Rules for Differentiation and Tangent Planes

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) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\vec{u} \times \vec{v})' = \vec{u'}(t) \times \vec{v}(t) + \vec{u}(t) \times \vec{v'}(t)} (Cross Product Rule).

g) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u}(f(t)) = f'(t) \vec{u'}(f(t))} (Chain Rule).

• Proof of a)

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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)

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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)

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{r}(t) = (x(t), y(t), z(t))} be a vector-valued function that traces the curve Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t \in I} . If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \| \vec{r}(t) \| = c} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} is a constant, then for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t \in I} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{r}(t) \perp \vec{r'}(t)} .

• Proof: We note that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{r}(t) \perp \vec{r'}(t)} . Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \blacksquare}

Licensing

Content obtained and/or adapted from:

• Derivative Rules for Vector-Valued Functions, mathonline.wikidot.com under a CC BY-SA license