Section 11.6: Vector Functions and Space Curves
~
Definition: A vector function is a function R(t)
= hf (t), g(t), h(t)i that assigns a unique
3
vector in R to every value of t in its domain. The functions f , g, and h are called the
~
component functions of R.
~ is the set of all values t ∈ R such that R(t)
~
Note: The domain of a vector function R
is
defined.
~
Example: Find the domain of R(t)
=
√
1
.
t − 2, ln(8 − t),
t−4
√
The component function t − 2 is defined for t ≥ 2. The component function ln(8 − t) is
1
~ is
defined for t < 8. The component function t−4
is defined for t 6= 4. So the domain of R
[2, 4) ∪ (4, 8).
Definition: The limit, derivative, and integral of a vector function are defined componentwise.
~
That is, if R(t)
= hf (t), g(t), h(t)i, then
D
E
~
lim R(t) = lim f (t), lim g(t), lim h(t) ,
t→a
t→a
t→a
t→a
~ 0 (t) = hf 0 (t), g 0 (t), h0 (t)i,
R
Z
Z
Z
Z
~
f (t)dt, g(t)dt, h(t)dt .
R(t)dt =
−1
−t 2t − 3
Example: Find lim e ,
, tan t .
t→∞
4t + 1
By definition,
2t − 3
1 π
−t 2t − 3
−1
−t
−1
lim e ,
, tan t = lim e , lim
, lim tan t = 0, ,
.
t→∞
t→∞
t→∞ 4t + 1 t→∞
4t + 1
2 2
~
Example: Find the derivative of R(t)
=
√
t
−t
4t2 + 3,
, e sin t .
ln t
By definition,
~ 0 (t) =
R
4t
ln t − 1 −t
√
,
, e (cos t − sin t) .
4t2 + 3 (ln t)2
~ 0 (t) = hsin t, − cos t, 2ti and R(0)
~
~
Example: If R
= h1, 1, 2i, find R(t).
By the Fundamental Theorem of Calculus,
Z
~
R(t) = hsin t, − cos t, 2tidt = h− cos t + C1 , − sin t + C2 , t2 + C3 i.
~
Then R(0)
= hC1 − 1, C2 , C3 i = h1, 1, 2i. So C1 = 2, C2 = 1, and C3 = 2. Thus,
~
R(t)
= h2 − cos t, 1 − sin t, t2 + 2i.
~
= hf (t), g(t), h(t)i is the set of all points
Definition: The graph of a vector function R(t)
3
(x, y, z) ∈ R such that x = f (t), y = g(t), and z = h(t). The graph is a curve in R3 called
a space curve.
~
Example: Sketch the space curve defined by R(t)
= hcos t, sin t, ti.
Let x = cos t, y = sin t, and z = t. Then
x2 + y 2 = cos2 t + sin2 t = 1.
Thus, the curve lies on a circular cylinder of radius 1 centered about the z-axis. Since z = t,
the curve moves upward along the cylinder as t increases. This curve is called a helix.
~
Figure 1: Graph of the helix defined by R(t)
= hcos t, sin t, ti.
~
Example: Sketch the space curve defined by R(t)
= ht cos t, t sin t, ti.
Let x = t cos t, y = t sin t, and z = t. Then
x2 + y 2 = t2 cos2 t + t2 sin2 t = t2 .
This curve is similar to a helix, but has an increasing radius.
~
Figure 2: Graph of the space curve defined by R(t)
= ht cos t, t sin t, ti.
~ 0 (t) is called the tangent vector since it lies on the tangent line to
Definition: The vector R
~
the space curve defined by R(t).
The vector
~ 0 (t)
R
T~ (t) =
~ 0 (t)||
||R
is called the unit tangent vector.
Example: Find parametric equations for the tangent line to the space curve defined by
√
~
R(t)
= h4 ln t, 8 t, t3 i at (0, 8, 1).
The tangent vector is
~0
R (t) =
4 4
2
, √ , 3t .
t
t
~ 0 (1) = h4, 4, 3i. The tangent line is
At the point (0, 8, 1), t = 1, and the tangent vector is R
defined by the parametric equations
x = 4t,
y = 8 + 4t,
z = 1 + 3t.
~ 1 (t) = h3t, t2 , t3 i and R
~ 2 (t) = hsin t, sin(5t), 4ti intersect
Example: The curves defined by R
at the origin. Find the angle of intersection.
The tangent vectors for these curves are
~ 10 (t) = h3, 2t, 3t2 i
R
~ 2 (t) = hcos t, −5 cos(5t), 4i.
R
~ 0 (0) = h3, 0, 0i and R
~ 0 (0) = h1, −5, 4i. If θ is the angle of intersection,
At the origin (t = 0), R
1
2
then
~ 20 (0)
~ 10 (0) · R
3
R
1
= √ =√ .
cos θ =
0
0
~ 2 (0)||
~ 1 (0)||||R
3 42
42
||R
Then
−1
θ = cos
1
√
42
≈ 81◦ .