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Calculus III (part 1): Vectors and 3-Dimensional Geometry
(by Evan Dummit, 2015, v. 2.25)
Contents
1
Vectors and 3-Dimensional Geometry
1.1
1
Functions of Several Variables and 3-Space
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . .
2
. . . . . . . . . . . . . . . . . . . . . . . .
4
1.2
Vectors and Scalars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.3
The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.4
The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.5
Lines and Planes in 3-Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
1.6
Vector-Valued Functions of One Variable, Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
1.7
1.1.1
Graphing Functions of 2 Variables: Surfaces and Level Sets
1.1.2
Graphing Functions of 3 Variables: Level Surfaces
1
Curves in 3-Space:
Velocity and Acceleration; Arclength; Unit Tangent, Normal, and Binormal
Vectors; Curvature and Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
14
Vectors and 3-Dimensional Geometry
•
f (x), we know how to produce the
y = f (x). We also know how to graph parametrically-dened functions, like the cycloid
x = t + sin(t), y = 1 + cos(t). And we even know, more or less, how to graph implicitly-dened functions, like
2
2
the circle x + y = 1.
We are very used to 2-dimensional planar geometry: given a function
graph of the curve
•
We now introduce 3-dimensional space (often called 3-space for short), and study the geometry of lines and
planes, as well as more general curves and surfaces. Along the way, we will introduce and study vectors, which
help to clarify a great deal of the concepts.
1.1
•
Functions of Several Variables and 3-Space
A function of several variables is, as the name indicates, a function that takes in several variables and outputs
a value associated to the inputs.
◦
◦
f (x, y) = x + y
Example: The function
Example: The function
takes in two values
g(x, y, z) = x y z
x y z2.
2
x
and
takes in three values
y
and outputs their sum
x, y ,
and
z,
x + y.
and outputs the product
√
l2 + w2 gives the length of a diagonal of a rectangle as a function of
w.
1
◦ Example: The function V (r, h) = πr2 h gives the volume of a (right circular) cone as a function of its
3
base radius r and its height h.
◦
Example: The function
the rectangle's length
•
l
d(l, w) =
and its width
As with functions of one variable, every function has a domain and a range: the domain is still the set of
input values, and the range is still the set of output values.
◦
For
f (x, y)
a function of two variables, the domain is now a subset of the 2-dimensional plane, rather
than a subset of the real line. Because of this, domains of functions of more than one variable can be
more complicated.
1
√
√
f (x, y) = x + y , the domain is the rst quadrant of the xy -plane, dened by the
inequalities x ≥ 0 and y ≥ 0.
p
◦ Example: For f (x, y) = x2 + y 2 − 1, the domain is the set of points in the xy -plane which satisfy
x2 + y 2 ≥ 1: this describes all the points of the plane except for those lying strictly inside the unit circle.
◦
•
Example:
Points in 3-space are represented by a triplet of numbers
above the
•
For
•
The new coordinate
z
represents height
In 3-space, we have a distance formula, which is just the Pythagorean Theorem applied twice: it says that
the distance between points
1.1.1
(x, y, z).
xy -plane.
(x1 , y1 , z1 )
and
(x2 , y2 , z2 )
is given by
p
(x1 − x2 )2 + (y1 − y2 )2 + (z1 − z2 )2 .
Graphing Functions of 2 Variables: Surfaces and Level Sets
There are two primary ways to visualize a function
◦
(x, y, z)
The rst way is to plot the points
∗
At the point
(x, y)
The second way is to plot the points
of
c),
∗
of two variables.
in 3-dimensional space satisfying
in the plane, the graph has the height
through the plane, the function
◦
f (x, y)
z = f (x, y)
z = f (x, y);
z = f (x, y).
so we see that as
will trace out a surface, called the graph of
(x, y) in the plane on the level
sets
(x, y) varies
f (x, y).
f (x, y) = c (for particular values
as implicit curves.
f (x, y) and a particular value of c, the points (x, y) satisfying f (x, y) = c
f . For a function of two variables these sets will generally be curves, so they
For a given function
are
called a level set of
are
also sometimes called level curves.
∗
If we graph many of these level curves together on the same axes, we will obtain a topographical
map of the function
•
Some example of simple graphs
◦
Example: The graph
∗
z=0
z = f (x, y)
is the
are given below.
xy -plane.
Note more generally that any equation of the form
(with not all of
•
f (x, y).
a, b, c
◦
Example: The graph
◦
Example: The graph
ax + by + cz = d
for some constants
a, b, c, d
zero) will give a plane. We will study planes in more depth in a later section.
z = x2 + y 2 is
p
z = x2 + y 2
a paraboloid (i.e., a parabolic dish).
is a right circular cone opening upward, with vertex at the origin.
Some examples of level sets are given below:
◦
Example: The level sets for the function
level set
2
2
x +y = c
just the single point
(for
c > 0)
(0, 0),
f (x, y) = x2 + y√2
is a circle with radius
and for
c<0
are circles in the plane. More specically, the
c
centered at
(0, 0).
For
c=0
the level set is
the level sets do not contain any points at all. The rst graph
2
is the level set
x2 + y 2 = 1;
the second graph contains level sets for
3
3
2
2
1
1
0
0
-1
-1
-2
-2
-3
-3
-3
◦
c = 1, 2, 3, · · · , 8, 9.
-2
-1
0
1
2
-3
3
-2
-1
0
1
2
3
f (x, y) = x2 − y 2 , along with the 3D
c = −4, −3, −2, −1, 0, and the second
Example: Here are two collections of level curves for the function
graph
z = f (x, y).
The rst set of level curves is
set of level curves is
f (x, y) = c
for
3
3
2
2
1
1
0
0
-1
-1
-2
-2
-3
f (x, y) = c
for
c = 0, 1, 2, 3, 4.
-3
-3
∗
-2
-1
0
1
2
-3
3
-2
-1
0
1
2
3
z = x2 − y 2 is called a hyperbolic paraboloid, or more colloquially, a saddle, since it
curves upward along the x-direction but downward along the y -direction. [The hyperbolic paraboloid
The graph of
is called that because it looks like a hyperbola in one cross-section, and a parabola in two others.]
•
2
f (x, y) = (x2 − y 2 )2 e−x
−3 ≤ x ≤ 3 and −3 ≤ y ≤ 3:
Example: Here are some level curves for the function
z = f (x, y),
both plotted on the region
−y 2
, along with a 3D graph of
3
2
1
0
-1
-2
-3
-3
-2
◦
-1
0
1
2
3
As one can see from comparing the plots, the level curves indicate where the function is changing in
value: many level curves grouped closely together indicates that the function is changing quickly so that
the function's graph will be steep, while having level curves grouped far apart means that the function's
graph will be fairly at.
