MATH 200-004 Multivariate Calculus Winter 2014Chapter 12.docx

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.
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
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.

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
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)
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
(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.
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)
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
∫
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
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 =
∫ 2π
0
||r′(t)|| dt =
∫ 2π
0

√
2 dt = 2
√
2π.
12.3.2 Arc Length as a Parameter
A single curve C can be represented by more then one vector
function. For instance, the twisted cubic
r1(t) = 〈t, t2, t3〉 1 ≤ t ≤ 2 (12.13)
can also be represented by the function
r2(u) = 〈eu, e2u, e3u〉 0 ≤ u ≤ ln 2 (12.14)
where the connection between the parameters t and u is given by
t = eu. We say that Equations (12.13)
and (12.14) are parametrizations of the curve C. If we use
Equation (12.12) to compute the arc length of
C using Equations (12.13) and (12.14), we get the same answer.
(Check this!) In general, when Equation
(12.12) is used to compute the length of any piecewise-smooth
curve, the arc length is independent of the
parametrization that is used.
Theorem 12.25 The arc length of a smooth curve is independent
of its parametrization.
Proof: Let r1 and r2 be two smooth parametrizations of a curve

C on the finite closed intervals [a, b] and
[α, β], respectively, where r1([a, b]) = r2([α, β]). Since r1 and
r2 are parametrizations of the same curve, it
follows that r1(b) = r2(β) and for every u ∈ [α, β], there exists a
unique t ∈ [a, b] such that r2(s) = r1(t).
Thus, there is an invertible function f : [α, β]→ [a, b] so that
r2(u) = r1(f(u)). Since r1 and r2 are smooth,
the Inverse Function Theorem and the Chain Rule dictate that f
is continuously differentiable on [α, β], and
r2
′(u) =
d
du
[r1(f(u))] = f
′(u)r′1(f(u)).
Since f is invertible and differentiable, f is strictly increasing,
and so f ′(u) > 0 on [α, β]. To finish the proof,
we have
∫ β
α
||r′2(u)|| du =
∫ β
α
||f ′(u)r′1(f(u))|| du =
∫ β
α
||r′1(f(u))||f ′(u) du

since f ′(u) is a positive scalar. Making the substitution t = f(u),
we have dt = f ′(u) du so that
∫ β
α
||r′1(f(u))||f ′(u) du =
∫ f(β)=b
f(α)=a
||r′1(t)|| dt,
both of which give the arc length of C.
Definition 12.26 Suppose that C is a piecewise-smooth curve
given by a vector function r(t), a ≤ t ≤ b,
and C is traversed exactly once as t increases from a to b. We
define its arc length function s by
s(t) =
∫ t
a
||r′(u)|| du (12.15)
Thus s(t) is the length of the part of C between r(a) and r(t).
(See Figure 12.6.) The point on C given by
r(a) is called the reference point . If we differentiate both sides
of Equation (12.15) using the Fundamental
Theorem of Calculus, we obtain
ds
dt

= ||r′(t)|| (12.16)
Figure 12.6: The Arc Length Function
Theorem 12.27 (a) If C is the graph of a smooth vector function
r, where t is a general parameter, and if
s is the arc length parameter for C defined by Formula (12.12),
then for every value of t the tangent
vector has length ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ drdt
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = dsdt (12.17)
(b) If C is the graph of a smooth vector function r, where s is an
arc length parameter, then for every value
of s the tangent vector to C has length ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ drds
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 1 (12.18)
(c) If C is the graph of a smooth vector function r, and if
||dr/dt|| = 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.
Proof:
(a) See the paragraph preceding Theorem 12.27.
(b) Let t = s in part (a).
(c) It follows from Formula (12.12) that the formula
s =

∫ t
t0
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ drdu
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ du
defines an arc length parameter for C with reference point r(0).
However ||dr/du|| = 1 by hypothesis, so
we can rewrite the formula for s as
s =
∫ t
t0
du = t− t0.
It is often useful to parametrize a curve with respect to arc
length because arc length arises naturally
from the shape of the curve and does not depend on a particular
coordinate system. If a curve r(t) is already
given in terms of a parameter t and s(t) is the arc length
function given by Equation (12.15), then we may
be able to solve for t as a function of s: t = t(s). Then the curve
can be reparametrized in terms of s by
substituting for t : r = r(t(s)). Thus, if s = 3 for instance, r(t(3))
is the position vector of the point 3 units
of length along the curve from its reference point.
Example 12.28 We wish to reparametrize the helix r(t) = cos ti
+ sin tj + tk with respect to arc length

measured from (1, 0, 0) in the direction of increasing t. The
reference point (1, 0, 0) corresponds to the
parameter t = 0. From Example 12.24, we have
ds
dt
= ||r′(t)|| =
√
2,
and so
s = s(t) =
∫ t
0
r′(u)|| du =
∫ t
0
√
2 du =
√
2t.
Therefore t = s/
√
2 and the required reparametrization is obtained by substituting
for t :

r(s) = cos
(
s√
2
)
i + sin
(
s√
2
)
j +
s√
2
k.
Example 12.29 A bug walks along the trunk of a tree following
a path modeled by the circular helix in
Example 12.28. The bug starts at the reference point (1, 0, 0)
and walks up the helix for a distance of 10
units. To find the bug’s final coordinates, it suffices to set s =
10 into our answer in Example 12.28 so that
its final coordinates are
(
cos
(
10√
2

