# Multivariate Calculus

MATH 200-004: Multivariate Calculus Winter 2014

Chapter 12: Vector-Valued Functions Extra Credit Scribe: Charles Burnette

Disclaimer: These notes have not been subjected to the usual scrutiny reserved for formal publications. They may be distributed outside this class only with my permission.

These notes cover some topics in addition to what you are expected to know about vector-valued functions. You will not be tested on material from sections 12.3 – 12.6. An extra credit quiz on the content of these notes is available at the end. Problems marked with a ? are ‘‘for your entertainment’’ and are not essential.

12.1 Introduction to Vector-Valued Functions

In general, a function is a rule that assigns to each element in the domain an element in the range. A vector-valued function , or vector function , is simply a function whose domain is a set of real numbers and whose range is a set of vectors. We are most interested in vector functions r whose values are three- dimensional vectors. This means that for every t in the domain of r there is a unique vector denoted by r(t). If f(t), g(t), and h(t) are the components of the vector r(t), then f, g, and h are real-valued functions called the component functions of r and we can write

r(t) = 〈f(t), g(t), h(t)〉 = f(t)i + g(t)j + h(t)k.

We use the letter t to denote the independent variable, because it represents time in most applications of vector functions.

Example 12.1 If

r(t) = ⟨ t3, ln(3− t),

√ t ⟩ ,

then the component functions are

f(t) = t3 g(t) = ln(3− t) h(t) = √ t.

By our usual convention, the domain of r consists of all values of t for which the expression r(t) is defined. The expressions t3, ln(3− t), and

√ t are all defined when 3− t > 0 and t ≥ 0. Therefore, the domain of r is

the interval [0, 3).

We now wish to develop a notion of what it means for a vector function r in 2-space or 3-space to approach a limiting vector L as t approaches a number a. That is, we want to define

12-1

12-2 Chapter 12: Vector-Valued Functions

Figure 12.1: Visualization of Limits for Vector Functions

lim t→a

r(t) = L. (12.1)

One way to motivate a reasonable definition of (12.1) is to position r(t) and L with their initial points at the origin and interpret this limit to mean that the terminal point of r(t) approaches the terminal point of L as t approaches a or, equivalently, that the vector r(t) approaches the vector L in both length and direction as t approaches a (Figure 12.1). Algebraically, this is equivalent to stating that

lim t→a ||r(t)− L|| = 0 (12.2)

(Figure 12.1). Thus, we make the following definition.

Definition 12.2 Let r(t) be a vector function that is defined for all t in some open interval containing the number a, except that r(t) need not be defined at a. We will write

lim t→a

r(t) = L

if and only if

lim t→a ||r(t)− L|| = 0.

Theorem 12.3 If r has a limit at a, then this limit is unique.

Proof: Suppose that r has limits L1 and L2 at a. Let � > 0 be given, but fixed. Then there exist δ1, δ2 > 0 such that ||r(t) − L1|| < �/2 whenever 0 < |t − a| < δ1 and ||r(t) − L1|| < �/2 whenever 0 < |t − a| < δ2. Hence, for each t such that 0 < |t− a| < min{δ1, δ2},

Chapter 12: Vector-Valued Functions 12-3

||L1 − L2|| = ∣∣∣∣((L1 − r(t)) + (r(t)− L2)∣∣∣∣ ≤ ||r(t)− L1||+ ||r(t)− L2|| ≤ �

2 + �

2 = �.

Yet � was arbitrary, and so ||L1 −L2|| is a nonnegative number that is smaller than every positive number �. It follows that ||L1 − L2|| = 0, and so L1 = L2, making the limit unique.

It is clear intuitively that r(t) will approach a limiting vector L as t approaches a if and only if the component functions of r(t) approach the corresponding components of L. This suggests the following.

Theorem 12.4 (a) If r(t) = 〈x(t), y(t)〉, then

lim t→a

r(t) = ⟨

lim t→a

x(t), lim t→a

y(t) ⟩

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 a.

(b) If r(t) = 〈x(t), y(t), z(t)〉, then

lim t→a

r(t) = ⟨

lim t→a

x(t), lim t→a

y(t), lim t→a

z(t) ⟩

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 a.

