Lesson 7: Vector-valued functions
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Transcript of Lesson 7: Vector-valued functions
Sections 10.1–2Vector-Valued Functions and Curves in Space
Derivatives and Integrals of Vector-ValuedFunctions
Math 21a
February 20, 2008
Announcements
I Problem Sessions:I Monday, 8:30, SC 103b (Sophie)I Thursday, 7:30, SC 103b (Jeremy)
I Office hours Wednesday 2/20 2–4pm SC 323.
Outline
Vector-valued functions
Derivatives of vector-valued functions
Integrals of vector-valued functions
Recall
If P and Q are two points in the plane, u =−→OP, and v =
−→OQ,
then the line through P and Q can be parametrized as
r(t) = tv + (1− t)u
This is a function whose domain is R and whose range is a subsetof R3 (the line).
Recall
If P and Q are two points in the plane, u =−→OP, and v =
−→OQ,
then the line through P and Q can be parametrized as
r(t) = tv + (1− t)u
This is a function whose domain is R and whose range is a subsetof R3 (the line).
DefinitionA vector-valued function or vector function is a function r(t)whose domain is a set of real numbers and whose range is a set ofvectors.
I We can split r(t) into its components
r(t) = f (t)i + g(t)j + h(t)k
Then f , g , and h are called the component functions of r.
I The range of r is a curve in R2 or R3.
Example
Given the plane curve described by the vector equation
r(t) = sin(t)i + 2 cos(t)j
(a) Sketch the plane curve.
(b) Find r′(t)
(c) Sketch the position vector r(π/4) and the tangent vectorr′(π/4).
r(t) = r(t) = sin(t)i + 2 cos(t)j
r′(t) = cos(t)i− 2 sin(t)j
t r(t)
0 2jπ/2 iπ −2j
3π/2 −i2π 2j
x
y
r(π/4)
r′(π/4)
Curves and functions
Example
Two particles travel along the space curves
r1(t) = 〈3t, 7t − 12, t2〉 r2(t) = 〈4t − 3, t2, 5t − 6〉
Do the particles collide?
AnswerYes. r1(3) = r2(3).
Curves and functions
Example
Two particles travel along the space curves
r1(t) = 〈3t, 7t − 12, t2〉 r2(t) = 〈4t − 3, t2, 5t − 6〉
Do the particles collide?
AnswerYes. r1(3) = r2(3).
Outline
Vector-valued functions
Derivatives of vector-valued functions
Integrals of vector-valued functions
Derivatives of vector-valued functions
DefinitionLet r be a vector function.
I The limit of r at a point a is defined componentwise:
limt→a
r(t) =⟨
limt→a
f (t), limt→a
g(t), limt→a
h(t)⟩
I The derivative of r is defined in much the same way as it isfor real-valued functions:
dr
dt= r′(t) = lim
h→0
r(t + h)− r(t)
h
Example
Given r(t) = 〈t, cos 2t, sin 2t〉, find r′(t).
Answer〈1,−2 sin 2t, 2 cos(2t)〉
Example
Given r(t) = 〈t, cos 2t, sin 2t〉, find r′(t).
Answer〈1,−2 sin 2t, 2 cos(2t)〉
FactIf r(t) = 〈f (t), g(t), h(t)〉, then
r′(t) =⟨f ′(t), g ′(t), h′(t)
⟩
Proof.Follow your nose:
r′(t) = limh→0
r(t + h)− r(t)
h
= limη→0
1
η[〈f (t + η), g(t + η), h(t + η)〉 − 〈f (t), g(t), h(t)〉]
= limη→0
1
η[〈f (t + η)− f (t), g(t + η)− g(t), h(t + η)− h(t)〉]
=
⟨limη→0
f (t + η)− f (t)
η, limη→0
g(t + η)− g(t)
η, limη→0
h(t + η)− h(t)
η
⟩=⟨f ′(t), g ′(t), h′(t)
⟩
FactIf r(t) = 〈f (t), g(t), h(t)〉, then
r′(t) =⟨f ′(t), g ′(t), h′(t)
⟩Proof.Follow your nose:
r′(t) = limh→0
r(t + h)− r(t)
h
= limη→0
1
η[〈f (t + η), g(t + η), h(t + η)〉 − 〈f (t), g(t), h(t)〉]
= limη→0
1
η[〈f (t + η)− f (t), g(t + η)− g(t), h(t + η)− h(t)〉]
=
⟨limη→0
f (t + η)− f (t)
η, limη→0
g(t + η)− g(t)
η, limη→0
h(t + η)− h(t)
η
⟩=⟨f ′(t), g ′(t), h′(t)
⟩
Example
Given the plane curve described by the vector equation
r(t) = sin(t)i + 2 cos(t)j
(a) Sketch the plane curve.
(b) Find r′(t)
(c) Sketch the position vector r(π/4) and the tangent vectorr′(π/4).
r(t) = r(t) = sin(t)i + 2 cos(t)j
r′(t) = cos(t)i− 2 sin(t)j
t r(t)
0 2jπ/2 iπ −2j
3π/2 −i2π 2j
x
y
r(π/4)
r′(π/4)
Example
Given the plane curve described by the vector equation
r(t) = sin(t)i + 2 cos(t)j
(a) Sketch the plane curve.
