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Calculus Volume 3

3.1 Vector-Valued Functions and Space Curves

Calculus Volume 33.1 Vector-Valued Functions and Space Curves
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  1. Preface
  2. 1 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 1.1 Parametric Equations
    3. 1.2 Calculus of Parametric Curves
    4. 1.3 Polar Coordinates
    5. 1.4 Area and Arc Length in Polar Coordinates
    6. 1.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Vectors in Space
    1. Introduction
    2. 2.1 Vectors in the Plane
    3. 2.2 Vectors in Three Dimensions
    4. 2.3 The Dot Product
    5. 2.4 The Cross Product
    6. 2.5 Equations of Lines and Planes in Space
    7. 2.6 Quadric Surfaces
    8. 2.7 Cylindrical and Spherical Coordinates
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  4. 3 Vector-Valued Functions
    1. Introduction
    2. 3.1 Vector-Valued Functions and Space Curves
    3. 3.2 Calculus of Vector-Valued Functions
    4. 3.3 Arc Length and Curvature
    5. 3.4 Motion in Space
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  5. 4 Differentiation of Functions of Several Variables
    1. Introduction
    2. 4.1 Functions of Several Variables
    3. 4.2 Limits and Continuity
    4. 4.3 Partial Derivatives
    5. 4.4 Tangent Planes and Linear Approximations
    6. 4.5 The Chain Rule
    7. 4.6 Directional Derivatives and the Gradient
    8. 4.7 Maxima/Minima Problems
    9. 4.8 Lagrange Multipliers
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  6. 5 Multiple Integration
    1. Introduction
    2. 5.1 Double Integrals over Rectangular Regions
    3. 5.2 Double Integrals over General Regions
    4. 5.3 Double Integrals in Polar Coordinates
    5. 5.4 Triple Integrals
    6. 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
    7. 5.6 Calculating Centers of Mass and Moments of Inertia
    8. 5.7 Change of Variables in Multiple Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Vector Calculus
    1. Introduction
    2. 6.1 Vector Fields
    3. 6.2 Line Integrals
    4. 6.3 Conservative Vector Fields
    5. 6.4 Green’s Theorem
    6. 6.5 Divergence and Curl
    7. 6.6 Surface Integrals
    8. 6.7 Stokes’ Theorem
    9. 6.8 The Divergence Theorem
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  8. 7 Second-Order Differential Equations
    1. Introduction
    2. 7.1 Second-Order Linear Equations
    3. 7.2 Nonhomogeneous Linear Equations
    4. 7.3 Applications
    5. 7.4 Series Solutions of Differential Equations
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 3.1.1. Write the general equation of a vector-valued function in component form and unit-vector form.
  • 3.1.2. Recognize parametric equations for a space curve.
  • 3.1.3. Describe the shape of a helix and write its equation.
  • 3.1.4. Define the limit of a vector-valued function.

Our study of vector-valued functions combines ideas from our earlier examination of single-variable calculus with our description of vectors in three dimensions from the preceding chapter. In this section we extend concepts from earlier chapters and also examine new ideas concerning curves in three-dimensional space. These definitions and theorems support the presentation of material in the rest of this chapter and also in the remaining chapters of the text.

Definition of a Vector-Valued Function

Our first step in studying the calculus of vector-valued functions is to define what exactly a vector-valued function is. We can then look at graphs of vector-valued functions and see how they define curves in both two and three dimensions.

Definition

A vector-valued function is a function of the form

r(t)=f(t)i+g(t)jorr(t)=f(t)i+g(t)j+h(t)k,r(t)=f(t)i+g(t)jorr(t)=f(t)i+g(t)j+h(t)k,
3.1

where the component functions f, g, and h, are real-valued functions of the parameter t. Vector-valued functions are also written in the form

r(t)=f(t),g(t)orr(t)=f(t),g(t),h(t).r(t)=f(t),g(t)orr(t)=f(t),g(t),h(t).
3.2

In both cases, the first form of the function defines a two-dimensional vector-valued function; the second form describes a three-dimensional vector-valued function.