Proof: We will only prove part (a), as the proof of part (b) is virtually identical. Suppose that lim t→a

x(t) = L1

and lim t→a

y(t) = L2. Then

lim t→a ||r(t)−〈L1, L2〉|| = lim

t→a

√ (x(t)− L1)2 + (y(t)− L2)2 =

√( lim t→a

[x(t)− a] )2

+ (

lim t→a

[y(t)− a] )2

= √

02 + 02 = 0

so that

lim t→a

r(t) = 〈L1, L2〉 = ⟨

lim t→a

x(t), lim t→a

y(t) ⟩

Conversely, suppose that lim t→a

r(t) = 〈L1, L2〉. Then

0 ≤ |x(t)− L1| = √ (x(t)− L1)2 ≤

√ (x(t)− L1)2 + (y(t)− L2)2 = ||r(t)− 〈L1, L2〉|| → 0

as t→ a. It follows by the Squeeze Theorem that lim t→a

x(t) = L1. The same reasoning begets lim t→a

y(t) = L2.

Limits of vector functions obey the same rules as limits of real-valued functions. This is a trivial consequence of Theorem 12.4, so we omit the proof.

12-4 Chapter 12: Vector-Valued Functions

Figure 12.2: C is traced out by the tip of a moving position vector r(t).

Proposition 12.5 Suppose u and v are vector functions that possess limits as t approaches a and let c be a scalar. Then the following properties hold.

(a) lim t→a

[u(t)± v(t)] = lim t→a

u(t)± lim t→a

v(t)

(b) lim t→a

cu(t) = c lim t→a

u(t)

(c) lim t→a

[u(t) •v(t)] = lim t→a

u(t) • lim t→a

v(t)

(d) lim t→a

[u(t)× v(t)] = lim t→a

u(t)× lim t→a

v(t)

Example 12.6 According to Theorem 12.4, if r(t) = (1 + t3)i + te–tj + sin t

t k, then

lim t→0

r(t) = [ lim t→0

(1 + t3) ] i + [ lim t→0

te–t ] j +

[ lim t→0

sin t

t

] k = i + k.

Definition 12.7 A vector function r is continuous at a if

lim t→a

r(t) = r(a). (12.3)

In the case where r is continuous on R, we will say that r is continuous everywhere, or just continuous.

In view of Definition 12.2, we see that r is continuous at a if and only if its component functions f, g,

and h are continuous at a. Additionally, for any real-valued function F (t), lim t→a

r(F (t)) = r (

lim t→a

F (t) ) .

There is a close connection between continuous vector functions and space curves. Suppose that f, g, and h are continuous real-valued functions on an interval I. Then the set C of all points (x, y, z) in space, where

x = f(t), y = g(t), z = h(t), (12.4)

and t varies throughout the interval I, is called a space curve . The equations in (12.4) are called parametric equations of C , and t is called a parameter . We can think of C as being traced out by a moving particle

Chapter 12: Vector-Valued Functions 12-5

whose position at time t is (f(t), g(t), h(t)). If we now consider the vector function r(t) = 〈f(t), g(t), h(t)〉, then r(t) is the position vector of the point P (f(t), g(t), h(t)) on C. Thus, any continuous vector function r defines a space curve C that is traced out by the tip of the moving vector r(t), as shown in Figure 12.2.

Example 12.8 Let us find a vector function that represents the curve of intersection of the cylinder x2 + y2 = 1 and the plane y + z = 2. The projection of C onto the xy-plane is the circle x2 + y2 = 1, z = 0. So we know that we can write

x = cos t, y = sin t, t ∈ R.

From the equation of the plane, we have

z = 2− y = 2− sin t.

So we can write parametric equations for C as

x = cos t, y = sin t, z = 2− sin t, t ∈ R.

The corresponding vector equation is

r(t) = cos ti + sin tj + (2− sin t)k.

This equation is called a parametrization of the curve C.

12.2 Calculus of Vector-Valued Functions

In this section, we will define derivatives and integrals of vector functions and discuss their properties.

12.2.1 Derivatives

The derivative of a vector function is defined in much the same way as for real-valued functions.

Definition 12.9 If r is a vector function, we define the derivative of r with respect to t to be the vector function r′ given by

r′(t) = lim h→0

r(t+ h)− r(t) h

. (12.5)

The domain of r′ consists of all values of t in the domain of r(t) for which the limit exists. The function r(t) is differentiable at t if the limit in (12.5) exists. All of the standard notations for derivatives still apply.

12-6 Chapter 12: Vector-Valued Functions

Figure 12.3: Geometric Interpretation of the Derivative

Theorem 12.10 If r is differentiable at a, then r is continuous at a.

Proof: We have r(t)− r(a) = r(t)− r(a) t− a

· (t− a)→ r′(a) · 0 = 0 as t→ a. Therefore lim t→a

r(t) = r(a).

