \(\mathbf{\hat{j}}\) are unit vectors, \(\mathbf{\hat{||i||}} = \mathbf{\hat{||j||}} = 1\). Therefore, the coordiantes of \(\mathbf{\hat{i}}\) at any time \(t\) are given by,
\[\begin{equation} (x_{i}, y_{i}, z_{i}) = (\cos \theta (t),\, \sin \theta (t),\, 0) \end{equation}\]\(z-\) coordinate is zero because \(\mathbf{\hat{i}}\) lies in the \(xy-\) plane.
Similarly, the coordinates of \(\mathbf{\hat{j}}\) are given by,
\[\begin{align} (x_{j}, y_{j}, z_{j}) &= (\cos\, (90 + \theta (t)),\, \sin\, (90 + \theta (t)),\, 0) \nonumber \\ &= (-\sin\, \theta (t), \cos\, \theta (t), 0) \end{align}\]Since we want to find the time derivatives of unit vectors, let’s differentiate \(\mathbf{\hat{i}}\) with respect to \(t\).
\[\begin{align} \frac{d\mathbf{\hat{i}}}{dt} &= (-\sin \theta(t)\, \dot{\theta},\, \cos\, \theta(t)\, \dot{\theta},\, 0) \nonumber \\ &= \dot{\theta}\, (-\sin \theta(t),\, \cos \theta(t),\, 0) \nonumber \\ &= \mathbf{\Omega}\, \mathbf{\hat{j}} \label{eq:didt} \end{align}\]Now, let’s take the cross product of \(\mathbf{\Omega}\) with \(\mathbf{\hat{i}}\) and observe something interesting. Since \(\mathbf{\Omega}\) is along the \(Z-\) axis, \(\mathbf{\Omega} = (0, 0, \Omega)\).
\[\begin{equation} \mathbf{\Omega} \times \mathbf{\hat{i}} = \begin{vmatrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ 0 & 0 & \Omega \\ 1 & 0 & 0 \end{vmatrix} = \mathbf{\Omega}\, \mathbf{\hat{j}} \end{equation} \label{eq:cross-product}\]From equations \eqref{eq:didt} and \eqref{eq:cross-product}, we can write,
\[\begin{equation} \frac{d\mathbf{\hat{i}}}{dt} = \mathbf{\Omega} \times \mathbf{\hat{i}} \end{equation}\]In general, for an arbitrary vector \(A\) of fixed magnitude and changing direction, the time derivative is given by,
\[\begin{equation} \boxed{ \frac{d\mathbf{A}}{dt} = \mathbf{\Omega} \times \mathbf{A} } \end{equation}\]
Reference
Wikipedia. Rotating reference frame. https://en.wikipedia.org/wiki/Rotating_reference_frame. Accessed on August 1, 2019.