Calculus III 2

Vector-Valued Functions

Calculus-III

Dr. Imran Rashid

Department of Mathematics

COMSATS Institute of Information Technology, Islamabad,

Pakistan

[email protected]

Calculus-III, BCE-III Spring 2013


Outline


Parametric Curves in 3−SpaceCalculus of Vector-Valued Functions

Change of Parameter; Arc Length

Unit Tangent, Normal, and Binormal Vectors

Curvature



Parametric Curves in 3−SpaceCalculus of Vector-Valued FunctionsChange of Parameter; Arc LengthUnit Tangent, Normal, and Binormal VectorsCurvature





Parametric Curves in 3−Space




Overview

When a body (or objet) travels through space, the equation

x = f (t), y = g(t), and z = h(t) that the body's coordinates as

functions of time serve as parametric equations for the body's

motion and path. With vector notation, we condense these into a

single equation

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

that gives the body's position as a vector function of time. For an

object moving in xy−plane, the component function h(t) is zero for

all time (that is, identically zero).




Overview

p The position vector r =⇀

OP of a particle moving through space is a function

of time.




Overview

The point (x , y , z) = (f (t), g(t), h(t)), t ∈ I , makes up the curve

in space that we call the particle's path.

The equations

x = f (t), y = g(t), and z = h(t)

parameterize the curve.

A curve in space can also be represented in vector form. The vector

r =⇀OP = f (t)i+ g(t)j+ h(t)k

from the origin to the particle's position P(f (t), g(t), h(t)) at time

t is the particle's position vector.




Overview

The functions f , g , and h are the component functions

(components) of the position vector.

We think of the particle's path as the curve traced by r during the

time interval I .

The equations

x = f (t), y = g(t), and z = h(t)

is a vector function or vector-valued function on a domain set D is

a rule that assigns a vector in space to each element in D.




Graphing Helix

The vector function r(t) = (cos t)i+ (sin t)j+ tk is de�ned for all

real values of t.

The upper half of the helix r = (cos t)i + (sin t)j + tk




Domain of Vector-Valued Function

Example: The component functions of

r(t) = 〈t, t2, t3〉 = ti+ t2j+ t3k

are

x(t) = t, y(t) = t2, z(t) = t3

The domain of r(t) is the set of allowable values for t. If the r(t) isde�ned in terms of component functions and the domain is not

speci�ed explicitly, then it will be understood that the domain is

the intersection of the natural domains of the component functions;

that is called the natural domain of r(t).




Domain of Vector-Valued Function

Example: Find the natural domain of

r(t) = 〈ln |t − 1|, et ,√t〉 = ln |t − 1|i+ et j+

√tk

Solution: The natural domains of the component functions

x(t) = ln |t − 1|, y(t) = et , z(t) =√t

are

(−∞, 1] ∪ (1,+∞), (−∞,+∞), [0,+∞)

respectively. The intersection of these sets is

[0, 1) ∪ (1,+∞)

. So the natural domain of r(t) is

0 ≤ t < 1 or t > 1




Vector form of a line segment

The line through the terminal point of r0 and is parallel to the

vector v:

r = r0 + tv.

The two-point vector form of a line is

r = r0 + t(r1 − r0) or r = (1− t)r0 + tr1.

The line segment traced from r0 to r1

r = (1− t)r0 + tr1 (0 ≤ t ≤ 1)




Vector form of a line segment

The line through the terminal point of r0 and is parallel to the

vector v:

r = r0 + tv.

The two-point vector form of a line is

r = r0 + t(r1 − r0) or r = (1− t)r0 + tr1.