)
, sin
(
10√
2
)
,
10√
2
)
.
Proposition 12.30 Let r(t) = r0 + tv be the vector equation of a
line that passes through the terminal point
r0 and is parallel to the vector v. The arc length parametrization
of the line that has reference points r0 and
the same orientation as r is
r(s) = r0 + s
(
v
||v||
)
. (12.19)
Proof: Note that r(0) = r0, the reference point, and dr/dt = v.
Hence

s =
∫ t
0
||r′(u)|| du =
∫ t
0
||v|| du = t||v||.
This implies that t = s/||v||, and the reparametrization now
follows.
Example 12.31 We wish to find the arc length parametrization
of the line
x = 2t+ 1, y = 3t− 2
that has the same orientation as the given line and uses (1, –2)
as the reference point. Observe that the line
is parallel to v = 2i + 3j. To find the arc length parametrization
of the line, we need only rewrite the given
equation using v/||v|| rather than v to determine the direction
and replace t by s. Since
v
||v||
=

2i + 3j√
13
=
2√
13
i +
3√
13
j,
it follows that the parametric equations for the line in terms of s
are
s =
2√
13
s+ 1, y =
3√
13
s− 2.
? A cable has radius r and length L and is wound around a spool
with radius R without overlapping. What
is the shortest length along the spool that is covered by the
cable?
12.4 Unit Tangent, Normal, and Binormal Vectors
In this section, we will discuss some of the fundamental
geometric properties of vector functions. Our work
here will have important applications to the study of motion
along a curved path in 2-space or 3-space and

to the study of the geometric properties of curves and surfaces.
12.4.1 Unit Tangent Vectors
Definition 12.32 Let C be the graph of a smooth vector function
r(t) so that the vector r′(t) is nonzero,
tangent to C, and points in the direction of increasing
parameter. Thus, by normalizing r′(t), we obtain a
unit vector
T(t) =
r′(t)
||r′(t)||
(12.20)
that is tangent to C and points in the direction of increasing
parameter t. We call T(t) the unit tangent
vector to C at t.
Figure 12.7: Vectors Orthogonal to T(t)
Example 12.33 Let r(t) = 2ti + t2j + t2k. Then r′(t) = 2i + 2tj +
2tk, and so the unit tangent vector to
graph of r(t) at the point where t = 2 is
T(2) =
r′(2)
||r′(2)||
=

2i + 4j + 4k√
36
=
2i + 4j + 4k
6
=
1
3
i +
2
3
j +
2
3
k.
12.4.2 Unit Normal Vectors
At a given point on a smooth curve r(t), there are multiple
vectors that are orthogonal to the unit tangent
vector T(t) (Figure 12.7). We single out one by observing that,
because ||T(t)|| = 1 for all t, we have
T(t) •T′(t) = 0 for all t by Corollary 12.18. So T′(t) is
orthogonal to T(t). Note that T′(t) is not necessarily
a unit vector, but if r′ is also smooth, we can obtain one that
points in the same direction as T′(t).
Definition 12.34

If r and r′ are smooth curves, then we define the principal unit
normal vector N(t) (or simply unit
normal) at t as
N(t) =
T′(t)
||T′(t)||
(12.21)
We can think of the normal vector as indicating the direction in
which the curve is turning at each point.
Example 12.35 Consider the circular helix r(t) = a cos ti+a sin
tj+ctk, where a > 0, c is a constant. Then
Figure 12.8: Unit Normal Pointing Towards the z-Axis
r′(t) = –a sin ti + a cos tj + ck
||r′(t)|| =
√
(–a sin t)2 + (a cos t)2 + c2 =
√
a2 + c2
T(t) =
r′(t)
||r′(t)||

