MATH 214: VECTOR-VALUED FUNCTIONS
1. Introduction
In this section we consider the following
Definition. A curve is a collection of points of the form
{(x, y, z) | x = f (t), y = g(t), z = h(t)}
for some functions f, g, h : R → R. The functions f, g, h are called the coordinate functions of the
−→
curve. Given a point P (f (t), g(t), h(t)) on the curve, the vector OP is called the position vector of
P . We usually denote
r(t) = hf (t), g(t), h(t)i
for the function that takes a value t to the position vector for P (f (t), g(t), h(t)).
Example 1. Lines, considered in the last section, are the simplest example of curves. The function
−→
r(t) = OP + tv
gives the position vector for a point on the line.
Example 2. Consider the set of points:
1
{(cos(t), sin(t), ) | t ∈ (π, 10π)}
t
this is a curve.
For a curve, the function r(t) takes a number t to a vector r(t) (the position vector associated to
the point on a curve). This is an example of a
Definition. A vector-valued function is a function that takes a number to a vector.
Regular functions R → R are also known as scalar-valued functions
1.1. Limits and Continuity. Limits of vector-valued functions are defined the same way as limits
of regular functions:
Definition. Say that r(t) approaches the limit L ∈ R3 as t approaches t0 if, for every > 0, there
exists a δ > 0 such that |t − t0 | < δ, |r(t) − L| < . We write this as
lim r(t) = L
t→t0
It doesn’t take much work (although it is not directly from the definition) to see that
Proposition 1.1. If r(t) = hf (t), g(t), h(t)i, then lim r(t) = h`1 , `2 , `3 i if and only if
t→t0
lim f (t) = `1 , lim g(t) = `2 , and lim h(t) = `3
t→t0
t→t0
t→t0
−1
Example 3. The limit of r(t) = h 1t sin(t), tan
(t)−t
, sin(π
t3
· t)i as t → 0 is h1, −1
3 , 0i.
The definition of continuity is also defined the same way for vector-valued functions as it is for
scalar-valued functions:
Definition. A vector valued function is continuous at a point t0 if lim r(t) = r(t0 ).
t→t0
A vector-valued function is continuous if and only if each of the component functions is continuous.
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MATH 214: VECTOR-VALUED FUNCTIONS
Example 4. The previous vector-valued function is not continuous at t0 = 0 because the first two
functions are not defined at t0 = 0.
t2
The vector-valued function r(t) = hln(t), cos(t), t−1
i is continuous for all t > 0 except t = 1.
1.2. Derivatives.
Definition. If r(t) = hf (t), g(t), h(t)i and f, g, h are differentiable then r(t) is differentiable and the
derivative of r(t) is defined as
r0 (t) = lim
∆t→0
r(t + ∆t) − r(t)
= hf 0 (t), g 0 (t), h0 (t)i
∆t
r(t) is smooth if it is differentiable and r0 (t) 6= h0, 0, 0i for any t. A curve is piecewise smooth if it
can be constructed by joining smooth curves together, end to end.
−1
Example 5. The curve h 1t sin(t), tan t3(t)−t , sin(π · t)i is smooth on any curve not containing t = 0,
and if we consider the curve given by
(
if t ≤ 0
h1 − t, −t
3 , ti
r(t) =
tan−1 (t)−t
1
h t sin(t),
, sin(π · t)i if t > 0
t3
The curve is continuous by construction and it is piecewise smooth, although it is not smooth at
t = 0.
If we assume that r(t) describes the position vector of a particle at time t, then we given the
following definitions:
Definition.
(1) The velocity vector of the particle is v =
(2) The speed of the particle is |v|
d2 r
(3) The acceleration of the particle is a = dv
dt = dt2
v
(4) The unit vector |v| is the direction of motion
dr
dt
Example 6. Consider a hang-glider rising in a warm air current whose motion is described by
r(t) = h3 cos(t), 3 sin(t), t2 i. Then
v(t) = h−3 sin(t), 3 cos(t), 2ti and a(t) = h−3 cos(t), −3 sin(t), 2i
q
√
This this means that the speed at t is |v| = 9 sin2 (t) + 9 cos2 (t) + 4t2 = 9 + 4t2 and the direction
sin(t) 3 cos(t)
2t
√
of motion at time t is h −3
, √9+4t2 , √9+4t
i.