The line segment traced from r0 to r1

r = (1− t)r0 + tr1 (0 ≤ t ≤ 1)Calculus-III, BCE-III Spring 2013



Calculus of Vector-Valued Functions




Limit

Let r(t) = f (t)i+ g(t)j+ h(t)k be a vector function and L a

vector. We say that r has limit L as t approaches t0 and we write

limt→t0

r(t) = L if and only if limt→t0

‖r(t)− L‖ = 0

Theorem: If r(t) =⟨x(t), y(t), z(t)

⟩= x(t)i+ y(t)j+ z(t)k, then

limt→t0

r(t) =⟨limt→t0

x(t), limt→t0

y(t), limt→t0

z(t)⟩

= limt→t0

x(t)i+ limt→t0

y(t)j+ limt→t0

z(t)k

provided the limits of the component functions exist. Conversely,

the limits of the component functions exist provided r(t)approaches a limiting vector as t approaches t0.




Continuity

De�nition: A vector valued function r(t) is continuous at a point

t = t0 in its domain if limt→t0

r(t) = r(t0). The function is continuous

at every point in its domain.

Note

I As in the case for real-valued functions, we say that r(t) iscontinuous on an interval I if it is continuous at each point of I

I A vector valued function r(t) is continuous at t = t0 if and

only if its component functions are continuous at t = t0.




Derivative

If r(t) is a vector-valued function, we de�ne the derivative of r with

respect to t to be the vector-valued function r′ given by

r′(t) = limh→0

r(t + h)− r(t)

h

The domain of r′ consists of all values of t in the domain of r(t) forwhich the limit exists.

The function r(t) is di�erentiable at t if the above limit exists.

r′(t) is a vector that is tangent to the curve C at the tip of r(t)and points in the direction of increasing parameter. The standard

notations for derivative of r(t) are

d

dt[r(t)],

dr

dt, r′(t), or r




Derivative



r′(t) = limh→0

r(t + h)− r(t)

h



r′(t) is a vector that is tangent to the curve C at the tip of r(t)and points in the direction of increasing parameter.

The standard


d

dt[r(t)],

dr

dt, r′(t), or r




Derivative



r′(t) = limh→0

r(t + h)− r(t)

h



r′(t) is a vector that is tangent to the curve C at the tip of r(t)and points in the direction of increasing parameter. The standard


d

dt[r(t)],

dr

dt, r′(t), or r




Derivative Geometrically




Geometric Interpretation of the Derivative

Suppose that C is the graph of a vector-valued function r(t) andthat r′(t) exists and is nonzero for a given value of t. If the vector

r′(t) is positioned with its initial point at the terminal point of the

radius vector r(t), then r′(t) is tangent to C and points in the

direction of increasing parameter.

Theorem: If r(t) is a vector-valued function, then r is di�erentiable

at t if and only if each of its component functions is di�erentiable

at t, in which case the component functions of r′(t) are the

derivatives of the corresponding component functions of r(t).




Derivative of a Vector Function

Let u and v be di�erentiable vector functions of t, C a constant

vector, k any scalar, and f any di�erentiable scalar function.

1. Constant Function Ruled

dtC = 0

2. Scalar Multiple Rule

d

dt[ku(t)] = ku′(t)

d

dt[f (t)u(t)] = f ′(t)u(t) + f (t)u′(t)

3. Sum Ruled

dt[u(t) + v(t)] = u′(t) + v′(t)




Derivative Rules

4. Di�erence Rule

d

dt[u(t)− v(t)] = u′(t)− v′(t)

5. Dot Product Rule

d

dt[u(t) · v(t)] = u′(t) · v(t) + u(t) · v′(t)

6. Cross Product Rule

d

dt[u(t)× v(t)] = u′(t)× v(t) + u(t)× v′(t)

7. Chain Ruled

dt[u(f (t))] = f ′(t)u′(f (t))




Tangent Lines

De�nition: Let P be a point on the graph of a vector-valued

function r(t), and let r(t0) be the radius vector from the origin to

P . If r′(t0) exists and r′(t0) 6= 0, then we call r′(t0) a tangent

vector to the graph of r(t) at r(t0), and we call the line through P

that is parallel to the tangent line to the graph of r(t) at r(t0)

