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Transcript of VECTOR FUNCTIONS 13. VECTOR FUNCTIONS Later in this chapter, we are going to use vector functions to...
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VECTOR FUNCTIONSVECTOR FUNCTIONS
13
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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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13.2Derivatives 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 tt
dt h→
+ −= =
r r rr
Equation 1
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The geometric significance
of this definition is shown as
follows.
DERIVATIVE
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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
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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 C at P is defined to be
the line through P parallel 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 t
t=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 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
=
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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
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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
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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 kT
r
j k
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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
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We have:
and
DERIVATIVES Example 2
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
DERIVATIVES Example 2
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).
DERIVATIVES Example 2
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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, π/2).
DERIVATIVES Example 3
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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›
DERIVATIVES Example 3
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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
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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
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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
f is a real-valued function
DIFFERENTIATION RULES Theorem 3
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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
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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
df t f t f t
dt
⋅ = ⋅ + ⋅
× = × + ×
=
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
DIFFERENTIATION RULES
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The proof of Formula 4 follows.
The remaining are left as exercises.
DIFFERENTIATION RULES
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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 ii
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 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'
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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
dtt 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).
DIFFERENTIATION RULES
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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).
DIFFERENTIATION RULES
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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 functions—except that
the integral is a vector.
INTEGRALS
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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
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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
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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)dt
a
b
∫⎛⎝⎞⎠i+ g(t)dt
a
b
∫⎛⎝⎞⎠ j+ h(t)dt
a
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) dt for 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 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