4
Chapter
CURVES
Force Fields
According
to Newton's laws of
is
accelerate
third law
F
velocity unless it
according to Newton's
motion, a particle will
subjected to forces.
line at constant
=
move
in
In that
a
straight
case
it will
(4.1)
ma
In this chapter we shall study the motion
m is the mass of the particle.
particles subjected to variable forces. That is, we must allow the possi
bility that the force applied to the particle depends upon its position (as in
gravitation) or even upon time (in the case of a variable electromagnet).
This gives rise to the notion of a field of force. A field of force will be given
in this way: at time t and position x a particle of unit mass will experience
Thus for each t0 we have associated a vector F(x, f0) to
a force F(x, t).
each point x. We can illustrate this as in Figure 4.1. Now, we have seen
that a particle of unit mass situated at x0 at time t
0, with a velocity v0
at t
0 will follow the path of motion determined by the given field of force
as the solution of the differential equation
where
of
=
=
f"(0
f '(0)
f(0)
=
F(f((), 0
=
v0
=
x0
306
Force Fields
Figure
path of motion
equation.
The
this
is
a curve
307
4.1
in space
given by the function
f which solves
Examples
Suppose a particle moves around the unit circle in
according to the function
1.
f(0
=
(cos
t, sin
f'0)
f"(0
we
=
=
plane
(4.2)
f)
What force field would account for this motion?
twice
the
Differentiating
find that
cost)
(-sin
t,
(-cos
t, -sin
0
=
-
f (0
particle is accelerating
magnitude (see Figure 4.2). This
Thus the
toward the
motion
can
origin
with constant
be accounted for
by
the force field
F(z,t)=
In
t
a
-z
fact, in the presence of this field, if
=
0
orthogonal
to its
circle centered at the
ential
f"(t)
f(0)
equation
=
=
-fit)
zo,f'(0)
=
izo
position
origin.
a
particle
has
a
velocity
at
time
vector, then it will continue to move in
We can see this by solving the differ
308
4
Curves
Figure 4.2
The solution of this
f(z)
=
z0 e"
which is
2.
z
=
equation
z0(cos
just (4.2)
t, sin
with z0
is
f)
=
1.
Suppose we are given in space a force field directed toward the
magnitude the distance from the z axis (Figure 4.3).
axis with
F(x.y,z) =-(x,y,0)
(Jf.y.z)
Figure 4.3
Fluid Flows
Here
again the force field is independent
F(x,y,
If
z,
of time and is
309
given by
f)= ~(x,y,0)
particle is at (1, 0, 0) with an initial velocity of (0, 1, a), what is
path of motion? We must solve this differential equation for
three unknown functions f(f)
(x(t), y(t), z(t))
a
its
=
f"(0
(x"(0, y"(t), z"(t))
=
x(0)
1, y(0)
=
x'(0)
=
0, y'(0)
=
The solution is
f(0
=
(cos
Thus, if a
is
positive,
0, z(0)
=
=
1, z'(0)
easily
f, sin t,
=
=
(x(f), X0, 0)
0
=
a
found to be
at)
0, the path of motion is
the
circle, of slope
path
a, and if
a <
circle in the
plane z 0. If a
upward spiral lying over the unit
the
0,
path followed is a downward spiral
of motion is
a
=
an
(Figure 4.4).
If
given a time-independent
R2,
R3,
graph sufficiently many
values of the field, it seems to be a broken line picture of a family of
curves.
In fact, there is a family of curves which fits the picture in
this sense: there is a curve through each point x which is tangent to
the vector F(x) at that point. These curves are called the lines of
force of the field and are found by solving the differential equation
3.
Time-independent fields.
force field
f '(f)
f(0)
=
=
on a
domain in
or
we
are
and
we
F(f(0)
x0
The solution of this differential
passing through
the
point
x0
equation
describes the line of force
.
Fluid Flows
of field of vectors arises in many other ways besides
as force fields.
Such an example which gives rise to a field is that of a fluid
in motion in a certain domain in R3. There are various ways of describing
0,
we
that flow. First of
idealize, by assuming that at the time t
The
general notion
all,
=
may
310
Curves
4
a
a >
=
0
a< 0
0
Figure
4.4
Then we can describe the flow by
a particle at each point x0 in R3.
describing the motion of each particle. The particle which is at x0 at time
t
0 follows a certain path which is given by a function f(x0 t). The
equations of motion are thus
there is
=
,
x
=
f(xo,0
position x at time t of the particle originally at x0 is f(x0 0particles are neither created nor destroyed; this amounts to
f(x0 0 is one-to-one and onto, and
asking that, for each t the function x0
Precisely,
We
the
assume
,
that
->
thus
can
be inverted.
x0
for
some
time
-
=
we can
,
also write
<b(x, t)
function <b.
1 was
So
(b(x, t).
Precisely,
the
original position
of the
particle
at
x
at
Fluid Flows
4.
Suppose
a
the vertical axis.
in
gas is rising at constant speed, and spiraling around
Thus the motion of particle is a helix as described
2.
Example
We do best to express this motion in cylindrical
be the (complex) coordinates in the plane (z
reie)
the height off the plane. Thus the path of motion described
coordinates: Let
and
311
w
z
=
the gas is
by
(z,w)
=
(zoeu,at
Thus the
time
t
+
w0)
(4.3)
particle originally at (z0 w0) will be
can certainly invert these
equations :
,
at z0
e",
w0 + at at
We
.
(z0, w0)
=
(ze-it, w-at)
(4.4)
Now, another
Let v(x, t)
way to describe a fluid flow is by its velocity.
velocity of the particle which is at position x at time t The field v is
called the velocity field of the flow. We can find the equations of motion
from the velocity field by solving the appropriate differential equation. For
the function f (x0 0 describes the motion of the particle originally at x0
The velocity of this particle at time t is f '(x0 0 and its position is f (x0 t).
be the
.
,
.
,
Thus
we
must have
f'(xo,0
f (x0
This
,
,
0)
equation
x0
can
5. Let
flow
v(f(xo,0, 0
=
=
us
be solved
find the
equations
(z',w')
=
uniquely.
are
velocity field of the gas flow in Example 4. The
(4.3). The velocity of the particle originally at x0 is
(iz0e",a)
velocity field we must write this as a
original position. We can
inversion (4.4), obtaining as velocity field
To find the
function of
at time t, rather than
do this
the
y(z, w)
=
by
position
means
of
(iz, a)
6. Suppose a fluid on the plane is spiraling
(Figure 4.5) according to this equation of flow
in toward the
origin
312
4
Curves
Figure 4.5
Here the
particle at time t 1 moves toward the origin so that its
argument is proportional to time elapsed, and its distance from the
origin is inversely proportional to time. Then
*
=
=
iz0eil
z0eu
Thus the
velocity
v(z,t)=
(--)z
The
I
field is
angular velocity is
decreases
as
1\
F-'=(,")Z(0
-
time goes
thus constant whereas the radial
7. Suppose now a fluid spiraled
velocity field was time independent,
v(z)
The
/'(0
/(0)
=
(i
=
=
of motion
0-i)/0)
z0
in toward the
for
example,
l)z
equations
velocity
on.
are
the solutions of
origin
so
that its
4.1
This
/(z)
Parametrization
of Curves
313
gives
=
e("1 + i)(
=
e-'ei'
In this case the distance from the
time (Figure 4.6).
origin decreases exponentially
with
We shall make a study of the geometry of paths of motion of single particles
and fluid flows, or families of motions, in this chapter. This study is a con
tinuation of analytic geometry, and begins the subject of differential geometry.
Figure 4.6
4.1
A
Parametrization of Curves
curve
in R" is
a
one-dimensional subset T of R".
be put into one-to-one correspondence with
We make this notion a little more precise.
set T can
a
This
means
line, in
a
that the
smooth way.
Definition 1. The image in R" of an interval under a continuously differ
entiable one-to-one function with a nowhere vanishing derivative is called a
C1 curve. If the function is A>times continuously differentiable we shall call
this
curve a
of the
curve.
Ck
curve.
The
particular
function is called
a
parametrization
314
Curves
4
Examples
8. The unit circle in R2 is
T:
z(f)
t, sin
parametrization:
teR
t)
(4.5)
( sin t, cos t) is never zero (the sine and cosine
simultaneously zero), this is a good parametrization.
could also parametrize the unit circle in this way :
Since
z'O)
never
We
z(t)
(cos
=
It has this
a curve.
(t,(l-t2)1'2)
=
but this
(4.6)
parametrization fails
is not differentiable there.
at t
=
other parts of the circle.
(0
=
(4.6)
does not
parametrize
of these failings
parametrize
z(0
=
the circle in the
care
cos t
Another
z(f)
=
where
z(t)
For
plane.
be written
=
=
example,
+ i sin t
curve
use
the
so on.
complex notation to
parametrization of
describe
curves
the circle
(4.5)
as
=
is the
eu
spiral :
ect
c
is
some
complex
number.
ett,eibt
or, in
polar notation,
r(t)
e"'
=
right half-plane,
of the lower semicircle, and
9. It is often convenient to
z(0
the
0, -(l-*2)1'2)
in the
can
cover
is,
t2)1'2, 0
-
will
takes
That
f2)1/2
+1, since the function (1
Notice that
the whole circle, but only the upper semicircle. Both
can be alleviated by introducing parametrizations which
z(0
are
=
0(0
=
bt
z
=
rew
Writing
c
=
a
+
ib, this becomes
4.1
Thus the modulus of
varies
z
is linear in t
(see Figures
10. The
T:
x(t)
x'0)
is
(sin t,
=
called
a
=
.
cos
right
(cos
never
11
curve
t,
circular
(4.7)
exponentially
4.8)
315
of Curves
with /, and the argument
4.7 and
0
t, sin t,
zero,
Parametrization
is
(4.7)
helix, is pictured in Figure 4.9.
Since
1)
a
valid
parametrization
The intersection of two
of the
curve.
cylinders with different axes is a curve
cylinders are both of radius 1 and
the
(see Figure 4.10). Suppose
one, Cx, has as axis the y axis, and
Then Cx has the equation
x2
+
and
y2
+
z2
=
the other,
C2 has
,
as
axis the
x
axis.
(4.8)
1
C2 has the equation
z2
=
(4.9)
1
z(t) =eu',Rea >0
Figure 4.7
316
4
Curves
zU)
=
e"',Rea
Figure
<
0
4.8
The intersection is, of course, the set of points where both
can be written x2
1
z2, y2 1 z2.
thus parametrize at least part of the curve by
hold and thus
x
=
f(0
(l-z2)1'2
=
=
y
=
(l-z2)1/2
((i-2)1/2,0-f2)1/2,0
Figure
4.9
-
or
=
-
equations
We
can
4.1
Parametrization
317
of Curves
Figure 4.10
Other parts will be found
f(0
f(0
on
this theme :
(-(l-f2)1/2,(l-f2)1/2,0
(M,(l-*2)1/2)
=
=
and
by variations
so on.
simpler parametrization
A
Then
we
fi(0
=
f2(0
=
is found
by
the substitution
have the two distinct branches of the intersection
(cos,
f,
(cos,
t,
cos
-
t, sin
cos t ,
x
=
cos f.
given by
t)
sin
t)
Implicitly Defined Curves
In the situation of the above
the
example,
(4.9).
Equations (4.8)
implicitly by
are given a collection
and
of
equations
such
as
say that the curve is given
More often than not, when we
we
these,
we can
determine, just by
318
4
Curves
working with them,
whether or not they do implicitly define a curve. Never
theless, the theoretical question remains: under what conditions can the
set defined by a collection of equations be parametrized as a curve ?
We
have
answered this
already
restate the conclusion
as a
question
fact about
in R2 in Theorem 2.14.
We shall
curves.
Proposition 1. Suppose that F is a differentiable real-valued function
defined in a neighborhood of (a0 b0) and F(a0, b0) 0 but dF(a0 b0) # 0.
=
,
,
Then the set
{(x,y)eN:F(x,y)
is
a curve
in
some
=
0}
(4.10)
neighborhood N of(a0 b0).
,
Proof. Since dFia0 b0) = 0, then either (SF/8x)(a0 b0) = 0 or i8F/Sy)iao b0)
Suppose the latter. Then, according to Theorem 2.16, there is an e > 0 and a
differentiable function g defined on the interval (a0
0
e, a0 + e) such that Fix, y)
if and only if y
Let /: (a 0
gix). In particular, gia0) b0
e, a0 + e) ->- R2 be
defined by /(f)
(f, git)). Then / parametrizes the set (4.10) near (a0 b0), and
clearly f'(t)
(1, g'it )) ^ 0. If instead (dF/8x)ia0 b0) # 0 we can give the same
argument merely by changing the roles of x and y.
,
,
,
= 0.
=
=
=
.
=
,
=
,
In
dimensions the situation is
higher
describe it in R3.
borhood
of a point
{p: Fiji)
is
-
p0
,
little
more complicated.
We shall
differentiable functions defined in a neigh
and VF(p0), VG(p0) are independent, then the set
If F,G
F(p0)
=
are
a
two
0, G(p)
-
through p0.
The verification of this fact is
G(p0)
=
0}
(4.1 1)
a curve
basically
another
of the fixed
point
assume that
algebra.
F(Vo) 0 G(p0). Since the vectors VF(p0), VG(p0) are independent, we
can change coordinates in R3 so that
That
VF(p0) E2 and VG(p0) E3
is, with respect to the new coordinates (x, y, z), dF/dx(p0) 0, dF/dy(p0)
1,
1. Now let
dF/dz(pQ) 0 and dG/dx(p0) 0, dG/dy(j,0) 0, dG/dz(p0)
Po
(xo yo zo) ; fr x near x0 we want to show that there are uniquely
theorem, complicated by
=
some more
linear
use
We first
=
=
=
.
=
=
=
=
=
=
=
>
,
determined y,
such that
0
G(x,
Newton's
method,
F(x,
Following
z
y,
z)
=
y,
we
z)
=
0
ask to find the fixed
point
in the y,
z
plane
4.1
of Curves
Parametrization
319
of the transformation
T(y, z)
dF
/
Our conditions
VF(p0)
borhood of p0
such that
T is
,
dG
[y + Yy iv*)
=
a
T(g(x), h(x))
Fix>
y>
E2 VG(p0)
=
=
,
contraction.
Z^>z
+
Tz
(x'
(Po)
\
y'
z))
E3 will guarantee that in some neigh
are unique y
g(x), z h(x)
Thus there
=
=
(g(x), h(x))
=
or
F(x, g(x), h(x))
Thus the function
=
/(0
0
=
G(x, g(x), h(x))
=
0, g(t), h(t)) parametrizes
the set
(4.11)
as a curve.
Examples
plane is the set ex+y y a curve? Let
1). Since dF/dx
F(x, y) ex+y y. Then VF(x, y) (ex+y, ex+y
0
is never zero, this is everywhere a curve and the equation ex+y
y
determines x as a function of y implicitly. dF/dy is zero when
0
The only point on the curve where ex+y
0.
x + y
y and x + y
a
function
as
to
find
is ( 1, 1), so at that point we cannot expect
y
12. At what
points
in the
=
=
-
=
-
=
=
=
of
x
=
x.