The tangent line to the graph of r(t) at r0 is given by

r = r0 + tv0




Tangent Lines

De�nition: Let P be a point on the graph of a vector-valued

function r(t), and let r(t0) be the radius vector from the origin to

P . If r′(t0) exists and r′(t0) 6= 0, then we call r′(t0) a tangent

vector to the graph of r(t) at r(t0), and we call the line through P

that is parallel to the tangent line to the graph of r(t) at r(t0)

The tangent line to the graph of r(t) at r0 is given by

r = r0 + tv0Calculus-III, BCE-III Spring 2013



Derivatives of Dot and Cross Products

Theorem: If r(t) is a vector-valued function and ‖r(t)‖ is constantfor all t, then

r(t) · r′(t) = 0

that is , r(t) and r′(t) are orthogonal vectors for all t.

Proof: Asd

dt[r1(t) · r2(t)] = r1(t) ·

dr2

dt+

dr1

dt· r2(t)

It follows with r1 = r2 = r that

d

dt[r(t) · r(t)] = r(t) · dr

dt+

dr

dt· r(t)

d

dt[‖r(t)‖2] = 2r(t) · dr

dt

But ‖r(t)‖2 is constant, so its derivative is zero. Thus

2r(t) · drdt

= 0




Antiderivatives and De�nite Integrals

An antiderivative for a vector-valued function r(t) is avector-valued function R(t) such that

R′(t) = r(t)

Using integral notation∫r(t)dt = R(t) + C

where C represents an arbitrary constant vector.

Like di�erentiation of a vector-valued function, antiderivative is also

performed component wise.




Rules of Integration

Theorem: Let r(t), r1(t), and r2(t) be vector-valued functions that

are continuous on the interval a ≤ t ≤ b, and let k be a scalar.

Then the following rules of integration hold:

(a)

∫b

a

kr(t)dt = k

∫b

a

r(t)dt

(b)

∫b

a

[r1(t) + r2(t)]dt =

∫b

a

r1(t)dt +

∫b

a

r2(t)dt

(c)

∫b

a

[r1(t)− r2(t)]dt =

∫b

a

r1(t)dt −∫

b

a

r2(t)dt




Change of Parameter; Arc Length




Smooth Parametrization

A curve represented by r(t) is smoothly parameterized by r(t), orthat r(t) is a smooth function of t if r′(t) is continuous andr′(t) 6= 0 for any allowable value of t.




Arc Length

The arc length L of parametric curve

x = x(t), y = y(t), z = z(t) (a ≤ t ≤ b)

is

L =

∫b

a

√(dxdt

)2+(dydt

)2+(dzdt

)2dt

that can have vector form by letting

r(t) = x(t)i+ y(t)j+ z(t)k

dr

dt=

dx

dti+

dy

dtj+

dz

dtk

∥∥∥drdt

∥∥∥ =

√(dxdt

)2+(dydt

)2+(dzdt

)2




Arc Length

The arc length L of parametric curve

x = x(t), y = y(t), z = z(t) (a ≤ t ≤ b)

is

L =

∫b

a

√(dxdt

)2+(dydt

)2+(dzdt

)2dt

that can have vector form by letting

r(t) = x(t)i+ y(t)j+ z(t)k

dr

dt=

dx

dti+

dy

dtj+

dz

dtk

∥∥∥drdt

∥∥∥ =

√(dxdt

)2+(dydt

)2+(dzdt

)2Calculus-III, BCE-III Spring 2013



Arc Length

Theorem: If C is the graph of a smooth vector-valued function r(t),then its arc length L from t = a to t = b is

L =

∫b

a

∥∥∥drdt

∥∥∥ dt




Using Arc Length as a Parameter

Step 1. Select an arbitrary point on the curve C to

serve as a reference point.

Step 2. Starting from the reference point, choose one

direction along the curve to be the positive direction

and the other to be the negative direction.

Step 3. If P is a point on the curve, let s be the signed

arc length along C from the reference point to P ,

where s is positive if P is in the positive direction from

the reference point and s is negative if P is in the

negative direction.





A parametric representation of a curve with arc length as the

parameter is called an arc length parametrization of the curve.