The geometric significance of this definition is shown in parts (a) and (b) of Figure 12.3. These illustrations show the graph C of r(t) (with its orientation) and the vectors r(t), r(t+ h), and r(t+ h)− r(t) for positive h and negative h. In both cases, the vector r(t+ h)− r(t) runs along the secant line joining the terminal points of r(t+ h) and r(t), but with opposite directions in the two cases. In the case where h > 0, the vector r(t+ h)− r(t) points in the direction of increasing t, and in the case where h < 0, it points in the opposite direction. However, in the case where h < 0, the direction gets reversed when we multiply by 1/h. So in both cases, the scalar multiple (1/h)[r(t+ h)− r(t)] points in the direction of increasing t and runs along the secant line. As h→ 0, it appears that this vector approaches a vector that lies on the tangent line.

The following theorem gives us a convenient method for computing the derivative of a vector function r : just differentiate each component of r.

Theorem 12.11 If r(t) is a vector function, then r is differentiable at t if and only if each of its component functions is differentiable at t, in which case the component functions of r′(t) are the derivatives of the corresponding component functions of r(t).

Proof: For simplicity, we give the proof in 2-space; the proof in 3-space is identical, except for the additional component. Assume that r(t) = 〈x(t), y(t)〉. Then

Chapter 12: Vector-Valued Functions 12-7

Figure 12.4: The Tangent Vector and the Tangent Line

r′(t) = lim h→0

r(t+ h)− r(t) h

= lim h→0

1

h [〈x(t+ h), y(t+ h)〉 − 〈x(t), y(t)〉]

= lim h→0

⟨ x(t+ h)− x(t)

h , y(t+ h)− y(t)

h

⟩ =

⟨ lim h→0

x(t+ h)− x(t) h

, lim h→0

y(t+ h)− y(t) h

⟩ = 〈x′(t), y′(t)〉.

Example 12.12 The derivative of r(t) = (1 + t3)i + te–tj + sin 2tk is

r′(t) = 3t2i + (1− t)e–tj + 2 cos 2tk.

Motivated by the discussion of the geometric interpretation of the derivative of a vector function, we make the following definition.

Definition 12.13 Let P be a point on the graph of a vector function r(t), and let r(t0) be the radial vector from the origin to P (Figure 12.6). If r′(t0) exists and r

′(t0) 6= 0, then we call r′(t0) the tangent vector to the graph of r(t) at r(t0), and we call the line P that is parallel to the tangent vector the tangent line to the graph of r(t) at r(t0).

Let r0 = r(t0) and v0 = r ′(t0). It follows that the tangent line to the graph of r(t) at r0 is given by the

vector equation

r = r0 + tv. (12.6)

12-8 Chapter 12: Vector-Valued Functions

Example 12.14 Suppose we want to find the tangent line to the helix to with parametric equations

x = 2 cos t, y = sin t, z = t

at the point (0, 1, π/2) is the line through the point. Its vector equation is r(t) = 〈2 cos t, sin t, t〉, so

r′(t) = 〈–2 sin t, cos t, 1〉.

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〉. The tangent line is the line through (0, 1, π/2) parallel to the vector 〈–2, 0, 1〉, so its parametric equations are

x = –2t, y = t, z = π

2 + t.

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 12.13 is

r′′(t) = 〈–2 cos t, – sin t, 0〉.

Definition 12.15 A curve given by a vector function r(t) on an interval I is called smooth if r′ is continuous and r′(t) 6= 0, except possible at the endpoints of I.

For instance, the helix in Example 12.13 is smooth because r′(t) is never 0.

Example 12.16 The semi cubical parabola r(t) = 〈1 + t3, t2〉 is not smooth, because

r′(t) = 〈3t2, 2t〉,

and so r′(0) = 〈0, 0〉 = 0. To see the behavior of r from a Calc I perspective, note that the parametric equations of r satisfy y = (x− 1)2/3 so that dy/dx = (2/3)(x− 1)–1/3 for x 6= 1, which is also the point at which t = 0. Observe that lim

x→1− (2/3)(x − 1)–1/3 = –∞ whereas lim

x→1+ (2/3)(x − 1)–1/3 = +∞. This tells us

that there is a sharp corner, called a cusp, at (1, 0). Any curve with this type of behavior—an abrupt change in direction—is not smooth.

A curve, such as the semi cubical parabola, that is made up of a finite number of smooth pieces is called piecewise smooth .

The next theorem shows that the differentiation formulas for real valued functions have their counterparts for vector-valued functions.

