1. **Problem statement:** Given a vector function $\mathbf{v}(t)$ and a constant vector $\mathbf{k}$, find the time derivatives of: (i) $|\mathbf{v}|^2$ (ii) $(\mathbf{v} \cdot \mathbf{k}) \mathbf{v}$ (iii) $[\mathbf{v}, \mathbf{v}, \mathbf{k}]$ (assuming this denotes the scalar triple product or a vector triple product, clarify if needed) --- 2. **Recall formulas and rules:** - The magnitude squared of a vector is $|\mathbf{v}|^2 = \mathbf{v} \cdot \mathbf{v}$. - The derivative of a dot product: $\frac{d}{dt}(\mathbf{a} \cdot \mathbf{b}) = \frac{d\mathbf{a}}{dt} \cdot \mathbf{b} + \mathbf{a} \cdot \frac{d\mathbf{b}}{dt}$. - The derivative of a scalar times a vector: $\frac{d}{dt}(f\mathbf{a}) = \frac{df}{dt}\mathbf{a} + f\frac{d\mathbf{a}}{dt}$. - The scalar triple product $[\mathbf{a}, \mathbf{b}, \mathbf{c}] = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$. - The derivative of a cross product: $\frac{d}{dt}(\mathbf{a} \times \mathbf{b}) = \frac{d\mathbf{a}}{dt} \times \mathbf{b} + \mathbf{a} \times \frac{d\mathbf{b}}{dt}$. --- 3. **Calculate each derivative:** (i) $\frac{d}{dt} |\mathbf{v}|^2 = \frac{d}{dt} (\mathbf{v} \cdot \mathbf{v}) = \mathbf{v}' \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{v}' = 2 \mathbf{v} \cdot \mathbf{v}'$ (ii) $\frac{d}{dt} \big((\mathbf{v} \cdot \mathbf{k}) \mathbf{v}\big) = \frac{d}{dt}(\mathbf{v} \cdot \mathbf{k}) \mathbf{v} + (\mathbf{v} \cdot \mathbf{k}) \mathbf{v}'$ Since $\mathbf{k}$ is constant, $\frac{d}{dt}(\mathbf{v} \cdot \mathbf{k}) = \mathbf{v}' \cdot \mathbf{k}$. So, $$ \frac{d}{dt} \big((\mathbf{v} \cdot \mathbf{k}) \mathbf{v}\big) = (\mathbf{v}' \cdot \mathbf{k}) \mathbf{v} + (\mathbf{v} \cdot \mathbf{k}) \mathbf{v}' $$ (iii) Assuming $[\mathbf{v}, \mathbf{v}, \mathbf{k}] = \mathbf{v} \cdot (\mathbf{v} \times \mathbf{k})$ (scalar triple product), note that $\mathbf{v} \times \mathbf{v} = \mathbf{0}$, so this is zero for all $t$. Therefore, $$ \frac{d}{dt} [\mathbf{v}, \mathbf{v}, \mathbf{k}] = \frac{d}{dt} 0 = 0 $$ If the notation means something else, please clarify. --- **Final answers:** (i) $\frac{d}{dt} |\mathbf{v}|^2 = 2 \mathbf{v} \cdot \mathbf{v}'$ (ii) $\frac{d}{dt} \big((\mathbf{v} \cdot \mathbf{k}) \mathbf{v}\big) = (\mathbf{v}' \cdot \mathbf{k}) \mathbf{v} + (\mathbf{v} \cdot \mathbf{k}) \mathbf{v}'$ (iii) $\frac{d}{dt} [\mathbf{v}, \mathbf{v}, \mathbf{k}] = 0$