Derivative Rules for Vector-Valued Functions

# Derivative Rules for Vector-Valued Functions

We will now look at a bunch of rules for differentiating vector-valued function, all of which are analogous to that of differentiating real-valued functions. We will not prove all parts of the following theorem, but the reader is encouraged to attempt the proofs.

 Theorem 1: Let $\vec{u}(t) = (x_1(t), y_1(t), z_1(t))$ and $\vec{v}(t) = (x_2(t), y_2(t), z_2(t))$ be vector-valued functions that are differentiable for $t$ in the interval $I$, let $k$ be a scalar, and let $f(t)$ be a real-valued function that is differentiable on $I$. Then: a) $( \vec{u}(t) + \vec{v}(t) )' = \vec{u'}(t) + \vec{v'}(t)$ (Sum Rule). b) $( \vec{u}(t) - \vec{v}(t) )' = \vec{u'}(t) - \vec{v'}(t)$ (Difference Rule). c) $(k \vec{u}(t))' = k \vec{u'} (t)$ (Scalar Multiple Rule). d) $(f(t) \vec{u}(t))' = f'(t) \vec{u}(t) + f(t) \vec{u'}(t)$ (Product Rule for Real-Valued and Vector-Valued Functions). e) $(\vec{u} \cdot \vec{v})' = \vec{u'}(t) \cdot \vec{v}(t) + \vec{u}(t) \cdot \vec{v'}(t)$ (Dot Product Rule). f) $(\vec{u} \times \vec{v})' = \vec{u'}(t) \times \vec{v}(t) + \vec{u}(t) \times \vec{v'}(t)$ (Cross Product Rule). g) $\vec{u}(f(t)) = f'(t) \vec{u'}(f(t))$ (Chain Rule).
• Proof of a)
(1)
\begin{align} \quad (\vec{u}(t) + \vec{v}(t))' = \lim_{h \to 0} \frac{[\vec{u}(t + h) + \vec{v}(t + h)] - [\vec{u}(t) + \vec{v}(t)]}{h} \\ \quad (\vec{u}(t) + \vec{v}(t))' = \lim_{h \to 0} \frac{[\vec{u}(t + h) - \vec{u}(t)] + [\vec{v}(t + h) - \vec{v}(t)]}{h} \\ \quad (\vec{u}(t) + \vec{v}(t))' = \lim_{h \to 0} \frac{\vec{u}(t + h) - \vec{u}(t)}{h} + \lim_{h \to 0} \frac{\vec{v}(t + h) - \vec{v}(t)}{h} \\ \quad (\vec{u}(t) + \vec{v}(t))' = \vec{u'}(t) + \vec{v'}(t) \quad \blacksquare \end{align}
• Proof of c)
(2)
\begin{align} \quad (k \vec{u}(t))' = \lim_{h \to 0} \frac{k\vec{u}(t + h) - k \vec{u}(t)}{h} \\ \quad (k \vec{u}(t))' = \lim_{h \to 0} k \frac{\vec{u}(t + h) - \vec{u}(t)}{h} \\ \quad (k \vec{u}(t))' = k \vec{u'}(t) \quad \blacksquare \end{align}
• Proof of d)
(3)
\begin{align} \quad (f(t)\vec{u}(t))' = \lim_{h \to 0} \frac{f(t + h)\vec{u}(t + h) - f(t)\vec{u}(t)}{h} \\ \quad (f(t)\vec{u}(t))' = \lim_{h \to 0} \frac{f(t+h)\vec{u}(t + h) - f(t+h)\vec{u}(t) + f(t+h)\vec{u}(t) - f(t) \vec{u}(t)}{h} \\ \quad (f(t)\vec{u}(t))' = \lim_{h \to 0} \frac{f(t+h) [ \vec{u}(t + h) - \vec{u}(t)] + \vec{u}(t) [f(t+h) - f(t)]}{h} \\ \quad (f(t)\vec{u}(t))' = \lim_{h \to 0} \frac{f(t+h) [ \vec{u}(t + h) - \vec{u}(t)]}{h} + \lim_{h \to 0} \frac{\vec{u}(t) [f(t+h) - f(t)]}{h} \\ \quad (f(t)\vec{u}(t))' = \lim_{h \to 0} f(t + h) \left ( \frac{\vec{u}(t + h) - \vec{u}(t)}{h} \right) + \lim_{h \to 0} \vec{u}(t) \left ( \frac{[f(t+h) - f(t)]}{h} \right) \\ \quad (f(t)\vec{u}(t))' = f(t) \vec{u'}(t) + \vec{u} f'(t) \\ \quad (f(t)\vec{u}(t))' = f'(t) \vec{u} + f(t) \vec{u'}(t) \quad \blacksquare \end{align}
 Theorem 2: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function that traces the curve $C$ for $t \in I$. If $\| \vec{r}(t) \| = c$ where $c$ is a constant, then for all $t \in I$, $\vec{r}(t) \perp \vec{r'}(t)$.
• Proof: We note that $\vec{r}(t) \cdot \vec{r}(t) = \| \vec{r}(t) \|^2 = c^2$. Taking the derivative of both sides of this equation and applying the dot product rule, we get that:
(4)
\begin{align} \frac{d}{dt} \vec{r}(t) \cdot \vec{r}(t) = \frac{d}{dt} c^2 \\ \vec{r}(t) \cdot \vec{r'}(t) + \vec{r'}(t) \cdot \vec{r}(t) = 0 \\ 2 \vec{r}(t) \cdot \vec{r'}(t) = 0 \\ \vec{r}(t) \cdot \vec{r'}(t) = 0 \end{align}
• Therefore $\vec{r}(t) \perp \vec{r'}(t)$. $\blacksquare$