Theorem 12.17 Suppose u and v are differentiable vector functions, k is a scalar, c is a constant vector, and f is a real-valued function. Then

Chapter 12: Vector-Valued Functions 12-9

(a) d

dt [c] = 0

(b) d

dt [ku(t)] = ku′(t)

(c) d

dt [u(t)± v(t)] = u′(t)± v′(t)

(d) d

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

(e) d

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

(f) d

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

(g) d

dt [u(f(t))] = f ′(t)u′(f(t))

Proof: Formulas (a)-(d) and (g) (the chain rule) can be proved by using Theorem 12.10 and the corresponding differentiation formulas for real-valued functions. Formulas (e) and (f) can be seen more readily by applying Definition 12.9 directly. The proof of (e) is included here due to its importance. Indeed

d

dt [u(t) •v(t)] = lim

h→0

u(t+ h) •v(t+ h)− u(t) •v(t) h

= lim h→0

u(t+ h) •v(t+ h)− u(t) •v(t) + u(t+ h) •v(t)− u(t+ h) •v(t) h

= lim h→0

[u(t+ h) •v(t+ h)− u(t+ h) •v(t)] + [u(t+ h) •v(t)− u(t) •v(t)] h

= lim h→0

[u(t+ h)− u(t)] •v(t) + u(t+ h) • [v(t+ h)− v(t)] h

= lim h→0

[( u(t+ h)− u(t)

h

) •v(t)

] + lim h→0

[ u(t+ h) •

( v(t+ h)− v(t)

h

)] =

( lim h→0

u(t+ h)− u(t) h

) • lim h→0

v(t) + lim h→0

u(t+ h) •

( lim h→0

v(t+ h)− v(t) h

) = u′(t) •v(t) + u(t) •v′(t).

The proof of Formula (f) is the same. Just replace each • with ×.

Corollary 12.18 If r is a differentiable vector function and ||r(t)|| is constant for all t, then

r(t) • r′(t) = 0, (12.7)

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

12-10 Chapter 12: Vector-Valued Functions

Figure 12.5: Orthogonality of r(t) and r′(t) for Curves on the Surface of a Sphere

Proof: Suppose that ||r(t)|| = c, where c is a constant. Since ||r(t)||2 = c2 and c2 is a constant, Formula (e) of Theorem 12.17 gives

0 = d

dt [||r(t)||2] d

dt [r(t) • r(t)] = r′(t) • r(t) + r(t) • r′(t) = 2r(t) • r′(t).

Thus r(t) • r′(t) = 0.

Geometrically, this result says that if a curve lies on a sphere with center the origin, then the tangent vector r′(t) is always orthogonal to the position vector r(t) (Figure 12.5).

12.2.2 Integrals

The definite integral of a continuous vector function r can be defined in much the same way as for real-valued functions expect that the integral is a vector. (We use the notation of Chapter 5.)

Definition 12.19 A continuous vector function r is said to be integrable on a finite closed interval [a, b] if the limit

lim max ∆tk→0

n∑ k=1

r(t∗k)∆tk

exists and does not depend on the choice of partitions or on the choice of the points x∗k in the subintervals. When this is the case, we denote the limit by

∫ b a

r(t) dt = lim max ∆tk→0

n∑ k=1

r(t∗k)∆tk (12.8)

Chapter 12: Vector-Valued Functions 12-11

which is called the definite integral of r from a to b. The numbers a and b are called the lower limit of integration and the upper limit of integration, respectively, and r(t) is called the integrand.

But then we can express the integral of r in terms of the integrals of its component functions f, g, and h. For example, if r(t) = 〈x(t), y(t)〉, then

∫ b a

r(t) dt = lim max ∆tk→0

n∑ k=1

r(t∗k)∆tk

= lim max ∆tk→0

n∑ k=1

〈x(t), y(t)〉∆tk

= lim max ∆tk→0

⟨ n∑ k=1

x(t)∆tk, y(t)∆yk

⟩

=

⟨ lim

max ∆tk→0

n∑ k=1

x(t)∆tk, lim max ∆tk→0

n∑ k=1

y(t)∆tk

⟩ =

⟨∫ b a

x(t) dt,

∫ b a

y(t) dt

⟩ .

This begets the following theorem.

Theorem 12.20 If r is a continuous vector function on a finite closed interval [a, b], then r is integrable on [a, b] if and only if each of its component functions is integrable on [a, b], in which case the components of∫ b a r(t) dt are the definite integrals of the corresponding component functions of r(t).