(b) Find r′(t)
(c) Sketch the position vector r(π/4) and the tangent vectorr′(π/4).
r(t) = r(t) = sin(t)i + 2 cos(t)j
r′(t) = cos(t)i− 2 sin(t)j
t r(t)
0 2jπ/2 iπ −2j
3π/2 −i2π 2j
x
y
r(π/4) r′(π/4)
Rules for differentiation
TheoremLet u and v be differentiable vector functions, c a scalar, and f areal-valued function. Then:
1.d
dt[u(t) + v(t)] = u′(t) + v′(t)
2.d
dt[cu(t)] = cu′(t)
3.d
dt[f (t)u(t)] = f ′(t)u(t) + f (t)u′(t)
4.d
dt[u(t) · v(t)] = u′(t) · v(t) + u(t) · v′(t)
5.d
dt[u(t)× v(t)] = u′(t)× v(t) + u(t)× v′(t)
6.d
dt[u(f (t))] = f ′(t)u′(f (t))
Leibniz rule for cross products
Let u = 〈f1(t), g1(t), h1(t)〉 and v = 〈f2(t), g2(t), h2(t)〉. The firstcomponent of u(t)× v(t) is
(u(t)× v(t)) · i = g1h2 − g2h1
Differentiating gives
(u(t)× v(t))′ · i = g ′1h2 + g1h′2 − g ′2h1 − g2h′1
= g ′1h2 − g2h′1 + g1h′2 − g ′2h1
= (u′(t)× v(t)) · i + (u(t)× v′(t)) · i=(u′(t)× v(t) + u(t)× v′(t)
)· i
Meet the Mathematician: Isaac Newton
I English, 1643–1727
I Professor at Cambridge(England)
I Philosophiae NaturalisPrincipia Mathematicapublished 1687
Meet the Mathematician: Gottfried Leibniz
I German, 1646–1716
I Eminent philosopher aswell as mathematician
I Contemporarily disgracedby the calculus prioritydispute
Smooth curves
Example
Which of the following curves are smooth? That is, which curvessatisfy the property that r′(t) 6= 0 for all t?
(a) r(t) = 〈t3, t4, t5〉(b) r(t) = 〈t3 + t, t4, t5〉(c) r(t) = 〈cos3 t, sin3 t〉
The first curve r(t) =⟨t3, t4, t5
⟩has r′(t) =
⟨3t2, 4t3, 5t4
⟩, and is
not smooth at t = 0.
x y
z
Projecting r(t) onto the yz-plane gives y = z4/5, which is notdifferentiable at 0.
If r(t) =⟨t3 + t, t4, t5
⟩, then r′(t) =
⟨3t2 + 1, 4t3, 5t4
⟩, which is
never 0.So this curve is smooth.
If r(t) =⟨cos3 t, sin3 t
⟩, then
r′(t) =⟨−3 cos2(t) sin(t), 3 sin2(t) cos(t)
⟩. This is 0 when
cos t = 0 or sin t = 0, i.e., when t = π/2, π, 3π/2, 2π.
x
y
Outline
Vector-valued functions
Derivatives of vector-valued functions
Integrals of vector-valued functions
Integrals of vector-valued functions
DefinitionLet r be a vector function defined on [a, b]. For each whole numbern, divide the interval [a, b] into n pieces of equal width ∆t.Choose a point t∗i on each subinterval and form the Riemann sum
Sn =n∑
i=1
r(t∗i ) ∆t
Then define∫ b
ar(t) dt = lim
n→∞Sn = lim
n→∞
n∑i=1
r(t∗i ) ∆t
= limn→∞
[n∑
i=1
f (t∗i ) ∆ti +n∑
i=1
g(t∗i ) ∆tj +n∑
i=1
h(t∗i ) ∆tk
]
=
(∫ b
af (t) dt
)i +
(∫ b
ag(t) dt
)j +
(∫ b
ah(t) dt
)k
Example
Given r(t) = 〈t, cos 2t, sin 2t〉, find∫ π
0r(t) dt
Answer ⟨π2
2, 0, 0
⟩
Example
Given r(t) = 〈t, cos 2t, sin 2t〉, find∫ π
0r(t) dt
Answer ⟨π2
2, 0, 0
⟩
FTC for vector functions
Theorem (Second Fundamental Theorem of Calculus)
If r(t) = R′(t), then ∫ b
ar(t) dt = R(t)
Proof.Let R(t) = 〈F (t),G (t),H(t)〉. To say that R′(t) = r(t) meansthat F ′ = f , G ′ = g , and H ′ = h. That and the componentwise
definition of
∫ b
ar(t) dt are all you need.
FTC for vector functions
Theorem (Second Fundamental Theorem of Calculus)
If r(t) = R′(t), then ∫ b
ar(t) dt = R(t)
Proof.Let R(t) = 〈F (t),G (t),H(t)〉. To say that R′(t) = r(t) meansthat F ′ = f , G ′ = g , and H ′ = h. That and the componentwise
definition of
∫ b
ar(t) dt are all you need.