Notice, that even though we cannot explicitly determine the function
=f(y) given implicitly by ex+y y, we can find its derivative. For
=
exp(/(y)
so
upon
+
y)
-
y
=
0
differentiating
we
have
1
y)(f'(y)
+
D
exp[-(/(y)
+
^)]-l
exp(/(y)
+
-
=
0
or
/'(y)
=
13.
F(x, y)
=
x
sin xy
-
cos
y
(4.12)
320
4
Curves
WF(x, y)
If
x >
of
x.
(sin
=
xy + xy
cos
1, dF/dy(x, y) # 0,
Differentiating (4.12)
sin xy +
cos(xy)(y
x
xy')
+
xy,
x2
cos
(4.12)
so
xy + sin
defines y
with respect to
+
y'
sin y
=
y)
implicitly
x we
as a
function
find
0
or
sin xy + xy
=
y
:
sin
14.
VF
F(x,
y +
y,
VF and VG
x3
3x2y
=
These
=
y2, G(x, y, z)
when
=
(yz,
=
xyz + ez.
xz, xy +
e2)
0
2y
xy + ez
become
and
0
+
dependent
equations
xy + ez
x3y
VG
xz
yz
xy
xy
2y, 0)
+
are
+
cos
x
z)
(3x2y, x3
=
cos
2
=
y
x3
or
3x2y
=
0
The first
=
x3
pair
+
has
2y
no
solutions, and the second pair
amounts to
x
=
0
0.
But the set F(x, y, z)
and y
0 never intersects
0, G(x, y, z)
this plane, so everywhere on that set F and G are independent. Thus
=
{(x,
is
y,
=
z): F(x,
a curve
y,
z)
=
G(x,
y,
z)
=
=
0}
in R3.
Comparison of Parametrizations
that a given curve admits many parametrizations, and
advantage to be able to single out a best possible one. In
the study of the motion of particles there is a distinguished parameter, that
of time. But as far as the geometric study is concerned we can take any
parametrization we care to, the only criterion being that of convenience.
Now,
we
have
it would be to
seen
our
4.1
Geometrically,
a
is that of length
as
Before
most convenient
measured from
considering
how to compare
see
Parametrization
parameter,
or
321
of Curves
along
measure,
the
curve
fixed
point.
particular parametrization by arc length, let us first
two different parametrizations.
Suppose T is a curve,
a
the
parametrized by
x
=
a<t<b
f(t)
continuously differentiable function with nonzero derivative defined
[a, /J] and taking values on the interval \a, b], then the
composed function / a also parametrizes T. That is, we can write T
as the image of
If
is
a
a
the interval
on
x
If
t
increases
sense
*<z<0
<700=/(<t(t))
=
does, then these
as f
of direction
the
along
two
curve
parametrizations
T.
This
sense
determine the
same
of direction is called
We know from calculus that the necessary and sufficient
condition for t, x to increase simultaneously along the curve is that a' > 0
We shall say that x is an orientation-preserving
on the interval [a, j8].
orientation.
parameter if this condition is satisfied, and otherwise x is orientation reversing.
On the other hand, if we started out with two different parametrizations
of
a curve
r:x=/(0
then there must exist
point
of
=
function
a
(4.13)
flf(t)
a
precisely
corresponds
correspondence
to
of T
The
t.
x
or
relating
one
For each
the two parameters.
value of t and
precisely
one
value
T-0(T)=/(t)-*f
defines the function
function of
t
and
Notice that,
the chain rule
we
given
a.
have
We shall
g{x)
the two
=
verify below
and
point
same
in the
g'(x) and/'O)
same
is
a
differentiable
so
that t
=
o(x),
we
have
by
(4-14)
are
collinear when t, x are the same points,
> 0, that is, when g,f induce the
direction when a'
orientation along T.
a
f(o(x)).
parametrizations,
g'(r)=f'(o(x))-o'(x)
Thus the vectors
that
322
Curves
4
Definition 2.
Let T be
a curve
parametrized by x =f(t), a<t<b.
The
unit tangent vector to T at /(f) is the vector
no
rS
l/'0)l
=
By the above remarks
parametrization
orientation.
For if
(since
a' >
we
have the two
/VOOVO)
l/VO)) r'OOl
_
determine the
x
we
parametrizations (4.13), then by (4.14)
0)
/'0)
Ifftol
when f,
that the unit tangent vector is the same no
choose so long as it induces the same
we see
matter what
same
fit)
l/'(0l
_
point
of T.
Examples
15. Consider the unit
z
=
circle, given parametrically by
e"
Then z'
=
Notice :
ie",
we
which is
have T
a
iz,
unit vector, soJ"= ieu.
so that the tangent vector is
orthogonal to
position vector.
More generally, consider the spiral z
eat, where a is a complex
number. Then z'
aeat, so the tangent vector is exp f(Im a + arg a)t.
Notice that the angle between the tangent vector and the
position
=
the
=
=
vector is
arg T
arg
Thus T,
z
z
=
we
ft
have
(l,2t,3t2)
a
make the
curve
(t, t2, t3)
=
arg
always
16. For the
x
=
same
in space
angle.
given by
4.1
Parametrization of Curves
323
so
T(0=(i
8f2l+9fr2(1'2t'3f2)
+
Now, here is the verification of the fact that
related
by
continuously differentiable
a
2.
Proposition
Let T be
a
curve,
two
parametrizations
are
function.
and
f: [a, b]->r,
g:
[a, /?]
->
T two
parametrizations of T. Then there is a continuously differentiable function <r
mapping [a, b~] one-to-one onto [a, /?] such that g(x) f(o(x))for all x e [a, /?],
=
andf(t)
=
g(o-l(t)), for
all
t e
[a, 6].
Proof. Let t g [a, /?]. Since / maps [a, b] one-to-one onto V there is precisely
t.
Then a is a well-defined
te[a, b] such that fit)=gir). Define <t(t)
function from [a, /3] to [a, b]. a is one-to-one. Suppose airL)
ct(t2). Then
one
=
=
9(ri) =/(o-(t0) =/(a(r2)) =^(r2)
Since g is one-to-one
a
fit)
maps
=
gir).
[a, /3]
we
Clearly,
must have
then
t
ti=t2.
For if t
[a, 6].
onto
=
6
[a, 6] there is
a
point
t g
[a, /?] such that
o-(t).
We
now have only to verify that a is a continuously differentiable function.
Let
[a, ft] and f0
o-(t0). Now / is a differentiable function at f0 and /'(f0) = 0.
Let /
(/,...,/) in coordinates. There is a / such that />(/<>) ^ 0. Then
/ is a real-valued continuously differentiable function of a real variable and
since /'j(fo) ^ 0, it is invertible.
That is, there is a function /; defined on the
/? is also continuously differ
t for f near f0
range of/ near f0 such that hifit))
entiable. Now, since /(o-(t)) =#(t), we have /j(<t(t)) =^/t), so
t0 e
=
=
=
.
^(t)=(/(o-(t))
Since
and /}
are
=
(/Io^)(t)
continuously differentiable
so
is
<j.
The
proposition is
proven.
requirement that the derivative of the parametrization is
nonzero we would in general not have such a good relationship between
different parametrizations. Notice that by the same argument the inverse
mapping <r-1 to o is also continuously differentiable. Since o-"1 <r(t) t,
for all f e (a, b), we must have, by the chain rule,
Without the
=
(ff-1)'W0)-ff'(0
=
i
324
4
Curves
is also
never zero.
If it is always positive, a is an increasing function
always negative o is a decreasing function of t. Notice that if
/ g are two parametrizations of a curve and they do reverse orientation,
then they will become compatible simply by negating one of the param
eters.
Thus if /is not compatible with g, then/: \_-b, a]
C defined by
f(t) =/( 0 certainly is.
T is a parametrization of a curve we shall call f(a) the left
If/: la, b~\
end point of T and f(b) the right end point.
so
cr'O)
of t; if
-
->
The
Line
Tangent
Now, let T be
at x0
is the
in R", and x0 a point
through x0 which best
a curve
straight
line
shall show that this is the line
through
T.
The tangent line to T
approximates the curve. We
on
the tangent vector and is
given by
this
equation
x
x0 +
=
tT(x0)
t e R
The tangent line at x0 can be
x0 and nearby points xx
through
be that line.
to xx
Now
-
x0
L(xx)
the
limiting position of lines
(Figure 4.11). Let L(xx)
Then Z,(xj) is the set of all vectors originating at x0 and parallel
Let /give a parametrization of T so that x0
/(f0), xx f(tx).
the set of points x such that x
x0 is parallel to
computed
on
T,
as
as
xx
->
x0
=
.
is
=
xi-x0=/01)-/0o)
But that is the
same as
the set of points
x
such that
fih) -fjtp)
h-t0
Figure
4.11
x
-
x0 is
parallel
to
Parametrization
4.1
Now Xt
fj
-
f0 is
/'00),
x0 is the
->
as
same as t x
Thus
/'(f0)-
L0u)
->
10
of Curves
325
and the limit of the difference quotient as
through x0 and parallel to
tends t0 tne lme
desired.
Examples
17. Consider the helix
f(0
(a
=
t,
cos
a
sin t,
(Figure 4.9), given by
the
parametrization
bt)
Then
f'O)
f is
(
=
a
T(0
w
a
sin t,
a cos
t,
b)
positive parametrization
=
r-2
(a2
+
T2TT72
b2)1'2
(~fl
sin *'
if
we
a cos
take for the unit tangent
'
b>>
(see Figure 4.12).
18. A
f(0
=
(e'
damped
cos
helix
t, e' sin t,
(Figure 4.13) parametrized by
bt)
Thus
f'O)
=
(e'(cos
t
-
sin
t), e'(sin
f + cos
Figure 4.12
t), b)
(4-15)
326
4
Curves
Figure
so we can
T(0
take
as
the tangent vector
=
(2e2t
+
b2)1'2
Notice that the
4.13
^' (-CS
'
Sin
~
^' e'(sin
* + C0S
^' ^
('4*16^
the unit
sphere swept out by the tangent
(Figure 4.14), and that the functions
(4.15) and (4.16) give two different parametrizations of this curve.
If we consider the parameter as t, then the moving point described
by (4.15) has no tangential acceleration, whereas in (4.16) it is acceler
ating exponentially.
is the
same
curve on
for both helices
"
19. A different helix is this
f(0
Here
T(0
(cos
=
we
t, sin t,
take
as
(1
+
e2()i/2
(Figure 4.15):
e')
tangent
=
one
(~sin
vector
'>
cos
'>
e<)
"
4.1
Parametrization
of Curves
327
Figure 4.14
This
(Figure 4.16).
to the
equator
as t
+
->
as t
again
->
(Notice
oo.
1
z(T(0)
-
oo
is
a
helix
on
and winds
the unit
rapidly
that
?1
as t
oo
-
=
(l
+
-2t\l/2
e~2<)
*
0
as t
-
Figure 4.15
-oo)
sphere
which tends
around the north
pole
328
4
Curves
Figure
20. The intersection of a
x2
y2
+
z2
(x-i)2
+
y2=i
+
=
x(0)
=
and
z(6) is
z(9) is
curve
lying
(1, 0, 0)
at
cross
We shall
i
+
above the xy
the
the
(2)
y(9)
=
\
shall restrict attention to the
plane.
Let
angle
us
9
first
as
parametrize
shown in the
point on the unit sphere lying above (x(6), y(9), 0),
positive square root of 1
(x(0))2 (y(8))2, which is
-
.
we can
9
parametrize this
Then
-
1
-
-
=Sm2
+
=
we
sin 9
curve
f(0)=(- -cos0,^sin0,sin^
f '(6)
cylinder (Figure 4.17)
parameter the
use as
i cos 9
/l-cos0\1/2
Thus,
a
Then
figure.
thus
sphere and
1
In order to avoid the
part of the
this curve.
4.16
sin 6,
cos
9,
cos
-1
with the function
4.1
and
take
we can
as
Parametrization
of Curves
329
tangent line
\1^2/
9\
(2
JTc^j {-^e^os9,oos^
(4.17)
Notice that this does not
parametrize T at the point (1, 0, 0), since
point corresponds
parametric values 0, 2n. In fact, T
is not a curve at the point (1, 0, 0) since it does not have a unique
tangent line: the limiting position to (4.17) as x-> (1, 0, 0) is either
(0,1, 1)/V2or(0, 1, -1)/V2!
this
to both
EXERCISES
1. Find
x2 +
a
parametrization for the
iy2 + z2
with the
x2 + z2
=
curve
of intersection of the
ellipsoid
1
cylinder
=
1
paraboliod z=x2 + y2 with the
1.
y2 + z2
3. At what points is the set defined (in polar coordinates in R2) by
1 a curve? Find a parametrization of the curve.
r(l + a cos 6)
2. Parametrize the intersection of the
unit
sphere
x2 +
=
=
Figure 4.17
330
4
Curves
Figure
4. Consider the
r
4.18
family of cardiods (Figure 4.18)
(l+c)_1(l +CCOS0)
=
(a) Describe the behavior of this family as c ranges between 0 and + oo
1, c
2, calculate the unit tangent vector to the curve as a
(b) For c
.
=
=
function of 6.
5. What is the tangent vector to the
for 6
=
1,2,
curve r
a cos
bd ?
Graph the curve
5,V2.
6. Calculate the tangent lines to the
(a)
f(f)=(e-'cosf, e-'sinf)
(b)
f(x)
(c)
f (f )
=
=
(X'Sinx)
(e''7+-rsin')
(x,sin-|
I e',
,
following curves :
at (1,0).
at
(1,0).
sin f I
at
(1
,
1
,
0).
at (2a, 0, 0).
x2+y2 + z2=4a2,ix-a)2 + y2=a2
f it)
at (0, 1, 0).
it, cos f, sin f)
at (1,0,1).
f(f)=(f2,l-f2,f)
Find the tangent line at the origin for these curves.
(a) ex+v-y-\=0
(b) cos xy y + 1
exy2
cos xy =0
(c) x2 + y3z + sin z 0
1
x2 + y2 + z2
x + y + z
(d) exp(sin (xy + z))
(d)
(e)
(f)
7.
=
=
=
=
=
=
PROBLEMS
1
A snail
deposits calcium at the leading edge of its shell in a direction
a fixed angle with the ray from the snail's center to the leading
edge. Show that this hypothesis explains the spiral form of a snail's shell.
2. Graph the curve r
(1 + d2Y\2 + 62) and compute its tangent
.
which makes
=
vector.
4.2
Arc
331
Length
Graph the curve in R3 given in spherical coordinates by r <?', 6
t,
Graph the curve on the unit sphere made by the tangent vector of
the given curve.
3.
Arc
4.2
=
Length
Let T be
Definition 3.
f:
[a, b~]
and
->
f(b0)
=
e'.
=
z
T.
Let
a <
oriented
an
a0 <
b0
<
b.
curve
be the least upper bound of all
to
positively parametrized by
length of T between f(a0)
Define the
sums
(4.18)
I llfft) -*('-.
over
all choices of
=
a0
t0
<
tx
points
10
<
<
,
tk
.
.
=
.
,
tk such that
b0
description. Approximate (Figure 4.19) the curve
joining a succession of points along T between a0
by a
of the lengths of the line segments is less than,
sum
Then
the
and b0
and approximates the length of the curve. Now, if the points t, and tt-t
are very close, then the vector f(t,)
f(tt-x) is approximately equal to
we get a sum
in
this
If
we
(4.18)
replace
f'O.X'i *i-i)-
This definition has this
"
broken line
"
.
-
-
(4.19)
El|f(*,)ll('i-'i-i)
which is
a
Riemann
sum
approximating
the
integral
/.&0
f
J
l|f'(OII
(4.20)
dt
fin
t
Figure 4.19
b
332
4
Curves
(4.18) to (4.19) admit a small error
by
large, we have no hold on the error
between (4.18) and (4.19). Nevertheless, we can, by being very careful,
justify that substitution and deduce that the limit of the lengths of the approxi
mating line segment curves is the integral (4.20).
Of course, the substitutions
taking us
from
term but since k may be very
term
Proposition 3. Let T be a curve parametrized by f : [_a, b~\
length ofT between f(a0) and f(b0) is given by the integral (4.20).
->
F.
The
Proof. We will use the fundamental theorem of calculus to show this. Let
sit) be the length of T between f (a0) and f (f). We shall show that s is a differen
tiable function of f, and /(f)
||f '(f) ||.
If S0 is any sum like (4.19) approximating the
Fix f0 > a0 and consider a f > f0
length of r between f (a0) and f (f0), then S0 + ||f (f ) f (fo) II is a sum like (4.19) for
the length between f (a0) and f (f ). Thus
=
.
-
S0+\\f(t)-f(to)\\<s(t)
Taking the
least upper bound
over
all such So
,
we
obtain the inequality
(4.21)
5(f0)+||f(f)-f(fo)ll<s(f)
(4.19) corresponding to a partition of the interval
one of the points in this partition.
For if not,
add it to the given partition, and get a still larger sum. Let f0 < fi <
t be the points of the partition between f0 and f.
Then
Now,
[a0 t ].
,
we can
< tk
=
suppose S is
S
like
a sum
We may suppose that t0 is
=
S0+
l!f(f,)-f0,-i)ll
i=i
where So is
a sum
s<sit0) +
^s(t0)+
corresponding
to the interval
[a0 t0].
,
Thus
2\\fit>)-Hti-M
2
f'
f'O) dt
"'i-l
'
<^s(t0) + 2
<
i
=
f
-Vi
||f '(Oil dt<s(t0) +
Since this is true for all such
sums
*0)^^o)+f'l|f'(OII*
S,
we
f ||f '(f)|| dt
Jt0
have
(4.22)
4.2
From
and
inequalities (4.21)
||f(f)-f(f0)ll
(4.22),
s(t)-s(t0)
<
As t->t0, both the left and
Thus
for f0 between
5
a0
Now, if T is
1
<
right
Length
333
obtain
f
,
,
tto
t-t0
||f'(fo)ll.
we
Arc
llf '0)11 dt
(4.23)
ends converge, since f is differentiable, to
Since this is valid
is differentiable at f0, and s'(t0)
||f'(f0)!l.
and b0 we have the desired conclusion.
=
,
parametrized by f : [a, b~\ T, we can consider arc
T.
along
Precisely, let s(t) be the length of the piece of
length
T from /(a) to /(f). Then, from the above proposition,
a curve
->
function
as a
5(f)
=
J'||f'(0ll
dt
parametrize T by arc length, and it induces
the same orientation
original parametrization. Thus g(s), for every
on
T
of
distance
s from a: g(s(t))
s is the point
f(0- If L 's the length of
Notice that
T parametrizes T.
T from a to b, g: [0, L]
Since
s'(t)
||f0)11
=
>
0,
as
we can
the
=
-*
f'O)
=
=
so
g'0(0) s'(t)
g'O(0)-llf'(0ll
that
8,(!(,,)=If|i=T(,)
Thus
g'OO
is the unit tangent to T at
g(s).
Examples
21. The circle x2 +
x
=
a cos
9
y
=
a
y2
sin 9
Thus
f(9)
f'(0)
=
=
(a
cos
9,
a
sin
9)
a(-sin0,cos0)
||f'(0)||=a
=
a2.
Parametrize this circle by
334
4
Curves
Thus
s
=
x
re
s(9)
=
ad9
a9
=
Jo
parametrization according to
The
ing
length is given by
arc
s
=
I
g(s)
I
=
s
a cos
,
a sin
(\ -sin-,
cos-
given by
substitut
given by
1
af
a
(a
=
is thus
-
22. Consider the helix of
f(0
length
s\1
.
-
The unit tangent vector is
T(s)=
arc
a9.
=
t,
cos
a
sin t,
Example
17
given by
bt)
Then
f'O)
sin f,
(-a
=
\\f'(t)\\=(a2
Thus
g(s)
s
=
=
(a
+
a cos
t,
bt)
b2)112
s(t) =((a2
cos
(fl2
+
b2)ll2)t,
+Sb2)X/2
,
a
and the
sin
(<j2
arc
+Sfo2)1/2
length parametrization
,
(fl2
+
fc2)1/2
The tangent vector is
1
T()
=
(fl2
23. The
+
b2)l/i(-a
curve
of
Sin
*>
Example
C0S
fc)
20 has the
f(0)=^ ^cos0,-sin0,sin-j
+
f>
parametrization
s)
is
4.2
and
we
Arc
Length
335
find
||f'(0)||=-i=(3 2cos0)i2
+
2sJ2
so
5(0)
=
-^= f (3
2^2 Jo
+ 2 cos
<b)1/2
deb
and the unit tangent vector is
given
as
T(0)=(3r!o70)1/2(-sin0'cos0'co4)
Equations of Motion
Now
shall consider in greater detail the equations of a particle in
Suppose a particle moves through R" along the path given by
we
motion.
The
at time t is
x'(f), and the acceleration is x"(f). These
describing the instantaneous change in the
motion (direction and magnitude) of the particle. The speed of the particle
is the rate at which the distance covered changes, and thus is the time deri
vative, ds/dt, of arc length. As we have seen above, this is the magnitude
of the velocity. Thus
x
=
are
x(f).
velocity
vector-valued functions
.
dx
.
velocity
dt
Now, it is instinctive
tangent
to
a
d2x
acceleration
=
=
-=
dt2
(4.24)
=
dt
decompose
to the curve, and
dx
ds
.
speed
=
dt
the acceleration vector into
component orthogonal
a
component
to the curve.
We write
_
=
aTi + aN IN
where T is the tangent vector and N is a unit vector orthogonal to T and lying
in the plane spanned by the velocity and acceleration vectors. N is called
the principal normal to the curve of motion, aT is the tangential acceleration
of the
particle,
and aN is the normal acceleration.
We
now
show how to
336
4
Curves
compute these components of the acceleration.
dx
Differentiate the
equation
ds
T
=
dt
dt
obtaining
d2x
d2s
ds dT
l?~dT2
+Jtli
_
Now
(4-25)
dT/dt is orthogonal to T,
<T, T>
=
since T is
a
unit vector.
Differentiate
1
We then have
<T, T>
Thus
we can
N
(Of
+
<T, T'>
2
<T, T>
-
=
0
dT/ds
(4.26)
dT/dt
(d 27)
-
course, the differentiation in
length as well as time.) Let k
path of motion. Then dT/ds
<fx
=
take for the normal vector the unit vector in the direction
dT/dt
=
=
=
(4.25) could have been with respect to arc
|| dT/ds || This is called the curvature of the
.
=
dh
dsdTds
dt2
Ttls It
jcN, and (4.25) becomes
_
~
dt2
.
Thus the
d2x
.
acceleration
=
^
d2s
d2^
(ds\:
/ds\2 N
T + k\
dr2T+idt)\
=
tangential acceleration is the rate of change of the speed, and
proportional to the curvature, or bending, of
normal acceleration is
the
the
curve.
d2s
aT
=
dT2
a
=
/ds\2
{Tt)K
(4-28)
4.2
Arc
337
Length
Examples
24.
Suppose a particle moves along
according to these equations
x
=
t- 1
y
=
2t
-
the
parabola
y
=
1
x2
t2
Then
x
=
(t-l,2t-t2)
l-O.Ki-0)
d^X -(0,-2)
dt
2
particle is determined by a downward vertical
acceleration of constant magnitude (perhaps due to gravity) (see
Figure 4.20). The speed of the particle is
Thus the motion of the
dx
=
It
Thus
mum
(1
+
we see
=
(1
+
4(1
-
speed is decreasing until time t
trajectory), and then increases.
that the
path
-
(4-29)
t)2)1/2
of the
height
vector to the
T
4(1
of motion is
OT1/20> 2(1
-
0)
Figure 4.20
=
1
(the maxi
The tangent
338
Curves
4
and
so
(4.30)
-[l+4(l2-,ff"0(1-"'-1)
The normal vector is the unit vector in this direction :
N
(1
=
+
4(1
-
f)2)"1/2(2(l
-
t),-l)
Now
dT
dT /ds
ds
dt/dt
(l+4(l-02)2
(2(1-0,-1)
2
N
[1
+
4(1
02]3/2
-
Thus the curvature of the
path
of motion is
2
K
~
(1
And
+
4(1
-
02)3/2
finally
d2s
aT
4(1
-
tds\2
0
_
""
=
dt2
The
+
4(1
t2))3/2
length of the trajectory from
dx
f
Jo
(1
-
dt
x
=
-
"
\dt)
1 to
x
2
K
~
(1
=
4(1
-
f)2)1/2
+ 1 is
dt=\Jq [1 + 4(1 -f)2l1/2 dt
(Rotation) (Figure 4.21). Suppose now that
according to the equations
25.
+
a
particle
rotates
around the unit circle
x
=
cos(e')
y
=
sin(e')
Then
x
=
(cos(e'), sin(e'))
dx
=
It
(4.31)
e'( sin(e'), cos(e'))
dht.
-j-%
dt
=
e'(-sin(e'), cos(e'))
-
e2'(cos(e'), sin(e'))
(4.32)
4.2
Arc
339
Length
Figure 4.21
Now
are
T
we
the tangent and normal
=
(
Thus
(4.31)
d2x
=
di
can
to
N
sin(e'), cos(e'))
-
from
already know, just
be written
the
=
-
geometric considerations,
path
(cos(e'), sin(e'))
as
e'T + e2tN
Thus the normal acceleration is the square of the
tion. From (4.31) we read
ds_
dx
dt~
dt
=
tangential
accelera
e'
thus
s
=
e'
and the curvature of the unit circle is 1
Notice, that
x
=
what
of motion:
any motion
on
the unit circle
(cos(/(0), sin(/(0)
.
can
be written in the form
340
Curves
4
Figure 4.22
where
f(t) represents
of the unit circle is 1,
acceleration
The
arc
=
length
as a
function of time.
Since the curvature
obtain for any circular motion
we
d2s
(ds\2^
I N
T + I
-j
dt2
\dt]
tangential acceleration
is the rate of
acceleration is the square of the
change of speed, and the normal
speed.
us consider the motion of an object down a slide (see
The slide will be represented by the curve T. Let
4.22).
Figure
z
z(f) x(f) + z'yO) be the equation of motion of the particle. The
acceleration is z"(t) ; according to Newton's laws
26. Now let
=
=
mz"
=
where
F
m
is the
mass
of the
object,
and F is the
sum
of the forces
One such force is the force due to
on the object.
gravity
which is mg, where g is the gravitational field. The other force is
the restraining force due to the curve. This force acts in a direction
normal to the curve, and has undetermined magnitude. (That is, its
acting
magnitude
<j>N, where
is determined
<h is
a
have
mz"
=
mg +
<j>N
only by
the
object.)
Let
scalar and N is the normal to the
us
call this force
curve.
Thus
we
4.2
Arc
Length
341
Now, since we know the path of motion, we need only determine the
tangential acceleration aT
By Equation (4.28), we have
.
dh
=
dt
2
aT
=
<z", T>
=
<flf, T>
(4.33)
where T is the tangent of the
parametrized by
then the
curve.
If
we
consider the
curve
as
length: z=f(s) is the equation of the curve,
tangent vector is f'(s). Then Equation (4.33) becomes
arc
dh
=
dt
2
<9, f(s)>
and the
speed
can
be found
with initial conditions
as
the solution to this differential
s(0) s'(0)
For specific examples, let us first
line (Figure 4.23) with equation
z(s)
=
where
Tit)
=
=
=
consider the
i +
s0
0
o
is constant, and the force due to
=
a
+ ib is
a
unit vector in the third
Figure 4.23
equation
0.
curve
to be
a
straight
quadrant (b > 0). Then
gravity is ig. The speed
342
4
Curves
Figure 4.24
is thus found
as
the solution of the differential
equation
d2s
2
s(0)
=
s'(0)
=
Thus
z
{-ig,a
=
s(t)
Z(t)
27.
z(s)
sin
i
s
=
(gbt2)/2
-gb
and the
equation
of motion is
(\gbt2)^
-
Suppose
=
ib}
0
=
=
=
+
+
the
now
curve
is
a
semicircle
(Figure 4.24)
/ cos s
Then
T(s)
=
and the
i sin
cos s
is the solution of the differential
speed
d2s
~2
s(0)
s
equation
.
=
=
ig,
\
s'(0)
cos 5
i sin
5>
=
g
sm 5
0
=
Rotating Plates
28. We
referring
can
describe the motion of
to the
center of the
angle
plate be
as a
rotating
chosen
this line makes at time t with
flat circular
plate by
through the
at time t
0 and let 0(f) be the angle
its original position. Then a point at
a
function of time.
=
Let
a
line
4.2
z0 at time f
z
z"(0
=
iz0 0 V<"
Thus the
343
of motion
The acceleration of
z0(9')2eie^
-
tangential
(4.34)
acceleration is
|zo|0"
and the radial acceleration
z0(9')2.
If there is
so
that the
provide
is
path
velocity is iz09'e'm, so its speed is |zo|0'.
point is found by differentiating further :
the
is
0 follows the
Length
z0em,)
=
Its
=
Arc
object of mass
object will follow
an
this force.
zo(0')2,
so
even
m on
the
plate,
a
force
the motion of the
mz"(t) is required
plate. Friction may
Notice that the central component of this force
no angular acceleration, friction must
if there is
do its job. The further the object is from the center, or the faster the
plate spins, the greater the force required. It is this principle which
explains the centrifuge, which settles precipitates in solution by
spinning the fluid.
29.