= –
a sin t√
a2 + c2
i +
a cos t√
a2 + c2
j +
c√
a2 + c2
k
T′(t) = –
a cos t√
a2 + c2
i− a sin t√
a2 + c2
j
||T′(t)|| =
√(
–
a cos t√
a2 + c2
)2
+
(

a sin t√
a2 + c2
)2
=
√
a2
a2 + c2
=
a√
a2 + c2
N(t) =
T′(t)
||T′(t)||
= – cos ti− sin tj = –(cos ti + sin tj)
Note that the k component of the unit normal N(t) is zero for
every value of t, so this vector always lies in a
horizontal plane, as illustrated in Figure 12.8. We leave it as an
exercise to show that this vector actually
always point toward the z-axis.
Our next objective is to show that for a nonlinear parametric
curve C in 2-space the unit normal always
points toward the concave side of C. For this purpose, let φ(t)
be the angle from the positive x-axis to T(t),
and let n(t) be the unit vector that results from rotating T(t)
counterclockwise through an angle of π/2
(Figure 12.9). Since T(t) and n(t) are unit vectors, it follows
that these vectors can be expressed as

T(t) = cosφ(t)i + sinφ(t)j (12.22)
and
n(t) = cos
(
φ(t) +
π
2
)
i + sin
(
φ(t) +
π
2
)
j = – sinφ(t)i + cosφ(t)j. (12.23)
Figure 12.9: Angle of Inclination Function φ(t)
Observe that on intervals where φ(t) is increasing the vector n(t)
points toward the concave side of C, and
on intervals where φ(t) is decreasing it points away from the
concave side (Figure 12.9).

Now let us differentiate T(t) by using Formula (12.22) and
applying the chain rule. This yields
dT
dt
=
dT
dφ
dφ
dt
= [(– sinφ)i + (cosφ)j]
dφ
dt
,
and thus from Formula (12.23)
dT
dt
= n(t)
dφ
dt
. (12.24)
Yet dφ/dt > 0 on intervals where φ(t) is increasing and dφ/dt < 0
on intervals where φ(t) is decreasing.
It thus follows from Formula (12.24) that dT/dt has the same

direction as n(t) on intervals where φ(t) is
increasing and the opposite direction on intervals where φ(t) is
decreasing. Therefore T′(t) points ‘‘inward’’
toward the concave side of the curve at all times, and hence so
does N(t). For this reason, N(t) is also called
the inward unit normal when applied to curves in 2-space.
In the case where r(s) is parametrized by arc length, the
procedures for computing T(s) and N(s) can
be simplified. For example, Theorem 12.27 says that if s is an
arc length parameter, then ||r′(s)|| = 1, and so
T(s) = r′(s) (12.25)
and
N(s) =
r′′(s)
||r′′(s)||
. (12.26)
Example 12.36 The circle of radius a with counterclockwise
orientation and centered at the origin can be
represented by the vector function
Figure 12.10: Unit Tangent and Unit Normal of a Circle
r(t) = a cos ti + a sin tj (0 ≤ t ≤ 2π)
We can interpret t as the angle in radian measure from the
positive x-axis to the radius vector. From basic

precalculus, this angle subtends an arc of length s = at on the
circle, so we can reparametrize the circle in
terms of s by substituting s/a for t. This yields
r(s) = a cos
( s
a
)
i + a sin
( s
a
)
j (0 ≤ s ≤ 2πa).
Hence
T(s) = r′(s) = – sin
( s
a
)
i + cos
( s
a
)
j
r′′(s) = –
1
a

cos
( s
a
)
i− 1
a
sin
( s
a
)
j
||r′′(s)|| =
√[
–
1
a
cos
( s
a
)]2
−
[
1
a
sin
( s

a
)]2
=
1
a
N(s) =
r′′(s)
||r′′(s)||
= – cos
( s
a
)
i− sin
( s
a
)
j
so N(s) points toward the center of the circle for all s (Figure
12.10). This makes sense geometrically and is
also consistent with our earlier observation that in 2-space the
unit normal points inward.

Figure 12.11: Right-Hand Rule Orientation of T, N, and B
12.4.3 Binormal Vectors
Definition 12.37 If C is the graph of a vector-valued function
r(t) in 3-space, then we define the binormal
vector to C at t to be
B(t) = T(t)×N(t) (12.27)
Note that if T and N are parametrized with respect to arc length
s, then
B(s) = T(s)×N(s) = r′(s)× r
′′(s)
||r′′(s)||
=
r′(s)× r′′(s)
||r′′(s)||
.
By the definition of the cross product, B is orthogonal to both T
and N and is oriented relative to T
and N by the right-hand rule. Moreover, B(t) is a unit vector
since
||B(t)|| = ||T(t)×N(t)|| = ||T(t)|| ||N(t)|| sin
(π
2
)
= 1.