•
Here are a few more graphs, which appear here for no particular reason:
◦
Example: The graph
z = x3 − 3xy 2
is called the monkey saddle, as it has three depressions rather than
the two for the regular saddle (one for each leg, and one for the tail).
√
◦
Example: The graph
z = e3−
x2 +y 2 /12
· cos
p
x2 + y 2
pool of water.
3
produces a surface that looks like ripples in a
1.1.2
•
Graphing Functions of 3 Variables: Level Surfaces
We can also represent some information graphically for functions
f (x, y, z)
of three variables. Unfortunately,
we cannot really produce proper graphs of these functions, since plotting a graph
w = f (x, y, z) would require
drawing a 4-dimensional picture.
•
However, we can still talk about level sets of functions of 3 variables these are the points
a relation
◦
f (x, y, z) = c:
Any graph
(x, y, z)
•
(x, y, z)
satisfying
these will (in general) give rise to level surfaces in 3-dimensional space.
z = g(x, y) is an example of a
z = g(x, y) are the same as
level surface for the function
with
those with
f (x, y, z) = g(x, y) − z :
the points
f (x, y, z) = 0.
f (x, y, z) = x2 + y 2 + z 2 are spheres centered at the origin: by
2
2
the distance formula, the points satisfying x + y + z = c (for c > 0) are precisely those which are at a
√
√
distance of
c from the origin but this is just another way of describing the sphere of radius c centered
2
2
2
at (0, 0, 0). A graph of the sphere x + y + z = 1 is below.
Example: The level surfaces for the function
2
•
Example: The level surfaces for the function
the
y -axis:
•
√
c from the y -axis but this is just another way of
y -axis. A graph of the cylinder x2 + z 2 = 1 is above.
a distance of
along the
f (x, y, z) = x2 + z 2 are (right circular) cylinders running along
x2 + z 2 = c (for c > 0) are those which are at
again, from the distance formula, points satisfying
Example: Let us examine the level surfaces for the function
◦
The set of points satisfying
describing a right circular cylinder oriented
f (x, y, z) = x2 + y 2 − z 2 .
x2 + y 2 − z 2 = c for c > 0 forms a surface called a hyperboloid of one sheet, so
named because the shape is a hyperbola in two cross-sections and an ellipse in the third, and the graph
is connected (one sheet).
∗
Remark: A hyperboloid of one sheet whose cross-sections are circles is an example of what is called
a ruled surface: through each point on the surface pass two lines which are contained in the surface.
Such surfaces can therefore be (physically) constructed using materials which are not curved. The
hyperboloid of one sheet, in particular, is a common design for cooling towers.
4
◦
and whose center point at
◦
x2 + y 2 − z 2 = 0 forms
the origin (0, 0, 0).
The set of points satisfying
The set of points satisfying
x2 + y 2 − z 2 = c
for
a (right circular) double cone whose axis is the
c<0
z -axis
forms a surface called a hyperboloid of two sheets,
so named because the shape is a hyperbola in two cross-sections and an ellipse in the third, and the
graph consists of two pieces (two sheets).
◦
1.2
•
c = 1, c = 0,
and
c = −1
respectively:
Vectors and Scalars
A vector is a quantity which has both a magnitude and a direction.
◦
•
Here are plots of the level surfaces for
This is in contrast to a scalar, which carries only a magnitude.
We denote the
where the
◦
ai
n-dimensional
vector from the origin to the point
(a1 , a2 , · · · , an )
as
v = ha1 , a2 , · · · , an i,
are scalars.
We use the angle brackets
h·i
rather than parentheses
a vector and the coordinates of a point in space.
(·)
so as to underscore the dierence between
We will, however, view coordinates of vectors and
coordinates of points as essentially interchangeable.
◦
We also write vectors in boldface (v, not
v ),
so that we can tell them apart from scalars. When writing
by hand, it is hard to dierentiate boldface, so the notation
•
~v
is also sometimes used.
The typical way to think of vectors is as directed line segments: the length of the line segment gives the
magnitude of the vector, and the direction the segment is pointing gives the direction of the vector.
◦
Note/Warning: Vectors are a little bit dierent from directed line segments, because we don't care where
a vector starts: we only care about the dierence between the starting and ending positions. Thus: the
directed segment whose start is
(2, 2) represent the
•
same vector,
(0, 0) and end is (1, 1) and the segment starting at (1, 1) and ending at
h1, 1i. This distinction is rarely necessary in most applications, however.
We can add vectors (provided they are of the same dimension!) in the obvious way, one component at a time:
if
v = ha1 , · · · , an i
◦
and
w = hb1 , · · · , bn i
then
v + w = ha1 + b1 , · · · , an + bn i.
Similarly we can 'scale' a vector by a scalar, one component at a time: if
r
is a scalar, then we have
r v = hra1 , · · · , ran i.
◦
Scaling a vector by a factor of
1/2,
for example, produces a new vector in the same direction, but with
half the length as the original.
◦
Scaling a vector by
−1 produces a new vector with the same length but pointing in the opposite direction
from the original vector.
•
Example: If
v = h−1, 2, 2i and w = h3, 0, −4i then 2w = h6, 0, −8i
2w = h−7, 2, 10i
.
5
, and
v+w = h2, 2, −2i
. Furthermore,v−
•
At this point, we will restrict ourselves to talking just about 2-dimensional space and 3-dimensional space.
Our primary reason for this is that most of the immediate applications of vectors (e.g., to physics) happen in
3-dimensional space.
•
It will be useful to have a way to denote the unit coordinate vectors of 3-dimensional space. So we denote
i = h1, 0, 0i, j = h0, 1, 0i,
and
k = h0, 0, 1i.
h3, 2, −5i
◦
Example: The vector
◦
Note: Some authors primarily use
is equal to
3i + 2j − 5k.
ijk-notation
when working with vectors. We will generally use angle
bracket notation, except in a few cases when there are useful mnemonics that are easier to remember
using
1.3
•
ijk-notation.
The Dot Product
Denition: We dene the norm (length, magnitude) of the vector
◦
v = h−1, 2, 2i
32 + 02 + (−4)2 = 5 .
Example: For
p
◦
If
r
and
(0, . . . , 0)
ha1 , . . . , an i is just the length
(a1 , . . . , an ).
p
||v|| = (−1)2 + 22 + 22 = 3 , and ||w|| =
w = h3, 0, −4i,
we have
√
r2 = |r|
||r v|| = |r| ||v||,
just by scaling
direction as
v.
v
u=
by 1 over its length. In other words, the vector
This vector
u
is sometimes called the normalization of
◦
Example: For
direction as
v = h−1, 2, 2i,
since
||v|| = 3,
we see that
u1 =
is a unit vector in the same
1 2 2
− , ,
3 3 3
is a unit vector in the same
v.
Example: For
direction of
v
||v||
v.
◦
w = h3, 0, −4i,
since
||w|| = 5,
we see that
u2 =
3
4
, 0, −
5
5
is a unit vector in the
w.