Suppose
now
we
have
angular velocity,
(Figure 4.25). Assuming
constant
a
curved circular
and there is
a
there is
friction,
no
ball of
plate spinning at
in the plate
mass m
we
can
describe the
motion of the ball in terms of the initial data.
spherical coordinates r, 0, z in R3,
equation z f(r). In Figure 4.25
given by
section of the plate. Let
Let
us use
the
r
=
r(t)
be the
of the
9
=
=
0(f)
z
=
so
that the
we
depict
the
angular velocity
a
plate is
planar
z(t)
equations of motion of the ball, and let a be
plate. Since there is no friction, the ball
Figure 4.25
rotates as does the
344
4
Curves
plate, then 0 0(0 a'- Since
plate we must have z(f) /(r(0),
=
=
=
x(t)
=
Thus
for all t.
we
on
the
have
(r(t)e"",f(r(t)))
equation of motion, and we must find, using Newton's laws,
r(t). Now the acceleration is
the
as
the ball is constrained to lie
the function
x"
=
((r"
a2r
-
+
2ir')ei, f'(r')2
+
(4.35)
f'r")
-(0, g) we have a force
g be the gravitational field, g
that which restrains the
is
another
There
due
to
force,
gravity.
mg
motion to the profile of the plate. This acts in a direction normal
Let cbN denote this
to the plate and has undetermined magnitude.
Letting
There is
force.
of
=
plate
a
third force
acting
on
the ball, due to the rotation
tangential to the circle on
and the direction of this force is
which the ball lies.
We shall denote this force
by
C.
Then, by
Newton's laws
<bN + C + mg
Let
=
(4.36)
mx"
equate coordinates.
us
it lies in the
normal to
Now, since N is normal
plane through the
the curve z
f(r).
=
z
axis and the ball
Thus N
=
to the
surface,
(the
plane)
(nxe"*', n2) and
rz
and is
-=-(f(r)y1
i
(since n2/nx
is the
slope
of the line
perpendicular to
the
curve z
=/(r)).
Since C is tangent to the circle on which the ball lies, C
(ceix', 0).
The magnitude c of C is yet to be determined.
Finally, g is vertical,
=
Using (4.35) and substituting
equations as a result:
(0, g).
so
g
we
have these three
(bnx
c
=
=
these values in
a2r)
m(r"
2ar'm
=
-mg +
<bn2=m(f"(r')2+f'r")
Thus, eliminating <b from the first and last equations,
r
(1
=
(4.36)
r(t) is
+
a
solution of the differential
f(r)2)r"
=
a2r
-
f(r)f"(r)(r')2
-
we
find that
equation
f'(r)g
(4.37)
4.2
For what kind of
down,
or
up
have r'
=
r"
a
released
0,
(4.37)
=
so
Length
345
will it be true that the ball will not move
We must
no matter what its position?
plate
once
Arc
becomes
<x2r=f'(r)g
Solving,
we
obtain /(r)
=
a
plate is a paraboloid
angular velocity so that it
Thus if the
(a2/2g)r2.
of revolution, we can rotate it at
will have this property.
suitable
Suppose we are given a field of force in space, and the initial
of
position and velocity of a particle. Then we can find the path
30.
example, suppose the force field is
the particle are
x, and the initial position and velocity of
F(x)
of this
solution
the
Then the path of motion is given by
x0 v0
differential equation:
motion of that
For
particle.
=
.
,
/"(0
/(0)
-At)
=
=
x0
/'(0)
=
v0
We know the solution; it is
f(t)
=
cos t
x0 + sin f
v0
path of the particle is an ellipse in the plane determined by
If x0 , v 0 are orthogonal the major and minor
the vectors x0 , v0
axis have lengths |x0|, \v0\ (see Figure 4.26). The velocity vector is
Thus the
.
/'(t)
=
and the
-sin
f
speed
x0 +
is the
cos t
v0
length of this
vector.
X"
Figure 4.26
346
4
Curves
Suppose we have a force field in the plane which is of the same
magnitude as the position vector, but orthogonal to it. Using
complex variables on the plane, the force field is given by
31
.
F(z)
Let
iz
=
iz
or
it is the former.
us assume
z0 and
velocity
v0
f'V)
i/(0
/(0)
=
Then,
.
=
Suppose
a
particle has initial position
by solving
the motion is found
f(P)
zo
=
v0
The solution is of the form
f(t)
=
Aea' + Be-'"
where
a
y/i
=
=
(1
i)/~j2.
+
We solve for
A,
B
by substituting
the
initial conditions,
/(0)
=
f'(0)
z0
=
A + B
=
=
v0
a(A
B)
-
Thus
f{t)
Suppose
f(t)
For
=
z0
=
<LZi5>
=
1
2-(e*<
large positive
z
=
,
=
i'0
+
e< +
0.
z +
fttP
e-
Then
e-<)
t, the second term is
negligible,
and the
curve
is very close to
fe"'
which
we know is an outgoing counterclockwise
spiral. For large negative
t, the second term e~"' is dominant and that gives an incoming clockwise
spiral. Thus the particle comes spiraling in from outer space and then at
time
came.
t
=
0 pauses for
a
breath and then goes
(See Figure 4.27.)
racing
back from whence it
4.2
Arc
Length
347
Figure 4.27
EXERCISES
8. Find
arc
length
as a
function of the parameter for each of the following
curves.
(a)
(b)
(c)
r(l +
r
=
a cos
1 + 2
The
6)
cos
curves
=
1
9
in Exercises
9. Parametrize these
curves
and 7(a).
length, and find the curvature
6(a)(b)(d)(f ),
according to
arc
and normal.
(a)
(b)
(c)
x2 +
y2
=
1 x2 + z2
curves
The
curve
10. Find
the
1
=
,
The
in Exercise
in Exercise
normal
and
.
8(a)(b).
(d) The curve of Example 22.
(e) The curve of Example 23.
6(a)(e).
tangential accelerations for these planar
motions:
(a)
(b)
1 1
Find the
.
space
zit)
x(f)
i)t
(c) z(t)
(1 + 2 cos t)e"
exp(l
t + e1'
f^,y(f)=f3
(d) zit)
normal and tangential accelerations of these
=
=
-
=
=
:
(a)
(b)
x(f)
x(t)
=
=
(f, sin f, sin t)
(e', e~',t2)
(c)
x(f)
=
f(sin t,
cos
f,
1)
motions in
348
4
Curves
PROBLEMS
4. The
graph of a differentiable function
Find the curvature
5. The
as a
function of
y
=/(x)
is
a curve
in the plane.
x.
graph of a differentiable Revalued function y =/(x),
z
=
gix) is
a
in space. Find its curvature as a function of x.
6. A skier has to negotiate a series of hills whose
curve
profile is the curve
(Figure 4.28). There are three forces acting on the skier:
that due to gravity, the restraining force of the hills, and a force due to
friction which is proportional to his velocity. Find the differential equation
describing his motion.
7. I shot an arrow into the sky at an initial velocity of 80 feet/second and
at an angle of tt/3 with the horizontal.
The gravitational field is vertical
downward with a magnitude of 32 feet/second2
The air drags the arrow
with a force of 0.05 times its velocity. Find the equation of motion, and
the curvature of the curve of motion (the arrow weighs one pound).
8. In Example 26, let k be the curvature of the slide.
Show that the
magnitude of the constraining force due to the slide is/= ids/dt)2K
<g, N>.
Find the differential equations which determine x(f ), yit). Write out these
equations when the slide is the curve y cos x.
9. Suppose we have a field of force in space given by F(x, y, z)
( y, x, z). Find the path of motion of a particle which at time f 0 is at
(1, 1, 1) with velocity (-1, -1, 1).
y
=
e'x
cos x
=
=
=
Figure 4.28
4.3
Local Geometry
349
of Curves
Figure 4.29
10.
Suppose
l)2 + z2
l,
by rotating the curve (x
(The surface is, in cylindrical coordinates,
1, Figure 4.29). A cyclist cycling around the track tends to
a race
track is formed
1 <z<0 around the
(r
z'
l)2 +
=
z
=
axis.
ride up the bank as he goes faster.
Explain that.
11. Water is at rest in a very large sink when a stopper is removed in the
bottom center of the sink.
An idealization of the
The water accelerates toward the hole.
follows.
particle of
water
are
due to
gravity and the
ensuing motion is as
acting on each
The forces
mass
of the fluid itself.
The
field due to the former is (0, 0, g) and the field due to the latter operates
Find
as if the particle were on an inclined plane with vertex at the hole.
the resultant force field. Find the differential
equation giving the
rate of
rotation around the hole.
a ball of unit mass over a hill whose profile is the curve
l.
What minimum initial speed is
x
1 to x
from
x2)
y=exp(
required to ensure that the ball maneuvers this hill?
13. Suppose we are given in space a force field which is directed toward
Find
the origin and so that its component in the z direction is always 1.
0 at the point
the path of motion of a particle which is at rest at time t
(1,1,1).
12. We must send
=
=
=
4.3
Local Geometry of Curves
physical problems discussed, that the higher-order
parametrizing a curve have some significance. In
this section we will discuss the higher-order invariants of a curve ; that is, those
concepts which depend only on the geometry and not on the particular
We have seen, from the
derivatives of
a
parametrization.
function
350
4
Curves
Let r be a curve in R".
For purposes of simplicity, we shall take T to be
parametrized by arc length by x x{s). If T is twice differentiable, the tangent
vector TOO
1 for
x'00 is a differentiable function. Since (T(s), (Ts)>
all s, we obtain through differentiation 2(T(s), T'(s)}
0. Thus at any
point T' is orthogonal to T.
=
=
=
=
Definition 4.
The normal line to T at x0
x(s0) is the line through x0
parallel to the vector T'(j0). The osculating (or tangent) plane to T at
x0 is the plane spanned by, the tangent and normal lines.
The name osculating plane is quite descriptive. This plane osculates in
the following sense.
=
and
Proposition 4. Let x0,xx, x2 be three points on the curve Y. If they are
noncollinear, they determine a plane. This plane has the osculating plane as
limiting position, as xx, x2 tend to x0
.
In order to determine the
Proof.
it suffices to find two
independent
limiting position of the plane through
vectors which
are
limits of vectors
on
x0
,
Xi, x2
the variable
plane. The easiest way to do this is to refer to the Taylor expansion of the arc
length parametrization. Suppose/: (a, b)^T parametrizes T with respect to arc
For simplicity we may assume x0 is the origin, 0.
x0
length, and/(0)
According
=
.
to Theorem 4. 1
fis)
where lim
=
eis)
we can
write
W)s + T'i0)s2 + eis)s2
=
(4.38)
0.
5-.0
Let
x0
is
,
Xl=f(si), X2=fis2). Since x0=/(0)=0, the plane tt(si,s2) through
plane spanned by the vectors /OO, /(s2). Now, for each su si TX?,).
-rrisi,s2). Now
xi, x2 is the
on
Ji-1/(*i)
Letting
st
=
7X0) + T\0)si + eis)si2
-^0, that
says that lim *r
V00
7"(0) is
=
on
the
limiting plane. Now,
to
find another vector on the limiting plane, we take an
appropriate combination of
f(si),f(s2) so as to dispose of the T(0)s term in the Taylor expansion (4.38). Thus, we
consider
s2
We
are
tend to
f(si)
-
sif(s2)
interested in
zero.
7"(0)
Let
T'(0)is2 si2
finding
us
2s3 +
=
some
take the
(25)
4s3
Sis22) + eisi)s2 Si2
-
eis2)sis22
vector of this form which has
special
-
-
case Si
Eis) 2s3
=
=
2s2
=
2s3iT'iO)
a
limit
2s; (4.39) becomes
+
2ei2s)
-
sis))
(4.39)
as
su s2
Local Geometry
4.3
Thus
T'iO) + 2ei2s
that T'iO) is
by 7X0) and 7"(0).
-
we see
A few remarks
defined.
vector is
line has
In
eis))
the
on
are
351
is on the plane spanned by fis) and /(2s).
Letting s - 0,
limiting plane. Thus the limiting plane is indeed spanned
in order.
particular,
of Curves
If
if the
0, then the osculating plane is not
a straight line, then the tangent
plane which is closest to Y, so a straight
T'(0)
constant, and there is
no
=
T is
curve
osculating plane anywhere. Conversely, if T and T' are always
collinear along Y, then T must be a straight line (Problem 14). Now, in the
case where T' and T are collinear at the point in question, but not always
collinear, it may happen that the plane through x0, x^ x2 of Proposition 3
has a limiting position as xx, x2
x0 and it may not (see Problem 14). In
the former case we shall consider the normal plane as defined by the limiting
position, and in the latter case, we shall say that the normal plane does not
exist. Generally speaking, such cases are pathological, and we shall exclude
no
-*
,
them from further discussion.
Observe that for
For
N
on
Y in
R2,
the
osculating plane
Then the normal vector N varies
will form
and the vectors
R2 along the
no
in
is
(of course) just R2.
we
(see Figure 4.30).
curve
curves
define the normal vector to Y at x0 as that unit vector
R2,
the normal line so that the sense of rotation T -> N is counterclockwise
curves
uniquely
curve
Definition 5.
save
Let Y be
normal vector to Y is
natural
(T, N)
(called the moving frame).
a
"
continuously along
the
orthonormal basis for
In R" for
n >
2 there is
normal vector, and thus we leave the
that it should vary continuously along Y.
determined choice for
choice undetermined
"
a
a
a
twice differentiable oriented
choice of unit vector
Figure 4.30
on
curve
in R".
The
the normal line which varies
352
4
Curves
continuously along Y. The curvature
such that T'(s)
k(s)N(s) along Y.
of Y is the scalar function of s,
k(s),
=
Examples
32. The circle in R2
x(s)
a cos
=
\
,
cos
-
\
,
orthogonal
/
I
=
sin
-
,
=
z(0)
z'(0)
=
=
(l
eVa
+
-
so
)
[cos(s/a), sin(s/a)1/a
33. The spiral r
parametrization is
=
and counterclockwise from T
a)
a
T'(s)
to
s\
s
cos
\
the circle of radius
z
-
af
a
The normal is
Then
s\
s
sin
=
N(s)
a)
a
I
T(s)
-I
a sin
-
(Figure 4.31)
=
a
=
=
N(s)/a,
so
the curvature
of
is a"1.
ee (in polar coordinates) (Figure 4.32).
e(1 + i)(,
Oe(1+i)e
Figure 4.31
The
4.3
Local
Geometry of Curves
353
T()
t/4
N()
Figure 4.32
SO
ee
ds
Te
-*m-Tl
Thus the tangent vector is
1 + i
T(9)-
e>8
The normal is
dl
=
dldl
d0 ds
ds
=
_
gi( + */4)
N(0)
=
iew+lvJ2e-e
Thus the curvature is
Here is
plane
a
e'<9+ 3*/4>.
proposition
=
Now,
J2e~ V<9+3*':/4)
given by k(9)
which
gives
an
=
yJ2e~e.
interpretation of curvature in the
easily computable. It says that
and sometimes makes the curvature
354
Curves
4
the curvature is the rate of rotation of the
moving
frame with respect to
arc
length.
Proposition 5. Let Y be a given plane curve. The curvature ofY
of rotation of the tangent with respect to arc length, that is,
k(s)
=
js
r(s)e"(,) in polar coordinates.
T(s)
?'<"< +*'2>, and
Then NO)
JPT1
Since T
'
is
a
unit vector,
J
_(gI(J)\
=
iQ'eW
_
ds
ds
k(s)
=
=
=
Thus
(arg TOO)
Let
Proof.