The parameter t can lie between two real numbers: atb.atb. Another possibility is that the value of t might take on all real numbers. Last, the component functions themselves may have domain restrictions that enforce restrictions on the value of t. We often use t as a parameter because t can represent time.

Example 3.1

Evaluating Vector-Valued Functions and Determining Domains

For each of the following vector-valued functions, evaluate r(0),r(π2),andr(2π3).r(0),r(π2),andr(2π3). Do any of these functions have domain restrictions?

  1. r(t)=4costi+3sintjr(t)=4costi+3sintj
  2. r(t)=3tanti+4sectj+5tkr(t)=3tanti+4sectj+5tk

Solution

  1. To calculate each of the function values, substitute the appropriate value of t into the function:
    r(0)=4cos(0)i+3sin(0)j=4i+0j=4ir(π2)=4cos(π2)i+3sin(π2)j=0i+3j=3jr(2π3)=4cos(2π3)i+3sin(2π3)j=4(12)i+3(32)j=−2i+332j.r(0)=4cos(0)i+3sin(0)j=4i+0j=4ir(π2)=4cos(π2)i+3sin(π2)j=0i+3j=3jr(2π3)=4cos(2π3)i+3sin(2π3)j=4(12)i+3(32)j=−2i+332j.

    To determine whether this function has any domain restrictions, consider the component functions separately. The first component function is f(t)=4costf(t)=4cost and the second component function is g(t)=3sint.g(t)=3sint. Neither of these functions has a domain restriction, so the domain of r(t)=4costi+3sintjr(t)=4costi+3sintj is all real numbers.
  2. To calculate each of the function values, substitute the appropriate value of t into the function:
    r(0)=3tan(0)i+4sec(0)j+5(0)k=0i+4j+0k=4jr(π2)=3tan(π2)i+4sec(π2)j+5(π2)k,which does not existr(2π3)=3tan(2π3)i+4sec(2π3)j+5(2π3)k=3(3)i+4(−2)j+10π3k=−33i8j+10π3k.r(0)=3tan(0)i+4sec(0)j+5(0)k=0i+4j+0k=4jr(π2)=3tan(π2)i+4sec(π2)j+5(π2)k,which does not existr(2π3)=3tan(2π3)i+4sec(2π3)j+5(2π3)k=3(3)i+4(−2)j+10π3k=−33i8j+10π3k.

    To determine whether this function has any domain restrictions, consider the component functions separately. The first component function is f(t)=3tant,f(t)=3tant, the second component function is g(t)=4sect,g(t)=4sect, and the third component function is h(t)=5t.h(t)=5t. The first two functions are not defined for odd multiples of π/2,π/2, so the function is not defined for odd multiples of π/2.π/2. Therefore, dom(r(t))={t|t(2n+1)π2},dom(r(t))={t|t(2n+1)π2}, where n is any integer.
Checkpoint 3.1

For the vector-valued function r(t)=(t23t)i+(4t+1)j,r(t)=(t23t)i+(4t+1)j, evaluate r(0),r(1),andr(−4).r(0),r(1),andr(−4). Does this function have any domain restrictions?

Example 3.1 illustrates an important concept. The domain of a vector-valued function consists of real numbers. The domain can be all real numbers or a subset of the real numbers. The range of a vector-valued function consists of vectors. Each real number in the domain of a vector-valued function is mapped to either a two- or a three-dimensional vector.

Graphing Vector-Valued Functions

Recall that a plane vector consists of two quantities: direction and magnitude. Given any point in the plane (the initial point), if we move in a specific direction for a specific distance, we arrive at a second point. This represents the terminal point of the vector. We calculate the components of the vector by subtracting the coordinates of the initial point from the coordinates of the terminal point.

