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 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}(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)

${\begin{aligned}\quad ({\vec {u}}(t)+{\vec {v}}(t))'=\lim _{h\to 0}{\frac {[{\vec {u}}(t+h)+{\vec {v}}(t+h)]-[{\vec {u}}(t)+{\vec {v}}(t)]}{h}}\\\quad ({\vec {u}}(t)+{\vec {v}}(t))'=\lim _{h\to 0}{\frac {[{\vec {u}}(t+h)-{\vec {u}}(t)]+[{\vec {v}}(t+h)-{\vec {v}}(t)]}{h}}\\\quad ({\vec {u}}(t)+{\vec {v}}(t))'=\lim _{h\to 0}{\frac {{\vec {u}}(t+h)-{\vec {u}}(t)}{h}}+\lim _{h\to 0}{\frac {{\vec {v}}(t+h)-{\vec {v}}(t)}{h}}\\\quad ({\vec {u}}(t)+{\vec {v}}(t))'={\vec {u'}}(t)+{\vec {v'}}(t)\quad \blacksquare \end{aligned}}$

• Proof of c)

${\begin{aligned}\quad (k{\vec {u}}(t))'=\lim _{h\to 0}{\frac {k{\vec {u}}(t+h)-k{\vec {u}}(t)}{h}}\\\quad (k{\vec {u}}(t))'=\lim _{h\to 0}k{\frac {{\vec {u}}(t+h)-{\vec {u}}(t)}{h}}\\\quad (k{\vec {u}}(t))'=k{\vec {u'}}(t)\quad \blacksquare \end{aligned}}$

• Proof of d)

${\begin{aligned}\quad (f(t){\vec {u}}(t))'=\lim _{h\to 0}{\frac {f(t+h){\vec {u}}(t+h)-f(t){\vec {u}}(t)}{h}}\\\quad (f(t){\vec {u}}(t))'=\lim _{h\to 0}{\frac {f(t+h){\vec {u}}(t+h)-f(t+h){\vec {u}}(t)+f(t+h){\vec {u}}(t)-f(t){\vec {u}}(t)}{h}}\\\quad (f(t){\vec {u}}(t))'=\lim _{h\to 0}{\frac {f(t+h)[{\vec {u}}(t+h)-{\vec {u}}(t)]+{\vec {u}}(t)[f(t+h)-f(t)]}{h}}\\\quad (f(t){\vec {u}}(t))'=\lim _{h\to 0}{\frac {f(t+h)[{\vec {u}}(t+h)-{\vec {u}}(t)]}{h}}+\lim _{h\to 0}{\frac {{\vec {u}}(t)[f(t+h)-f(t)]}{h}}\\\quad (f(t){\vec {u}}(t))'=\lim _{h\to 0}f(t+h)\left({\frac {{\vec {u}}(t+h)-{\vec {u}}(t)}{h}}\right)+\lim _{h\to 0}{\vec {u}}(t)\left({\frac {[f(t+h)-f(t)]}{h}}\right)\\\quad (f(t){\vec {u}}(t))'=f(t){\vec {u'}}(t)+{\vec {u}}f'(t)\\\quad (f(t){\vec {u}}(t))'=f'(t){\vec {u}}+f(t){\vec {u'}}(t)\quad \blacksquare \end{aligned}}$

Theorem 2: Let ${\vec {r}}(t)=(x(t),y(t),z(t))$ 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