3.2 Calculus of vector-valued functions (Page 3/11)

Calculus volume 3 Page 3 / 11

□

Now for some examples using these properties.

Using the properties of derivatives of vector-valued functions

Given the vector-valued functions

r (t) = (6 t + 8) i + (4 t^{2} + 2 t - 3) j + 5 t k

and

u (t) = (t^{2} - 3) i + (2 t + 4) j + (t^{3} - 3 t) k,

calculate each of the following derivatives using the properties of the derivative of vector-valued functions.

$\frac{d}{d t} [r (t) \cdot u (t)]$
$\frac{d}{d t} [u (t) \times u^{'} (t)]$

We have $r^{'} (t) = 6 i + (8 t + 2) j + 5 k$ and $u^{'} (t) = 2 t i + 2 j + (3 t^{2} - 3) k .$ Therefore, according to property iv.:

$\begin{matrix} \frac{d}{d t} [r (t) \cdot u (t)] & = r^{'} (t) \cdot u (t) + r (t) \cdot u^{'} (t) \\ = (6 i + (8 t + 2) j + 5 k) \cdot ((t^{2} - 3) i + (2 t + 4) j + (t^{3} - 3 t) k) \\ + ((6 t + 8) i + (4 t^{2} + 2 t - 3) j + 5 t k) \cdot (2 t i + 2 j + (3 t^{2} - 3) k) \\ = 6 (t^{2} - 3) + (8 t + 2) (2 t + 4) + 5 (t^{3} - 3 t) \\ + 2 t (6 t + 8) + 2 (4 t^{2} + 2 t - 3) + 5 t (3 t^{2} - 3) \\ = 20 t^{3} + 42 t^{2} + 26 t - 16 . \end{matrix}$
First, we need to adapt property v. for this problem:

$\frac{d}{d t} [u (t) \times u^{'} (t)] = u^{'} (t) \times u^{'} (t) + u (t) \times u″ (t) .$

Recall that the cross product of any vector with itself is zero. Furthermore, $u″ (t)$ represents the second derivative of $u (t) :$

$u″ (t) = \frac{d}{d t} [u^{'} (t)] = \frac{d}{d t} [2 t i + 2 j + (3 t^{2} - 3) k] = 2 i + 6 t k .$

Therefore,

$\begin{matrix} \frac{d}{d t} [u (t) \times u^{'} (t)] & = 0 + ((t^{2} - 3) i + (2 t + 4) j + (t^{3} - 3 t) k) \times (2 i + 6 t k) \\ = | \begin{matrix} i & j & k \\ t^{2} - 3 & 2 t + 4 & t^{3} - 3 t \\ 2 & 0 & 6 t \end{matrix} | \\ = 6 t (2 t + 4) i - (6 t (t^{2} - 3) - 2 (t^{3} - 3 t)) j - 2 (2 t + 4) k \\ = (12 t^{2} + 24 t) i + (12 t - 4 t^{3}) j - (4 t + 8) k . \end{matrix}$

Got questions? Get instant answers now!

Given the vector-valued functions $r (t) = cos t i + sin t j - e^{2 t} k$ and $u (t) = t i + sin t j + cos t k,$ calculate $\frac{d}{d t} [r (t) \cdot r^{'} (t)]$ and $\frac{d}{d t} [u (t) \times r (t)] .$

$\frac{d}{d t} [r (t) \cdot r^{'} (t)] = 8 e^{4 t}$

$\begin{array}{l} \frac{d}{d t} [u (t) \times r (t)] \\ = - (e^{2 t} (cos t + 2 sin t) + cos 2 t) i + (e^{2 t} (2 t + 1) - sin 2 t) j + (t cos t + sin t - cos 2 t) k \end{array}$

Got questions? Get instant answers now!

Tangent vectors and unit tangent vectors

Recall from the Introduction to Derivatives that the derivative at a point can be interpreted as the slope of the tangent line to the graph at that point. In the case of a vector-valued function, the derivative provides a tangent vector to the curve represented by the function. Consider the vector-valued function $r (t) = cos t i + sin t j .$ The derivative of this function is $r^{'} (t) = - sin t i + cos t j .$ If we substitute the value $t = π / 6$ into both functions we get

r (\frac{π}{6}) = \frac{\sqrt{3}}{2} i + \frac{1}{2} j and r^{'} (\frac{π}{6}) = - \frac{1}{2} i + \frac{\sqrt{3}}{2} j .

The graph of this function appears in [link] , along with the vectors $r (\frac{π}{6})$ and $r^{'} (\frac{π}{6}) .$

This figure is the graph of a circle represented by the vector-valued function r(t) = cost i + sint j. It is a circle centered at the origin with radius of 1, and counter-clockwise orientation. It has a vector from the origin pointing to the curve and labeled r(pi/6). At the same point on the circle there is a tangent vector labeled “r’(pi/6)”. — The tangent line at a point is calculated from the derivative of the vector-valued function $r (t) .$

Notice that the vector $r^{'} (\frac{π}{6})$ is tangent to the circle at the point corresponding to $t = π / 6 .$ This is an example of a tangent vector to the plane curve defined by $r (t) = cos t i + sin t j .$

Definition

Let C be a curve defined by a vector-valued function r, and assume that $r^{'} (t)$ exists when $t = t_{0} .$ A tangent vector v at $t = t_{0}$ is any vector such that, when the tail of the vector is placed at point $r (t_{0})$ on the graph, vector v is tangent to curve C. Vector $r^{'} (t_{0})$ is an example of a tangent vector at point $t = t_{0} .$ Furthermore, assume that $r^{'} (t) \neq 0 .$ The principal unit tangent vector at t is defined to be

T (t) = \frac{r^{'} (t)}{‖ r^{'} (t) ‖},

provided $‖ r^{'} (t) ‖ \neq 0 .$

The unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. To calculate a unit tangent vector, first find the derivative $r^{'} (t) .$ Second, calculate the magnitude of the derivative. The third step is to divide the derivative by its magnitude.

Finding a unit tangent vector

Find the unit tangent vector for each of the following vector-valued functions:

$r (t) = cos t i + sin t j$
$u (t) = (3 t^{2} + 2 t) i + (2 - 4 t^{3}) j + (6 t + 5) k$

$\begin{matrix} First step: & r^{'} (t) & = & −sin t i + cos t j \\ Second step: & ‖ r^{'} (t) ‖ & = & \sqrt{{(− sin t)}^{2} + {(cos t)}^{2}} = 1 \\ Third step: & T (t) & = & \frac{r^{'} (t)}{‖ r^{'} (t) ‖} = \frac{−sin t i + cos t j}{1} = −sin t i + cos t j \end{matrix}$
$\begin{matrix} First step: & u^{'} (t) & = & (6 t + 2) i - 12 t^{2} j + 6 k \\ Second step: & ‖ u^{'} (t) ‖ & = & \sqrt{{(6 t + 2)}^{2} + {(−12 t^{2})}^{2} + 6^{2}} \\ = & \sqrt{144 t^{4} + 36 t^{2} + 24 t + 40} \\ = & 2 \sqrt{36 t^{4} + 9 t^{2} + 6 t + 10} \\ Third step: & T (t) & = & \frac{u^{'} (t)}{‖ u^{'} (t) ‖} = \frac{(6 t + 2) i - 12 t^{2} j + 6 k}{2 \sqrt{36 t^{4} + 9 t^{2} + 6 t + 10}} \\ = & \frac{3 t + 1}{\sqrt{36 t^{4} + 9 t^{2} + 6 t + 10}} i - \frac{6 t^{2}}{\sqrt{36 t^{4} + 9 t^{2} + 6 t + 10}} j + \frac{3}{\sqrt{36 t^{4} + 9 t^{2} + 6 t + 10}} k \end{matrix}$

Got questions? Get instant answers now!

Questions & Answers

what are components of cells

ofosola Reply

twugzfisfjxxkvdsifgfuy7 it

Sami

58214993

Sami

what is a salt

John

the difference between male and female reproduction

John

what is computed

IBRAHIM Reply

what is biology

IBRAHIM

what is the full meaning of biology

IBRAHIM

what is biology

Jeneba

what is cell

Kuot

425844168

Sami

what is biology

Inenevwo

what is cytoplasm

Emmanuel Reply

structure of an animal cell

Arrey Reply

what happens when the eustachian tube is blocked

Puseletso Reply

what's atoms

Achol Reply

discuss how the following factors such as predation risk, competition and habitat structure influence animal's foraging behavior in essay form

Burnet Reply

cell?

Kuot

location of cervical vertebra

KENNEDY Reply

What are acid

Sheriff Reply

define biology infour way

Happiness Reply

What are types of cell

Nansoh Reply

how can I get this book

Gatyin Reply

what is lump

Chineye Reply

what is cell

Maluak Reply

what is biology

Maluak

what is vertibrate

Jeneba

what's cornea?

Majak Reply

what are cell

Achol

Got questions? Join the online conversation and get instant answers!

Jobilize.com Reply

<< Chapter < Page Page > Chapter >>

Practice Key Terms 5

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 3' conversation and receive update notifications?

Ask

	Summer kitchens By Heather McAvoy Start Quiz
	Young Economist MCQ Test By Robert Murphy Start Test
	1 Arts Society: Theater 1 By Jonathan Long Start Quiz
	NCE Ch 08 Appraisal By Anh Dao Start Quiz
	14 Biology 14 DNA Structure and Function MCQ By OpenStax Start Quiz
	9 Neuroanatomy 09 The Auditory System By Stephen Voron Start Quiz
	3 AP 03 Cellular Level of Organization Essay By OpenStax Start Flashcards
©flickr: Dave	Infection Control By Cath Yu Start Quiz
	Vocabulary for "A Rose for Emily" By Bonnie Hurst Start Quiz
	Socialism Command Economy By Robert Murphy Start Test