Note that a given curve will generally have in�nitely many di�erent

arc length parameterizations, since the reference point and

orientation can be chosen arbitrarily.





Example: Find the arc length parametrization of the circle

x2 + y2 = a2 with counterclockwise orientation and (a, 0) as thereference point.

Solution: x = a cos t, y = a sin t (0 ≤ t ≤ 2π)

s = at or t =s

a

x = a cos(s

a), y = a sin(

s

a) (0 ≤ s ≤ 2πa)




Change of Parameter

A change of parameter in a vector-valued function r(t) is asubstitution t = g(τ) that produces a new vector-valued function

r(g(τ)) having the same graph as r(t), but possibly, traceddi�erently as the parameter τ increases.




Change of Parameter

Example: Find a change of parameter t = g(τ) for the circle

r(t) = cos ti+ sin tj (0 ≤ t ≤ 2π)

such that

(a) the circle is traced counterclockwise as τ increases over the

interval [0, 1];(b) the circle is traced clockwise as τ increases over the interval

[0, 1].

Solution (a): If we choose g to be an increasing function, then it

will follow from the relationship t = g(τ) that t increases when τincreases, ensuring that the circle will be traced counterclockwise as

τ increases. We also want to choose g so that t increases from

0 to 2π as τ increases from 0 to 1. A simple choice of g that

satis�es all of the criteria is the linear function t = g(τ) = 2πτ ,which is the desired change of parameter.




Change of Parameter

Example: Find a change of parameter t = g(τ) for the circle

r(t) = cos ti+ sin tj (0 ≤ t ≤ 2π)

such that

(a) the circle is traced counterclockwise as τ increases over the

interval [0, 1];(b) the circle is traced clockwise as τ increases over the interval

[0, 1].

Solution (a): If we choose g to be an increasing function, then it

will follow from the relationship t = g(τ) that t increases when τincreases, ensuring that the circle will be traced counterclockwise as

τ increases. We also want to choose g so that t increases from

0 to 2π as τ increases from 0 to 1. A simple choice of g that

satis�es all of the criteria is the linear function t = g(τ) = 2πτ ,which is the desired change of parameter.Calculus-III, BCE-III Spring 2013



Change of Parameter

The resulting representation of the circle in terms of τ is

r(g(τ)) = cos 2πτ i+ sin 2πτ j (0 ≤ τ ≤ 1)

Solution (b): To ensure that the circle is traced clockwise, we will

choose g to be a decreasing function such that t decreases from

2π to 0 as τ increases from 0 to 1. A simple choice of g that

achieves this is the linear function t = g(τ) = 2π(1− τ). Theresulting representation of the circle in terms of τ is

r(g(τ)) = cos(2π(1− τ))i+ sin(2π(1− τ))j (0 ≤ τ ≤ 1)

r(g(τ)) = cos 2πτ i− sin 2πτ j (0 ≤ τ ≤ 1)




Change of Parameter

When making a change of parameter t = g(τ) in a vector-valued

function r(t), it will be important to ensure that the new

vector-valued function r(g(τ)) is smooth if r(t) is smooth. The

following theorem establishes the condition under which it happens.

Theorem (Chain Rule): Let r(t) be a vector-valued function that is

di�erentiable with respect to t. If t = g(τ) is a change of

parameter in which g is di�erentiable with respect to τ , thenr(g(τ)) is di�erentiable with respect to τ and

dr

dτ=

dr

dt

dt

dτ




Change of Parameter

A change of parameter t = g(τ) in which r(g(τ)) is smooth if r(t)is smooth is called a smooth change of parameter.

It follows that t = g(τ) will be a smooth change of parameter if dt

dτ

is continuous and dt

dτ 6= 0 for all values of τ , since these conditions

imply that dr

dτ is continuous and nonzero if dr

dtis continuous and

nonzero.

→ If dt

dτ > 0 for all τ then it is a positive change of parameter.

→ If dt

dτ < 0 for all τ then it is a negative change of parameter.

A positive change of parameter preserves the orientation of a

parametric curve, and a negative change of parameter reverses it.




