VECTOR FUNCTIONSVECTOR FUNCTIONS

13

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.

13.2Derivatives and Integrals

of Vector Functions

In this section, we will learn how to:

Develop the calculus of vector functions.

VECTOR FUNCTIONS

The derivative r’ of a vector function

is defined in much the same way as for

real-valued functions.

DERIVATIVES

if this limit exists.

DERIVATIVE

0

( ) ( )'( ) lim

h

d t h tt

dt h→

+ −= =

r r rr

Equation 1

The geometric significance

of this definition is shown as

follows.

DERIVATIVE

If the points P and Q have position

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

the vector r(t + h) – r(t).

This can therefore be regarded as a secant vector.

SECANT VECTOR

PQuuur

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

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

The tangent line to C at P is defined to be

the line through P parallel to the tangent

vector r’(t).

TANGENT LINE

We will also have occasion to consider

the unit tangent vector:

UNIT TANGENT VECTOR

'( )( )

| '( ) |

tT t

t=rr

The following theorem gives us

a convenient method for computing

the derivative of a vector function r:

Just differentiate each component of r.

DERIVATIVES

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

DERIVATIVES Proof

0

0

0

0 0 0

'( )

1lim [ ( ) ( )]

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

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

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

t

t

t

t t t

t

t t tt

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

t t t

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

t t

Δ →

Δ →

Δ →

Δ → Δ → Δ →

= +Δ −Δ

= ⎡ +Δ +Δ +Δ − ⎤⎣ ⎦Δ+Δ − +Δ − +Δ −

=Δ Δ Δ

+Δ − +Δ − +Δ −=

Δ Δ

r

r r

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

f t g t h tΔ

=

a. Find the derivative of

r(t) = (1 + t3) i + te–t j + sin 2t k

b. Find the unit tangent vector at the point

where t = 0.

DERIVATIVES Example 1

According to Theorem 2, we differentiate

each component of r:

r’(t) = 3t2 i + (1 – t)e–t j + 2 cos 2t k

DERIVATIVES Example 1 a

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 kT

r

j k

For the curve ,

find r’(t) and sketch the position vector r(1)

and the tangent vector r’(1).

DERIVATIVES Example 2

( ) (2 )t t t= + −r i j

We have:

and

DERIVATIVES Example 2

1'( )

2

1'(1)

2

tt

= −

= −

r i j

r i j

The curve is a plane curve.

Elimination of the parameter from

the equations , y = 2 – t gives:

y = 2 – x2, x ≥ 0

DERIVATIVES Example 2

x t=

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

at the origin.

The tangent vector r’(1)

starts at the

corresponding point

(1, 1).

DERIVATIVES Example 2

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, π/2).

DERIVATIVES Example 3

The vector equation of the helix is:

r(t) = ‹2 cos t, sin t, t›

Thus,

r’(t) = ‹–2 sin t, cos t, 1›

DERIVATIVES Example 3

The parameter value corresponding to

the point (0, 1, π/2) is t = π/2.

So, the tangent vector there is:

r’(π/2) = ‹–2, 0, 1›

DERIVATIVES Example 3

The tangent line is the line through

(0, 1, π/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 tπ

=− = = +

Example 3

The helix and the tangent line in

the Example 3 are shown.

DERIVATIVES

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

The next theorem shows that

the differentiation formulas for real-valued

functions have their counterparts for

vector-valued functions.

DIFFERENTIATION RULES

Suppose:

u and v are differentiable vector functions

c is a scalar

f is a real-valued function

DIFFERENTIATION RULES Theorem 3

Then,

DIFFERENTIATION RULES

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

2. [ ( )] ( )

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

dt t t t

dt

dc 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

DIFFERENTIATION RULES Theorem 3

( )

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

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

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

dt t t t t t

dt

dt t t t t t

dt

df t f t f t

dt

⋅ = ⋅ + ⋅

× = × + ×

=

u v u' v u v'

u v u' v u v'

u u'

This theorem can be proved either:

Directly from Definition 1

By using Theorem 2 and the corresponding differentiation formulas for real-valued functions

DIFFERENTIATION RULES

The proof of Formula 4 follows.

The remaining are left as exercises.

DIFFERENTIATION RULES

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 ii

t t

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

f t g t=

= + +

=∑

u v

So, the ordinary Product Rule gives:

FORMULA 4 Proof

3

1

3

1

3

1

3 3

1 1

[ ( ) ( )] ( ) ( )

[ ( ) ( )]

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

'( ) ( ) ( ) '( )

( ) ( ) ( ) ( )

i ii

i ii

i i i ii

i i i ii i

d dt t f t g t

dt dt

df t g t

dt

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'

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

then r’(t) is orthogonal to r(t) for all t.

DIFFERENTIATION RULES Example 4

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

dtt t t t

t t

= ⋅

= ⋅ + ⋅= ⋅

r r

r r r r

r r

Thus,

r’(t) ∙ r(t) = 0

This says that r’(t) is orthogonal to r(t).

DIFFERENTIATION RULES

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).

DIFFERENTIATION RULES

The definite integral of a continuous vector

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

way as for real-valued functions—except that

the integral is a vector.

INTEGRALS

However, then, we can express

the integral of r in terms of the integrals

of its component functions f, g, and h

as follows.

We use the notation of Chapter 5.

INTEGRALS

INTEGRALS

*

1

* *

1 1

*

1

( )

lim ( )

lim ( ) ( )

( )

b

a

n

in

i

n n

i in

i i

n

ii

t dt

t t

f t t g t t

h t t

→ ∞=

→ ∞= =

=

= Δ

⎡⎛ ⎞ ⎛ ⎞= Δ + Δ⎢⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣

⎤⎛ ⎞+ Δ ⎥⎜ ⎟⎝ ⎠ ⎦

∫

∑

∑ ∑

∑

r

r

i j

k

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)dt

a

b

∫⎛⎝⎞⎠i+ g(t)dt

a

b

∫⎛⎝⎞⎠ j+ h(t)dt

a

b

∫⎛⎝⎞⎠k

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) dt for indefinite integrals (antiderivatives).

INTEGRALS

]b

a(t) ( ) ( ) ( )

b

adt t b a= = −∫ r R R R

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

where: C is a vector constant of integration

INTEGRALS Example 5

r(t)dt∫ = 2costdt∫( )i+ sintdt∫( ) j+ 2tdt∫( )k

=2sinti−cost j+ t2k +C

2 2 200

2

( ) [2sin cos ]

24

t dt t t tπ π

π

= − +

= + +

∫ r i j k

i j k