A vector is considered to be in standard position if the initial point is located at the origin. When graphing a vector-valued function, we typically graph the vectors in the domain of the function in standard position, because doing so guarantees the uniqueness of the graph. This convention applies to the graphs of three-dimensional vector-valued functions as well. The graph of a vector-valued function of the form r(t)=f(t)i+g(t)jr(t)=f(t)i+g(t)j consists of the set of all (t,r(t)),(t,r(t)), and the path it traces is called a plane curve. The graph of a vector-valued function of the form r(t)=f(t)i+g(t)j+h(t)kr(t)=f(t)i+g(t)j+h(t)k consists of the set of all (t,r(t)),(t,r(t)), and the path it traces is called a space curve. Any representation of a plane curve or space curve using a vector-valued function is called a vector parameterization of the curve.

Example 3.2

Graphing a Vector-Valued Function

Create a graph of each of the following vector-valued functions:

  1. The plane curve represented by r(t)=4costi+3sintj,r(t)=4costi+3sintj, 0t2π0t2π
  2. The plane curve represented by r(t)=4cost3i+3sint3j,r(t)=4cost3i+3sint3j, 0t2π0t2π
  3. The space curve represented by r(t)=costi+sintj+tk,r(t)=costi+sintj+tk, 0t4π0t4π

Solution

  1. As with any graph, we start with a table of values. We then graph each of the vectors in the second column of the table in standard position and connect the terminal points of each vector to form a curve (Figure 3.2). This curve turns out to be an ellipse centered at the origin.
    t r(t)r(t) t r(t)r(t)
    0 4i4i ππ −4i−4i
    π4π4 22i+322j22i+322j 5π45π4 −22i322j−22i322j
    π2π2 3j3j 3π23π2 −3j−3j
    3π43π4 −22i+322j−22i+322j 7π47π4 22i322j22i322j
    2π2π 4i4i
    Table 3.1 Table of Values for r(t)=4costi+3sintj,r(t)=4costi+3sintj, 0t2π0t2π
    This figure is a graph of an ellipse centered at the origin. The graph is the vector-valued function r(t)=4cost i + 3sint j. The ellipse has arrows on the curve representing counter-clockwise orientation. There are also line segments inside of the ellipse to the curve at different increments of t. The increments are t=0, t=pi/4, t=pi/2, t=3pi/4.
    Figure 3.2 The graph of the first vector-valued function is an ellipse.
  2. The table of values for r(t)=4costi+3sintj,r(t)=4costi+3sintj, 0t2π0t2π is as follows:
    t r(t)r(t) t r(t)r(t)
    0 4i4i ππ −4i−4i
    π4π4 22i+322j22i+322j 5π45π4 −22i322j−22i322j
    π2π2 3j3j 3π23π2 −3j−3j
    3π43π4 −22i+322j−22i+322j 7π47π4 22i322j22i322j
    2π2π 4i4i
    Table 3.2 Table of Values for r(t)=4costi+3sintj,r(t)=4costi+3sintj, 0t2π0t2π

    The graph of this curve is also an ellipse centered at the origin.
    This figure is a graph of an ellipse centered at the origin. The graph is the vector-valued function r(t)=4cost^3 i + 3sint^3 j.
    Figure 3.3 The graph of the second vector-valued function is also an ellipse.
  3. We go through the same procedure for a three-dimensional vector function.
    t r(t)r(t) t r(t)r(t)
    0 4i4i ππ −4j+πk−4j+πk
    π4π4 22i+22j+π4k22i+22j+π4k 5π45π4 −22i22j+5π4k−22i22j+5π4k
    π2π2 4j+π2k4j+π2k 3π23π2 −4j+3π2k−4j+3π2k
    3π43π4 −22i+22j+3π4k−22i+22j+3π4k 7π47π4 22i22j+7π4k22i22j+7π4k
    2π2π 4i+2πk4i+2πk
    Table 3.3 Table of Values for r(t)=costi+sintj+tk,r(t)=costi+sintj+tk, 0t4π0t4π

    The values then repeat themselves, except for the fact that the coefficient of k is always increasing (Figure 3.4). This curve is called a helix. Notice that if the k component is eliminated, then the function becomes r(t)=costi+sintj,r(t)=costi+sintj, which is a unit circle centered at the origin.
    This figure is the graph of a helix in the 3 dimensional coordinate system. The curve represents the function r(t) = cost i + sint j + tk. The curve spirals in a circular path around the vertical z-axis and has the look of a spring. The arrows on the curve represent orientation.
    Figure 3.4 The graph of the third vector-valued function is a helix.

