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) $({\vec {u}}\times {\vec {v}})'={\vec {u'}}(t)\times {\vec {v}}(t)+{\vec {u}}(t)\times {\vec {v'}}(t)$ (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 $C$ 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:

${\begin{aligned}{\frac {d}{dt}}{\vec {r}}(t)\cdot {\vec {r}}(t)={\frac {d}{dt}}c^{2}\\{\vec {r}}(t)\cdot {\vec {r'}}(t)+{\vec {r'}}(t)\cdot {\vec {r}}(t)=0\\2{\vec {r}}(t)\cdot {\vec {r'}}(t)=0\\{\vec {r}}(t)\cdot {\vec {r'}}(t)=0\end{aligned}}$

• Therefore ${\vec {r}}(t)\perp {\vec {r'}}(t)$. $\blacksquare$

Licensing

Content obtained and/or adapted from:

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