If we have two vectors, we now know how to nd their lengths. But another thing we might want to know
about two vectors is the angle
•
since we can just factor
from each term under the square root.
From any nonzero vector we can nd a unit vector (that is, a vector of length 1) in the same direction of
v
•
(a1 )2 + · · · + (an )2
to the point
is a scalar, we can see immediately from the denition that
out a
•
p
This is just an application of the distance formula: the norm of the vector
of the line segment joining the origin
◦
v = ha1 , . . . , an i as ||v|| =
θ
between them. This motivates the denition of the dot product:
Denition: The dot product of two vectors
v1 · v2 = a1 b1 + a2 b2 + · · · + an bn
◦
v1 = ha1 , . . . , an i
and
v2 = hb1 , . . . , bn i
is dened to be the scalar
.
Note: The dot product of two vectors is a scalar,
not
a vector!
(For this reason, the dot product is
sometimes called the scalar product of two vectors.)
•
◦
Example: The dot product
h1, 2i · h3, 4i
◦
Example: The dot product
h−1, 2, 2i · h3, 0, −4i
is
(1)(2) + (3)(4) = 14
is
.
(−1)(3) + (2)(0) + (2)(−4) = −11
.
The dot product obeys several very nice properties reminiscent of standard multiplication. For any vectors
v, v1 , v2 , w,
and any scalar
r,
we can verify the following properties directly from the denition:
◦
The dot product is commutative:
v·w =w·v
◦
The dot product distributes over addition:
◦
Scalars can be factored out of a dot product:
.
(v1 + v2 ) · w = (v1 · w) + (v2 · w)
(r v) · w = r (v · w)
6
.
.
.
◦
2
v · v = ||v||
The dot product of a vector with itself is the square of the norm:
•
There is a nice relation between the dot product and the angle between two vectors:
•
Theorem (Dot Product): For vectors
.
v1 and v2 forming an angle θ between them, we have v1 · v2 = ||v1 || ||v2 || cos(θ)
◦
Using the Dot Product Theorem, we can compute the angle between two vectors.
◦
Proof: To prove this statement, we use the Law of Cosines in the triangle formed by
which states that
2
2
2
||v2 − v1 || = ||v1 || + ||v2 || − 2 ||v1 || ||v2 || cos(θ).
v1 , v2 ,
and
v2 − v1 ,
Now we know that the square of the
norm is the dot product of a vector with itself so we can apply this and the other dot product properties
to see that
2
||v2 − v1 ||
=
(v2 − v1 ) · (v2 − v1 )
=
(v2 · v2 ) − (v1 · v2 ) − (v2 · v1 ) + (v1 · v1 )
2
2
= ||v2 || − 2(v1 · v2 ) + ||v1 || .
Now by comparing to the Law of Cosines expression, we see that everything cancels and leaves us precisely
with the result we wanted.
•
√ v = 2, 1, 3
and
√ w = 0, 3, 1 .
√
√
√
v · w = (2)(0) + (1)( 3) + ( 3)(1) = 2 3,
and
||v|| =
Example: Compute the angle between the vectors
◦
We compute
q
√
√
22 + 12 + ( 3)2 = 2 3
and
q√
||w|| = ( 3)2 + 02 + 12 = 2.
◦
Then by the Dot Product Theorem, the angle
meaning that
•
1
π
−1
θ = cos
=
2
3
θ
between the vectors satises
.
Example: Compute the angle between the vectors
v = h2, 2, −1i
◦ p
We compute v · w = (2)(3) + (2)(4) + (−1)(0) = 14,
32 + 02 + (−4)2 = 5.
◦
Then by the Dot Product Theorem, the angle
θ = cos−1
•
14
15
√
√
2 3 = 2 · 2 3 · cos(θ),
θ
and
w = h3, 4, 0i.
p
||v|| = 22 + 22 + (−1)2 = 3
and
between the vectors satises
and
||w|| =
14 = 3 · 5 · cos(θ),
so
.
Using the Dot Product Theorem, we can see that the sign and magnitude of the dot product is roughly
measuring whether the vectors are pointing in the same direction:
◦
v2 are nonzero vectors, then both ||v1 || and ||v2 || are positive, so by the theorem above, the
v1 · v2 = ||v1 || ||v2 || cos(θ) will have the same sign as cos(θ).
π
◦ If 0 ≤ θ < , the dot product v1 · v2 will be positive. Furthermore, the smaller θ is, the larger the value
2
of v1 · v2 will be. Thus, a large positive value for the dot product indicates that the vectors are pointing
If
v1
and
value
in roughly the same direction.
◦
Inversely, if
v1 · v2
π
< θ ≤ π,
2
the dot product
v1 · v2
will be negative, and the larger
θ
is, the larger negative
will be. Thus, a large negative value for the dot product indicates that the vectors are pointing
in roughly opposite directions.
◦
•
We have a special name for the case where the angle is
π
:
2
Denition: We say two vectors are orthogonal if their dot product is zero.
◦
From the Dot Product Theorem, since
the angle between them is
π
,
2
cos
π
2
= 0,
we see that two nonzero vectors are orthogonal if
which is to say, if they are perpendicular.
7
.
◦
Example: The vectors
h2, −1, 4i and h3, 2, −1i are orthogonal, since their dot product is (2)(3)+(−1)(2)+
(4)(−1) = 0.
◦
Remark: Since the dot product of the zero vector with any vector is zero, by our denition above, the
zero vector is orthogonal to every vector.
•
Another basic question about vectors is: given a vector
direction of
v,
and how much of
w
is orthogonal to
v
v?
and another vector
w,
how much of
w
is in the
(This is a problem that often arises in Newtonian
physics: one often needs to separate the vector representing a force into orthogonal components.)
◦
In other words: we want to write
value of the scalar
w = av + y,
where
y
is orthogonal to
v.
The goal is to determine the
a.
y = w −av, and y is orthogonal to v, taking the dot product with v gives 0 = y ·v = (w −av)·v =
w · v − a(v · v).
w·v
.
◦ Solving for a gives a =
v·v
w · v
v.
• Denition: The vector projection of w onto the nonzero vector v is the vector Projv (w) =
v·v
•
◦
Since
◦
The vector projection of
w
onto
v
Example: Find the vector projection of
gives the piece of
w = h3, 4, −2i
w
in the direction of
v.
onto the coordinate vectors i,
j,
and
k.
3
w·i
i = i = 3i = h3, 0, 0i.
i·i
1
w·j
4
◦ Next, Projj (w) =
j = j = 4j = h0, 4, 0i.
j·j
1
−2
w·k
k=
k = −2k = h0, 0, −2i.
◦ Finally, Projk (w) =
k·k
1
◦ These results are consistent with our intuition about what the vector projection should be: for example,
the component of h3, 4, −2i in the direction of h1, 0, 0i should logically be h3, 0, 0i, and this is what we
◦
We have Proji (w)
=
got.