You may notice that the graphs in parts a. and b. are identical. This happens because the function describing curve b is a so-called reparameterization of the function describing curve a. In fact, any curve has an infinite number of reparameterizations; for example, we can replace t with 2t2t in any of the three previous curves without changing the shape of the curve. The interval over which t is defined may change, but that is all. We return to this idea later in this chapter when we study arc-length parameterization.

As mentioned, the name of the shape of the curve of the graph in Example 3.2c. is a helix (Figure 3.4). The curve resembles a spring, with a circular cross-section looking down along the z-axis. It is possible for a helix to be elliptical in cross-section as well. For example, the vector-valued function r(t)=4costi+3sintj+tkr(t)=4costi+3sintj+tk describes an elliptical helix. The projection of this helix into the x,y-planex,y-plane is an ellipse. Last, the arrows in the graph of this helix indicate the orientation of the curve as t progresses from 0 to 4π.4π.

Checkpoint 3.2

Create a graph of the vector-valued function r(t)=(t21)i+(2t3)j,r(t)=(t21)i+(2t3)j, 0t3.0t3.

At this point, you may notice a similarity between vector-valued functions and parameterized curves. Indeed, given a vector-valued function r(t)=f(t)i+g(t)j,r(t)=f(t)i+g(t)j, we can define x=f(t)x=f(t) and y=g(t).y=g(t). If a restriction exists on the values of t (for example, t is restricted to the interval [a,b][a,b] for some constants a<b),a<b), then this restriction is enforced on the parameter. The graph of the parameterized function would then agree with the graph of the vector-valued function, except that the vector-valued graph would represent vectors rather than points. Since we can parameterize a curve defined by a function y=f(x),y=f(x), it is also possible to represent an arbitrary plane curve by a vector-valued function.

Limits and Continuity of a Vector-Valued Function

We now take a look at the limit of a vector-valued function. This is important to understand to study the calculus of vector-valued functions.

Definition

A vector-valued function r approaches the limit L as t approaches a, written

limtar(t)=L,limtar(t)=L,

provided

limta||r(t)L||=0.limta||r(t)L||=0.

This is a rigorous definition of the limit of a vector-valued function. In practice, we use the following theorem:

Theorem 3.1

Limit of a Vector-Valued Function

Let f, g, and h be functions of t. Then the limit of the vector-valued function r(t)=f(t)i+g(t)jr(t)=f(t)i+g(t)j as t approaches a is given by

limtar(t)=[limtaf(t)]i+[limtag(t)]j,limtar(t)=[limtaf(t)]i+[limtag(t)]j,
3.3

provided the limits limtaf(t)andlimtag(t)limtaf(t)andlimtag(t) exist. Similarly, the limit of the vector-valued function r(t)=f(t)i+g(t)j+h(t)kr(t)=f(t)i+g(t)j+h(t)k as t approaches a is given by

limtar(t)=[limtaf(t)]i+[limtag(t)]j+[limtah(t)]k,limtar(t)=[limtaf(t)]i+[limtag(t)]j+[limtah(t)]k,
3.4

provided the limits limtaf(t),limtag(t)andlimtah(t)limtaf(t),limtag(t)andlimtah(t) exist.

In the following example, we show how to calculate the limit of a vector-valued function.

