Khanh: Created page with "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..."

2021-11-12T02:04:36Z

Created page with "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..."

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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.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
:'''Theorem 1:''' Let <math>\vec{u}(t) = (x_1(t), y_1(t), z_1(t))</math> and <math>\vec{v}(t) = (x_2(t), y_2(t), z_2(t))</math> be vector-valued functions that are differentiable for <math>t</math> in the interval <math>I</math>, let <math>k</math> be a scalar, and let <math>f(t)</math> be a real-valued function that is differentiable on <math>I</math>. Then:
:'''a)''' <math>( \vec{u}(t) + \vec{v}(t) )' = \vec{u'}(t) + \vec{v'}(t)</math> (Sum Rule).
:'''b)''' <math>( \vec{u}(t) - \vec{v}(t) )' = \vec{u'}(t) - \vec{v'}(t)</math> (Difference Rule).
:'''c)''' <math>(k \vec{u}(t))' = k \vec{u'} (t)</math> (Scalar Multiple Rule).
:'''d)''' <math>(f(t) \vec{u}(t))' = f'(t) \vec{u}(t) + f(t) \vec{u'}(t)</math> (Product Rule for Real-Valued and Vector-Valued Functions).
:'''e)''' <math>(\vec{u} \cdot \vec{v})' = \vec{u'}(t) \cdot \vec{v}(t) + \vec{u}(t) \cdot \vec{v'}(t)</math> (Dot Product Rule).
:'''f)''' <math>(\vec{u} \times \vec{v})' = \vec{u'}(t) \times \vec{v}(t) + \vec{u}(t) \times \vec{v'}(t)</math> (Cross Product Rule).
:'''g)''' <math>\vec{u}(f(t)) = f'(t) \vec{u'}(f(t))</math> (Chain Rule).
</blockquote>

*'''Proof of a)'''
<div style="text-align: center;"><math>\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}</math></div>

*'''Proof of c)'''
<div style="text-align: center;"> <math>\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}</math></div>

*'''Proof of d)'''
<div style="text-align: center;"><math>\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}</math></div>

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
'''Theorem 2:''' Let <math>\vec{r}(t) = (x(t), y(t), z(t))</math> be a vector-valued function that traces the curve <math>C</math> for <math>t \in I</math>. If <math>\| \vec{r}(t) \| = c</math> where <math>c</math> is a constant, then for all <math>t \in I</math>, <math>\vec{r}(t) \perp \vec{r'}(t)</math>.
</blockquote>

*'''Proof:''' We note that <math>\vec{r}(t) \cdot \vec{r}(t) = \| \vec{r}(t) \|^2 = c^2</math>. Taking the derivative of both sides of this equation and applying the dot product rule, we get that:

<div style="text-align: center;"><math>\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}</math></div>

*Therefore <math>\vec{r}(t) \perp \vec{r'}(t)</math>. <math>\blacksquare</math>

== Licensing ==
Content obtained and/or adapted from:
* [http://mathonline.wikidot.com/derivative-rules-for-vector-valued-functions Derivative Rules for Vector-Valued Functions, mathonline.wikidot.com] under a CC BY-SA license

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Khanh: Created page with "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..."