•
1.4
•
Example: Find the vector projection of
w · v
◦
We have Projv (w)
◦
As a sanity check, note that
=
v·v
v=
w = h6, −3, 0i
onto the vector
v = h2, −2, −1i.
(6)(2) + (−3)(−2) + (0)(−1)
18
v=
v = 2v = h4, −4, −2i
22 + (−2)2 + (−1)2
9
w − Projv (w) = h2, 1, 2i, and indeed,
h2, 1, 2i · h2, −2, −1i = (2)(2) + (1)(−2) + (2)(−1) = 0.
this vector is orthogonal to
.
v,
because
The Cross Product
In addition to the dot product, we have another type of product dened for vectors in 3-space, called the cross
product:
•
v1 = hx1 , y1 , z1 i and v2 = hx2 , y2 , z2 i is dened to be the
v1 × v2 = hy1 z2 − y2 z1 , z1 x2 − z2 x1 , x1 y2 − x2 y2 i . It is orthogonal to both v1 and v2 .
Denition: The cross product of
◦
vector
Important Note: The cross product is only dened for vectors with 3 components, and outputs another
vector with 3 components. In contrast, the dot product is dened for vectors of any length, and outputs
a scalar.
◦
A way to remember the cross product formula (aside from memorization) is the determinant formula
i
v1 × v2 = det x1
x2
dard unit vectors:
j
y1
y2
k
z1
z2
y1
=
y2
x1
z1 i
−
x2
z2
i = h1, 0, 0i, j = h0, 1, 0i,
and
8
x1
z1 j
+
x2
z2
k = h0, 0, 1i.
y1 k
y2 , where
i, j, k
are the stan-
∗
It may seem a little unusual to have vectors inside a determinant, but it works out to the correct
answer. Don't forget the minus sign on the middle term!
•
A central property of the cross product
◦
v1 × v2
is that it is orthogonal both to
To verify this, we can just evaluate the dot products
v1
and to
v2 .
v1 · (v1 × v2 ) and v2 · (v1 × v2 ) and check that they
are both zero.
◦
For example, we have
v1 · (v1 × v2 ) = x1 (y1 z2 − y2 z1 ) + y1 (z1 x2 − z2 x1 ) + z1 (x1 y2 − x2 y2 ),
which some
algebra will conrm is equal to zero.
•
v1 = h1, 1, 3i and v2 = h2, −1, 1i, nd v1 × v2 and v2 × v1 and verify that v1 × v2
v2 .
i
j k 1 3 1 1 1 3 k = h4, 5, −3i .
j+
i−
◦ First, v1 × v2 = 1 1 3 = 2 −1 2 1 −1 1 2 −1 1 i
j k 2 −1 2 1 −1 1 k = h−4, −5, 3i .
j+
i−
◦ Also, v2 × v1 = 2 −1 1 = 1 1 1 3 1 3 1 1 3 Example: If
to both
◦
•
v1
is orthogonal
and
To check: we have v1 · (v1 × v2 ) = h1, 1, 3i · h4, 5, −3i = 4 + 5 − 9 = 0,
h2, −1, 1i · h4, 5, −3i = 8 − 5 − 3 = 0.
v2 · (v1 × v2 ) =
and also
Here are a few other algebraic properties of the cross product:
◦
Unlike the dot product, the cross product is NOT commutative! Indeed, we can see from the denition
(and the example above) that
we see that
◦
v×v =0
for any
◦
in this denition,
The cross product does have a distributive property, like with the dot product: it is fairly easy to check
from the denition that
◦
v1 × v2 = −(v2 × v1 ) . In particular, if we set v1 = v2
vector v, where 0 = h0, 0, 0i denotes the zero vector.
(v1 + v2 ) × w = (v1 × w) + (v2 × w)
Scalars can also be factored out of a cross product:
.
(r v) × w = r (v × w) = v × (r w)
The cross product is also not associative. It does satisfy the so-called Jacobi identity:
.
u × (v × w) + v ×
(w × u) + w × (u × v) = 0.
•
Theorem (Cross Product): If
where
◦
A
θ
is the angle between
is the area of the parallelogram formed by
Proof: We just need to show that
||v1 || ||v2 || cos(θ)
∗
v1
v1
and
and
2
v2 ,
then
||v1 × v2 || = ||v1 || ||v2 || sin(θ) = A
,
v2 .
2
2
2
||v1 × v2 || + (v1 · v2 ) = ||v1 || ||v2 ||
, because we know that
v1 · v2 =
from the Dot Product Theorem.
2
2
(y1 z2 − y2 z1 ) + (z1 x2 − z2 x1 ) +
(x1 y2 − x2 y2 ) + (x1 x2 + y1 y2 + z1 z2 ) is equal to (x1 )2 + (y1 )2 + (z1 )2 · (x2 )2 + (y2 )2 + (z2 )2 .
∗ When we expand the rst thing, we get each of the 9 possible square terms (1 42 )2 where and
4 are each one of x, y , or z , and the cross terms like 2x1 x2 y1 y2 will all cancel out.
∗ We get exactly the same sum of 9 square terms when we expand the second thing. So they are equal
To check this we multiply everything out. So we need to see that
2
2
and we're done.
∗
For the statement about the area, it is a standard fact of geometry that the area of the triangle with
sides
◦
•
v1
and
v2
is
1
||v1 || ||v2 || sin(θ).
2
The parallelogram's area is twice this.
Remark: This quite nice property is one reason we chose the denition we did for the cross product.
Example: Find the area of the triangle whose vertices are the points
◦
By the Cross Product Theorem, the area of the triangle is
representing two sides of the triangle.
9
A=
A(1, −1, 2), B(2, −3, 1),
1
||v × w||,
2
where
v
and
and
w
C(2, 2, 2).
are vectors
◦ We
can take v = B − A = h1, −2, −1i and w = C − A = h0, 5, 1i.
i
j
k 1 −2 −1 = −2 −1 i − 1 −1 j + 1 −2 k = h3, −1, 5i.
5
0 5 0 1
1
0 5
1 ◦
1.5
•
Hence the area of the triangle is
1
1p 2
||h3, −1, 5i|| =
3 + (−1)2 + 52 =
2
2
We then compute
v×w =
√
35
2
.
Lines and Planes in 3-Space
Among the many useful applications of vectors is that they can be used to give simple descriptions of lines
and planes in 3-space.
◦
In the 2-dimensional plane, the general equation of a line is
a
and
b
ax + by = d for some constants a, b, d (with
ax + by + cz = d (with a, b, c not all
not both zero). The 3-dimensional version of this would be
zero), which will describe a plane.
◦
x = x(t),
(x(t), y(t), z(t)) will
To describe lines in 3-space, we instead need to describe them as parametric curves, in the form
y = y(t), z = z(t)
for some functions
x(t), y(t), z(t).