Example 3.3

Evaluating the Limit of a Vector-Valued Function

For each of the following vector-valued functions, calculate limt3r(t)limt3r(t) for

  1. r(t)=(t23t+4)i+(4t+3)jr(t)=(t23t+4)i+(4t+3)j
  2. r(t)=2t4t+1i+tt2+1j+(4t3)kr(t)=2t4t+1i+tt2+1j+(4t3)k

Solution

  1. Use Equation 3.3 and substitute the value t=3t=3 into the two component expressions:
    limt3r(t)=limt3[(t23t+4)i+(4t+3)j]=[limt3(t23t+4)]i+[limt3(4t+3)]j=4i+15j.limt3r(t)=limt3[(t23t+4)i+(4t+3)j]=[limt3(t23t+4)]i+[limt3(4t+3)]j=4i+15j.
  2. Use Equation 3.4 and substitute the value t=3t=3 into the three component expressions:
    limt3r(t)=limt3(2t4t+1i+tt2+1j+(4t3)k)=[limt3(2t4t+1)]i+[limt3(tt2+1)]j+[limt3(4t3)]k=12i+310j+9k.limt3r(t)=limt3(2t4t+1i+tt2+1j+(4t3)k)=[limt3(2t4t+1)]i+[limt3(tt2+1)]j+[limt3(4t3)]k=12i+310j+9k.
Checkpoint 3.3

Calculate limt−2r(t)limt−2r(t) for the function r(t)=t23t1i+(4t+3)j+sin(t+1)π2k.r(t)=t23t1i+(4t+3)j+sin(t+1)π2k.

Now that we know how to calculate the limit of a vector-valued function, we can define continuity at a point for such a function.

Definition

Let f, g, and h be functions of t. Then, the vector-valued function r(t)=f(t)i+g(t)jr(t)=f(t)i+g(t)j is continuous at point t=at=a if the following three conditions hold:

  1. r(a)r(a) exists
  2. limtar(t)limtar(t) exists
  3. limtar(t)=r(a)limtar(t)=r(a)

Similarly, the vector-valued function r(t)=f(t)i+g(t)j+h(t)kr(t)=f(t)i+g(t)j+h(t)k is continuous at point t=at=a if the following three conditions hold:

  1. r(a)r(a) exists
  2. limtar(t)limtar(t) exists
  3. limtar(t)=r(a)limtar(t)=r(a)

Section 3.1 Exercises

1.

Give the component functions x=f(t)x=f(t) and y=g(t)y=g(t) for the vector-valued function r(t)=3secti+2tantj.r(t)=3secti+2tantj.

2.

Given r(t)=3secti+2tantj,r(t)=3secti+2tantj, find the following values (if possible).

  1. r(π4)r(π4)
  2. r(π)r(π)
  3. r(π2)r(π2)
3.

Sketch the curve of the vector-valued function r(t)=3secti+2tantjr(t)=3secti+2tantj and give the orientation of the curve. Sketch asymptotes as a guide to the graph.

4.

Evaluate limt0eti+sinttj+etk.limt0eti+sinttj+etk.

5.

Given the vector-valued function r(t)=cost,sint,r(t)=cost,sint, find the following values:

  1. limtπ4r(t)limtπ4r(t)
  2. r(π3)r(π3)
  3. Is r(t)r(t) continuous at t=π3?t=π3?
  4. Graph r(t).r(t).
6.

Given the vector-valued function r(t)=t,t2+1,r(t)=t,t2+1, find the following values:

  1. limt−3r(t)limt−3r(t)
  2. r(−3)r(−3)
  3. Is r(t)r(t) continuous at x=−3?x=−3?
  4. r(t+2)r(t)r(t+2)r(t)
7.

Let r(t)=eti+sintj+lntk.r(t)=eti+sintj+lntk. Find the following values:

  1. r(π4)r(π4)
  2. limtπ/4r(t)limtπ/4r(t)
  3. Is r(t)r(t) continuous at t=t=π4?t=t=π4?

Find the limit of the following vector-valued functions at the indicated value of t.

8.

limt4t3,t2t4,tan(πt)limt4t3,t2t4,tan(πt)

9.

limtπ/2r(t)limtπ/2r(t) for r(t)=eti+sintj+lntkr(t)=eti+sintj+lntk

10.

limte−2t,2t+33t1,arctan(2t)limte−2t,2t+33t1,arctan(2t)

11.

limte2tln(t),lntt2,ln(t2)limte2tln(t),lntt2,ln(t2)

12.

limtπ/6cos2t,sin2t,1limtπ/6cos2t,sin2t,1

13.

limtr(t)limtr(t) for r(t)=2eti+etj+ln(t1)kr(t)=2eti+etj+ln(t1)k

14.

