Study Sheet (13.3)
Arc Length and Curvature
In this section, we will find the arc length of a space curve.
We already know how to find the length L of a curve C given in the form y = f (x) , a ≤ x ≤ b .
Arc Length Formula
If f ' is continuous on [a, b], then the length of the curve y = f (x) , a ≤ x ≤ b is:
L =
Now suppose that C can also be described by the parametric equations
Now we define the length of a plane curve with
dx
= f '(t) > 0 .
x = f (t), y = g(t), ∂ ≤ t ≤ β where
dt
Then
L=
Plane Curve Length
For the case where f’ and g’ are continuous, we arrived at the following formula:
L=
SPACE CURVE LENGTH:
The length of a space curve is defined in exactly the same way.
Suppose that the curve has the vector equation
r(t) = <f(t), g(t), h(t)>, a ≤ t ≤ b
Equivalently, it could have the parametric equations
x = f(t), y = g(t), z = h(t)
where f’, g’ and h’ are continuous.
Space Curve Length
If the curve is traversed exactly once as t increases from a to b, then it can be shown that its
length is:
L =
ARC LENGTH
Notice that Space Curve Length can be put into the more
compact form
L=
That is because:
§ For plane curves r(t) = f(t) i + g(t) j,
§
r '(t ) = f '(t ) i + g '(t ) j =
2
2
[ f '(t )] + [g '(t )]
For space curves r(t) = f(t) i + g(t) j + h(t) k, r '(t ) = f '(t )i + g '(t ) j + h '(t )k
=
2
2
2
[ f '(t )] + [g '(t )] + [h '(t )]
Five steps to find the length of the space curve over a given interval:
Step 1: Set up the integral.
Step 2: Find the derivative of the vector.
Step 3: Substitute into the formula.
Step 4: Simplify.
Step 5: Integrate and evaluate.
Example 1: Find the length of the arc of the circular helix with vector equation
r(t) = cos t i + sin t j + t k
If
the
curve
exactly
from the point (1, 0,is0)traversed
to the point
(1, 0, once
2π). as t increases
from a to b, then it can be shown
that its length is:
ARC LENGTH: A single curve C can be represented by more than one vector function.
The twisted cubic
r1(t) = <t, t 2, t 3>
1≤t≤2
could also be represented by the function
r2(u) = 0 ≤ u ≤ ln 2
The connection between the parameters t and u is given by t = eu.
PARAMETRIZATION:
We say that above equations are parametrizations of the curve C.
b
If we were to use Equation L =
∫ r '(t) dt
to compute the length of C using the above equations, we
a
b
would get the same answer. In general, it can be shown that, when Equation L =
∫ r '(t) dt is used to
a
compute arc length, the answer is independent of the parametrization that is used.
Now, we suppose that C is a curve given by a vector function
r(t) = f(t) i + g(t) j + h(t) k
a≤t≤b
where:
§ r’ is continuous.
§ C is traversed exactly once as t increases from a to b.
ARC LENGTH FUNCTION:
We define its arc length function s by:
s (t ) =
Thus, s(t) is the length of the part of C between r(a) and r(t).
If we differentiate both sides of the above equation using Part 1
of the Fundamental Theorem of Calculus (FTC1), we obtain:
It is often useful to parametrize a curve with respect to arc length.
§ This is because arc length arises naturally from the shape of the curve and does not
depend on a particular coordinate system.
If the curve is traversed exactly once as t increases from a to b, then it can be
If a curve r(t)
is already given in terms of a parameter t and s(t) is the arc length function given by
shown
Equation 6,
then
may be
t = t(s)
that
itswe
length
is: able to solve for t as a function of s:
Then, the curve can be reparametrized in terms of s by substituting for t:
r = r(t(s))
Thus, if s = 3 for instance, r(t(3)) is the position vector of the point 3 units of length along the curve
from its starting point.
Example 2: Reparametrize the helix
r(t) = cos t i + sin t j + t k
with respect to arc length measured from (1, 0, 0) in the direction of increasing t.
SMOOTH PARAMETRIZATION:
A parametrization r(t) is called smooth on an interval I if:
§
r’ is continuous.
§
r’(t) ≠ 0 on I.
SMOOTH CURVE:
A curve is called smooth if it has a smooth parametrization.
§
A smooth curve has no sharp corners or cusps.