As
t
varies, the set of points
form a curve: the goal is to choose the functions so that the curve is a line.
•
P1 = hx1 , y1 , z1 i and P2 = hx2 , y2 , z2 i, the points hx, y, zi on the line l
through P1 and P2 are given parametrically by hx, y, zi = P1 + t v , as t varies through the real numbers,
and v = P2 − P1 = hx2 − x1 , y2 − y1 , z2 − z1 i.
Proposition: Given distinct points
◦
Proof: There is a unique line between two points, by the axioms of geometry. So we just need to check
that this is a line, and that it goes through
∗
P1
and
P2 .
l explicitly is tells us that x = x1 + t(x2 − x1 ), y = y1 + t(y2 − y1 ), and
z = z1 + t(z2 − z1 ), and these are all linear equations. So it's a line.
∗ We see l goes through P1 because at t = 0 we get P1 . Similarly, at t = 1 we get P2 . So we're done.
◦
The parametric equation for
Note: This procedure works to nd the parametrization of a line in a space of any dimension, not just
3-space.
◦
Remark: We call the vector
v = P2 − P1 the direction vector for the line l: it tells us in which direction
P1 in the sum P1 + t v species which, of all possible lines in that direction,
the line is moving. The term
is the line we want.
•
Example: Find a parametrization of the line through the points
◦
First, we nd the direction vector: we get
◦
Then the line is given parametrically by
◦
We can write this out explicitly, as
the points
•
(x, y, z)
(1, 2, 3)
and
(−1, 2, −1).
v = h−1, 2, −1i − h1, 2, 3i = h−2, 0, −4i.
hx, y, zi = h1 − 2t, 2, 3 − 4ti
x = 1 − 2t, y = 2, z = 3 − 4t.
As
t
.
ranges through the real numbers,
will range along the desired line.
Proposition: The plane dened by
ax+by +cz = d
words, every line lying in this plane is orthogonal to
n = ha, b, ci. In other
n = ha, b, ci,
(x0 , y0 , z0 ), and its equation is
is orthogonal to its normal vector
ha, b, ci.
Conversely, given a nonzero vector
there is a unique plane normal to that vector passing through a given point
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0.
◦
Proof: Suppose
l
is a line in the plane. All we need to show is that its direction vector is orthogonal to
n.
∗
v = P2 − P1 , where both of the points P2 = hx2 , y2 , z2 i
hx1 , y1 , z1 i lie in the plane.
∗ Then P1 · n = a x1 + b y1 + c z1 = d since P1 lies in the plane, and similarly P2 · n = d.
∗ But then we have v · n = P2 · n − P1 · n = d − d = 0, which is exactly what we wanted.
So say the direction vector is
10
and
P1 =
∗
For the converse statement, clearly if
cz = ∗
for some value of
,
n = ha, b, ci
then the equation of the plane must be
ax+by+
by the previous proposition.
But if we are given a point that lies in the plane, we can plug in to see that
= ax0 + by0 + cz0 ,
and so we have uniquely determined the equation of the plane, and hence the plane. We can rewrite
the equation as
◦
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0,
so this is the desired plane.
Remark: This proposition says that a plane is specied by its normal vector along with a point that the
plane passes through.
h2, 3, −1i passing through (2, 2, 7) has equation 2(x − 2) + 3(y − 2) − (z − 7) = 0
•
Example: The plane with normal vector
•
Using these two results we can solve a wide variety of problems involving lines and planes:
•
Example: Find an equation for the plane containing the vectors
through the point
◦
v1 = h1, 1, 1i
and
v2 = h1, 2, 3i
and passing
P = (1, −1, 1).
The normal vector to the plane is orthogonal to both
v1
and
v2 ,
so we can nd it by taking the cross
product.
◦
We get
∗
◦
Sanity check: we have
j
1
2
k
1
3
1
=
2
1
1 i − 1
3 x − 2y + z = 4
1
1 j + 1
3 1 k = h1, −2, 1i.
2 n·v1 = (1)(1)+(−2)(1)+(1)(1) = 0 and n·v2 = (1)(1)+(−2)(2)+(1)(3) = 0.
Now we see that the equation for the plane is
as
•
i
n = v1 × v2 = 1
1
1(x − 1) − 2(y + 1) + 1(z − 1) = 0,
which we can rewrite
.
P1 = (3, 0, −1), P2 = (1, 2, 2)
Example: Find an equation for the plane passing through the three points
and
P3 = (−2, 1, 4).
◦
We need the plane's normal vector
take their cross product to get
◦
We have
◦
Then
∗
◦
n.
To nd it, we need to nd two vectors lying in the plane, and then
v1 = P2 − P1
Sanity check:
n · v1 = (7)(−5) + (−5)(1) + (8)(5) = 0
and
7(x − 3) − 5(y − 0) + 8(z + 1) = 0,
or
7x − 5y + 8z = 13
For an extra error check, we verify that all three points do lie in this plane: we have
Example: Parametrize the line of intersection of
◦
v2 = P3 − P1 .
n · v2 = (7)(−2) + (−5)(2) + (8)(3) = 0.
8(−1) = 13, 7(1) − 5(2) + 8(2) = 13, and 7(−2) − 5(1) + 8(4) = 13.
•
and
v1 = P2 − P1 = h−2, 2, 3i and v2 = P3 − P1 = h−5, 1, 5i.
−2 2 −2 3 2 3 k = h7, −5, 8i.
j+
i−
n = v1 × v2 = −5 1 −5 5 1 5 Thus the equation of the plane is
∗
n.
We can get two such vectors as
x − y + 2z = 3
and
.
7(3) − 5(0) +
Hence this is the correct equation.
2x + y − z = 0.
The normal vector to each plane will be orthogonal to the line of intersection. Therefore, we can get the
direction vector of the line by taking cross product of the two planes' normal vectors.
◦
We have
∗
◦
n1 = h1, −1, 2i
Sanity check:
and
n2 = h2, 1, −1i.
Then the cross product is
v · n1 = −1(1) + 5(−1) + 3(2) = 0
and
v = n1 × n2 = h−1, 5, 3i.
v · n2 = −1(2) + 5(1) + 3(−1) = 0.
Now we need to nd a point in both planes (since we need a point on the line). We try looking for one
with
x = 0:
this requires
−y + 2z = 3
and
y − z = 0.
Solving yields
y = z = 3,
so
(0, 3, 3)
is in both
planes and thus on the line l.
◦
•
Applying the line parametrization formula gives
l : hx, y, zi = h−t, 5t + 3, 3t + 3i
.
Another common problem is to compute the distance from a point to a plane. It turns out there is a nice
formula for this distance:
•
Proposition:
The distance from the point
P = (x0 , y0 , z0 )
|ax0 + by0 + cz0 − d|
√
.
a2 + b2 + c2
11
to the plane
ax + by + cz = d
is equal to
.