2
9+4t2
When is the direction of the glider’s motion orthogonal to the glider’s acceleration? When is it
parallel?
v
:
Both of these questions require us to look at a · |v
a·
v
−3 sin(t) 3 cos(t)
2t
= h−3 cos(t), −3 sin(t), 2i · h √
,√
,√
i
|v|
9 + 4t2
9 + 4t2
9 + 4t2
9 cos(t) sin(t) − 9 cos(t) sin(t) + 4t
√
=
9 + 4t2
4t
√
=
9 + 4t2
4t
thus the two vectors are orthogonal when t = 0 and they are parallel when √9+4t
= |a| =
2
which has no solutions (i.e. the acceleration is never parallel to the direction of motion).
√
13,
MATH 214: VECTOR-VALUED FUNCTIONS
3
1.3. Differentiation Rules. All of the following rules are consequences of the single-variable differentiation rules for f (t), g(t) and h(t) as well as the definitions:
d
Theorem 1.1.
(1) dt
(f (t) · r(t)) = f 0 (t)r(t) + f (t)r0 (t)
d
(2) dt (c1 r1 (t) + c2 r2 (t)) = c1 r10 (t) + c2 r20 (t)
d
r(t) · s(t) = r0 (t) · s(t) + r(t) · s0 (t)
(3) dt
d
(4) dt [r(t) × s(t)] = r0 (t) × s(t) + r(t) × s0 (t) NOTE THE ORDER
d
(5) dt
r(f (t)) = f 0 (t)r0 (f (t))
sin(t) 1
Exercise. Find the derivative of r(t) = h cos(t)
t ,
t , ti
Exercise. Use the derivative rules to find
d
dt u(t)(v(t)
× w(t))
1.4. Integrals of Vector Functions.
Definition. A differentiable vector-valued function R(t) is an antiderrivative of r(t) if R0 (t) = r(t).
Note that finding vector-valued antiderrivatives breaks down to finding three scalar-valued antiderrivatives. We also use similar definitions to the scalar case:
Definition. The indefinite integral of r(t) is
Z
r(t)dt = R(t) + C
for any antiderivative R(t) and any constant vector C.
Definition. If the components of r(t) are integral over [a, b] we define the definite integral of r(t)
to be
Z b
Z b
Z b
Z b
r(t)dt =
f (t)dti +
g(t)dtj +
h(t)dtk
a
a
a
a
Note that the fundamental theorem of calculus still holds for vector-valued integrals.
Exercise. From the example of the glider, starting with the assumption that a(t) = h−3 cos(t), −3 sin(t), 2i
and r(0) = h3, 0, 0i, v(0) = h0, 3, 0i to find the position vector as a function of time.
2. Arc Length and the Unit Tangent Vector T
We’ve seen arc length for curves in the xy-plane in polar coordinates, now we look at lengths of
curves in 3d-space:
Definition. The length of a smooth curve r(t) = x(t)i + y(t)j + z(t)k from r(a) to r(b) is
s
Z b 2 2 2
Z b
dx
dy
dz
+
+
dt =
|v(t)|dt
dt
dt
dt
a
a
Example 7. Suppose that we wrap a string around a cylinder of radius 2 in such a way that each
time we complete one revolution, we move up by the same amount. How much string would we need
to wrap the string 3 times?
Example 8. The cylinder is described by x2 + y 2 = 4 and z is allowed to be any value. If we
consider the curve r(t) = h2 cos(t), 2 sin(t), ti, then this curve satisfies the conditions in the question.
