with
A vector function is a function of the form
with
The vector function
is continuous at
if
are continuous at
Theorem
is continuous at
if and only if for each
there exists
such that
for
Proof.
If
then
If
then
A vector function
is differentiable at
if
are differentiable at
In this case
This is so since
for
for
for
for
for
Proof.
Example If
is differentiable and
show that the scalar function
is differentiable and
Proof. Note that
Since the square root function
is differentiable for
it follows that the composite
is differentiable. Differentiating the identity
we obtain
so we have
Example Let
and
Then
Proof.
It follows from CAB minus BAC formula that this expression equals
If
is continuous on the interval
![$[a,b],$](vectorfn__69.png)
we define
for
for
Proof. Set
Then it follows from Cauchy-Schwarz inequality that
If
the conclusion holds. If
then
Let the position of the curve be given by
Its velocity vector is
and its acceleration vector is
Assume that
The unit tangent vector is
Suppose that
The principal normal vector is
Since
it follows that
This shows that the principal normal vector is perpendicular to the unit
tangent vector.
The arc length of
for
![$u\in [t_0,t]$](vectorfn__94.png)
is given by
It follows from the Fundamental Theorem of Calculus that
is differentiable and
for all
The scalar function
is called the speed of
Thus
It follows that
for all
i.e.,
is an increasing function. Therefore the inverse function
of
exists. (Thus
is the unique
such that
)
The position vector
may be regarded as a function of the arc length
Consider the composition
Taking the derivative of
with respect to
The curvature of a curve is the magnitude of the change in
the tangent
per unit length:
Since
it follows that
and so
Thus the curvature may be computed as function of
Example Let
be a positive constant. Find the curvature of the curve
We have
so
Consequently
The tangent vector
satisfies
When this identity is differentiated with respect to
we obtain
Therefore
At a point where
is nonzero, the principal unit normal is given by
Thus
Since the velocity
satisfies
the acceleration
satisfies
In this way, the quantity
is expressed as the tangential component of acceleration while the
quantity
is expressed as the normal component of acceleration. Since
the tangential component of acceleration is
The computation
and the fact
show that
Therefore, the normal component of acceleration may be expressed as
Suppose that the plane curve is given by
Regard it as a space curve
its curvature equals
Suppose that the plane curve is the graph
of the function
It can be regarded as the parametric curve
Hence its curvature equals
Geometric Interpretation
the
angle that the tangent makes with the x-axis
Since
it follows that
so
It follows from the chain rule that
Hence
The reciprocal
of the curvature is called the radius of curvature. A point
at a distance
from the curve in the direction of principal normal is called the
center of curvature. The circle with center at center of
curvature and radius
is called the osculating circle (also called the circle
of curvature.) Thus the position vector of the center of curvature is
Show that the center of curvature
of a plane curve
is given by
Let
be that tangent and
the normal of a curve. The vector
is called the binormal. These vectors satisfy
Differentiating both sides of the relation
with respect to
we have:
Therefore
Differentiating both sides of the relation
with respect to
we have:
Therefore
It follows that the vector
is perpendicular to both
and
we may therefore write
for some scalar function
which is known as the torsion of the curve. We now show how
this function may be computed as function of
Differentiating
both sides of
with respect to
we obtain
Since
and since
it follows that
Also we have
Therefore
Consequently
Differentiating
with respect to
we have
Therefore
can be expressed as
for some scalars
To determine these scalars we have
and
Therefore
Summery:
