Chap12_Sec2 [Compatibility Mode]

7/23/2019 Chap12_Sec2 [Compatibility Mode]

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VECTOR FUNCTIONS


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VECTOR FUNCTIONS

Later in this chapter, we are going to use

vector functions to describe the motion of

planets and other objects through space.

Here, we prepare the way by developing

the calculus of vector functions.


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12.2

Derivatives and Integrals

of Vector Functions

In this section, we will learn how to:

Develop the calculus of vector functions.

VECTOR FUNCTIONS


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The derivative r of a vector function

is defined in much the same way as for

real-valued functions.

DERIVATIVES


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if this limit exists.

DERIVATIVE

0

( ) ( )'( ) lim

h

d t h t t

dt h

+ = =

r r rr

Equation 1


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The geometric significance

of this definition is shown asfollows.

DERIVATIVE


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If the points Pand Q have position

vectors r(t) and r(t + h), then represents

the vector r(t + h) r(t).

This can thereforebe regarded as

a secant vector.

SECANT VECTOR

PQuuur


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If h > 0, the scalar multiple (1/h)(r(t + h) r(t))

has the same direction as r(t + h) r(t).

As h 0, it appears

that this vector

approaches a vector

that lies on the

tangent line.

DERIVATIVES


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For this reason, the vector r(t) is called

the tangent vector to the curve defined by r

at the point P,

provided:

r(t) exists

r(t) 0

TANGENT VECTOR


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The tangent line to Cat Pis defined to be

the line through Pparallel to the tangent

vector r(t).

TANGENT LINE


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We will also have occasion to consider

the unit tangent vector:

UNIT TANGENT VECTOR

'( )( )| '( ) |

tT tt

= rr


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The following theorem gives us

a convenient method for computing

the derivative of a vector function r:

Just differentiate each component of r.

DERIVATIVES


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If r(t) = f(t), g(t), h(t) = f(t) i + g(t)j + h(t) k,

where f, g, and h are differentiable functions,

then:

r(t) = f(t), g(t), h(t)

= f(t) i + g(t)j + h(t) k

DERIVATIVES Theorem 2


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DERIVATIVES Proof

0

0

0

0 0 0

'( )

1lim [ ( ) ( )]

1lim ( ), ( ), ( ), ( ), ( ), ( )

( ) ( ) ( ) ( ) ( ) ( )lim , ,

( ) ( ) ( ) ( ) ( ) ( )lim , lim , lim

t

t

t

t t t

t

t t t

t

f t t g t t h t t f t g t h tt

f t t f t g t t g t h t t h tt t t

f t t f t g t t g t h t t h t

t t

= +

= + + +

+ + + =

+ + + =

r

r r

'( ), '( ), '( )t

f t g t h t

=


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a. Find the derivative of

r(t) = (1 + t3) i + tetj + sin 2t k

b. Find the unit tangent vector at the point

where t= 0.

DERIVATIVES Example 1


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According to Theorem 2, we differentiate

each component of r:

r(t) = 3t2 i + (1 t)etj + 2 cos 2t k

DERIVATIVES Example 1 a


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As r(0) = i and r(0) =j + 2k, the unit tangent

vector at the point (1, 0, 0) is:

DERIVATIVES Example 1 b

'(0) 2

(0) | '(0) | 1 4

1 2

5 5

+

= = +

= +

r j k

T r

k


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For the curve ,

find r(t) and sketch the position vector r(1)

and the tangent vector r(1).


( ) (2 )t t t= + r i j


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We have:

and


1'( )2

1'(1)

2

tt

=

=

r i j

r i j


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The curve is a plane curve.

Elimination of the parameter from

the equations , y = 2 t gives:

y = 2 x2, x 0


x t=


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The position vector r(1) = i +j starts

at the origin.

The tangent vector r(1)

starts at the

corresponding point

(1, 1).



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Find parametric equations for the tangent line

to the helix with parametric equations

x = 2 cos t y = sin t z = t

at the point (0, 1, /2).



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The vector equation of the helix is:

r(t) = 2 cos t, sin t, t

Thus,

r

(t) =

2 sin t, cos t, 1



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The parameter value corresponding to

the point (0, 1, /2) is t= /2.

So, the tangent vector there is:

r(/2) = 2, 0, 1



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The tangent line is the line through

(0, 1, /2) parallel to the vector 2, 0, 1.

So, by Equations 2 in Section 12.5,

its parametric equations are:

DERIVATIVES

2 12

x t y z t

= = = +

Example 3


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The helix and the tangent line in

the Example 3 are shown.

DERIVATIVES


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Just as for real-valued functions,

the second derivative of a vector function r

is the derivative of r, that is, r = (r).

For instance, the second derivative

of the function in Example 3 is:

r(t) =2 cos t, sin t, 0

SECOND DERIVATIVE


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The next theorem shows that

the differentiation formulas for real-valued

functions have their counterparts for

vector-valued functions.

DIFFERENTIATION RULES


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Suppose:

u and v are differentiable vector functions

c is a scalar

fis a real-valued function

DIFFERENTIATION RULES Theorem 3


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Then,


1. [ ( ) ( )] ( ) ( )

2. [ ( )] ( )

3. [ ( ) ( )] '( ) ( ) ( ) ( )

dt t t t

dt

d c t c t dt

df t t f t t f t t

dt

+ = +

=

= +

u v u' v'

u u'

u u u'

Theorem 3


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DIFFERENTIATION RULES Theorem 3

( )

4. [ ( ) ( )] ( ) ( ) ( ) ( )

5. [ ( ) ( )] ( ) ( ) ( ) ( )

6. [ ( ( ))] '( ) ( ) (Chain Rule)

dt t t t t t

dt

dt t t t t t

dt

d f t f t f tdt

= +

= +

=

u v u' v u v'

u v u' v u v'

u u'


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This theorem can be proved either:

Directly from Definition 1

By using Theorem 2 and the corresponding

differentiation formulas for real-valued functions



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The proof of Formula 4 follows.

The remaining are left as exercises.



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Let

u(t) = f1(t), f2(t), f3(t)

v(t) = g1(t), g2(t), g3(t)

Then,

FORMULA 4 Proof

1 1 2 2 3 3

3

1

( ) ? ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )i i

i

t t

f t g t f t g t f t g t

f t g t=

= + +

=

u v


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So, the ordinary Product Rule gives:

FORMULA 4 Proof

3

1

3

1

3

1

3 3

1 1

[ ( ) ( )] ( ) ( )

[ ( ) ( )]

[ '( ) ( ) ( ) '( )]

'( ) ( ) ( ) '( )

( ) ( ) ( ) ( )

i i

i

i i

i

i i i i

i

i i i i

i i

d dt t f t g t

dt dt d

f t g tdt

f t g t f t g t

f t g t f t g t

t t t t

=

=

=

= =

=

=

= +

= +

= +

u v

u' v u v'


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Show that, if |r(t)| = c(a constant),

then r(t) is orthogonal to r(t) for all t.

DIFFERENTIATION RULES Example 4


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Since

r(t) r(t) = |r(t)|2 = c2

and c2 is a constant,

Formula 4 of Theorem 3 gives:

DIFFERENTIATION RULES Example 4

0 [ ( ) ( )]

'( ) ( ) ( ) '( )2 '( ) ( )

dt t

dt

t t t t

t t

=

= + =

r r

r r r r

r r


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Thus,

r(t) r(t) = 0

This says that r(t) is orthogonal to r(t).



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Geometrically, this result says:

If a curve lies on a sphere with center

the origin, then the tangent vector r(t) is

always perpendicular to the position vector r(t).



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The definite integral of a continuous vector

function r(t) can be defined in much the same

way as for real-valued functionsexcept that

the integral is a vector.

INTEGRALS


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However, then, we can express

the integral of rin terms of the integrals

of its component functions f, g, and h

as follows.

We use the notation of Chapter 5.

INTEGRALS


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INTEGRALS

*

1

* *

1 1

*

1

( )

lim ( )

lim ( ) ( )

( )

b

a

n

in

i

n n

i in

i i

n

i

i

t dt

t t

f t t g t t

h t t

=

= =

=

=

= +

+

r

r

i j

k


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Thus,

This means that we can evaluate an integral of a vector

function by integrating each component function.

INTEGRALS

r(t)dta

b

= f(t)dta

b

i + g(t)dta

b

j + h(t)dta

b

k


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We can extend the Fundamental Theorem

of Calculus to continuous vector functions:

Here, R is an antiderivative of r, that is, R(t) = r(t).

We use the notation r(t) dtfor indefinite integrals(antiderivatives).

INTEGRALS

]b

a(t) ( ) ( ) ( )

b

adt t b a= = r R R R


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If r(t) = 2 cos t i + sin tj + 2t k, then

where:

C is a vector constant of integration

INTEGRALS Example 5

r(t)dt = 2cos t dt( )i + sint dt( )j + 2t dt( )k= 2sin ti cos tj + t2k + C

2

2 200

2

( ) [2sin cos ]

2

4

t dt t t t

= +

= + +

r i j k

i j k

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Chap12_Sec2 [Compatibility Mode]