If \mathbf{a}=3 t \mathbf{i}-t^2 \mathbf{j} \text { and } \mathbf{b}=2 t^2 \mathbf{i}+3 \mathbf{j}, verify
(a) \frac{\mathrm{d}}{\mathrm{d} t}(\mathbf{a} \cdot \mathbf{b})=\mathbf{a} \cdot \frac{\mathrm{d} \mathbf{b}}{\mathrm{d} t}+\frac{\mathrm{d} \mathbf{a}}{\mathrm{d} t} \cdot \mathbf{b} (b) \frac{\mathrm{d}}{\mathrm{d} t}(\mathbf{a} \times \mathbf{b})=\mathbf{a} \times \frac{\mathrm{d} \mathbf{b}}{\mathrm{d} t}+\frac{\mathrm{d} \mathbf{a}}{\mathrm{d} t} \times \mathbf{b}
Also
\frac{\mathrm{d} \mathbf{a}}{\mathrm{d} t}=3 \mathbf{i}-2 t \mathbf{j} \quad \frac{\mathrm{d} \mathbf{b}}{\mathrm{d} t}=4 t \mathbf{i}So,
\begin{aligned} \mathbf{a} \cdot \frac{\mathrm{d} \mathbf{b}}{\mathrm{d} t}+\mathbf{b} \cdot \frac{\mathrm{d} \mathbf{a}}{\mathrm{d} t} & =\left(3 t \mathbf{i}-t^2 \mathbf{j}\right) \cdot(4 t \mathbf{i})+\left(2 t^2 \mathbf{i}+3 \mathbf{j}\right) \cdot(3 \mathbf{i}-2 t \mathbf{j}) \\ & =12 t^2+6 t^2-6 t=18 t^2-6 t \end{aligned}We have verified \frac{\mathrm{d}}{\mathrm{d} t}(\mathbf{a} \cdot \mathbf{b})=\mathbf{a} \cdot \frac{\mathrm{d} \mathbf{b}}{\mathrm{d} t}+\frac{\mathrm{d} \mathbf{a}}{\mathrm{d} t} \cdot \mathbf{b}.
\begin{aligned}(b)\ & \mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 t & -t^2 & 0 \\ 2 t^2 & 3 & 0 \end{array}\right| \\ &=\left(9 t+2 t^4\right) \mathbf{k} \\ & \frac{\mathrm{d}}{\mathrm{d} t}(\mathbf{a} \times \mathbf{b})=\left(9+8 t^3\right) \mathbf{k} \end{aligned}Also,
\begin{aligned} \mathbf{a} \times \frac{\mathrm{d} \mathbf{b}}{\mathrm{d} t} & =\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 t & -t^2 & 0 \\ 4 t & 0 & 0 \end{array}\right| \\ & =4 t^3 \mathbf{k} \\\frac{\mathrm{d} \mathbf{a}}{\mathrm{d} t} \times \mathbf{b} & =\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -2 t & 0 \\ 2 t^2 & 3 & 0 \end{array}\right| \\ & =\left(9+4 t^3\right) \mathbf{k} \end{aligned}and so
\mathbf{a} \times \frac{\mathrm{d} \mathbf{b}}{\mathrm{d} t}+\frac{\mathrm{d} \mathbf{a}}{\mathrm{d} t} \times \mathbf{b}=4 t^3 \mathbf{k}+\left(9+4 t^3\right) \mathbf{k}=\left(9+8 t^3\right) \mathbf{k}=\frac{\mathrm{d}}{\mathrm{d} t}(\mathbf{a} \times \mathbf{b})as required.