1.
r(s)
is the rate
__
Q'eH6)+nl2
0'(s).
=
Examples
34. The helix
f(0
=
has
(a
=
t,
length
arc
TO)
cos
(a2
+
sin t,
a
s
=
(a2
bt)
+
b2)ll2t,
and tangent vector
b2y1/2(-a sin(a2
+
b2)~1/2s,
a
cos(a2
b2)~1(-a cos(a2
+
b2y1/2s,
-a
+
b2y1/2s, b)
Thus
T(s)
(a2
=
Thus N
=
+
(-cos
t, sin t,
0)
sin(a2
+
b2y1/2s, 0)
and
a
k
=
a' + b2
Observe that the normal line to the helix
axis of the helix.
35. Consider the
x(0
=
(cos
curve
t, sin t, sin
30
(Figure 4.33)
always points
toward the
4.3
Local
Geometry of Curves
355
Figure 4.33
Then
x'O)
(-sin t,
=
cos
t,
ds
||x'(OI|
=
-
T(f)
=
(1
=
+ 9
(1
+ 9
cos2
cos
3f)
cos2 3f)1/2
30"1/2(-sin
t,
cos
f,
cos
3f)
Computing
dT_dTaj_
ds
dt ds
=
-(1
+ 9
cos2 3f)"2(10
and the curvature is the
cos
t, sin 9f,3
length of this
cos
3f
cos
2f + sin 3f
sin2f)
vector.
Now, let us make one final remark about a curve in the plane. It is
completely determined, up to Euclidean motions, by its curvature. Thus, for
example, the only curve of constant curvature is a circle. This is, as we
shall see,
an
easy consequence of Picard's existence theorem for differential
equations.
Theorem 4.1.
Let
k(s)
be
a
continuous function
the origin. There is a curve Y whose
another curve parametrized by arc length
curvature, then
a
curvature
on
ofs in some interval I about
function is k(s). If Y' is
the interval I which has the
Euclidean motion will move Y'
onto
Y.
same
356
4
Curves
First
Proof.
curvature.
Euclidean
shall
verify the uniqueness. Let r be a curve with the given
x(s) be its arc length parametrization. We may apply a
motion (translation and rotation) so that x(0) is the origin and T(0) is
we
Let
x
the vector Ei.
=
Now
we
show there is only
one curve
with these
The
properties.
proof depends on the observation that the normal is rigidly attached to the tangent;
that is, its motion along the curve is completely determined by the tangent.
In
/e'e<9+"/2)
fact, writing T(s)=ems\ we have N(i) =e'<8(s+"'2). Thus N'
0V<9+")
T.
Now the system of differential equations
=
=
=
T(s)
has
T is
k(s)N(s)
=
N(s)
only one solution subject
unique, so
x(s)
f
=
Jo
(4.40)
-k(s)T(s)
=
to the initial conditions
T(0)
=
Ei, N(0)
=
E2
Thus
.
da
T(<r)
uniquely determined by the given conditions. Thus there is only one r
given curvature.
We now turn to the question of the existence of a plane curve with given curva
ture.
Again, by the fundamental theorem on differential equations, there exists a
solution of the system (4.40) subject to the initial conditions T(0)
E2
Ei, N(0)
If (T(s), N(s)) is the solution, then
is also
with the
=
=
.
x
defines
=
x(s)
=
'o
T(ct) da
plane curve r.
k(s)N(s), so k(s)
show that x'(s)
T(s) is
x"(s)
We must show that
a
=
=
Now, let f (s)
f (0)
=
f '(s)
N'(s)
=
=
-
=
-
iE2
=
=
N(0)
Ei
=
=
is
arc
-
=
iT(s).
1E1
-K-k(s)T(s))
=
=
-
as
so
<T(j), T(s)}
T(s) has
constant
we
must
E2
kCs)N(s)
i/c(i)N(j) =-k(s)T(s)
By the uniqueness, T
=
=
For then
Then
Thus T, N also solve the given initial value problem.
N.
Thus N
/T, so N T. It follows that
N
length along r.
5 is arc length
unit vector.
iN(s), N(s)
-iN'(j)
f'T'Cr)
a
j
To show that
is the curvature.
=
2<T(i), T'(*)>
length.
=
2k(sKT(s), N(s)>
Since T(0) =Ei, it is
a
=
0
unit vector.
=
T,
4.3
Local Geometry
357
of Curves
PROBLEMS
14. Show that if T :
collinear,
15. (a)
x
=
x
x(s) is a curve in R3 and T(s), T(s)
straight line.
be given by
then T is
Let r
=
T(0), T'(0)
origin.
(b) Let
\x3
.
*W
=
=
collinear, but
are
T has
an
osculating plane
at the
ifx<0
(o
ifx>0
Show that the
x
everywhere
(x, x3, x3)
Show that
,
are
a
curve
V
given by
(x, -g(x),g(-x))
does
have
osculating plane at the origin.
the sphere x2 + y2 + z2
1.
Show that r is
an arc of a great (i.e., diametric) circle if and only if the normal to T is
always collinear with the position vector.
17. Show that a curve is a straight line if all its tangents are parallel.
18. Three noncollinear points in R2 determine a circle.
If, for the
purposes of this exercise, we consider a straight line as a circle (of infinite
radius) we may assert that any three points determine a circle. Suppose T
is a curve in R through p0
Following the kind of reasoning on pages 324
and 325, define the osculating circle to T at p0 and find its equation in terms
not
16. Let r be
an
a curve on
=
2
.
of
a
parametrization
of r.
19. The radius of the
osculating circle is called the radius of
curvature.
Show that it is k~\
20. If the
a
straight
osculating circle
to F is
always
a
straight line, deduce that
T is
line.
osculating circle at a general point of an ellipse.
osculating circle at a general point of a parabola.
that if the osculating circle to a curve is always a circle of
21. Find the
22. Find the
23. Show
radius R, the
curve
is
a
circle of radius R.
given parametrically by
Suppose
Show that the curvature is given by
T is
24.
k
=
x'y"
-
y'x"
=
25. Show that
arc
length by
x
=
x(s),
y
=
[{x")2 + (y")2]"2
a curve
in the
plane of
constant curvature is
a
circle.
y(s).
358
4
Curves
Figure 4.34
26.
V:
Suppose
x=t(s) is
a curve
distance between f (s) and f (s +
circle.
27.
Suppose f is
t)
is
with this property: for every /, the
s.
Show that T is a
independent of
nonnegative function of a real variable with the
graph of /between 0 and x is proportional
to the arc length of that graph.
Find the curve.
28. Find the curve V with the property that at any point p the angle
between the tangent to r at p and the tangent to the ellipse
property that the
E: x2 +
2y2
=
a
area
under the
l
at the
point of intersection of E with the ray through p is constant.
x
t(s) be a planar curve. Suppose we have a string along
T with one end point at x0
If we unwind the string tautly and without
stretching, the end point will follow a curve E, called an evolute of T
(Figure 4.34). If 5 measures arc length from x
f(0), the curve E is
f (s) + sf'(s).
Find the evolutes to (a) the unit circle
parametrized by x
ell + n', (c) the parabola y
(b) the spiral z
x2, (d) an ellipse.
30. If we rotate a cylinder of water about its axis, the surface of the water
does not remain a plane.
What shape does it take and why ?
29. Let T:
=
.
=
=
=
=
4.4
4.4
Curves in
Curves in
359
Space
Space
T is a curve in space.
Let x0 e I\ and suppose T and N are the
and
normal
T
at
A
third unit vector orthogonal to both T
to
tangent
x0
and N will serve to provide a natural frame within which to discuss the
Suppose
.
behavior of the
is chosen
curve
near
x0
that the
.
T
This vector B, called the binormal to the
-> N -> B forms a
right-handed frame (see
triple
Figure 4.35). In this section we
trihedron along the curve, much as
plane curves.
curve
so
shall
we
use
this frame, called the moving
used the tangent and normal to
study
The three vectors T, N, B determine three planes : the tangent {or osculating)
plane is spanned by T and N, the normal plane is spanned by N and B, and
the
plane spanned by
of the
T and B is called the
Now the
rectifying plane.
cur
have seen, the rate of rotation, with respect to
arc length, of the tangent line in the
osculating plane. In three dimensions
there is another important intrinsic function on the curve.
Since B is a
vature
unit vector
<B, T>
=
0,
curve
on
we
<B', T>
Since T"
=
is,
as we
Y, <B', B>
0.
=
Thus B' lies in the
osculating plane.
Since
have
+
<B, T'>
kN, <B, T">
=
=
0
k
<B, N>
=
0, thus also <B\ T>
be collinear with N.
Figure 4.35
=
0
so
B' must
360
4
Curves
Definition 6.
B'
=
The torsion
of
i
a
curve
T is that function such that
tN.
-
The torsion measures the torque, that is, the twisting of the osculating
plane about the tangent line. That is, since the binormal is orthogonal to
the osculating plane, the change in the binormal reflects adequately the
change in the osculating plane. The Taylor development of the binormal
in a neighborhood of a point x0
x(0) is
=
B(s)
=
B(0)
t(0)N(0>
-
+
b(s)
(considering only first-order terms) the binormal at x(s) has moved
t(0) s toward the normal. Thus if t(0) > 0, the osculating plane has
twisted in the right-handed sense about the tangent line. At a point where
t
0, the osculating plane pauses ; it may or may not change its direction
of rotation about the tangent line. If x
0, the osculating plane remains
fixed along the curve; it follows that the curve lies on this plane.
Thus
=
=
t
Let F be
Proposition 6.
0 along T.
a curve
in
R3.
T is
a
plane
curve
if and only if
=
If T is
Proof.
a plane curve, let n be the plane containing T.
The tangent
always lie on n, so the binormal is always the unit vector orthogonal
and normal to T
to n.
Thus the binormal is constant,
On the other
hand,
r
=
0.
so
B'
Let
=
0, thus
t
=
0.
x(s) be the parametrization of T by
arc length.
Since t
0, B'
0, so B is constant along T. If for some s0 x(so)is
not on the plane through x(0) and orthogonal to B, then
suppose
x
=
=
=
,
<xC$o)-x(0),B>^0
Let
6(s)
since B
6(0)
=
be the function
=
B(j)
0, 6(s0)
x(0), B>. Then 6'(s)
<T(s), B> which is zero
orthogonal to T(s). Thus d(s) is constant. Since
0 also contradicting (4.41).
for all
=
(4.41)
s
<x(s)
-
=
and is
The fundamental formulas of space curve theory
We can now easily derive them.
N', B' with T, N, B.
Theorem 4.2.
T'=
(Frenet-Serret Formula)
kN
N'=-kT
B'
=
+ tB
-
tN
are
those
relating 1",
Curves in
4.4
The first and the third
Proof.
N is
N
=
unit vector,
a
<xT +
<N, B>
=
a
p
,8b;
0.
=
=
we
<N', N>
must verify a
are
0,
=
=
just the definitions of k, t, respectively.
so N' lies in the rectifying plane.
j8
-k,
=
But that follows from
t.
361
Space
Since
Write
<N, T>
=
0,
For
<N', T>
<N', B>
=
=
-<N, T'>
<N, B'>
-
=
=
-k
-(-T)
=
t
Examples
36. The circular helix:
x(t)
(a
=
cos
We have
T(,)
t,
sin t,
a
bt)
already computed
that
s
=
ct, where
c
=
(a2
+
b2)112,
and
l(-asinQ,flCos(^)
=
kN(s)
=
-5
I
a
cosl-J,
a
sinl-l,
01
thus
K
B
=
^N=-(cosQ,sing),o)
=
T
_TN
thus
=
t
N
x
B'
=
=
=
-
I
-tsinl-J, b cosl-J,
-a
J
iI(-bcosQ,^sing),0)
b/c2.
37. Let C be a curve in the xy plane, and let T be a curve of constant
slope lying over the curve C (see Figure 4.36). Thus if T is the tangent
Let b be that constant. Then T has the
to T, <T, E3> is constant.
parametrization
x(0
=
where
WO, y(t), 6(0)
(x(t), y(t)) parametrizes
C.
We may
assume
the parameter
362
4
Curves
Figure
is
length along
arc
x'
=
C.
4.36
Then
(x',y',b)
so
ds
=
-
Thus
T
||x'||
s
=
=
(1
x')2
+
+
{y'f
b2)1/2t
=
(1
+
^1,2
+
b2)1'2
and the
=
(1
tangent
+
b2)1'2
to T is
(*'. /.
Thus
kN
=
T'
=
(TTW7l(^/'.0)
Now if kc is the curvature of C, since
( /, x') is its normal vector, so
(x\y")=Kc(-y',x')
(*', /)
is its tangent vector,
4.4
Curves in
363
Space
Thus
Kq
kN=
{1
+
b2yl2(-y',x',o)
so
K
N
{1+Cb2)m
=
=
(-/,x',0)
Then
B
=
T
N
x
(1
+
^2)i/2 i-bx',-
(l
+
b2)i/2(-bx',-by',l)
=
=
by', (x')2
+
(y'f)
Differentiating,
1
~TN
B'
=
fojc
=
(i
+
b2Y'2
{-bx"> -by"> 0)
=
(i
+
b4>2 {~y'' x'' 0)
Thus
bKc
(1
+
b2)1'2
Local Behavior
of a
Curve
now make a close study of the local behavior of a curve relative
moving trihedron. Let T be a sufficiently differentiate curve, par
a < s < a.
We may perform a
ametrized by arc length by x
x(s),
Euclidean transformation so that x(0)
0, T(0)
E^ N(0) E2 B(0) E3
Expanding \(s) in a Taylor series, we obtain
We shall
to the
=
=
=
=
=
,
s2
x(s)
=
x(0)
+
x'(0)s
+
x"(0)
-
2
s3
+
x'"(0)
-
+
6
e(s3)
(4.42)
Now
x'
=
T, x"
=
kN, x*
=
jc'N + kN'
=
k'N +
k(-kT
.
+
tB)
364
4
Curves
Evaluating these
x(s)
In
=
at
sEi
zero
+
and
substituting into (4.42),
s2E2
we
+
E2
Ei +.
2ooo
-
-
-
obtain
E3
+
e(s3)
coordinates,
x
=
s(s3)
+
s
6
y
z
=
'iL s2
+ t
s3
=
6
s3
+
o
+
(53)
e(53)
Thus for small values of s, the
the equations
given
looks like the cubic
curve
curve
given
by
KX
y
Figure
=
z
r
4.37 is
a
=
picture
,
x3
of this
-
zz
4
3
curve
for
X
,
'3
K
jc>
# 0 the curve always passes through its
but lies on one side of its rectifying plane.
as kx
Figure 4.37
Notice that, so long
osculating and normal planes,
0,
x >
0.
4.5.
Varying
a
Curve in the Plane
365
Now, just as the curvature determines plane curves up to a Euclidean
motion, space curves are so determined by the curvature and torsion. The
proof of this fact is by the same kind of application of Picard's existence
and uniqueness theorem as we used in the case of the plane. We shall leave
the verification to the interested reader.