◦
The shortest vector connecting
P
to the plane will be in the same direction as the normal vector to the
plane. In theory, we could use this information to parametrize the line joining
P
intersection of the line and plane, and nally compute the distance we seek.
However, there is a less
to the plane, nd the
messy way using vector projections.
◦
Proof: Let
∗
Q
be any point on the plane
Then the vector
the normal
v
connecting
n.
P
ax + by + cz = d
and let
n
be the normal vector to the plane.
to the plane is given by the vector projection of
onto
|n · w|
.
||n||
||n||
∗ From our earlier
√ results, we can take n = ha, b, ci: then n · w = n · P − n · Q = (ax0 + by0 + cz0 ) − (d),
a2 + b2 + c2 . (The fact that n · Q = d is just a restatement of the fact that Q lies in
and ||n|| =
the plane ax + by + cz = d.)
|n · w|
|ax0 + by0 + cz0 − d|
√
∗ Then ||v|| =
=
, as claimed.
||n||
a2 + b2 + c2
∗
•
v = Projn (w) =
n · w
n·n
Example: Find the distance from the point
◦
•
Explicitly,
The formula gives
Then the length of
P = (1, 2, 4)
P = (1, 2, 4)
v
to the plane
|−6|
|1 · 1 + 2 · 2 − 2 · 4 − 3|
p
= √ = 2
2
2
2
9
1 + 2 + (−2)
Example: Find the distance from the point
◦
n.
is
||v|| =
|n · w|
w = Q−P
2
x + 2y − 2z = 3.
.
to the plane
z = 0.
0x + 0y + z = 0.
|0 · 1 + 0 · 2 + 1 · 4 − 0|
|4|
√
◦ Then the formula gives
=
= 4.
1
02 + 02 + 12
◦ This result agrees with our intuition: the plane z = 0 is the horizontal xy -plane,
any point to this plane is simply the absolute value of its z -coordinate.
•
The plane can be written as
Example: Find the distance between the planes
◦
◦
||n|| =
2x + y + 2z = 1
and
and so the distance of
2x + y + 2z = 7
These two planes are parallel, since they visibly have the same normal vector.
To compute the distance between them, we can pick any point on one plane and compute its distance to
the other plane.
◦
2x + y + 2z = 1.
|2 · 0 + 1 · 1 + 2 · 0 − 7|
|−6|
√
◦ Then the formula gives the distance as
= √ = 2.
9
22 + 12 + 22
◦ Remark: By the same argument, the distance between the two planes ax+by +cz = d1
|d2 − d1 |
d2 is equal to √
.
a2 + b2 + c2
•
The point
(0, 1, 0)
lies on
Example: Find the distance between the lines l1
◦
and
ax+by +cz =
: hx, y, zi = h4t, 1 − t, 2 − ti and l2 : hx, y, zi = h2 + 2s, 3, 5 − si.
Imagine the two lines l1 and l2 as sitting inside two parallel planes. The distance between the lines will
then be the same as the distance between the planes, which we can compute by nding the distance of
any arbitrary point in one plane to the other plane (using the point-to-plane distance formula).
◦
In order to nd these two parallel planes, we just need to nd their (common) normal vector
will be orthogonal to the direction vectors of both lines. So to compute
n,
n,
which
we can simply take the cross
product of the direction vectors.
◦
n = h4, −1, −1i×h2, 0, −1i = h1, 2, 2i, meaning that the planes are of the form x+2y+2z = d
d.
◦ To get the plane containing l1 , choose any point on l1 . Picking t = 0 gives (0, 1, 2), and so we get
d1 = 0 + 2 · 1 + 2 · 2 = 6, so the rst plane is x + 2y + 2z = 6.
◦ Setting s = 0 in the equation for l2 gives the point (2, 3, 5), which lies in the second plane.
|2 + 2 · 3 + 2 · 5 − 6|
√
=
◦ Then using the point-to-plane distance formula gives the required distance as
12 + 22 + 22
|12|
√ = 4.
9
We compute
for some value of
12
1.6
•
Vector-Valued Functions of One Variable, Derivatives
Denition: A vector-valued function of one variable
components is a function of the parameter
r(t)
is a function whose output is a vector, each of whose
t.
√
r(t) = t2 , 2t , r(t) = ht, t, ti, r(t) = cos(t2 ), e2t , tan−1 ( t2 + 1) .
◦
Examples:
◦
We have already encountered a few vector-valued functions when we studied lines in 3-space:
the
parametrization of a line is an example of a vector-valued function.
•
We can add and scalar-multiply vector-valued functions in the same manner as normal vectors.
◦
and
◦
t
2
2
t
r1 (t) = e2 , cos(t), t − 1 and r2 (t) = t, 0, −t we have r1 (t) + r2 (t) = he + t, cos(t), −1i
2r2 (t) = 2t, 0, −2t .
Example: For
r(t) = hx(t), y(t)i and r(t) = hx(t), y(t), z(t)i,
t and output a vector with 2 or 3 coordinates. These functions trace
We will primarily be interested in vector functions of the form
which have a single input parameter
out parametric curves in 2 or 3-dimensional space (respectively).
◦
Later, we will also be interested in vector functions of the form
input parameters
•
s
and
t.
If we graph a vector-valued function
◦
r(s, t) = hx(s, t), y(s, t), z(s, t)i,
Example: The curve given by
r(t) = hx(t), y(t)i
as
t
r(t) = hcos(t), sin(t)i
varies, we will obtain a curve in the plane.
traces around the unit circle as
counterclockwise, and completes one revolution after a time interval of
◦
for two
These functions, in general, will describe surfaces in 3-dimensional space.
t
varies. It traces
2π .
Remark: Vector-valued functions in the plane are often studied in single-variable calculus (where they
are usually termed parametric curves).
•
If we graph a vector-valued function
◦
r(t) = hx(t), y(t), z(t)i
A parametric curve in 3-space is just the set of points
as
t
varies, we will obtain a curve in space.
(x(t), y(t), z(t))
for some functions
x(t), y(t),
and
z(t).
◦
Reminder: We are thinking of functions giving parametric curves interchangeably with vector functions:
(x(t), y(t), z(t)) as output from a function, we can equally well think of getting
hx(t), y(t), z(t)i.
instead of getting a point
the vector output
•
Example: The curve given by
◦
r(t) = ht, t, ti
is a line passing through the origin with direction vector
More generally, as we have already seen, for any constants
r(t) = (a1 t + b1 , a2 t + b2 , a3 t + b3 ) will
the direction of the vector ha1 , a2 , a3 i.
a1 , a2 , a3
and
h1, 1, 1i.
b1 , b2 , b3 , the curve given by
(b1 , b2 , b3 ) and pointing in
be a line passing through the point
r(t) = (sin(t), cos(t), t) is a helix wrapping around the cylinder x2 + y 2 = 1.