The question is then answered by calculating (using the fact that three rotations is 6π radians):
Z 6π q
Z 6π √
√
2
2
4 cos (t) + 4 sin (t) + 1dt =
5dt = 6π 5
0
0
In the last section, we defined the direction of motion for a smooth curve r(t) to be T =
v
|v| . From this point on we call T the unit tangent vector of r(t).
r 0 (t)
|r 0 (t)|
=
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MATH 214: VECTOR-VALUED FUNCTIONS
Example 9. In the hang glider example from last section, we already calculated
−3 sin(t) 3 cos(t)
2t
T (t) = h √
,√
,√
i
9 + 4t2
9 + 4t2
9 + 4t2
3. Curvature and the Normal Vector
3.1. Curves of constant length. Assume that |r(t)| = c for some constant c and any t in our
d
d
interval (or t ∈ R for infinite curves). Then dt
|r(t)| = dt
c = 0. From our differentiation rules, we
get:
d
d
|r(t)| = r(t) · r(t) = r0 (t) · r(t) + r(t) · r0 (t) = 2r0 (t) · r(t) = 0
dt
dt
this means that r0 (t) · r(t) = 0, i.e. r0 (t) and r(t) are orthogonal.
3.2. Unit Normal Vector. Recall that T (t) has constant length 1. Using the last section, we can
therefore say that
T·
dT
=0
dt
i.e. T and its derivative are orthogonal.
Definition. The principal unit normal vector N to r(t) is given by
N=
1
dT
|dT /dt| dt
Definition. We define the curvature of our curve r(t) to be κ =
1
|v|
dT .
dt
The value of κ determines how sharply the vector T is turning as s advances through the curve.
√
2
3
√
2
2t
2t
Example 10. For the example r(t) = h 2t
2 , 3 − 1, 2 i, we have calculated already that |v| =
√
2
2t 1 + t and so
√
√
2
t
2
,√
, √
i
T =h √
2
2
2 1+t
1 + t 2 1 + t2
Next we calculate that
√
√
dT
− 2·t
1
− 2·t
=h
3 ,
3 ,
3 i
dt
2 (1 + t2 ) 2 (1 + t2 ) 2 2 (1 + t2 ) 2
so
p
dT 1
1
2
=
3
dt (1 + t2 ) 2 t + 1 = 1 + t2
This gives that κ =
1
3
2t(1+t2 ) 2
and
√
√
− 2t
1
− 2t
√
√
√
N =h
,
,
i
2 1 + t2
1 + t2 2 1 + t2
Notice that the curvature κ increases to infinity as t → 0, which reflects the fact that the curve is
not smooth at t = 0 (you can think of the cusp as the tangent vector moving ‘infinitely fast’)
MATH 214: VECTOR-VALUED FUNCTIONS
5
4. Unit Binormal Vector and Torsion
Definition. The unit binormal vector to a smooth curve r(t) is B(t) = T (t) × N (t).
B is called a unit vector because |B| = |T ||N | sin( π2 ) = 1.
Recall from the definition of the cross product that B(t) is perpendicular to T (t) and N (t).
The three vectors form what is known as an orthonormal basis of R3 , i.e. all three vectors are
perpendicular to each other and they all have length 1. The vectors T (t), N (t) and B(t) form what
is known as the Frenet frame (or the TNB frame).
Using the cross product rule of derivatives:
d
dT
dN
d
B = (T × N ) =
×N +T ×
dt
dt
dt
dt
dB
and the fact that N is in the same direction as dT
dt (by definition of N ), we get that dt is perpendicular
dN
to T (it is also perpendicular to dt , but we don’t care).
Since B is a unit vector, it has constant length, and so B is perpendicular to dB
dt . Because T, B
dB
and N form an orthonormal basis, this gives that dB
is
parallel
to
N
,
say
=
−τ
|v|N .
dt
dt
Definition. τ is called the torsion function associated to the curve r(t).
1
dB
Notice that τ = − |v|
dt · N . It is possible to show (but we won’t here) that:
τ=
(v(t) × a(t)) · a0 (t)
|v(t) × a(t)|2
as long as v(t) × a(t) 6= h0, 0, 0i.