Theorem 4.3
is
a
curve
space
interval
of
Given continuous functions f, g defined in an interval I there
Y: x
\(s) given parametrically by arc length in some sub=
I such that
K(s)=f(s)
Y is
x(s)
=
g(s)
up to Euclidean motions in
unique
R3.
PROBLEMS
31. Show that
through
a
a curve
in R3 is
plane
a
curve
if all its tangent
planes
pass
given point.
32. Show that
33. Let T be
a curve
a curve
in R3 is
a
plane
curve
if its binormal is constant.
in the plane and let y be the intersection of the cylinder
T with the cone x2 + y2
Find the curvature of T in terms
z, z > 0.
of that of y. What is the torsion of y?
34. Let T be a curve in space, and y its projection onto the xy plane.
What is the relation between the curvature and torsion of T and the curva
=
over
ture of
y?
Suppose that T is the intersection of the surface z y2 in R3, with the
plane ax + by 0. What is the curvature of r at the origin ?
x2 + 2y2 with the plane
36. Let T be the intersection of the surface z
35.
=
=
=
ax
+
by
=
0.
What
37. Let T be
are
given
the curvature and torsion of T ?
in R3
by
x
=
Show that
the tangent lines to T.
gonal to those tangent lines is
x
4.5
=
x(s) + (c
Varying
a
s)T(s)
for
x(s).
Let S be the surface swept out
a curve on
which is
by
everywhere ortho
given by
some
constant
c
Curve in the Plane
A family of curves in the plane is a collection of curves {Yc}, as c range
through some set, usually of n-tuples of numbers. It is to be understood that
the curves of the family vary smoothly ; although we shall not make this idea
precise. For example, if x(t, c) are functions defined for real t and c lying
366
4
Curves
in
set
S, then the equations
some
x
value of
More
(4.43)
y(t, c)
=
y
each curve in the family is found by fixing a
Equations (4.43) as the explicit form of the family.
often, a curve is determined by a relation between x, y and a family
be given by an equation
determine
could
x(t, c),
=
a
of
family
curves:
We refer to
c.
F(x, y,c)
=
0
(4.44)
which, for fixed c gives the relation determining a curve. We refer to (4.44)
Since it does not refer to any particular
as the implicit form of the family.
of
the
individual
members, this form is particularly useful.
parametrization
The "constant" c which picks out the member of the family usually ranges
through some set in R" : in which case we refer to the family ((4.43), (4.44))
as an n-parameter family of curves.
Examples
38. A
ax
+
by
straight line in
+
c
the
plane
is
given by
the
equation
0
=
(4.45)
Thus the set of all
straight lines is given by (4.45) implicitly as a
3-parameter family of curves. If instead, we write down the slopeintercept form of a straight line,
y
mx
=
then
we
+ b
(4.46)
exhibit this
family explicitly
as a
2-parameter family
of curves.
and consider the
family of
(x(t),
39. Let
x
x(t)
=
be the
y
equation
tangent lines
y(t))
y
=
=
y(t)
of
a curve
to Y.
The
Y in the
equation
plane,
of the line tangent to Y at
is
^
+
y'(t)
x\t)(X
~
X(0)
(4"47)
4.5
This is the
Varying
a
Curve in the Plane
explicit form then of a 1 -parameter family.
367
(The parameter
is*.)
40. Consider the
x
=
The
cos t
=
y
family
case
sin
where Y is the circle
t
of tangent lines to Y is
given by
the
equation
cos t
y
=
t
sin t
This
y
sin
simplifies
x
=
(x
cos
0
(4.48)
to
cot t +
esc
t
can make this appear even more palatable by
the parameter of the family. Letting c
cot t
so
becomes
+
-(1
c2)1/2/c,
(4.48)
We
as
=
,
XC
a
-
(i
+
1 -parameter
c2yi2
family
taking
we
find
cot t
esc t
=
(4.49)
of lines.
41. Suppose a hoop is rolling along a horizontal line (see Figure
4.38). This collection of positions of the hoop forms a 1-parameter
family of circles where the point of tangency with the horizontal (the
Figure
4.38
4
Curves
Figure
axis) is taken
family is thus
x
(x-c)2
+
42. The
is
a
to be the
(y-l)2
family
1 -parameter
4.39
parameter.
The
of circles tangent to both the x axis and the y axis
family of curves (Figure 4.39). We take for the
parameter the point of tangency of the
-
It is
c)2
+
easily
(y- r)2
seen
considerations.
for the
l
=
is the radius of the cth circle, then the
(x
implicit equation
=
curve
equation
with the
of the
x
axis.
family
is
If
r
clearly
r2
that
r
Thus
=
c; this follows from
the
elementary geometric
family is implicitly described by this
equation
(x
-
c)2
+
(y- c)2
=
c2
(4.50)
43. The family of circles of radius 1 tangent to the parabola y
x2
(Figure 4.40). We can take as the parameter the x coordinate c of
the points of tangency. The center of the circle is on the line per
pendicular to the parabola at (c, c2). Thus if (r, s) are the coordinates
=
4.5
of the center of the cth
s-c2
(r
These
+
(s- c2)2
equations
=
C
Curve in the Plane
have
we
have the solution
+
_
S~C
Thus the
1
2
(l+4c2)1/2
(1+4C2)1'2
implicit equation
for this
family
of circles is
(x-c-(TT^)2 (^c2 (JTi?F)2
+
44. Let T be
to
T.
369
1
2c
T
a
-h(r-c)
=
c)2
-
circle,
Varying
If T is
a curve
given
+
in the
as a
We seek the
plane.
function of
1
=
arc
family of tangents
length by x \(s), then the
=
lines
g(w)
=
x(s)
form the
+
uT(s)
family
(4.51)
of tangents to Y with
s as
Figure 4.40
Suppose now
family at every point.
particular tangent line
parameter.
that y is a curve which is orthogonal to this
If h(s) is the point of intersection of y with the
370
4
Curves
x(s), then y is parametrized by x h(s). h(s) is then of the
form (4.51) with a particular choice u(s) of u.
Writing then h(s)
x(s) + u(s)T(s), and differentiating, we obtain
(4.51)
at
=
=
h'(s)
(1
=
Since
h'(s)
is tangent to y and thus,
1
u'(s) 0. Thus
=
x(s)
+
family
given by
=
eis
These
The
+
are
to the
=
(s+ c)T(j)
The
x
by assumption, orthogonal to T,
u(s) s + c. So the family of
tangent lines to Y is given by
=
orthogonal
curves
is
u(s)T(s)
+
must have
we
x
u'(s))T(s)
-
i(s
of
+
just
(4.52)
curves
c)eis
=
orthogonal
[1
+
i(s
+
to
the tangents to the circle
z
=
e"
c)]e's
the evolutes of the circle.
Differential Equation of a Family
A differential
V'
=
determines
equation
Fix, y)
a
1 -parameter
family
of curves, if the function Fis decent enough.
c there is a unique solution of the initial
For, under such conditions, for each
value
problem
y'
=
F(x, y)
The solution
y(x0)
=
c
be written y =f(x, c), which can be considered as either
explicit,
implicit form of the family. Now, it is usually true that a
1 -parameter family of curves is the family of solutions of some differential
equation, and we would often like to find that differential equation.
the
can
or
Suppose, for example, that y f(x, c) is the equation of a given 1-parameter
family. If y y(x) is one particular curve (i.e., y(x) =/(*, c0) for some
fixed c0), then these two equations must hold
=
=
y
=
f(x,c)
y'
=
dJ-(x,c)
4.5
for
value of
Varying
a
Curve in the Plane
371
(i.e., c c0). It may be possible to eliminate the param
equations, thus obtaining a relation between x, y, y'
which must be satisfied; this is the differential equation of the family. For it is
a differential equation which must be valid for each member of the family,
and this is a differential equation which determines the family.
More generally, suppose the family is given implicitly by
some
eter
c
F(x, y,c)
If
x
is
a c
=
c
=
from these two
x(t),
y
0
=
y(t) parametrizes
=
of the
one
curves
in the
family,
then there
such that
F(x(t), y(t), c)
identically in
=
(4.53)
Differentiating
t.
now
with respect to t,
we
have
dF
dF
ox
0
(x,
c)x'
y,
+
oy
(x,
y,
c)y'
=
0
(4.54)
we can eliminate c from Equations (4.53) and (4.54), the result will be a
relation between x, y, x', y' which must be satisfied for each curve in the
family and thus is the differential equation of the family. Of course, if x
is the parameter along the curve, and y
y(x) is its equation, (4.54) becomes
If
=
8l(x,y,c) + d-f(x,y,c)^
dx
oy
=
0
(4.55)
ox
Examples
45. Consider the
y2
-
ex
=
-
c
=
with respect to
-
parabolas (Figure 4.41)
lyy'x
x
(considering
0
Thus the differential
y2
of
0
Differentiating
2yy'
family
=
0
equation
of the
family
is
y
as a
function of
x),
372
4
Curves
Figure 4.41
or,
y
excluding
2y'x
-
=
the
curve
family y
(as
already know).
equation
=
cex is
we
=
exp(-
0,
=
0
46. The
y
y
x
The
given by
family y
the differential
=
ecx is
equation y' y
given by the differential
=
j
(Clairaut's Equation). Let y f(x) give a curve in the plane,
and consider the family of lines tangent to that curve. That family
is given implicitly (taking the x coordinate of the point of tangency as
the parameter) by this equation,
47.
y=f(x)
=
+
f'(c)(x-c)
Now, upon differentiation
/ =/'(c)
(4.56)
we
find
(4.57)
4.5
To say that
(4.57)
y'. Then,
equation
y
=
we can
amounts to
y'x
where
upon
eliminate
that
saying
eliminating
Varying
c
from the
we can
we
a
Curve in the Plane
373
pair of Equations (4.56) and
(4.57) for c as a function of
as differential equation, the
solve
obtain
h{y')
+
(4.58)
the
h(y') represents
y'.
considered
expression f(c) f'(c)c,
as
a
function of
Thus
Equation (4.58), known
of the differential
solutions
y
equation
of
as
a
Clairauts'
family
equation,
is the
form
general
of lines tangent to
a
curve.
Its
are
=
+
ex
h(c)
Notice that the
given curve y=/(x) also solves Equation (4.58) (because
(4.56) and (4.57) which hold under the substitution y f(x)).
singular solution of the equation.
it is derived from
It is called the
48. The
implicit
y
=
c2
=
family
y
+
2c(x
c)
-
y
=
x2 has the
we
=
lex
obtain
-
c2
y' 2c. Thus c \y'', so
equation of the family,
=
=
we can
eliminate
to obtain this differential
=
/x-Ky')2
49. The
family
implicitly by
xc
y
Then
y'
=
yx
of lines tangent to the circle x2 +
(1-c2)1'2
=
c, so the Clairaut
(1
y
parabola
form
Differentiating,
c
of lines tangent to the
-
(/fy2
r
equation
of the
family
is
y2
=
1 is
given
374
4
Curves
Family Orthogonal
to a Given
Family
given family of curves. We propose to find a family
G of curves everywhere orthogonal to F. Thus, if p is a point in the
plane, and Y is the curve in F through p with tangent T1; and y is the
0.
curve in G through p with tangent T2 we must have <T1; T2>
Suppose the family F is given by the differential equation (Figure 4.42)
50. Let F be
a
=
a(x, y)x'
+
b(x, y)y'
=
(4.59)
0
Thus, since (x', y') is the tangent field
with
to
(a(x,y), b(x,y)) (for <Tl5 (a, b))
differential
*'
equation
for the
family
=
we
0
must have
by (4.59)).
T2 collinear
Thus
the
G is
y'
(4.60)
=
a(x, y)
F,
b(x, y)
Figure 4.42
4.5
51. Find the
xy
=
Varying
family orthogonal
a
Curve in the Plane
to the
family
of
family
is
+
375
hyperbolas
c
The differential
equation
of this
yx'
xy'
=
0.
Thus the
differential equation of the orthogonal family is
x'
y'
y
x
xx'
or
is
yy'
=
0 which
integrates
to
x2
y2
-
52. The family orthogonal to the family
given by the differential equation
1
V'
y
-2x
(here
x
is the parameter,
2,y2
+
*
=
y
2xx' +
=
This
1).
of parabolas in
integrates
Example
=
45
to
(4.61)
family
which makes
The differential
yy'
x'
c.
c
53. Find the
(4.61).
so
=
equation
an
of the
angle of n/4 with
family (4.61) is
the
family
0
The
family orthogonal to this family has tangent collinear with 2x + iy,
family we seek has tangent collinear with this vector rotated
by n/4. Thus the tangent field is collinear with ei("/4)(2x + iy), or,
what is the same, (1 + i)(2x + iy)
2x
y + i(2x + y).
Thus, the
differential equation is
thus the
=
2x
y
2x + y
Envelopes
we have been studying have the property that there
(or curves) which is not a member of the family but bounds the
family (see Figures 4.39-4.41). Similarly, for a family of lines tangent to a
Many of the families
is
a curve
376
4
Curves
given curve,
an envelope.
the
First of all,
x
=
c, y
exists
=
curve
bounds the
We want to
c, y
we can
find
a
bounding
curve
is called
for families.
families do not admit
some
=
Such
family.
how to find
envelopes
envelopes. Clearly,
x2 + c do not admit envelopes. However, if
it by the present techniques.
see
the families
an
envelope
Definition 7.
Let F be a family of curves in the plane. A curve Y is an
envelope for the family F, if through every point p in r there goes a curve in
F which is
tangent
Suppose
that
F(x, y,c)
a
to Y at p.
family
is
given implicitly by
0
=
and that the
curve Y: y =f(x) is an envelope of this family.
Then, for every
there
is
a
x0
c(x0) such that the curve C corresponding to F(x, y, c(x0)) 0
is tangent to Y at (x, f(x0)). Thus we must have
=
F(x0,f(x0), c(xo))
and since the
=
0
(4.62)
C has the tangent direction
curve
(l,/'(*o))>
we
must
have, by
(4.54),
dF
dF
+
-fa (*o /(*o)> c(x0)) y (*o f(x0), c(x0))/'(x0)
,
,
Differentiating (4.62)
dF
-dx
with respect to x0
8F
+
we
=
0
(4.63)
also find
dF
+
Comparing (4.63)
and
(4.64)
<4-64>
=
TyfiX) TcC'(X)
we
have
as a
result
dF
dc
Thus if
(x0 f(x0), c(x0)) c'(x0)
(x, y)
,
is
on
the
=
evenlope Y,
0
(4.65)
there is
dF
F(x,y,c)
=
0
oc
(x,y,c)
=
0
a c
such that
4.5
and
we
eliminate
can
of Y.
equation
x
also hold
3F
^
=
a
Curve in the Plane
from this
c
Notice that from
F(x, y,c)
Varying
0
pair of equations to
(4.64), the equations
obtain
an
377
implicit
dF
+
_
dx
dy
/
=
0
Y. Eliminating c from
equation of the family, so
differential equation.
this
on
differential
pair we obtain once again
envelope must also satisfy
the
the
this
Examples
54. Find the
(x
of
-
c)2
+
(y
Example
2(x
c)
of
-
c)2
+
-2(x
c)
-
or
-
family
1
(4.66)
x
=
obtain
c we
42.