We can see that as t increases, the x and y parts just trace around a unit circle at constant speed, while z
increases at constant speed. Here is a plot of the curve winding around the cylinder, for 0 ≤ t ≤ 8π :
•
Example: The curve given by
•
Example: The curve given by
r(t) = hcos(t), sin(t), cos(t)i is an ellipse. We can see that this curve is
z = x with the cylinder x2 + y 2 = 1 (above).
ellipse by observing that it is the intersection of the plane
13
an
•
The next question to ask is: can we take derivatives of vector-valued functions? The answer is yes:
•
Denition: The derivative of the vector function
r(t)
r(t + h) − r(t)
h→0
h
r0 (t) = lim
is given by
, provided the
limit exists.
◦
Note the extreme similarity of this denition with the denition of the derivative of a (scalar) function
f (t),
of one variable
◦
which reads
For r(t) = hx(t), y(t), z(t)i,
hx0 (t), y 0 (t), z 0 (t)i.
∗
f (t + h) − f (t)
.
h
f 0 (t) = lim
h→0
then by applying the denition of the derivative, we see that
r0 (t) =
More generally, we can see that taking the derivative of a vector function is the same thing as
dierentiating each component.
◦
•
Example: For
r(t) = et , cos(t), t2 − 1 ,
we have
r0 (t) = et , − sin(t), 2t
.
By the limit part of the denition, we see that the derivative of a vector function tells us in which direction
t.
the curve is moving at time
◦
More specically,
r0 (t)
is a vector tangent to the graph of
Compare to the situation of the derivative of a (scalar) function of one variable
f (t):
r(t).
the derivative
f 0 (t)
gives the slope of the tangent line.
•
We can also take higher derivatives: for example,
◦
•
◦
◦
r(t) = et , cos(t), t2 − 1 , we have r0 (t) = het , − sin(t), 2ti, so r00 (t) = et , − cos(t), 2
r000 (t) = et , sin(t), 0 , and so on.
Example: For
The rst derivative
r0 (t)
The second derivative
The magnitude
and
0
00
gives the velocity of the particle at time
r (t)
||r (t)||
t.
gives the acceleration of the particle at time
t.
of the velocity vector gives the speed of the particle.
Derivatives of vector functions satisfy rules strongly reminiscent of the product rule with regard to the dot
d
[r1 · r2 ] = r1 · (r02 ) + (r01 ) · r2
dt
and cross products:
◦
and
d
[r1 × r2 ] = r1 × (r02 ) + (r01 ) × r2
dt
These properties can be veried by expanding out the dot and cross products of
[r2 (t + h) − r2 (t)],
1.7
and so forth.
The derivatives of vector-valued functions have the same physical interpretations as we are used to:
◦
•
r0 (t + h) − r0 (t)
,
h→0
h
r00 (t) = lim
.
[r1 (t + h) − r1 (t)]
with
applying the limit denition of derivative, and simplifying.
Curves in 3-Space: Velocity and Acceleration; Arclength; Unit Tangent, Normal,
and Binormal Vectors; Curvature and Torsion
•
A particle traveling on a parametric curve
r(t) = hx(t), y(t), z(t)i
in 3-space possesses a number of associated
vectors and scalar quantities.
◦
The velocity of the particle is given by
tangent line to the curve
v(t) = r0 (t).
◦
The acceleration of the particle is given by
◦
The arclength of the curve between
ˆ
t = t1
a(t) = v0 (t) = r00 (t).
and
t = t2
t2
||v(t)|| dt.
t1
◦
The arclength
s
The velocity vector is the direction vector for the
r(t).
obeys the relation
ds
= ||v(t)||.
dt
14
is given by
s=
´ t2 q
t1
2
2
2
[x0 (t)] + [y 0 (t)] + [z 0 (t)] dt =
•
Example: Find the velocity and acceleration of a particle traveling along the path
Also, nd the arclength of the path from
•
D
v(t) = r0 (t) =
et ,
◦
The velocity is
◦
For the arclength, rst we compute
◦
Then the arclength is
s=
´1
0
t=0
√
to
2, e−t
E
√
r(t) = 2 + et , 2 + t 2, 3 − e−t .
t = 1.
, and the acceleration is
a(t) = v0 (t) =
et , 0, −e−t
.
q
√
√
(et )2 + ( 2)2 + (e−t )2 = e2t + 2 + e−2t = et + e−t .
||v(t)|| =
1
[et + e−t ] dt = [et − e−t ]|t=0 = e − e−1
.
When working with curves in 3-space (especially in physics), it is often useful to use normalized tangent
vectors, rather than just the velocity and acceleration vectors. There are three standard normalized tangent
vectors: the unit tangent vector
◦
T(t) =
The unit tangent vector
∗
T(t),
the unit normal vector
v(t)
||v(t)||
N(t),
and the unit binormal vector
B(t).
measures the direction in which the particle is moving.
By the denition, the unit tangent vector satises
||T(t)|| = 1,
and
T(t)
is in the same direction as
r0 (t).
∗
Thus, it is a unit vector, in the same direction as the velocity (or tangent) vector. (Whence the
name.)
◦
The unit normal vector
N(t) =
T0 (t)
||T0 (t)||
, orthogonal to the unit tangent vector, measures roughly the
direction that forces are pulling the particle.
∗
∗
||N(t)|| = 1, as long as T0 (t) 6= 0.
T(t) · T(t) = 1, taking the derivative with
By the denition, the unit normal vector satises
N(t) is orthogonal to T(t), since
respect to t
T(t) · T0 (t) + T0 (t) · T(t) = 0. But since the dot product is symmetric, this says T (the unit
0
tangent vector) and T (the normal vector) have dot product 0. In other words, the (unit) normal
To see that
gives
vector is orthogonal to the tangent vector.
◦
The unit binormal vector
B(t) = T(t) × N(t) =
v(t) × a(t)
||v(t) × a(t)||
, orthogonal to the tangent and normal
vectors, gives the direction perpendicular to the particle's plane of motion.
∗
By its denition via the cross product, the binormal vector is orthogonal to both the (unit) normal
and (unit) tangent vectors, and since those vectors have length 1, so does
∗
B.
Thus, we see that the unit tangent, unit normal, and unit binormal vectors are all orthogonal to each
other provided that the normal and binormal vectors are dened (which is not always guaranteed
because of the division by
◦
||T0 (t)||
in the denition of
N).
There are several dierent planes associated to these normalized tangent vectors:
∗
The osculating plane is the plane containing
T(t)
and
N(t),
passing through the point
r(t).
It is
sometimes called the tangent plane to the curve.
∗
∗
•
The normal plane is the plane containing
N(t) and B(t), passing through the point r(t).
T(t) and B(t), passing through the point r(t).