Thus {T(t),N(t),B(t)} is a set of mutually orthogonal unit
vectors. (More can be said actually. Like i,
j, and k, the vectors T(t), N(t), and B(t) determine a right-
handed coordinate system in 3-space (Figure
12.11). We will see why shortly.)
Example 12.38 Consider the circular helix r(t) = a cos ti + a sin
tj + ctk, where a > 0, c is a constant.
In Example 12.35, we found that T(t) = (–a sin t/
√
a2 + c2)i + (a cos t/
√
a2 + c2)j + (c/
√
a2 + c2)k and
N(t) = –(cos ti + sin tj). The binormal vector is
B(t) = T(t)×N(t) = 1√
a2 + c2
∣ ∣ ∣ ∣ ∣ ∣
i j k
–a sin t a cos t c
– cos t – sin t 0
∣ ∣ ∣ ∣ ∣ ∣ = c sin t√a2 + c2 i− c cos t√a2 + c2 j + a
2
√
a2 + c2

k.
Figure 12.12: The Frenet Frame
Not only are T(t), N(t), and B(t) mutually orthogonal unit
vectors, but each vector is related to the
other two by the right-hand rule. Indeed, using the formula a×
(b× c) = (a • c)b− (a •b)c,
T×N = B
N×B = N× (T×N) = (N •N)T− (N •T)N = ||N||2T− 0N = T
B×T = B× (N×B) = (B •B)N− (B •N)B = ||B||2N− 0B = N
The coordinate system determined by T(t), N(t), and B(t) is
called the TNB-frame or Frenet frame.
Whereas the xyz-coordinate system determined by i, j, and k
remains fixed, the Frenet frame changes as
its origin moves along the curve C (Figure 12.12).
At each point on a smooth curve given parametric equations r(t)
= 〈x(t), y(t), z(t)〉 in 3-space, the
vectors T(t), N(t), and B(t) determine three mutually
perpendicular planes that pass through the point
(x(t), y(t), z(t)). They are the TB-plane (called the rectifying
plane), the TN-plane (called the osculating
plane), and the NB-plane (called the normal plane) (Figure
12.13). The osculating plane is especially
important as it is the plane that comes closest to containing the
part of the curve near (x(t), y(t), z(t)). (For
a plane curve, the osculating plane is simply the plane that
contains the curve.) The name comes from the

Latin osculum, meaning ‘‘kiss.’’
Example 12.39 Let us find the equations of the rectifying,
osculating, and normal planes of the helix in
Example 12.38 at the point P (0, a, cπ/2). Note that r(t) = a cos
ti + a sin tj + ctk passes through P when
t = π/2. The rectifying plane at P contains the vectors T(π/2)
and B(π/2), so its normal vector is N(π/2).
So
N(t) = –(cos ti + sin tj)⇒ N
(π
2
)
= –j.
So an equation of the rectifying plane is
–(y − a) = 0 or y = a.
The osculating plane at P contains the vectors T(π/2) and
N(π/2), so its normal vector is B(π/2). So
B(t) =
c sin t√
a2 + c2
i− c cos t√
a2 + c2
j +
a2√
a2 + c2

k⇒ B
(π
2
)
=
c√
a2 + c2
i +
a2√
a2 + c2
k.
Figure 12.13: The Rectifying, Osculating, and Normal Planes
A simpler normal vector is 〈c, 0, a2〉, so an equation of the
osculating plane is
cx+ a2
(
z − π
2
)
= 0 or z = –
c

a2
x+
π
2
Lastly, the normal plane at P has normal vector r′(π/2) = 〈–a, 0,
c〉, so an equation of it is
–ax+ c
(
z − π
2
)
= 0 or z =
a
c
x+
π
2
.
12.5 Curvature
If C is a smooth curve, then we can use the derivative dT/ds to
measure the ‘‘bendiness’’ of a curve. We
formalize this idea by considering the curvature of a curve. The
curvature of C at a given point is a measure

of how quickly the curve changes direction at that point.
Specifically, we define it to be the magnitude of the
rate of change of the unit tangent vector with respect to arc
length. (We use arc length so that the curvature
is independent of the parametrization.)
12.5.1 Definition and Formulas for Curvature
Definition 12.40 The curvature of a smooth curve in 2-space or
3-space that is parametrized by arc
length is
κ(s) =
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ dTds
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ||r′′(s)|| (12.28)
where T is the unit tangent vector.
We can even express N(s) and B(s) in terms of curvature:
N(s) =
r′′(s)
||r′′(s)||
=
r′′(s)
κ(s)
=

1
κ(s)
dT
ds
(12.29)
B(s) =
r′(s)× r′′(s)
||r′′(s)||
=
r′(s)× r′′(s)
κ(s)
. (12.30)
Is this a good definition for curvature though? Let us put it to
the test.
Proposition 12.41 All lines have zero curvature.
Proof: Recall from Proposition 12.30 that a line in 2-space or 3-
space can be parametrized in terms of arc
length as
r(s) = r0 + su
where the terminal point of r0 is a point on the line and u is a
unit vector parallel to the line. Since r0
and u are constant, their derivatives with respect to s are zero,
and hence r′(s) = u and r′′(s) = 0. Thus
κ(s) = ||r′′(s)|| = 0.

