Section 13.2: Derivatives and Integrals of Vector Functions

The next theorem shows that the differentiation formulas for real-valued functions have their counterparts for vector-valued functions.

Theorem 3 Suppose \(\mathbf{u}\) and \(\mathbf{v}\) are differentiable vector functions, c is a scalar, and f is a real-valued function. Then 1. \(\frac{d}{dt}[\mathbf{u}(t) + \mathbf{v}(t)] = \mathbf{u}'(t) + \mathbf{v}'(t)\) 2. \(\frac{d}{dt}[c\mathbf{u}(t)] = c\mathbf{u}'(t)\) 3. \(\frac{d}{dt}[f(t)\mathbf{u}(t)] = f'(t)\mathbf{u}(t) + f(t)\mathbf{u}'(t)\) 4. \(\frac{d}{dt}[\mathbf{u}(t) \cdot \mathbf{v}(t)] = \mathbf{u}'(t) \cdot \mathbf{v}(t) + \mathbf{u}(t) \cdot \mathbf{v}'(t)\) 5. \(\frac{d}{dt}[\mathbf{u}(t) \times \mathbf{v}(t)] = \mathbf{u}'(t) \times \mathbf{v}(t) + \mathbf{u}(t) \times \mathbf{v}'(t)\) 6. \(\frac{d}{dt}[\mathbf{u}(f(t))] = f'(t)\mathbf{u}'(f(t))\) (Chain Rule)

This theorem can be proved either directly from Definition 1 or by using Theorem 2 and the corresponding differentiation formulas for real-valued functions. The proof of Formula 4 follows; the remaining formulas are left as exercises.

PROOF OF FORMULA 4 Let \[ \mathbf{u}(t) = \langle f_1(t), f_2(t), f_3(t) \rangle \quad \mathbf{v}(t) = \langle g_1(t), g_2(t), g_3(t) \rangle \] Then \[ \mathbf{u}(t) \cdot \mathbf{v}(t) = f_1(t)g_1(t) + f_2(t)g_2(t) + f_3(t)g_3(t) = \sum_{i=1}^3 f_i(t)g_i(t) \] so the ordinary Product Rule gives \[ \frac{d}{dt}[\mathbf{u}(t) \cdot \mathbf{v}(t)] = \frac{d}{dt} \sum_{i=1}^3 f_i(t)g_i(t) = \sum_{i=1}^3 \frac{d}{dt}[f_i(t)g_i(t)] \] \[ = \sum_{i=1}^3 [f_i'(t)g_i(t) + f_i(t)g_i'(t)] \] \[ = \sum_{i=1}^3 f_i'(t)g_i(t) + \sum_{i=1}^3 f_i(t)g_i'(t) \] \[ = \mathbf{u}'(t) \cdot \mathbf{v}(t) + \mathbf{u}(t) \cdot \mathbf{v}'(t) \]