The rectifying plane is the plane containing
r(t) = h3 sin t, 5 cos t, 4 sin ti.
p
||v(t)|| = 9 cos2 t + 25 sin2 t + 16 cos2 t =
Example: Find the unit tangent, normal, and binormal vectors for the curve
0
◦ p
We have v(t) = r (t) = h3 cos t, −5 sin t, −4 cos ti, so we get
25 cos2 t + 25 sin2 t = 5.
v(t)
3
4
◦ Therefore, T(t) =
=
cos t, − sin t, cos t .
||v(t)||
5
5
r
3
4
9
16
0
0
◦ Next, we have T (t) = − sin t, − cos t, − sin t , so ||T (t)|| =
sin2 t + cos2 t +
sin2 t =
5
5
25
25
p
sin2 t + cos2 t = 1.
15
◦
◦
•
Therefore,
Finally,
N(t) =
T0 (t)
=
||T0 (t)||
3
4
− sin t, − cos t, − sin t
5
5
.
i
j
k
3
4
cos t
B(t) = T(t) × N(t) = 5 cos t − sin t
5
3
4
− sin t − cos t − sin t
5
5
= 4 , 0, − 3
5
5
Another quantity associated to the motion of a particle in space is the curvature
.
κ(t) =
||T0 (t)||
||v(t) × a(t)||
=
3
||v(t)||
||v(t)||
which measures how much the path of the particle is curving.
◦
Note that the curvature is always nonnegative, since both the numerator and denominator are nonnegative.
◦
Example: Find the curvature of the line
r(t) = ht, 2t, 2ti.
∗
We have
v(t) = h1, 2, 2i
so
||v(t)|| = 3.
0
curvature of this line is always equal to
◦
Then
T(t) =
and
T0 (t) = h0, 0, 0i.
Thus, the
T(t) is always constant, implying that ||T0 (t)|| =
Conversely, from essentially the same observation, the only curves whose curvature is zero everywhere
are lines, since a curve with curvature zero must have
◦
.
More generally, any line has curvature 0, since for a line
0.
1 2 2
, ,
3 3 3
Example: Find the curvature of the circle
T(t)
be constant.
r(t) = hr cos t, r sin t, 0i.
∗
∗
Then
v(t) = h−r sin t, r cos t, 0i so ||v(t)|| = r.
T(t) = h− sin t, cos t, 0i and so T0 (t) = h− cos t, − sin t, 0i.
∗
Then
||T0 (t)|| = 1,
We have
so we get
κ(t) =
1
r
.
1
.
r
◦
More generally, for any circle of radius
◦
From these examples, we can see that a small curvature indicates that the path is close to a line segment,
r,
the curvature is
while a large curvature means the path is turning sharply.
◦
For a general curve, the quantity
1
κ(t)
is called the radius of curvature of the curve: it gives the radius
of the tangent circle to the curve. (For a circle of radius
•
r,
r.)
the radius of curvature is also
The acceleration of a particle can be decomposed into a component in the direction of the unit tangent vector
T and a component in the direction of the unit normal vector N.
(This decomposition is very frequently used
in physical applications.)
◦
The tangential and normal components of acceleration are
give the decomposition
∗
aT =
d
||v||
dt
and
2
aN = κ ||v||
, and they
a = aT T + aN N.
aN , observe that v = ||v|| T, and then dierd ||v||
dT
d ||v||
entiate both sides using the Product Rule to get a =
T + ||v||
=
T + ||v|| ||T0 (t)|| N;
dt
dt
dt
d
2
then aT =
||v|| and aN = ||v|| ||T0 (t)|| = κ ||v|| .
dt
∗ From a = aT T+aN N, one has v×a = ||v|| T×(aT T + aN N) = ||v|| aT (T × T)+||v|| aN (T × N) =
v×a
||v|| aN B. Then one can immediately derive the formulas B =
and ||v × a|| = ||v|| aN =
||v × a||
2
κ ||v|| .
◦
To see that
a = aT T + aN N for these values of aT
Note that although
||v||
aT
and
can be positive, negative, or zero,
are nonnegative.
16
aN
is always nonnegative, because both
κ
and
,
•
The nal quantity associated to a curve's motion is the torsion
B0 (t) · N(t)
1
τ =−
=
2
||v(t)||
||v × a||
0
x
00
x
000
x
y0
y 00
y 000
z0
z 00
z 000
which measures how sharply the path of the particle is twisting out of its plane of motion.
◦
The torsion is the least intuitive of all of the quantities we have listed. (Note that it is the only one to
involve the third derivatives of the coordinate functions.)
◦
Another relation satised by the torsion, which may be easier to use, is
◦
By observing that
ds
= ||v||,
dt
we can rewrite the above relation as
Frenet-Serret formulas: the other two say
◦
•
dN
= τ B − κT
ds
. This is one of the three
.
dT
dN
formula follows from the denition of κ, while the
formula can be proven by writing
ds
ds
dN
dB
dT
N = B × T and then taking the derivative with respect to s: one obtains
=
×T+B×
=
ds
ds
ds
−τ (N × T) + B × (κN) = τ B − κT.
We have
∗
Hence
T=
Next we compute
∗
r(t) = hsin(3t), cos(3t), 4ti,
v(t) = h3 cos(3t), −3 sin(3t), 4i,
◦
and
dB
= −τ N
ds
1 dB
.
||v|| dt
The
Example: For the curve
◦
dT
= κN
ds
−τ N =
T, N, B, κ, τ, aT , aN .
q
||v|| = 9 cos2 (3t) + 9 sin2 (3t) + 16 = 5.
nd
so that
3 cos(3t) 3 sin(3t) 4
,−
,
.
5
5
5
dT
9 sin(3t) 9 cos(3t)
= −
,−
,0
dt
5
5
N = h− sin(3t), − cos(3t), 0i .
i
j
3 sin(3t)
3 cos(3t)
get B = T × N = −
5
5
− sin(3t)
− cos(3t)
so that
dT 9
dt = 5 .
Hence
◦
Then we
◦
The curvature is given by
◦
We have
dB
=
dt
◦
Finally, we have
◦
As a sanity check, we
κ=
9
||T0 (t)||
9/5
=
=
||v(t)||
5
25
12 sin(3t) 12 cos(3t)
,−
,0 ,
−
5
5
aT =
so
k
4
5
0
4 cos(3t) 4 sin(3t) 3
,−
,−
=
5
5
5
.
.
dB
·N
12
=
τ = − dt
||v||
25
d
2
[5] = 0 and aN = κ ||v|| = 9 .
dt
compute a(t) = h−9 sin(3t), −9 cos(3t), 0i,
.
and indeed
a(t) = 0 T + 9 N.
Well, you're at the end of my handout. Hope it was helpful.
Copyright notice: This material is copyright Evan Dummit, 2013-2015. You may not reproduce or distribute this
material without my express permission.
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