The curvature is easier to compute if it is expressed in terms of
the parameter t instead of s, so we use
the Chain Rule to write
dT
dt
=
dT
ds
ds
dt
and κ(s) =
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ dTds
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ dT/dsds/dt
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ .
But ds/dt = ||r′(t)|| from Equation 12.16, so
κ(t) =
||T′(t)||
||r′(t)||
. (12.31)
Proposition 12.42 The curvature of a circle of radius a is 1/a.
Proof: We can take the circle to have center the origin, and then
a parametrization is
r(t) = a cos ti + a sin tj.

Therefore r′(t) = –a sin ti + a cos tj and ||r′(t)|| = a. So T(t) =
r(t)/||r′(t)|| = – sin ti + cos tj and T′(t) =
– cos ti− sin tj. This gives ||T′(t)|| = 1, so using Equation 12.29,
we have
κ(t) =
||T′(t)||
||r′(t)||
=
1
a
.
The result of Proposition 12.42 shows that small circles have
large curvature and large circles have small
curvature, in accordance with our intuition. This along with the
fact that lines have zero curvature show
that our definition of curvature is a reasonable one.
Although Formula (12.29) can be used in all cases to compute
the curvature, the formula given by the
following theorem is often more convenient to apply.
Theorem 12.43 The curvature of the curve given by the smooth
vector function r is
κ(t) =
||r′(t)× r′′(t)||
||r′(t)||3

. (12.32)
Proof: Since T(t) = r′(t)/||r′(t)||, we have
r′(t) = ||r′(t)||T(t), (12.33)
so the Product Rule gives
r′′(t) = ||r′(t)||′T(t) + ||r′(t)||T′(t). (12.34)
Yet, from the fact that N(t) = T′(t)/||T′(t)|| and Equation 12.29,
we have
T′(t) = ||T′(t)||N(t) and ||T′(t)|| = κ(t)||r′(t)||
so that T′(t) = κ(t)||r′(t)||N(t). Substituting this into Equation
(12.31) yields
r′′(t) = ||r′(t)||′T(t) + κ(t)||r′(t)||2N(t) (12.35)
Thus, from (12.31), (12.33), and the fact that the cross product
of a vector with itself is 0,
r′(t)× r′′(t) = ||r′(t)||T(t)× (||r′(t)||′T(t) + κ(t)||r′(t)||2N(t))
= ||r′(t)|| ||r′(t)||′(T(t)×T(t)) + κ(t)||r′(t)||3(T(t)×N(t)) =
κ(t)||r′(t)||3B(t)
Since B(t) is a unit vector,
||r′(t)× r′′(t)|| =
∣ ∣ ∣ ∣ κ(t)||r′(t)||3B(t)∣ ∣ ∣ ∣ = κ(t)||r′(t)||3.

Formula (12.30) now follows.
Note that the proof above provides us with a way to express
B(t) and, consequently, N(t) directly in
terms of r(t).
Corollary 12.44 If r is a smooth curve, then
B(t) =
r′(t)× r′′(t)
||r′(t)× r′′(t)||
(12.36)
and
N(t) =
r′(t)× (r′′(t)× r′(t))
||r′(t)|| ||r′′(t)× r′(t)||
. (12.37)
Proof: From the computations in the proof to Theorem 12.43,
B(t) =
r′(t)× r′′(t)
κ(t)||r′(t)||3
=
r′(t)× r′′(t)
(||r′(t)× r′′(t)||/||r′(t)||3)||r′(t)||3
=
r′(t)× r′′(t)

||r′(t)× r′′(t)||
,
and so
N(t) = B(t)×T(t)
=
r′(t)× r′′(t)
||r′(t)× r′′(t)||
× r
′(t)
||r′(t)||
= –
r′(t)× (r′(t)× r′′(t))
||r′(t)|| ||r′(t)× r′′(t)||
=
r′(t)× [–(r′(t)× r′′(t))]
||r′(t)|| ||r′(t)× r′′(t)||
=
r′(t)× (r′′(t)× r′(t))
||r′(t)|| ||r′(t)× r′′(t)||
Example 12.45 Let us find the curvature of the twisted cubic
r(t) = 〈t, t2, t3〉 at a general point and at
(0, 0, 0).
r′(t) = 〈1, 2t, 3t2〉
r′′(t) = 〈0, 2, 6t〉

||r′(t)|| =
√
1 + 4t2 + 9t4
r′(t)× r′′(t) =
∣ ∣ ∣ ∣ ∣ ∣
i j k
1 2t 3t2
0 2 6t
∣ ∣ ∣ ∣ ∣ ∣ = 6t2i− 6tj + 2k
||r′(t)× r′′(t)|| =
√
36t4 + 36t2 + 4 = 2
√
9t4 + 9t2 + 1.
Figure 12.14: The Osculating Circle and the Center of Curvature
Theorem 12.43 then gives
κ(t) =
||r′(t)× r′′(t)||
||r′(t)||3
=
2