Finding Arc Length Parameterizations

Theorem: Let C be the graph of a smooth vector-valued function

r(t) and let r(t0) be any point on C . Then the following formula

de�nes a positive change of parameter from t to s, where s is an

arc length parameter having r(t0) as its reference point

s =

∫t

t0

∥∥∥drdu

∥∥∥ duIn component form

s =

∫t

t0

√(dxdu

)2+(dydu

)2+(dzdu

)2du




Finding Arc Length Parameterizations

Example: Find the arc length parametrization of the circular helix

r = cos ti+ sin tj+ tk

that has reference point r(0) = (1, 0, 0) and the same orientation as

the given helix.

Example: A bug walks along the trunk of a tree following a path

modeled by the circular helix in last example. The bug starts at the

reference point (1, 0, 0) and walks up the helix for a distance of 10

units. what are the bug's �al coordinates?




Properties of Arc Length Parameterizations

Theorem:

(a) If C is the graph of a smooth vector-valued function r(t), wheret is a general parameter, and if s is the arc length parameter for C

de�ned by formula above, then for every value of t the tangent

vector has length ∥∥∥drdt

∥∥∥ =ds

dt

(b) If C is the graph of a smooth vector-valued function r(s),where s is an arc length parameter, then for every value of s the

tangent vector to C has length∥∥∥drds

∥∥∥ = 1




Properties of Arc Length Parameterizations

(c) If C is the graph of a smooth vector-valued function r(t), and if∥∥∥drdt

∥∥∥ = 1

for every value of t, then for any value of t0 in the domain of r, the

parameter s = t − t0 is an arc length parameter that has its

reference point at the point on C where t = t0.




Unit Tangent, Normal, and Binormal Vectors




Unit Tangent Vectors

If C is the graph of a smooth

vector-valued function r, then

the vector r′ is nonzero, tan-

gent to C , and points in the di-

rection of increasing parameter.

Thus by normalizing r′ we ob-

tain a unit vector

T(t) =r′(t)

‖r′(t)‖

that is tangent to C and points

in the direction of increasing pa-

rameter. T(t) is called unit tan-

gent vector to C at point t.Calculus-III, BCE-III Spring 2013



Unit Normal VectorsIf a vector-valued function r(t) has constant norm, then r(t) andr′(t) are orthogonal vectors. In particular, T(t) has constant norm1, so T(t) and T′(t) are orthogonal vectors. This implies that

T′(t) is perpendicular to the tangent line to C at t, that is T′(t) isnormal to C at t. It follows that if T′(t) 6= 0, and normalizing

T′(t), we obtain a unit vector

N(t) =T′(t)

‖T′(t)‖that is normal to C and points in the same direction as T′(t). We

call N(t) the principal unit normal vector to C at t , or more

simply, the unit normal vector. Observe that the unit normal vector

is de�ned only at points where T′(t) 6= 0. Unless stated otherwise,

we will assume that this condition is satis�ed. In particular, this

excludes straight lines.Calculus-III, BCE-III Spring 2013



T and N for curves parameterized by arc length

In case where r(s) is parameterized by arc length, we compute of

unit tangent T(s) and the unit normal vector N(s) as:

T(s) = r′(s)

N(s) =r′′(s)

‖r′′(s)‖

Example: The circle of radius a with counterclockwise orientation

and centered at the origin can be represented by the vector-valued

function

r = a cos ti+ a sin tj (0 ≤ t ≤ 2π)

Parameterize this circle by arc length and �nd T(s) and N(s).





In case where r(s) is parameterized by arc length, we compute of

unit tangent T(s) and the unit normal vector N(s) as:

T(s) = r′(s)

N(s) =r′′(s)

‖r′′(s)‖Example: The circle of radius a with counterclockwise orientation

and centered at the origin can be represented by the vector-valued

function

r = a cos ti+ a sin tj (0 ≤ t ≤ 2π)

Parameterize this circle by arc length and �nd T(s) and N(s).




