CALCULUS III: LECTURE 7
Today, we discuss arc length and curvature.
Relevant Book Sections: 13.3.
1. Motivating Example
A few lectures ago, I tried to emphasize the following point: “there is no best parameterization for a curve
or surface.”
I lied. For many curves, there is something like a best parameterization. (For a surface, or a higher
dimensional object (for example, like a 3-plane in 4 ), this is not true. We will see why this is so in a few
minutes.) Sometimes it’s even computable.
This best parameterization for a curve (really, there are two–we will see in a second) is given in terms of
arclength. Let’s work through an example to get some sense. We will define the concepts we need as we go.
Consider the following parameterized curve
R
r:
R
→
R3
t 7→ (t3 + 1, −t3 , 3t3 − 1)
or, if you prefer, we can write it as
r(t) = (1, −1, 3)t3 + (1, 0, −1)
What does the image of this parameterization look like? That is, what curve is this describing? We can try
to graph it. (Draw a picture.) What do we end up seeing, just looking at the equation: there is a given
vector (1, 0, −1), and then a direction vector (1, −1, 3) that doesn’t change, and just gets scaled. That is, this
is the usual parameterization of a line, but rescaled somehow. So let’s compare the weird parameterization,
which I will now call
rweird (t) = (1, −1, 3)t3 + (1, 0, −1)
to the usual parameterization we would write down for a line going through (1, 0, −1) with direction (1, −1, 3).
rusual (t) = (1, −1, 3)t + (1, 0, −1)
Now suppose I ask you to write down a tangent vector to this curve. Remember from yesterday that we can
do so by differentiating. Applying this to the usual parameterization, we find
r0usual (t) = (1, −1, 3)
which jives well with what we know intuitively: the tangent vector to a line at a point should be a non-zero
scalar multiple of the the direction vector. But what if differentiate the weird parameterization?
r0weird (t) = (1, −1, 3)(3t2 )
which is a non-zero scalar times the direction vector at all times t except one: at t = 0, r0weird (0) = 0 which
is troubling.
What happening? The parameterization rweird is describing a particle moving along the given line, but
with varying speed, and at time t = 0, the particle stops for a moment, then resumes.
But, if we are only interested in the curve the particle travels on, and not in the particle itself, then
this parameterizations rweird is not well suited for our purposes: at the point (1, 0, −1) it even “forgets” its
tangent vector.
How do we resolve this? Well, to be honest, we don’t: we usually don’t deal with such weird parameterizations of perfectly good curves. So instead, we make the definition
R
Definition 1.1. A parameterization r of a curve C in n is said to be C 1 (the book calls such a thing
smooth, but this conflicts with notation is all other math references) if r is differentiable with continuous
derivative r0 , and r0 (t) 6= 0. We say a curve C in n is C 1 if it admits a C 1 parameterization.
R
1
CALCULUS III: LECTURE 7
2
Remark 1.2. Not every parameterization of a smooth curve is smooth. For example, look above: the line we
are dealing with is smooth, but the parameterization rweird isn’t smooth: r0weird (0) = 0
We prefer to only look at smooth parameterizations of smooth curves. But here’s the question: how can
we find a reparameterization of a smooth curve given a non-smooth parameterization?
2. Arclength
We leave this example behind for a moment to discuss a necessary concept, namely arclength. Suppose
we are given a space curve C with differentiable parameterization r(t) = (x1 (t), ..., xn (t)) (which also has
continuous derivative). (Draw a picture, leave it on the board since we will modify it.) Let’s assume for
simplicity that the curve does not double back on itself. That is, let us assume that given a point P on the
curve with position vector p, that there do not exist two distinct t1 , t2 so that r(t1 ) = r(t2 ) = p.
We want a good way to compute the length of the curve between two points r(a) and r(b). What do we
do? Well, if we know that if a and b were close to one another, then the distance between the points r(a)
andr(b) is a good approximation to the length of the curve: since differentiable means locally looks like a
line, when a and b are close, the length of the curve between the two points r(a) and r(b) is essentially the
straight line distance |r(b) − r(a)|. (Draw different picture, zoomed in.)
So, if we want to approximate the arclength, what can we do? We can break the interval [a, b] into
k-pieces, i.e. we can partition it into
a = t0 < t1 < t2 < ... < tk = b
and then the arclength is pretty well approximated by (Draw in on the first picture the partition and the
vectors r(ti+1 ) − r(ti ).)
|r(t1 ) − r(t0 )| + |r(t2 ) − r(t1 )| + ... + |r(tk ) − r(tk−1 )| =
k
X
|r(ti ) − r(ti−1 )|
i=1
(Make sure everyone is comfortable with the summation notation.) What does this mean in terms of
coordinates? Each term in the sum is
p
(x1 (ti ) − x1 (ti−1 ))2 + (x2 (ti ) − x2 (ti−1 ))2 + ... + (xn (ti ) − xn (ti−1 ))2
|r(ti ) − r(ti−1 )| =
s
(x1 (ti ) − x1 (ti−1 ))2
(x2 (ti ) − x2 (ti−1 ))2
(xn (ti ) − xn (ti−1 ))2
=
+
+
...