√
9t4 + 9t2 + 1
(9t4 + 4t2 + 1)3/2
.
At the origin, where t = 0, the curvature is κ(0) = 2.
12.5.2 Radius of Curvature
The circle that lies in the osculating plane of a smooth curve C
at a point P, has the same tangent as C
at P, lies on the concave side of C (toward which N points), and
has radius ρ = 1/κ (the reciprocal of the
curvature) is called the osculating circle (or the circle of
curvature) of C at P. (Figure 12.14) It is the
circle that best describes how C behaves near P ; it shares the
same tangent, normal, and curvature at P.
The radius of ρ of the osculating circle at P is called the radius
of curvature at P, and the center of the
circle is called the center of curvature at P. (Figure 12.14)
12.5.3 An Interpretation of Curvature in 2-Space
A useful geometric interpretation of curvature in 2-space can be
obtained by considering the counterclockwise
angle of elevation φ from the positive x-axis to T. Since T is a
unit vector, we can express T in terms of φ as
T(φ) = cosφi + sinφj.
Thus dT/dφ = – sinφi + cosφj. By the Chain Rule, dT/ds =
(dT/dφ)(dφ/ds), and so we obtain
κ(s) =

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ dTds
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ dTdφ dφds
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ dφds
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ dTdφ
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ dφds
∣ ∣ ∣ ∣ √(– cosφ)2 + sin2φ = ∣ ∣ ∣ ∣ dφds
∣ ∣ ∣ ∣ .
Figure 12.15: The Osculating Circle and the Center of Curvature
In summary, we have shown that
κ(s) =
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ dφds
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ , (12.38)
which tells us that curvature in 2-space can be interpreted as the
size of the rate of change of φ with respect
to s. (Figure 12.15) The greater the curvature, the more rapidly
φ changes with s, and vice-versa. In the
case of a line, the angle φ is constant and, consequently, κ(s) =
|dφ/ds| = 0, which is consistent with the fact
that all lines have zero curvature.
For the special case of a plane curve with equation y = f(x), we
choose x as the parameter and write
r(x) = xi + f(x)j. Then r′(x) = i + f ′(x)j and r′′(x) = f ′′(x)j.

Hence
r′(x)× r′′(x) = (i + f ′(x)j)× f ′′(x)j = f ′′(x)(i× j) + f ′(x)f ′′(x)(j×
j) = f ′′(x)k.
We also have ||r′(x)|| =
√
1 + [f ′(x)]2, and so, by Theorem 12.43,
κ(x) =
|f ′′(x)|
[1 + (f ′(x))2]3/2
(12.39)
Example 12.46 Consider the parabola y = x2. Since y′ = 2x and
y′′ = 2, Formula (12.37) gives
κ(x) =
|y′′|
[1 + (y′)2]3/2
=
2
(1 + 4x2)3/2
.
Observe that κ(x)→ 0 as x→ ±∞. This corresponds to the fact
that the parabola flattens out as x→ ±∞.

Now the curvature of the parabola at the origin is κ(0) = 2. So
the radius of curvature at the origin is
1/κ = 1/2. Given that y′(0) = 0 so that the tangent to the
parabola is horizontal at the origin, the normal to
the parabola at the origin is thus vertical and points upward (the
concave side of the parabola). As a result,
the center of curvature is (0, 1/2). The equation of the
osculating circle is therefore
x2 +
(
y − 1
2
)2
=
1
4
.
? Let’s consider the problem of designing a railroad track to
make a smooth transition between sections of
straight track. Existing track along the negative x-axis is to be
joined smoothly to a track along the line
y = 1 for x ≥ 1. Find a polynomial P = P (x) of degree 5 such
that the function F defined by
F (x) =
1 if x ≥ 1

is continuous and has continuous slope and continuous
curvature.
12.6 Motion Along a Curve
In this section, we show how the ideas of tangent and normal
vectors and curvature can be used in physics
to study the motion of an object, including its velocity and
acceleration, along a space curve. After this, the
reader will be prepared to understand Newton’s derivation of
Kepler’s laws of planetary motion.
12.6.1 Velocity, Acceleration, Speed, Displacement, and
Distance Traveled
Definition 12.47 If r(t) is the position function of a particle
moving along a curve in 2-space or 3-space, then
the instantaneous velocity, instantaneous acceleration, and
instantaneous speed of the particle at
time t are
velocity = v(t) = r′(t)
acceleration = a(t) = v′(t) = r′′(t)
speed = ||r′(t)|| = ds
dt
.
The displacement of the particle over the time t1 ≤ t ≤ t2 is
∆r = r(t2)− r(t1).
The displacement vector, which describes the change in position
during the time interval, can be obtained