Describe the curve defined by the vector-valued function r(t)=(1+t)i+(2+5t)j+(−1+6t)k.r(t)=(1+t)i+(2+5t)j+(−1+6t)k.

Find the domain of the vector-valued functions.

15.

Domain: r(t)=t2,tant,lntr(t)=t2,tant,lnt

16.

Domain: r(t)=t2,t3,32t+1r(t)=t2,t3,32t+1

17.

Domain: r(t)=csc(t),1t3,ln(t2)r(t)=csc(t),1t3,ln(t2)

Let r(t)=cost,t,sintr(t)=cost,t,sint and use it to answer the following questions.

18.

For what values of t is r(t)r(t) continuous?

19.

Sketch the graph of r(t).r(t).

20.

Find the domain of r(t)=2eti+etj+ln(t1)k.r(t)=2eti+etj+ln(t1)k.

21.

For what values of t is r(t)=2eti+etj+ln(t1)kr(t)=2eti+etj+ln(t1)k continuous?

Eliminate the parameter t, write the equation in Cartesian coordinates, then sketch the graphs of the vector-valued functions.

22.

r(t)=2ti+t2jr(t)=2ti+t2j (Hint: Let x=2tx=2t and y=t2.y=t2. Solve the first equation for x in terms of t and substitute this result into the second equation.)

23.

r(t)=t3i+2tjr(t)=t3i+2tj

24.

r(t)=2(sinht)i+2(cosht)j,t>0r(t)=2(sinht)i+2(cosht)j,t>0

25.

r(t)=3(cost)i+3(sint)jr(t)=3(cost)i+3(sint)j

26.

r(t)=3sint,3costr(t)=3sint,3cost

Use a graphing utility to sketch each of the following vector-valued functions:

27.

[T] r(t)=2cost2i+(2t)jr(t)=2cost2i+(2t)j

28.

[T] r(t)=ecos(3t),esin(t)r(t)=ecos(3t),esin(t)

29.

[T] r(t)=2sin(2t),3+2costr(t)=2sin(2t),3+2cost

30.

4x2+9y2=36;4x2+9y2=36; clockwise and counterclockwise

31.

r(t)=t,t2;r(t)=t,t2; from left to right

32.

The line through P and Q where P is (1,4,−2)(1,4,−2) and Q is (3,9,6)(3,9,6)

Consider the curve described by the vector-valued function r(t)=(50etcost)i+(50etsint)j+(55et)k.r(t)=(50etcost)i+(50etsint)j+(55et)k.

33.

What is the initial point of the path corresponding to r(0)?r(0)?

34.

What is limtr(t)?limtr(t)?

35.

[T] Use technology to sketch the curve.

36.

Eliminate the parameter t to show that z=5r10z=5r10 where r2=x2+y2.r2=x2+y2.

37.

[T] Let r(t)=costi+sintj+0.3sin(2t)k.r(t)=costi+sintj+0.3sin(2t)k. Use technology to graph the curve (called the roller-coaster curve) over the interval [0,2π).[0,2π). Choose at least two views to determine the peaks and valleys.

38.

[T] Use the result of the preceding problem to construct an equation of a roller coaster with a steep drop from the peak and steep incline from the “valley.” Then, use technology to graph the equation.

39.

Use the results of the preceding two problems to construct an equation of a path of a roller coaster with more than two turning points (peaks and valleys).

40.
  1. Graph the curve r(t)=(4+cos(18t))cos(t)i+(4+cos(18t)sin(t))j+0.3sin(18t)kr(t)=(4+cos(18t))cos(t)i+(4+cos(18t)sin(t))j+0.3sin(18t)k using two viewing angles of your choice to see the overall shape of the curve.
  2. Does the curve resemble a “slinky”?
  3. What changes to the equation should be made to increase the number of coils of the slinky?
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