+
(ti − ti−1 )
(ti − ti−1 )2
(ti − ti−1 )2
(ti − ti−1 )2
And so we get that the arclength is approximately
s
k
X
(x1 (ti ) − x1 (ti−1 ))2
(x2 (ti ) − x2 (ti−1 ))2
(xn (ti ) − xn (ti−1 ))2
length(C) ≈
+
+ ... +
(ti − ti−1 )
2
2
(ti − ti−1 )
(ti − ti−1 )
(ti − ti−1 )2
i=0
Now, as the partition gets finer and finer, we find that each term inside the square root
2
2
xj (ti ) − xj (ti−1 )
dxj
→
ti − ti−1
dt
while
(ti − ti−1 ) → dt
and the sum becomes an integral. That is, we find
ˆ bq
length(C) =
x01 (t)2 + x02 (t)2 + ... + x0n (t)2 dt
a
ˆ
b
|r0 (t)|dt
=
a
This should make sense physically: r(t) is the position of a particle moving in space, r0 (t) is its velocity,
|r (t)| is its speed, and to find the total distance the particle has travelled from time a to time b, we should
integrate the speed from a to b. This should also explain why we asked that the curve not double back on
itself: if it did so, we would be double counting.
We have (essentially) shown the following theorem
0
CALCULUS III: LECTURE 7
3
Rn given by parameterization r(t), from point r(a) to r(b),
Theorem 2.1. The arclength of a curve C in
is given by the integral
ˆ
b
|r0 (t)|dt
L = length(C) =
a
as long as the curve does not double back on itself.
We also define the arclength function from r(a) to r(t) by the formula
ˆ t
|r0 (u)|du
s(t) =
a
We will often just call this the arclength function, with the starting time a implicitly understood.
Example 2.2. Let’s compute the arclength function for the curve given by
r(t) = (a cos t, a sin t)
with a > 0. Clearly this is the circle centered at the origin of radius a. We compute
r0 (t) = (−a sin t, a cos t)
and so
|r0 (t)| =
p
a2 sin2 t + a2 cos2 t =
Therefore
ˆ
√
a2 = a
t
s(t) =
adu = at
0
so in particular, at time t = 2π, we find the parameterization has gone around the circle once, and travelled
a distance of 2πa. That is, we just checked that the circumference of a circle of radius a is given by the well
known formula 2πa.
Why do we care about arclength? Arclength does not depend on parameterization.
Example 2.3. Consider the following two parameterizations of the same curve
r(t) = ti + t2 j + t3 k
and
u(t) = et i + e2t j + e3t k
Note that these are parameterizing the same curve: this is since u(t) = r(et ). (Draw a picture of the curve.)
Let’s compute the arclength of this curve between the points (1, 1, 1) and (2, 4, 8) for both parameterizations.
We see that the first parameterization will go from 1 to 2, since r(1) = (1, 1, 1) and r(2) = (2, 4, 8). On the
other hand, the other parameterization will go from 0 to ln 2, since u(0) = (1, 1, 1) and u(ln 2) = (2, 4, 8).
So we compute
r0 (t)
=
(1, 2t, 3t2 )
u0 (t)
=
(et , 2e2t , 3e3t )
and so
|r0 (t)| =
p
1 + 4t2 + 9t4
|u0 (t)| =
p
e2t + 4e4t + 9e6t = et
Then, when we do the integral
ˆ ln 2
ˆ
0
|u (t)|dt
0
et
0
ˆ
=
1
so the two arclengths are equal.