by integrating the velocity function from t1 to t2 :
Figure 12.16: Equations of Motion
∫ t2
t1
v(t) dt =
∫ t2
t1
r′(t) dt = r(t)|t2t1 = r(t2)− r(t1) = ∆r.
It follows from Equation (12.12) that we can find the distance s
traveled by a particle over a time interval
t1 ≤ t ≤ t2 by integrating the speed over that interval, since
s =
∫ t2
t1
||r′(t)|| dt =
∫ t2
t1
||v(t)|| dt.
Nothing here yet is really new. We have only translated the
content of sections 12.2 and 12.3 into the
language of kinematics.

12.6.2 Normal and Tangential Components of Acceleration
When we study the motion of a particle, it is often useful to
resolve the acceleration into two components,
one in the direction of the tangent and the other in the direction
of the normal.
Theorem 12.48 If a particle moves along a smooth curve C in 2-
space or 3-space, then at each point on
the curve, the velocity and acceleration vectors can be written
as
v =
ds
dt
T (12.40)
and
a =
d2s
dt2
T + κ
(
ds
dt
)2
N (12.41)

where s is an arc length parameter for the curve, and T, N, and κ
denote the unit tangent, unit normal and
curvature at a point.
Proof: Recall that
T(t) =
r′(t)
||r′(t)||
=
v
ds/dt
and so
v =
ds
dt
T.
If we differentiate both sides of this equation with respect to t
along with the fact that N = T′/||T′||,
κ = ||T′||/||r′||, and ||r′(t)|| = ds/dt, we get
a =
d
dt

(
ds
dt
T
)
=
d2s
dt2
T +
ds
dt
dT
dt
=
d2s
dt2
T +
ds
dt
||T′||N
=
d2s
dt2

T +
ds
dt
κ||r′||N = d
2s
dt2
T + κ
(
ds
dt
)2
N.
The coefficients of T and N are commonly denoted by
aT =
d2s
dt2
(12.42)
and
an = κ
(
ds
dt

)2
, (12.43)
in which case Formula (12.41) is expressed as
a = aTT + aNN (12.44)
In this formula, the scalars aT and aN are called the tangential
scalar component of acceleration and
the normal scalar component of acceleration, and the vectors
aTT and anN are called the tangential
vector component of acceleration and the normal vector
component of acceleration.
Let us look at what Formula (12.41) says. The first thing to
notice is that the binormal vector B is
absent. No matter how an object moves through space, its
acceleration always lies in the TN-plane (the
osculating plane). This makes sense, because T gives the
direction of motion and N points in the direction
the curve is turning. Next notice that the tangential component
of acceleration is d2s/dt2 = d(ds/dt)/dt,
the rate of change of speed, and the normal component of
acceleration is κ(ds/dt)2, the curvature times the
square of the speed. For tangibility, think of a passenger in a
car. If a car speeds up rapidly, then your body
is thrown back against the backrest of the seat. A sharp turn in a
road means a large value of the curvature
κ, so the component of the acceleration perpendicular to the
motion is large and the passenger is thrown
against a car door, toward the outside of the curve. High speed

around the turn has the same effect. In fact,
if you double your speed, aN is increased by a factor of 4.
Although we have expressions for the tangential and normal
components of acceleration in Formula
(12.41), it is desirable to have expressions that only depend on
r, r′, and r′′.
space or 3-space, then
aT =
v •a
||v||
, (12.45)
aN =
||v × a||
||v||
, (12.46)
and
κ =
||v × a||
||v||3
(12.47)
Proof: The formula for κ is just the one from Theorem 12.43
restated in terms of v and a. Taking the dot
product of v = (ds/dt)T with a as given by Formula (12.41)
gives

v •a =
ds
dt
T •
(
d2s
dt2
T + κ
(
ds
dt
)2
N
)
=
ds
dt
d2s
dt2
(T •T) + κ
(
ds
dt

)2
(T •N) = ||v||aT
since ds/dt = ||v||, d2s/dt2 = aT , T •T = ||T||2, as T is a unit
vector, and T •N = 0, as they are orthogonal.
Hence aT = v •a/||v||. Finally,
an = κ
(
ds
dt
)2
=
||v × a||
||v||3
||v||2 = ||v × a||
||v||
.
Example 12.50 A particle moves with position function r(t) =
〈t2, t2, t3〉. Then
v(t) = 〈2t, 2t, 3t2〉
a(t) = 〈2, 2, 6t〉
||v(t)|| =
√

