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Lesson 7: Vector-valued functions

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We define vector-valued functions, whose image is a curve in the plane or space. We also show how to differentiate and integrate them
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Page 1: 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.

Page 2: Lesson 7: Vector-valued functions

Outline

Vector-valued functions

Derivatives of vector-valued functions

Integrals of vector-valued functions

Page 3: Lesson 7: 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).

Page 4: Lesson 7: 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).

Page 5: Lesson 7: Vector-valued functions

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.

Page 6: Lesson 7: Vector-valued functions

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).

Page 7: Lesson 7: Vector-valued functions

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)

Page 8: Lesson 7: Vector-valued functions

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).

Page 9: Lesson 7: Vector-valued functions

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).

Page 10: Lesson 7: Vector-valued functions

Outline

Vector-valued functions

Derivatives of vector-valued functions

Integrals of vector-valued functions

Page 11: Lesson 7: 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

Page 12: Lesson 7: Vector-valued functions

Example

Given r(t) = 〈t, cos 2t, sin 2t〉, find r′(t).

Answer〈1,−2 sin 2t, 2 cos(2t)〉

Page 13: Lesson 7: Vector-valued functions

Example

Given r(t) = 〈t, cos 2t, sin 2t〉, find r′(t).

Answer〈1,−2 sin 2t, 2 cos(2t)〉

Page 14: Lesson 7: Vector-valued functions

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)

Page 15: Lesson 7: Vector-valued functions

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)

Page 16: Lesson 7: Vector-valued functions

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).

Page 17: Lesson 7: Vector-valued functions

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)

Page 18: Lesson 7: Vector-valued functions

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).

Page 19: Lesson 7: Vector-valued functions

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)

Page 20: Lesson 7: Vector-valued functions

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))

Page 21: Lesson 7: Vector-valued functions

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

Page 22: Lesson 7: Vector-valued functions

Meet the Mathematician: Isaac Newton

I English, 1643–1727

I Professor at Cambridge(England)

I Philosophiae NaturalisPrincipia Mathematicapublished 1687

Page 23: Lesson 7: Vector-valued functions

Meet the Mathematician: Gottfried Leibniz

I German, 1646–1716

I Eminent philosopher aswell as mathematician

I Contemporarily disgracedby the calculus prioritydispute

Page 24: Lesson 7: Vector-valued functions

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〉

Page 25: Lesson 7: Vector-valued functions

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.

Page 26: Lesson 7: Vector-valued functions

If r(t) =⟨t3 + t, t4, t5

⟩, then r′(t) =

⟨3t2 + 1, 4t3, 5t4

⟩, which is

never 0.So this curve is smooth.

Page 27: Lesson 7: Vector-valued functions

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

Page 28: Lesson 7: Vector-valued functions

Outline

Vector-valued functions

Derivatives of vector-valued functions

Integrals of vector-valued functions

Page 29: Lesson 7: 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

Page 30: Lesson 7: Vector-valued functions

Example

Given r(t) = 〈t, cos 2t, sin 2t〉, find∫ π

0r(t) dt

Answer ⟨π2

2, 0, 0

Page 31: Lesson 7: Vector-valued functions

Example

Given r(t) = 〈t, cos 2t, sin 2t〉, find∫ π

0r(t) dt

Answer ⟨π2

2, 0, 0

Page 32: Lesson 7: Vector-valued functions

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.

Page 33: Lesson 7: Vector-valued functions

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.


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