Binormal Vectors in 3−Space

If C is the graph of a vector-valued function r(t) in 3-space, then

we de�ne the binormal vector to C at t to be

B(t) = T(t)×N(t)

It follows from properties of the cross product that B(t) isorthogonal to both T(t) and N(t) and is oriented relative to T(t)and N(t) by the right-hand rule.

Moreover, T(t)×N(t) is a unit vector since

‖T(t)×N(t)‖ = ‖T(t)‖ ‖N(t)‖ sin(π2) = 1

Thus, {T(t),N(t),B(t)} is a set of three mutually orthogonal unit

vectors.




Planes Determined by T, N and B




Coordinate Sysytem Determined by T, N and B

The coordinate system determined by T(t),N(t), and B(t) isright-handed in the sense that each of these vectors is related to

the other two by the right-hand rule:

B(t) = T(t)×N(t), N(t) = B(t)×T(t), T(t) = N(t)× B(t)

This coordinate system is called the TNB-frame or sometimes the

Frenet frame in honor of the French mathematician Jean Frederic

Frenet (1816− 1900) who pioneered its application to the study of

space curves.

Typically, the xyz−coordinate system determined by the unit

vectors i, j, and k remains �xed, whereas the TNB−frame changes

as its origin moves along the curve C .




TNB−frame along the curve C

The binormal B(t) can be expressed directly in terms of r(t) as

B(t) =r′(t)× r′′(t)

‖r′(t)× r′′(t)‖In the case where the parameter is arc length it can be expressed in

terms of r(s) as

B(s) =r′(s)× r′′(s)

‖r′′(s)‖






B(t) =r′(t)× r′′(t)

‖r′(t)× r′′(t)‖

In the case where the parameter is arc length it can be expressed in

terms of r(s) as

B(s) =r′(s)× r′′(s)

‖r′′(s)‖






B(t) =r′(t)× r′′(t)

‖r′(t)× r′′(t)‖In the case where the parameter is arc length it can be expressed in

terms of r(s) as

B(s) =r′(s)× r′′(s)

‖r′′(s)‖Calculus-III, BCE-III Spring 2013



Curvature




What is Curvature




What is Curvature

De�nition: If C is a smooth curve in 2-space or 3-space that is

parameterized by arc length, then the curvature of C , denoted by

κ = κ(s) (κ is Greek �kappa�), is de�ned by

κ(s) =∥∥∥dTds

∥∥∥ = ‖r′′(s)‖

Observe that κ(s) is a real-valued function of s, since it is the

length of dT

dsthat measures the curvature. In general, the curvature

will vary from point to point along a curve.

Example: Show that the circle of radius a parameterized by arc

length

r(s) = a cos( sa

)i+ a sin

( sa

)j (0 ≤ s ≤ 2πa)

has a constant curvature.




What is Curvature

De�nition: If C is a smooth curve in 2-space or 3-space that is

parameterized by arc length, then the curvature of C , denoted by

κ = κ(s) (κ is Greek �kappa�), is de�ned by

κ(s) =∥∥∥dTds

∥∥∥ = ‖r′′(s)‖

Observe that κ(s) is a real-valued function of s, since it is the

length of dT

dsthat measures the curvature. In general, the curvature

will vary from point to point along a curve.

Example: Show that the circle of radius a parameterized by arc

length

r(s) = a cos( sa

)i+ a sin

( sa

)j (0 ≤ s ≤ 2πa)

has a constant curvature.Calculus-III, BCE-III Spring 2013



Formulas for Curvature

Theorem: If r(t) is a smooth vector-valued function in 2-space or

3-space, then for each value of t at which T′(t) and r′′(t) exist, thecurvature κ can be expressed as

(a) κ(t) =‖T′(t)‖‖r′(t)‖

(b) κ(t) =‖r′(t)× r′′(t)‖‖r′(t)‖3

Example: Find κ(t) for the circular helix

x = a cos t, y = a sin t, z = ct ; where a > 0.



Thank you for attention!


Calculus III 2

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