8t2 + 9t4.
Therefore
aT (t) =
v(t) •a(t)
||v(t)||
=
8t+ 18t2√
8t2 + 9t4
.
Since
v(t)× a(t) =
∣ ∣ ∣ ∣ ∣ ∣
i j k
2t 2t 3t2
2 2 6t
∣ ∣ ∣ ∣ ∣ ∣ | = 6t2i− 6t2j,
aN (t) =
v(t)× a(t)
||v(t)||
=
6
√

2t2√
8t2 + 9t4
.
Since
T(t) =
v(t)
||v(t)||
=
⟨
2t√
8t2 + 9t4
,
2t√
8t2 + 9t4
,
3t2√
8t2 + 9t4
⟩
,
the tangential vector component of acceleration at time t is
aTT =
⟨
2t(8t+ 18t2)

8t2 + 9t4
,
2t(8t+ 18t2)
8t2 + 9t4
,
3t2(8t+ 18t2)
8t2 + 9t4
⟩
.
Since a = aTT + aNN, we can simply calculate the normal
vector component of acceleration as
aNN = a− aTT =
⟨
2− 2t(8t+ 18t
2)
8t2 + 9t4
, 2− 2t(8t+ 18t
2)
8t2 + 9t4
, 6t− 3t
2(8t+ 18t2)
8t2 + 9t4

⟩
Finally
κ(t) =
||v(t)× a(t)||
||v(t)||3
=
6
√
2t2
(8t2 + 9t4)3/2
.
In the case where ||a|| and aT are known, there is a useful
alternative to Formula (12.46) for aN that does
not require the calculation of a cross product.
space or 3-space, then
aN =
√
||a||2 − a2T .
Proof: Since a = aTT + aNN and aTT and aNN are orthogonal
vectors (since T and N themselves are
orthogonal), it follows by the Pythagorean Theorem for vectors

that
||an||2 = ||aTT||2 + ||aNN||2 = |aT |2||T||2 + |aN |2||N||2 = a2T +
a2N .
The result now follows.
You are now prepared to understand one of the great
accomplishments of calculus. Read section 12.7 in
the textbook to see how Isaac Newton used the material of this
chapter to prove Kepler’s laws of planetary
motion. You will not be disappointed!
References
[ABD12] H. Anton, I. Bivens, and S. Davis, Calculus: Early
Transcendentals, 10th Edition, Brooks/Cole,
2005, pp. 841--888.
[S05] I. Stewart, Calculus: Concepts & Contexts, 3rd Edition,
Wiley and Sons, 2012, pp. 694--722.
Extra Credit Quiz Math 200-004, Multivariate Calculus, Winter
2014
Name:
Given that this is a take-home quiz, it is expected that your
work will be extremely well-organized and
immaculate. I reserve the right to deduct points for failing to
meet these expectations.
Problem 1. Reparametrize the curve

r(t) =
(
2
t2 + 1
− 1
)
i +
2t
t2 + 1
j
with respect to arc length measured from the point (1, 0) in the
direction of increasing t. Express the
reparametrization in its simplest form. What can you conclude
about the curve? (5 points)
Problem 2. A particle moving through 3-space has velocity
vector
v(t) = 3et sin(t)i + 4et sin(t)j + 5et cos(t)k.
Find its speed, unit tangent vector T, principal unit normal
vector N, binormal vector B, curvature, scalar
and vector tangential components of acceleration, and scalar
and vector normal components of acceleration
as functions of time t. (5 points)
Communication Channel and Context Matrices

COMM/400 Version 5
1
University of Phoenix Material
Communication Channel and Context Matrices
Part I – Communication Channel Matrix
Fill in descriptions of the characteristics and examples, pros,
cons, and recommended etiquette of each communication
channel.
Communication Channel Matrix
Communication channel
Characteristics and examples
Pros
Cons
Etiquette for managers and staff
Personal e-mail
Company-wide e-mail
Phone call
Teleconference

Virtual meeting or web conference
Face-to-face meeting
Part II – Communication Context Matrix
Recommend and provide justification for the appropriate
communication channel you would use in the following
contexts. In your justification, explain whether the channels
may vary according to company size or culture.
Communication Context Matrix
Situation
Recommended channel
(specify the type of intrapersonal, interpersonal, public, mass,
or computer-mediated context channel)
Justification
Publicizing a change in employee benefits
Handling a conflict situation between virtual teams
Handing a conflict situation between a manager and an
employee
Detailing a new procedure to a small number of employees
Training a team on a new software program

Explaining a new process to the company
Announcing promotions
Announcing the termination of a dangerous employee
Announcing a major reorganization
Announcing a major layoff cycle
References

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