§
When the tangent vector turns, it does so continuously.
If C is a smooth curve defined by the vector function r, recall that the unit tangent vector T(t) is given
by:
§
This indicates the direction of the curve.
You can see that T(t) changes direction:
§
Very slowly when C is fairly straight.
§
More quickly when C bends or twists more sharply.
CURVATURE:
An important use of the arc length parameter is to find curvature.
Curvature is the measure of how sharply the curve bends.
We can calculate curvature by calculating the magnitude of the rate of change of the unit tangent
vector T with respect to the arc length s.
T2
T3
T1
Definition of Curvature
Let C be a smooth curve (in the plane or in space) given by r(s), where s is the arc
length parameter. The curvature K at s is given by:
The curvature is easier to compute if it is expressed in terms of the parameter t instead of s.
So, we use the Chain Rule (Theorem 3 in Section 13.2, Formula 6) to write:
d T d T ds
d T d T / dt
=
and κ =
=
dt
ds dt
ds
ds / dt
However,
ds
= r '(t) , so we have
dt
Example 3: Show that the curvature of a circle of radius a is 1/a.
If the curve is traversed exactly once as t increases from a to b, then it can
be shown
that its length is:
The result of Example 3 shows—in accordance with our intuition—that:
§ Small circles have large curvature.
§ Large circles have small curvature.
We can see directly from the definition of curvature that the curvature of a straight line is always 0
because the tangent vector is constant.
k(t) =
T '(t)
can be used in all cases to compute the curvature. Nevertheless, the following formula is
r '(t)
often more convenient to apply.
Formula for Curvature:
The curvature of the curve given by the vector function r is:
Example 4: Find the curvature of the twisted cubic r(t) = <t, t2, t3> at a general point (0, 0, 0).
For the special case of a plane curve with equation y = f(x), we choose x as the parameter and write:
r(x) = x i + f(x) j
Then,
r’(x) = iis+ traversed
f’(x) j andexactly
r’’(x) =once
f’’(x) jas t increases
If the curve
Since i x j = k and from
j x j =a0,towe
have:
b, then it can be shown
r’(x)
x r’’(x)is:= f’’(x) k
that its length
2
We also have r '(x) = 1+ [ f '(x)] and
k(x) =
Example 5: Find the curvature of the parabola y = x2 at the points (0, 0), (1, 1), (2, 4).
If the curve is traversed exactly once as t
increases from a to b, then it can be shown
that its length is:
NORMAL AND BINORMAL VECTORS:
At a given point on a smooth space curve r(t), there are many vectors that are orthogonal to the unit
tangent vector T(t).
NORMAL VECTORS:
We single out one by observing that, because |T(t)| = 1 for all t,
we have T(t) · T’(t) by Example 4 in Section 13.2. So, T’(t) is orthogonal to T(t).
§
Note that T’(t) is itself not a unit vector.
However, if r’ is also smooth, we can define the principal unit normal
vector N(t) (simply unit normal) as: .
We can think of the normal vector as indicating the direction in which the curve is turning at each
point.
BINORMAL VECTOR:
The vector
B(t) = T(t) x N(t)
is called the binormal vector. It is perpendicular to both T and N and is also a unit vector.
Example 6: Find the unit normal and binormal vectors for the circular helix
r(t) = cost i + sin t j + t k
NORMAL & BINORMAL VECTORS:
The figure illustrates Example 6 by showing the vectors
T, N, and B at two locations on the helix.
Example 7: Find the equations of the normal plane and osculating plane of the helix in Example 6 at
the point P(0, 1, π/2)
NORMAL & OSCULATING PLANES
The figure shows the helix and the osculating plane in Example 7.
Example 8
Find and graph the osculating circle of the parabola y = x2 at the origin.
§ From Example 5, the curvature of the parabola at the origin is ĸ(0) = 2.
§ So, the radius of the osculating circle at the origin is 1/ĸ = ½ and its center is (0, ½).
SUMMARY
We summarize the formulas for unit tangent, unit normal and binormal vectors, and curvature.
T(t ) =
r '(t )
r '(t )
κ=
N (t ) =
T '(t )
T '(t )
B(t ) = T(t ) × N(t )
T '(t )
r '(t ) × r ''(t )
dT
=
=
3
ds
r '(t )
r '(t )