ln 2
=
2
p
p
1 + 4e2t + 9e4t
1 + 4e2t + 9e4t dt
ˆ
p
2
4
1 + 4t + 9t dt =
1
2
|r0 (t)|dt
CALCULUS III: LECTURE 7
4
This is essentially the proof in general. If there are two different parameterizations of a curve, one given
by r(t), the other given by u(t), then there is a function f : → so that u(t) = r(f (t)), and we compute
ˆ b
ˆ bq
|u(t)|dt =
u01 (t)2 + ... + u0n (t)2 dt
R R
a
a
ˆ
b
=
q
(x01 (f (t))f 0 (t))2 + ... + (x0n (f (t))f 0 (t))2 dt
q
x01 (f (t))2 + ... + x0n (f (t))2 f 0 (t)dt
a
ˆ
=
b
a
ˆ
f (b)
=
q
ˆ
f (b)
|r0 (t)|dt
x01 (t)2 + ... + x0n (t)2 dt =
f (a)
f (a)
2.1. The motivating example, continued. Let’s come back to our original example. We had
rweird (t) = (1, −1, 3)t3 + (1, 0, −1)
Let’s compute what the arclength from (1, 0,
√−1) to rweird (t) is as function of time t. We find, since
r0weird (t) = (1, −1, 3)(3t2 ), that |r0weird (t)| = 3 11t2 . So we compute, and find that
ˆ t √
√
s(t) =
3 11u2 du = 11t3
0
Note that this function has an inverse. That is, it is of such nice form that we can actually write t as a
function of s, namely
t = s1/3 111/6
So if we want to, we can use this expression to reparameterize rweird in terms of the parameter s, arclength,
and subsituting back in we find the expression
1
3
1
(1, −1, 3)(s1/3 111/6 )3 + (1, 0, −1) = ( √ , − √ , √ )s + (1, 0, −1)
11
11
11
This is now called the arclength parameterization of the curve.
1
1
3
rarclength (s) = ( √ , − √ , √ )s + (1, 0, −1)
11
11
11
Note that in this example, we have recovered something that looks much like our rusual but, in fact, is
even better: the direction vector is now a unit vector, so this is really the parameterization of the curve
given by a particle moving along the curve at constant speed 1, so that after time 1 the particle has moved
distance 1.
2.2. Arclength Parameterization. This is essentially what we did above, but let’s just say is more formally. Given a curve C with parameterization r(t), one can find the arclength parameterization of C as long
as the arclength function (which we only define in this way when r is a parameterization that doesn’t double
back on itself)
ˆ t
s(t) =
|r0 (u)|du
a
is invertible, i.e. if we can write t = t(s) as a function of s. (Remark: Since arclength is always nondecreasing, we are very happy as long as the parameterization r(t) doesn’t stop for some time.) In this case,
we define the arclength parameterization by
rarclength (s) = r(t(s))
we often abuse notation and just write r(s) instead of the bulky rarclength (s), keeping implicit in the notation
of using s intead of the customay t that we have reparameterized by arclength.
Example 2.4. Reparameterize the helix r(t) = (cos t, sin t, t) with respect to arclength measured from
(1, 0, 0) in the direction of increasing t.
We see that (1, 0, 0) corresponds to t = 0, so we find the arclength function by first computing
r0 (t) = (− sin t, cos t, 1)
CALCULUS III: LECTURE 7
5
p
√
|r0 (t)| = sin2 t + cos2 t + 1 = 2
ˆ t
ˆ t√
√
0
2 = 2t
s(t) =
|r (u)|du =
√
so in fact we can solve t = s/ 2, and so
0
0
s
s
s
r(s) = (cos √ , sin √ , √ )
2
2
2
2.3. Remarks. Why did I say that there were two possible “best” parameterizations? Well, you could
do arclength in either direction–the direction of positive t, or of negative t. And what happens in higher
dimension? Well, there aren’t just two directions for the parameters t1 , t2 , ..., tk instead, there are lots
of possible “directions” to choose for the parameterization, so you have no hope of something which is
distinguished over other directions.
If this section makes no sense, that’s ok. I just wanted to say a few words justifying why I called arclength
parameterization the “best.”
3. Curvature
What should curvature of a smooth curve C with smooth parameterization r(t) be intuitively? Well,
recall yesterday when we defined the unit normal vector N(t), we did it by first taking
n(t) = T0 (t)
and that this was some normal vector whose length measured loosely how much the parameterization curved
(i.e. how quickly the unit tangent vector changed direction).
But this depends on parameterization r! We can make this length independent of parameterization if we
work only with a parameterization which is intrinsic to the curve. That is, we should differentiate not with
respect to time, but with respect to arclength (or, equivalently, reparameterize the curve w.r.t arclength and
then differentiate with respect to the new parameter). So we define
Definition 3.1. The curvature of a smooth curve is
dT κ = ds where T is the unit tangent vector.
We can manipulate this formula a little to make it easier to work with. Namely if we notice that
dT ds
dT
=
dt
ds dt
Then by taking magnitudes, we find
|dT/dt|
1
1
κ=
= 0 T0 (t) = 0
n(t)
|ds/dt|
s (t)
|r (t)|
So this relates the curvature to the normal vector n(t) we described yesterday.
We can even determine another formula for the curvature, given only in terms of r. However, we stop
here for today, and resume tomorrow.