=
l)2
(y
of the
=
1,
c)
-
+ y
=
(-j)2
+
this in
(-*)2
=
=
2c
(4.67)
we
obtain
(*+>')2
or
2xy
=
0
Thus the
envelopes
y
=
2,
y
=
0.
(4.67)
c
Substituting
or
c2
or
x
to find
family
We must eliminate
2(y
c
c
envelopes
(y- c)2
Example
=
of the
We differentiate with respect to
0
55. Find the
(x
l)2
41.
=
Eliminating
-
envelopes
are x
=
0,
y
=
0.
c
from this
equation
and
4
378
Curves
56. Find the
y
=
x2 sin
of the
envelopes
ex
Differentiation with respect
0
=
ex2
family
to
c
yields
cos ex
or
n
c
=
0,
cx
=
gives y 0 which fails as an envelope.
x2 (Figure 4.43).
envelopes y
n/2, 3tt/2 yields
The condition
ex
=
=
Xy'
This is
y
=
c
=
0
=
the
57. Find the
y
2>n
-,,...
ex
+
a
(1
+
envelope
of the
family given by
{y')2)
Clairaut
+ 1 +
=
equation
and has the solution
c2
Figure 4.43
But
4.5
Differentiation with respect to
0
=
x
+ 2c
or
c
=
Varying
a
379
Curve in the Plane
yields
c
-
2
Thus the
envelope
3x2
y
=
of this
family
is the
curve
1
-+l
EXERCISES
equations for these families of curves:
l
(c) xec"
0
(d) x sin y + c sin x 0
xy
12. Find the differential
(a)
(b)
(e)
(f )
xyc
=
sin xy
l
=
a cos
=
=
1
yec('+
1
sin(x + y+c) + cos(x + y + c)
13. Find the implicit form of the family given by these differential
equations :
l
(c) (y')2 + ^
(a) xy'-yx'=0
l
(d) y + y'x + siny=0
(b) x' + yy'
\y
(e) y'(sec x tan x)
14. Find the implicit form and the differential equation of the family of
=
=
=
=
=
circles with center
on
the y axis and tangent to the
x
axis.
family of ellipses with foci at (-1, 0), (0, 1).
Find the family of curves orthogonal to the family in
15. Find the
16.
Exercise
14;
Exercise 15.
17. Find the
family orthogonal
to the families of Exercises
12(a), (b),
(f), 13(b), (d).
18. Find the
family making
an
angle of tt/3 with the family of
Exercises
12(a), (b), (c).
19. Find the envelopes of the families of Exercises
20. Find the envelopes of these families:
(c) 13e
(a) y sin(x-c)2
12(a), (b), (c), (d), (e).
=
(d)
family of cardiods
sin ad.
The family r
136
(b)
(e)
(f )
The
y
r
=
=
e*sincx
(1
+
c)_1(l
+
c cos
6).
=
PROBLEMS
38. Find the
orthogonal
to
family of evolutes of the parabola
y
=
x2.
Find the
family
this family of evolutes.
cee.
39. Find the family orthogonal to the family of spirals r
a
tall
feet
building
slips
originally leaning against
40. A ladder 10
of
curves which are the trajectories of the
the
Find
family
4.44).
=
(Figure
points on
the ladder.
380
4
Curves
\\\\\\\\\\\\\\\\\\\\\\\\\V\\\VVVX\V
Figure
41
Find the
.
family
of
4.44
trajectories of the points
horizontal plane.
42. A line segment of length 2 has its
ball
rolling
on
the circumference of
a
on a
of
Find the
trajectories
parabola (Figure 4.45).
43. A ball of unit
points
mass
x0
on
endpoints
on
the segment
is at the end of
a
the
as
parabola
y
=
x1
it slides along the
string of unit length attached
vertical bar rotating at constant angular velocity. Find the
0
of
motion
of the ball assuming its position and velocity at time t
path
Find the trajectory of any point on
to be (1, 0, 0), (1, 0, 1), respectively.
to the
top of
a
=
the
string.
44. Find the
length
with
family of curves swept
endpoints along the curve
out
xy
by the midpoints of bars of given
1 in the first quadrant.
=
Figure 4.45
4.6
Vector Fields and Fluid Flows
We have
come across
vector fields several times
now
mass
of
a
study
noninterreacting particles.
want to
already : the gradient of a
field of forces, are all vector fields. We
such fields in connection with fluid flows : motions of a
function, the gravitational field,
4.6
Vector Fields and Fluid Flows
381
Figure 4.46
A vector field is
in R",
a
a
a
function which
vector field defined
on
assigns
R", usually considered
vector in
on
D in R" is
U, but interpreted pictorially
as
as
to each
point in a given domain
given point. Thus,
based at the
nothing more than
in Figure 4.46.
an
Revalued function
Examples
body in
58. A
Suppose
there is
a
space sets up a field of gravitational attraction.
body of unit mass situated at the origin. According
to Newton's laws another
body
at the
origin
squared.
distance
with
body of unit
a
force
gravitational
vector field defined on
or, in
,
rectangular
_
to
field of
R3
-
a body
{0} by
given
the inverse of the
at
"2
situated at
a point p by a
(see Figure 4.47).
the origin is the
coordinates
(x,
.
is attracted to the
represent this attraction
origin and of length ||p ||
We
vector directed toward the
Thus the
mass
proportional
y,
z)
v(x,y,z)--(x2 + y>2 + z2)3/2
family of curves, we
tangents to the family (Figure 4.48).
gents to the family of circles x2 + y2
59. Given
a
may consider the field of unit
In particular the field of tan
=
c2 is defined
on
R2
-
{0, 0},
4
Curves
Figure 4.47
and is
given by
}
The
R
-
(x2
family
+
y2f'2
of unit tangents to the
[0, 0} by
T(x, y)
family
(x, y)
=
(x2
+
y2)1'2
Figure 4.48
of rays is defined
on
4.6
If we
it
are
given
a vector
field tangent to a
Suppose then that
field
is
v
v on a
domain D in R", the
a
in D such that
v(x) is tangent to Y
parametrizes the curve Y.
so we must have f '(0 an(l v(f(0) collinear.
of the differential equation
curve
at each
function which
f
then f
of the
'
(0
arise: Is
point
x on
Let f be
Y.
a
Then f '(0 is tangent to Y at f(0
In particular then, if f is a solution
v(f(0)
=
parametrizes a curve tangent
preceding section
dx
,
to the
given field.
In the
terminology
,
--v(x)
is the
questions
383
of curves, and if so, can we discover the curves ?
given vector field in the domain D, and T is a
family
a
Vector Fields and Fluid Flows
0
=
(parametric) differential equation of
the
family of
curves
tangent
to
the vector field.
60.
Suppose v(x, y)
differential
X'
x(0)
=
y(0)
x0
=
Xoe'
We
can
y-cx2
y
(4.68)
2y
=
The solution is
x
(x, 2y). (Figure 4.49.) Then the family of
field v is given parametrically by this
equation:
y'
x
=
=
to the vector
tangent
curves
=
y0
given by
=
(4.69)
y0e2t
write this
family
of curves
implicitly
as
=0
(taking the constant
of parabolas.
c as
y0 Xo
2).
Thus the
family
we
Another way to find the implicit equation of the
one equation in (4.68) by the other:
dy
_
dx
This
dyjdt
curve
2y
_
dx/dt
x
we can
solve
directly by separation
seek is
of variables.
a
system
is to divide
384
4
Curves
-^
Figure
4.49
4.6
61. Let
dy
v(x, y)
dyjdt
x
=
(x
y=
Now let
us
equation
is
y=x + y
or
general solution
+ cex
-(x+ l)
consider
of fluid motion
Then the differential
1).
385
+ y
Tx=jxidr~r
which has the
4- y,
Vector Fields and Fluid Flows
a
fluid in motion in
written
a
domain D in R".
The
equations
0
We suppose that at time t
there is a particle of fluid at each point x0 in D. The position of that particle
The equation of motion
at the subsequent time t is denoted by <j)(x0 t).
are
follows.
as
=
,
then is
x
For
a
=
fixed x0
,
is at x0 at time
xo
(4-7)
<Kxo,0
=
the
curve
/
0.
=
described
Thus
we are
of the
by (4.69) is the path
assuming that
particle which
(4.71)
<Kxo,0)
no two particles can ever occupy the same position
Then for each t, the function <f>(x0 t) is one-to-one and
be inverted : there is also a function i/^(x, 0 which describes
It is also assumed that
at the same time.
thus
can
the f
=
,
always
0 position of the particle
x
=
,
at the time t
~St
only
at time t such that
x
if
xo
=
<A(x, 0
Given the fluid motion described
1 0 is the vector field
Definition 8.
velocity
if and
</>(x0 0
at
by Equation (4.70) its
=
t=t0
situated at the
point
x
=
0(xo *o)>
If tne vector field is
independent
of
time, we say that the fluid motion is a steady flow.
Thus the velocity field of a flow at time t0 and
v(x, t0)
of the
particle
dent of the time,
river of constant
which is at
x
at that time.
point x is the velocity
velocity is indepen
is steady. The flow in a
If the
particular particle, the flow
volume is determined by the shape of the river bed,
or
the
and thus
386
4
Curves
tends to be
steady, whereas the flow of clouds in the sky is time dependent.
steady, then the path lines (the curves described by (4.70)) are
the curves of the family tangent to the velocity field. If the velocity field is
time dependent, then these tangent families (called the lines of force) vary
with time and have little to do with the paths of individual particles. This is
easy to see.
Suppose the flow
If the flow is
x
<Kxo,0
=
(4-72)
velocity field v(x, t).
equation
has the
Then the
path lines (4.72)
are
the solutions of
the differential
^
x(0)
v(x(0,0
=
The lines of force at time
t
=
=
r0
(4.73)
x0
are
the solutions of the
equation
dx
-
These
=
v(x(r), t0)
the
are
same
all t, that is, if and
x(0)
differential
only
=
(4.74)
x0
equations if and only
steady.
if
v(x, 0
=
v(x, r0)
for
if the flow is
Examples
62. Consider the flow in R2
x
=
x0e'
=
y
given by Equation (4.67):
y0e2'
(4.75)
Then
x*=x0e'
y'
=
2y0e2'
(4.76)
Thus the
velocity at time t of the particle originally at (x0 y0) is
(x0 e', 2y0 e2'). To find the velocity field we must solve (4.75) for
x0 y0 in terms of x, t and substitute. Thus (4.76) becomes
,
,
x'
=
so
the
x
y'
=
velocity
2y
field is
v(x, y)
=
(x, 2y)
and the flow is
steady.
Vector Fields and Fluid Flows
4.6
63. Consider the flow in R3
x
x'
=x0 + 1
=
1
y
y'
Thus the
\(x,y, z)
=
=
z'
2r
velocity
=
y0 + t2
=
387
given by
z
=
z0 +
t3
3i2
field
(l,2t, 3f2)
independent of position but is time dependent. In fact, the path
lines are independent of position and are just translates of the twisted
cubic (Figure 4.50). It is as if all of space were being rigidly trans
lated along the line curve y
x2,z x3. Notice that since the
t
time
field
at
t0 is a constant field, the lines of
any given
velocity
force are straight lines.
is
=
=
=
64.
x
=
x0 + t, y
=
y0(l
+
t),
z
=
8x
=
~di
so
the
(1, y0,z0e')
velocity
field is
v(x,y,z)=^l,^-^,zj
Figure 4.50
z0e'.
Then
388
4
Curves
the flow is not
The lines of force at time t
steady.
=
t0
are
the solutions
of
x'
=
so
x
1
is the
=
?
l + f0
y'=
y
=
=
y0
explyj
quite different
65. The flow is
x
z
=
family
x0 + t
which is
z'
x0e'
y
from the
z
family
z0e'
=
of
path
lines.
given by
y0e~'
=
x0(e'
+
e~')
-
z
=
z0
e2t
-
x0(e'
-
e2t)
(4.77)
x'
=
x0e'
y' =-y0e~'
+
x0e'
z'
x0e~'
+
2z0e2t-x0(et-e2t)
=
(4.78)
The Equations (4.77) are linear in x0 y0 z0 so we may solve for these
in terms of x, y, z. Doing so, and substituting the result in '(4.78),
we obtain the velocity field of the flow,
,
v(x,
y,
z)
=
(x,
2x
This flow is time
It is
y,2z
+
,
,
x)
independent,
or
steady.
immediate consequence of the uniqueness assertion of Picard's
a flow is
completely determined by its velocity field. For the
flow equation is the solution of the initial value problem (4.73), which is
an
theorem that
unique. Notice also that the existence part of Picard's theorem asserts that
always is a flow associated with a given velocity field (which is sufficiently
smooth).
The last remark we care to make at this time (we shall continue the study
of fluid flows in Chapter 8) is that in the case of a steady flow, the particles
follow one another along a fixed family of paths (whereas in general each
particle determines its own path). These are of course the lines of force.
there
4.6
What
show is this : If two
we must
at different times
there
are
s0
(of course),
,
xt occupy the
same
same
paths.
position
That is, if
</>(xi, 'i)
=
>
curves
T0 :
are
they
x0
follow the
^ such that
,
<K*o so)
then the
particles
then
389
Vector Fields and Fluid Flows
x
=
(j>(x0 ,0
I\ :
x
=
<Kxi, 0
the same, except for parametrization. The following proposition proves
more.
It makes explicit the relation between the two parametriza-
this, and
tions.
Proposition 7.
(x0 s0), (xu st),
,
(j)(x0 s0)
Suppose
=
x
<j)(x0,t)
describes
a
steady flow.
Iffor
some
have
we
(j)(xu st)
=
,
then
(l>(xo,So
In
particular,
+
xt
t)
=
=
<t)(xi,s1
(j)(x0
,
s0
-
+
t)
for all
t
(4.79)
s^).
proof is simply that the two functions in (4.79) solve the same
problem. Let v(x) be the velocity field of the flow (by assumption v is
independent). Consider these functions
Proof.
The
initial value
time
f(0
=
0(xo ,50
+ /)
We have
f0(0)
=
fi(0)
Since
^(xo,0=v(#Xo,r))
ot
fx(t)
=
#xi, s,
+
t)
(4.80)
390
4
Curves
for all x, t,
have
we
S
dia
(r)
-
=
<f>(x0
,
*>
+ f)
=
y(<t>(x0
,
so
+ 0)
=
v(fo(0)
dU
(0
Thus f0
same
,
=
v(f.(0)
fi solve the
value at 0.
Planetary
same
Thus f0
first-order differential
=
equation
and by
(4.80) have the
fi identically.
Motion
chapter with a study of the classical equations of planetary
study first requires these simplifications. We assume all
action is in a plane, and that the only force acting is that due to the sun's
gravitational field. These simplifications approximate the true situation with
For the other forces acting on the body are gravitational
enormous accuracy.
forces due to other celestial bodies, which are either too far away or too small,
relative to the sun, to make a substantial contribution. According to
Newton's laws, the acceleration of a body due to the gravitational field is
proportional to the field. The motion is thus completely determined by
this force and an initial position and velocity. For if s
s(t) is the equation
of motion, then s is the solution of an initial value problem ;
We conclude this
motion.
This
=
s(0)
=
s'(0)
s0
=
s"(t)
=
v0
kF(s(t))
where F is the
given
force field.
Our purpose here is to describe the motion of planets in terms of an
observed position and velocity. If we locate the sun at the origin, then the
gravitation
Thus,
we
force field is
must
z(0)
z'(0)
w
given (in complex notation) by
explicitly
=
z0
=
v0
solve this system
(4.81)
wois
4.6
Vector Fields and Fluid Flows
The best way to solve this is by
z(t) r(t)e,ei'\ Then differentiating,
=
z'
=
z-
and
r'ew
=
=
r"eie
+
+
2i9'r'eie
+
id"rew
(d')2r
r9")
i(26'r'
+
+
which reduces to this system
r"
(6')2r
-
The second
either &'
natives.
^
=
In
r
or
(4.83)
-
%
r2
(4.84)
obtain
=
^
r
(equating
real and
20V + r6"
=
0
imaginary parts) :
(4.85)
6-=(ln6'y
t)
0' is
one case
=
reads
=
=
0
(e')2reie
'
-
r
equation
2(lnr)'
so
=
Write
becomes
we
-
coordinates.
i8"reie -(9')2reie
Multiplying through by e~'e,
r"
polar
(4.82)
2ie'r'eie
+
Equation (4.81)
Z"
of
have
iO're'e
+
r"eie
means
we
391
proportional
to
r~2.
We have then these two alter
9 is constant, in which case the planet approaches the
line. In the other case, the planet rotates around the
a straight
angular velocity inversely proportional to the square of the distance
from the sun (the closer it is to the sun the faster it rotates around it). Notice
constant is possible, so that an
also that the solution r
constant, 9'
The
motion
of
constant angular velocity.
admissible path is that of circular
the
circle
gets larger.
angular velocity decreases as
h,
We proceed now to the full solution of (4.84). We already have 9'r2
From (4.84), we obtain
a constant (determined by the initial conditions).
sun
along
sun
at an
=
=
=
pie
1
i
r2
h
h
392
Thus
4
Curves
we can
'
z
=
Where C
peim is
+
=
arbitrary
an
i9'rew
=
Multiply through by
B'r
to obtain
1h e'e + C
=
r'eie
integrate
i eie
ie
e
-+p sin(co
again using 9'r2
h
=
rl-
+ p
we
have
imaginary parts :
9)
-
=
Now, using (4.82)
pe"
+
and equate
h
Once
constant.
h,
obtain this
we
implicit
relation between
r
and 9:
9)\
sin(a)
or
(4-86)
r-l+p/,sin(a)-0)
The constants p, h, co are to be determined by the initial conditions. Equation
(4.86) is the polar form of the equation of a conic with one focus at the origin.
If pA < 1, it is an ellipse; ph
1, a parabola; and ph > 1, a hyperbola.
These are then the possible paths of motion of a planet, or comet, around the
=
sun.
EXERCISES
21. Find the
(a)
(b)
(c)
(d)
22. Find
(a)
(b)
(c)
(d)
family of curves tangent
v(x, y)
(x, -y)
y(x, y)
(-y,x)
v(x, y, z)
(-x, -y, z)
1 z)
v(x, y, z) =(x,
to the
given
vector fields:
=
=
=
,
a
field of vectors tangent to these families:
e(1+c'"
z
=
z
=
x
=
x
=
ec+"
let,
x0
y
=
+ t,y
1
-
=
(cO2
e'y0
,
z
=
sin
t
4.7
393
Summary
23. Find the
velocity field of these flows:
(e~'xo ,y0 + t, e"'z0)
(x0(l + 0, yo(l + t), z0 + t2(x02 + y02))
x(t) (x0, ya + t, z0 cos t)
x(0 =e-'(x0,y0,z0cost)
24. Find the flow with the given velocity field :
(a) v(0
r(x,y, z)
(c) Exercise 21(b).
(b) y(t)
t(y,x,l)
(d) Exercise 21(c).
25. Is there a steady flow whose path lines are the trajectories
particles at (x0 y0 0) at time t 0 in the flow in Exercise 23(b) ?
(a)
(b)
(c)
(d)
x(t)
x(0
=
=
=
=
=
of the
=
,
,
PROBLEMS
45. Under what conditions
velocity field of a flow are the lines of
paths of motion ?
46. Consider a flow which spirals around the line L : x
z at constant
y
angular velocity, whose distance from the origin increases exponentially
with time and whose distance from L decreases exponentially with time.
Find the velocity field of the flow.
47. If we are given a family of curves in the plane we may consider the
tangent field of the family as well as its differential equation and the tangent
field of the orthogonal family as well as its differential equation. How are
force at all times the
same as
on
the
the
=
=
all these formulas related ?
4.7
Summary
The
where
image in R" of an interval under a one-to-one C1
vanishing derivative is called a curve. If Y is a
function with
curve
function
x
=
f(0
a<t<b
the variable t is called the parameter of the
x
is another
t
%if)
=
parametrization
=
mapping
a < t <
p
of the curve, there is
a
one-to-one function
a(x)
the interval
g(t)
If
curve.
=
i(c(x))
[a, j3]
onto the interval
cc^x<p
[a, b~\
such that
a no
given by
the
394
4
If a'
>
0
Curves
is
(a
orientation
on
An oriented
increasing)
Y.
say that the parameters t,
we
This notion divides all
is
curve
for which
one
one
x
induce the
same
into two classes.
parametrizations
classes, the well-oriented
of these
parameters is chosen.
If F is a differentiate function of two variables such that VF is never zero,
then the equation F(x, y)
0 defines a curve implicitly. For we can find a
=
parametrization
x=f(t)
for the set
=
y
F(x, y)
=
g(t)
0.
Similarly,
if F, G
of three variables such that VFand VG
F(x, y,z)
implicitly
If T is
x
the
defines
z)
y,
=
two differentiable
are
everywhere independent
functions
in the set
0
in R3.
curve
with
parametrization
a
a<t<b
f(0
length of Y
G(x,
a curve
oriented
an
=
0
=
are
between
upper bound of all
and
i(a)
i(t)
for
a < t <
b is defined to be the least
sums
Iiifc)-fa,-i)ii
;=i
over
all choices of
a
If
s(t)
=
t0
<
t^
points t0
<
<
,
tk
.
.
=
.
,
tk such that
t
is this
number, the function s s(t), a<t<b gives a parametrization
parametrization by arc length, s is the solution of the
differential equation
of T.
=
This is the
*'(0=l|f'(OII
s(a) 0
=
The unit tangent to a curve Y: x
x(s) is the vector T(s) x'(s). The
tangent line is the line through f(s) spanned by this vector. The unit normal
=
to the curve is a choice of unit vector
In two dimensions N is chosen
wise.
N(j) lying
on
the line spanned by T(s).
-* N is
counterclock
that the rotation T
so
In three dimensions the T
=
-
N
plane is called
the
osculating plane.
4.7
The unit binormal'is the vector B
orthornormal basis : B
ential
equations,
T'
=
T
x
so
N.
that the basis T
->
N
->
395
Summary
B is
This frame is determined
a
right-handed
by
these differ
the Frenet-Serret formulas:
kN
=
N'= -kT+ tB
B'= -tN
The scalar functions k, x, the curvature and torsion respectively of the curve
The curvature k is the angular
are defined by the first and third equations.
osculating plane and the torsion is the angular
osculating plane about the tangent. A curve in R3 is uniquely
determined (but for Euclidean motions) by its curvature and torsion. A
curve in R2 is uniquely determined (but for Euclidean motion) by its curvature.
If x
f(0 is the equation of motion of a particle, we call the curve de
scribed by this function the path of motion, ds/dt is the speed, f '(0 is the
velocity and f "(0 is the acceleration of the particle. The acceleration vector
lies in the osculating (T
N) plane. We can write
velocity
velocity
of the tangent in the
of the
=
-
acceleration
=
aTT
+
aNN
where aT is the tangential acceleration and aN the normal acceleration.
equations hold:
d2s
aT
where
k
dT2
(ds\2
\Jt)K
path of motion.
plane is a collection
of
equations
pair
is the curvature of the
family of curves
through some set. A
A
x
=
a
=
These
=
x(t, c)
in the
y
family of
functional equation
determines
a
F(x, y,c)
=
=
+
{Yc}
curves
as c
ranges
y(t, c)
curves.
This is the
explicit form
of the
family.
A
o
also determines a family. This is the
of solutions of a differential equation
a(x, y)x'
of
b(x, y)y'
=
0
implicit form
of the
family.
The set
(4.87)
396
4
forms
Curves
Fix,
is the
c
of
family
a
c)
y,
curves
in the
plane.
If
0
=
implicit form
(4.88) and
(4.88)
of a
family its differential equation is found by eliminating
from
'
>-o
dx
If
is the differential
(4.87)
gonal
dy
to the
family
b(x, y)x'
Fis
+
A vector field in
a
equation of a family F, the family
given by the differential equation
a(x, y)y'
domain U
The vector associated to
that
x
The
ortho
0
<=
R" is
an
Rn-valued function defined in U.
in U is
particular point
given by the function
depicted
as
originating
at
,
properties :
<p(x0 0) x0
<)> has continuous partial
=
,
.
For each t the function
curves x
velocity v(x, 0
V(X, 0
If
curves
<p(x0 0
=
with these
(i)
(ii)
(iii)
a
A fluid flow is
point.
=
of
x
derivatives.
=
<|)(x0 t)
,
is invertible.
<|>(x0 0 x0 fixed, are the paths of motion of the flow.
particle at x at time t is the velocity field of the flow
=
,
The
of the
=
"^ (X0 0 |x
,
=
Kxo, 0
is
independent of t, the flow is steady. The velocity field of a flow
completely determines the flow, for the paths of motion are obtained by
solving the differential equation
v
dx
=
^
x(0)
v(x,0
=
x0
When the flow is
particles
on
the
steady the paths of motion do not change with time,
path remain on the same path.
same
and
4.7
397
Summary
FURTHER READING
In addition to the
bibliography
mention these excellent texts
D.
Struik, Lectures
on
at the end of
Reading, Mass., 1950.
H. Guggenheimer, Differential Geometry,
R. T.
Chapter 3,
differential geometry :
Classical Differential Geometry,
we
should also
on
Addison-Wesley,
McGraw Hill, New York, 1963.
Seeley, Calculus, Scott-Foresman, Glenview, 111., 1967 has a
gravitational attraction from Kepler's laws.
deriva
tion of Newton's law of
MISCELLANEOUS PROBLEMS
48. Suppose that y is
a
closed
curve
in the
plane which lies outside the
unit disk and encircles the origin. Show that the length of y is at least 2tt.
49. Suppose that y is a closed curve lying completely inside the unit disk
with the property that it crosses every ray
a priori bound on the length of y ?
Suppose that
50.
y is
a curve as
once
and only
once.
Is there
an
described in Problem 49, whose curvature
Is there now a bound to the length of y ?
is bounded by 1
5 1 A pendulum consists of a body of mass m hanging on a rope of length
If the mass is displaced from the vertical and let
L which is fixed at one end.
.
.
go it will
swing along
the circle of radius L.
Find the differential
equation
of the motion.
Suppose a particle is moving along the curve of Example 20 at con
speed. Find the speed of its projection onto the xy plane.
53. Suppose that a particle moves along the right circular cone according
the equation
52.
stant
to
x
=
(cos
Find the
x
=
t, sin t,
2t)
equation
of motion of the
projection of
the
particle
on
the
plane
l.
54. A horse is
x< +
2y2
=
running around the elliptical track
1
1 and a floodlight
speed. There is a wall along the line y
the
wall.
Find the
on
shadow
the
horse's
casts
which
at the point (0, 1)
the
shadow.
of
motion
of
equation
55. A man six feet tall walks at constant speed along a straight line
passing directly beneath a street lamp 12 feet off the ground. Find the
equation of motion of the head of the man's shadow cast by the street lamp.
56. A loose foot bridge of length L hangs across a chasm of width W
(L > W). A man appears at one entrance on a pair of roller skates.
at constant
=
398
4
Curves
Suddenly he lets go and begins skating down the bridge. Assuming the
only forces acting on him are those due to gravity and the restraining forces
of the bridge, find the differential equation governing his motion.
57. Why does a river going around a curve wear out the far bank and
deposit silt along the near bank directly after the curve ?
58. Suppose a disk of radius r rolls with constant speed (at the center)
along a disk of radius R in the plane. Find the equation of motion of a
typical point on the circumference of the smaller disk.
59. Find the differential equation of the motion of a ball rolling in a
parabolic dish (with profile y x2) starting at rest at some point other than
=
the center.
60.
Assuming that the population of the organisms on a given remote
can you say anything about the eigenvalues of the
island remains bounded,
biotic matrix ?
61. Find the curvature and torsion of these
curves
in R3:
cos u, 3)
(u sin u, 1
x
(sin u, 1 + cos u, sin u)
62. Let x
x(s) be the equation of a curve y in R3 whose tangent vector
traces
out
a circle on the sphere.
Show that y is a helix.
T(s)
63. A general helix is a curve lying on a surface of revolution z =f(r)
which cuts the curves z =f(r), 8
constant at a fixed angle.
Show that
the ratio k/t is constant on a helix.
64. Find the curve on the xy plane onto which a helix on a cone projects.
65. Let yi and y2 be two space curves for which we have a point for point
correspondence such that the line joining corresponding points is the normal
line to both curves. Show that the line segment between corresponding
points has constant length.
66. Let x
x(0 be an i?"-valued function of a real variable which is
-times continuously differentiable. Then the image of y is a curve in R".
The Frenet-Serret frame of y is the orthonormal set obtained by applying
(a)
(b)
x
=
=
=
=
=
the Gram-Schmidt process to the vectors
x'(0,x*(/),...,x"(0
(a)
(b)
Show that for
n
=
Show that if there
Serret frame at every
sion k.
3, the Frenet-Serret frame is T, N, B.
are
point, the
only k independent
curve
lies in
a
vectors in the Frenet-
linear
subspace of dimen
Suppose that the Frenet-Serret frame Ti, T2
T is a basis.
representing the vectors dT,jds,
dTJds in this
basis is skew-symmetric. These are the generalized Frenet-Serret formulas.
(c)
67. Find the Frenet-Serret formulas for the
x
=
(cos
in/?4.
/, sin t, t,
.
,
Show that the matrix
2t)
.
curve
.
.
..,
,
4.7
68.
law of
Kepler's laws of planetary motion (from which
gravitational attraction) are these :
planet the ray from the sun
equal times.
equal
II. The path of motion of each planet is
I. For each
areas
one
is
Summary
Newton derived his
to the
planet
sweeps out
in
an
ellipse with the
sun
at
focus.
III. The square of the time period required to make
proportional to the cube of the major axis of the ellipse.
of
399
proportionality
In the text
we
is the
same
have derived
one
revolution
This constant
for all the planets.
Kepler's second law from Newton's laws.
Now derive Kepler's first and third laws.