As with differentiation, many of the rules for integrating real-valued functions have analogs for vector functions. We omit the proof of the following proposition, as each fact follows trivially from Theorem 12.20

Proposition 12.21 Let u and v be continuous vector functions on [a, b], and let k be a scalar. Then

(a)

∫ b a

ku(t) dt = k

∫ b a

u(t) dt

(b)

∫ b a

[u(t)± v(t)] dt = ∫ b a

u(t) dt± ∫ b a

v(t) dt

We can extend the Fundamental Theorem of Calculus to continuous vector functions as follows:

∫ b a

r(t) dt = R(t)

∣∣∣∣∣ b

a

= R(b)−R(a), (12.9)

where R is an antiderivative of r, that is, R′(t) = r(t). We use the notation ∫ r(t) dt for indefinite

integrals (antiderivatives).

Example 12.22 If r(t) = 2 cos ti + sin tj + 2tk, then

12-12 Chapter 12: Vector-Valued Functions

∫ r(t) dt =

(∫ 2 cos t dt

) i +

(∫ sin t dt

) j +

(∫ 2t dt

) k = 2 sin ti− cos tj + t2k + C,

where C is a vector constant of integration, and

∫ π/2 0

r(t) dt = [ 2 sin ti− cos tj + t2k

]π/2 0

= 2i + j + π2

4 k

Unfortunately, our standard intuitive understanding of the definite integral as the ‘‘net signed area under a curve’’ makes no sense here, since any reasonable interpretation of ‘‘signed area’’ would be scalar, whereas the integral of a vector function is always a vector. Our interpretation of vector functions as position functions fares no better, since while we have interpretations for the integrals of velocity and acceleration, we do not have a physical interpretation of the integral of position. Nevertheless, we can still relate some important aspects of a space curve with vector-valued integration. The first of which is the displacement vector from r(a) to r(b), that is, the vector starting at the point on the curve corresponding to t = a and ending at the point corresponding to t = b, which can be determined by integrating the velocity function:

[ displacement vector from t = a to t = b

] =

∫ b a

r′(t) dt = r(b)− r(a).

Note that this is just the Fundamental Theorem of Calculus applied to a vector function. We investigate another application in the next section.

12.3 Change of Parameter; Arc Length

We observed in the past that a curve in 2-space or 3-space permit multiple parametrizations. Sometimes it is desirable to change the parameter for a parametric curve to a different parameter that is better suited for the problem at hand. In this section, we will investigate issues associated with changes of parameter, and we will show that arc length plays a special role in parametric representations of curves.

12.3.1 Arc Length from the Vector Viewpoint

In Calc II, we define the length of a plane curve with parametric equations x = f(t), y = g(t), a ≤ t ≤ b, as the limit of lengths of inscribed triangulations and, for the case where f ′ and g′ are continuous on [a, b], we arrived at the formula

L =

∫ b a

√( dx

dt

)2 +

( dy

dt

)2 dt. (12.10)

The length of a smooth space curve is defined in exactly the same way.

Definition 12.23 Suppose that a smooth space curve C has the vector equation r(t) = 〈f(t), g(t), g(t)〉, a ≤ t ≤ b, where f ′, g′, and h′ are continuous on [a, b]. If C is traversed exactly once as t increases from a to b, then its arc length is

Chapter 12: Vector-Valued Functions 12-13

L =

∫ b a

√( dx

dt

)2 +

( dy

dt

)2 +

( dz

dt

)2 dt. (12.11)

Notice that both of the arc length formulas (12.10) and (12.11) can be put into the more compact form

L =

∫ b a

||r′(t)|| dt (12.12)

because for plane curves r(t) = 〈x(t), g(t)〉,

||r′(t)|| = ∣∣∣∣∣∣∣∣⟨dxdt , dydt

⟩∣∣∣∣∣∣∣∣ = √(

dx

dt

)2 +

( dy

dt

)2 ,

whereas for space curves r(t) = 〈x(t), y(t), z(t)〉,

||r′(t)|| = ∣∣∣∣∣∣∣∣⟨dxdt , dydt , dzdt

⟩∣∣∣∣∣∣∣∣ = √(

dx

dt

)2 +

( dy

dt

)2 +

( dz

dt

)2 .

Example 12.24 For the circular helix with vector equation r(t) = cos ti + sin tj + tk, we have

||r′(t)|| = √

(– sin t)2 + cos2t+ 1 = √

2.

The arc from (1, 0, 0) to (1, 0, 2π) is described by the parameter interval 0 ≤ t ≤ 2π, and so, from Formula (12.12), we have

L =

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