Find the derivative of $\mathbf{u}(t)\times \mathbf{v}(t)$ if $$\mathbf{u}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+t\mathbf{k}\qquad \hbox{and}\qquad \mathbf{v}(t)=t\mathbf{i}+\ln t\mathbf{j}+\mathbf{k}$$

Solution The derivatives of $\mathbf{u}$ and $\mathbf{v}$ are $$\mathbf{u}^{\prime} (t)=-\!\sin t\mathbf{i}+\cos \dot{t}\mathbf{j}+\mathbf{k}\qquad \hbox{and}\qquad \mathbf{v}^{\prime}(t) =\mathbf{i}+\dfrac{1}{t} \mathbf{j}$$

We use the cross product formula. $$\begin{eqnarray*} \left[ \mathbf{u}(t)\times \mathbf{v}(t)\right] ^{\prime} &=&\mathbf{u}^{\prime} (t)\times \mathbf{v}(t)+\mathbf{u}(t)\times \mathbf{v}^{\prime} (t) \notag \\[6pt] &=&\left\vert \begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} &\mathbf{j} &\mathbf{k} \\ -\sin t &\cos t &1 \\ t &\ln t &1 \end{array} \right\vert +\left\vert \begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} &\mathbf{j} &\mathbf{k} \\ \cos t &\sin t &t \\ 1 &\dfrac{1}{t} &0 \end{array} \right\vert \notag \\[6pt] &=&\big[ ( \cos t-\ln t) \mathbf{i}+(\sin t+t)\mathbf{j} -[(\sin t)(\ln t)+t\cos t]\mathbf{k}\big] \\[6pt] &&+\left[ -\mathbf{i}+t\mathbf{ j}+\left( \frac{1}{t}\cos t-\sin t\right) \!\;\mathbf{k}\right] \notag \\[6pt] &=&(\cos t-\ln t-1)\mathbf{i}+(\sin t+2t)\mathbf{j}\\[6pt] &&+\left[ \frac{1}{t}\cos t-\sin t-(\sin t)(\ln t)-t\cos t\right] \!\;\mathbf{k} \notag \end{eqnarray*}$$