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) $[\mathbf{v}, \mathbf{v}, \mathbf{k}]$ (the scalar triple product) --- 2. **Recall formulas and rules:** - The time derivative of a scalar function $f(t)$ is $\frac{d}{dt}f(t)$. - The time derivative of a vector function $\mathbf{v}(t)$ is $\frac{d}{dt}\mathbf{v}(t) = \mathbf{v}'(t)$. - For dot product: $\frac{d}{dt}(\mathbf{a} \cdot \mathbf{b}) = \mathbf{a}' \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{b}'$. - For scalar multiplication: $\frac{d}{dt}(f \mathbf{v}) = f' \mathbf{v} + f \mathbf{v}'$ where $f$ is scalar. - For scalar triple product $[\mathbf{a}, \mathbf{b}, \mathbf{c}] = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$, derivative is: $$\frac{d}{dt}[\mathbf{a}, \mathbf{b}, \mathbf{c}] = \mathbf{a}' \cdot (\mathbf{b} \times \mathbf{c}) + \mathbf{a} \cdot (\mathbf{b}' \times \mathbf{c}) + \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}')$$ Since $\mathbf{k}$ is constant, $\mathbf{k}'=\mathbf{0}$. --- 3. **Step (i): Derivative of $[\mathbf{v}]^2$** Note $[\mathbf{v}]^2 = \mathbf{v} \cdot \mathbf{v}$. $$\frac{d}{dt}[\mathbf{v}]^2 = \frac{d}{dt}(\mathbf{v} \cdot \mathbf{v}) = \mathbf{v}' \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{v}' = 2 \mathbf{v} \cdot \mathbf{v}'$$ --- 4. **Step (ii): Derivative of $(\mathbf{v} \cdot \mathbf{k}) \mathbf{v}$** Let $f = \mathbf{v} \cdot \mathbf{k}$ (scalar), then: $$\frac{d}{dt}[(\mathbf{v} \cdot \mathbf{k}) \mathbf{v}] = f' \mathbf{v} + f \mathbf{v}'$$ Calculate $f'$: $$f' = \frac{d}{dt}(\mathbf{v} \cdot \mathbf{k}) = \mathbf{v}' \cdot \mathbf{k} + \mathbf{v} \cdot \mathbf{k}' = \mathbf{v}' \cdot \mathbf{k} + 0 = \mathbf{v}' \cdot \mathbf{k}$$ So: $$\frac{d}{dt}[(\mathbf{v} \cdot \mathbf{k}) \mathbf{v}] = (\mathbf{v}' \cdot \mathbf{k}) \mathbf{v} + (\mathbf{v} \cdot \mathbf{k}) \mathbf{v}'$$ --- 5. **Step (iii): Derivative of scalar triple product $[\mathbf{v}, \mathbf{v}, \mathbf{k}]$** Recall: $$[\mathbf{v}, \mathbf{v}, \mathbf{k}] = \mathbf{v} \cdot (\mathbf{v} \times \mathbf{k})$$ Derivative: $$\frac{d}{dt}[\mathbf{v}, \mathbf{v}, \mathbf{k}] = \mathbf{v}' \cdot (\mathbf{v} \times \mathbf{k}) + \mathbf{v} \cdot (\mathbf{v}' \times \mathbf{k}) + \mathbf{v} \cdot (\mathbf{v} \times \mathbf{k}')$$ Since $\mathbf{k}'=\mathbf{0}$: $$= \mathbf{v}' \cdot (\mathbf{v} \times \mathbf{k}) + \mathbf{v} \cdot (\mathbf{v}' \times \mathbf{k}) + 0$$ Note that $\mathbf{v} \times \mathbf{k}$ is a vector, and the scalar triple product is antisymmetric. Also, the scalar triple product with two identical vectors is zero: Since $[\mathbf{v}, \mathbf{v}, \mathbf{k}] = 0$ always (because two vectors are the same), its derivative is also zero. Alternatively, the two terms cancel each other: $$\mathbf{v}' \cdot (\mathbf{v} \times \mathbf{k}) = - \mathbf{v} \cdot (\mathbf{v}' \times \mathbf{k})$$ So sum is zero. --- **Final answers:** (i) $$\frac{d}{dt}[\mathbf{v}]^2 = 2 \mathbf{v} \cdot \mathbf{v}'$$ (ii) $$\frac{d}{dt}[(\mathbf{v} \cdot \mathbf{k}) \mathbf{v}] = (\mathbf{v}' \cdot \mathbf{k}) \mathbf{v} + (\mathbf{v} \cdot \mathbf{k}) \mathbf{v}'$$ (iii) $$\frac{d}{dt}[\mathbf{v}, \mathbf{v}, \mathbf{k}] = 0$$