√
Example 11. For the curve r(t) = h
√
√
2t2 2t3
2 , 3
− 1,
√ 2
2t
2 i
√
√
2
t
, √1+t
, √ 2 i
we know T (t) = h 2√1+t
2
2 2 1+t2
√ 2t , √ 1
√ 2t i This gives that
and N (t) = h 2−
, −
1+t2
1+t2 2 1+t2
√
√
− 2
2
B(t) = T (t) × N (t) = h
, 0,
i
2
2
To calculate τ , we’ll use the formula given above. First, we need to calculate a(t) and a0 (t):
√
√
a(t) = v 0 (t) = h 2, 4t, 2i and a0 (t) = v 00 (t) = h0, 4, 0i
√
√
this gives that v(t) × a(t) = h−2 2t2 , 0, 2 2t2 i =
6 h0, 0, 0i, so
√
√
(v(t) × a(t)) · a0 (t)
0(−2 2t2 ) + 4 · 0 + 0(2 2t2 )
0
τ=
=
=
=0
|v(t) × a(t)|2
16t4
16t4
We could have seen that τ = 0 in the last example by using the fact that B is constant, so dB
ds = 0.
In general, the geometric meaning of the Frenet frame is this:
• The unit tangent vector gives the direction of motion on the curve
• The unit normal vector tells us which direction the motion is changing in (i.e. the direction
of the velocity changes in the direction of the unit normal vector) and the curvature tells us
how quickly the motion is changing
• The torsion tells us how much the motion is moving out of the T N plane (i.e. which way
the acceleration is changing) and the binormal unit vector gives us the direction that the
motion is changing in.
In situations where we do have to calculate the curvature, we can use the formula (which we
won’t derive here):
|v × a|
κ=
|v|3
notice that when v is parallel to a, we get that κ = 0.
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MATH 214: VECTOR-VALUED FUNCTIONS
√
√
2t2 2t3
2 , 3
2
− 1, 2t
2 i, we have that
p
2t
2 + 4t2
d
d
and aN = κ|v(t)|2 = √
aT = |v(t)| = 2t 1 + t2 = √
dt
dt
1 + t2
1 + t2
so the decomposition of the acceleration into normal and tangential components is given by:
√
√
√
√
√
√
2
t
2
2t
− 2t
1
− 2t
2 + 4t2
√
√
√
√
√
√
√
√
a(t) = h 2, 4t, 2i =
h
,
,
i+
h
,
,
i
1 + t2 2 1 + t2
1 + t2 2 1 + t2
1 + t2 2 1 + t2
1 + t2 2 1 + t2
Example 12. For r(t) = h
5. Summary of Formulae
I list the formulae needed for curves in space here for reference:
v
T = |v|
1 dT κ = |v|
dt =
N=
|v×a|
|v|3
dT /dt
|dT /dt|
B =T ×N
τ=
(v(t)×a(t))·a0 (t)
|v(t)×a(t)|2
1
= − |v|
dB
dt
·N
If we plug in v(t) = hx0 (t), y 0 (t), z 0 (t)i as well as the definitions for acceleration and a0 (t) then the
formula for the the torsion can be restated as
0
0
0 x
00 y00 z00 x
000 y000 z000 x
y
z τ=
|v × a|2
where the numerator is a regular determinant.
Example 13. Calculate the Frenet frame, the curvature and the torsion for the helix r(t) =
ha cos(t), a sin(t), bti
sin(t) a cos(t)
√
Solution. First, v(t) = h−a sin(t), a cos(t), bi so T = h −a
, √a2 +b2 , √a2b+b2 i and
a2 +b2
dT
−a cos(t) −a sin(t)
= h√
,√
, 0i
dt
a2 + b2
a 2 + b2
so dT
dt =
√ a
a2 +b2
and N = h− cos(t), − sin(t), 0i. The binormal is
b sin(t) −b cos(t)
a
B = T × N = h√
,√
,√
i
2
2
2
2
2
a +b
a +b
a + b2
the curvature is given by
1
a
a
√
= 2
2
2
2
a
+
b2
+b
a +b
To calculate the torsion, first we calculate v × a:
κ= √
a2
v × a = h−ab sin(t), −ab cos(t), a2 i
so |v × a|2 = a2 (a2 + b2 ) and
0
0
0 x
00 y00 z00 x
000 y000 z000 x
y
z =
τ=
|v × a|2
−a sin(t) a cos(t)
−a cos(t) −a sin(t)
a sin(t) −a cos(t)
a2 (a2 + b2 )
b 0
0
=
a2 b
b
= 2
+ b2 )
a + b2
a2 (a2