CALCULUS III: LECTURE 8
Today, we talk a little more about curvature and discuss some physical applications of all this, such as
Kepler’s laws of planetary motion.
Relevant Book Sections: 13.3., 13.4
1. Curvature Again
Recall the definition of the curvature κ of a curve C with C 1 (smooth) parameterization given by r(t)
(remember that if C is a C 1 curve but you are given a non-C 1 parameterization r, you can find a smooth
parameterization by reparameterizing by arclength!):
Definition 1.1. The curvature of a smooth curve is at a given point r(t) is given by
dT κ=
ds where T is the unit tangent vector.
Using the chain rule, we manipulated this formula so that it became
κ(t) =
1
T0 (t)
|r0 (t)|
So this describes the curvature of C at the point r(t) without ever mentioning arclength.
One can go even further, and find an even more explicit formula for curvature. However, this formula
only works for curves C in three-dimensional space 3 .
Proposition 1.2. Let C be a smooth curve in
R
3
κ(t) =
R
with given smooth parameterization r(t). Then
|r0 (t) × r00 (t)|
|r0 (t)|3
To see why this formula is true, we simply compute the right hand side of the equation. We note since
r0 (t) = |r0 (t)|T(t) = ds
dt T(t), that
d2 s
ds
r00 (t) = 2 T(t) + T0 (t)
dt
dt
and so
2
ds
d s
ds 0
0
00
T(t) ×
r (t) × r (t) =
T(t) + T (t)
dt
dt2
dt
2
ds
d s
ds
ds 0
=
T(t) ×
T(t) +
T(t) ×
T (t)
dt
dt2
dt
dt
Note that since the cross product of a vector with itself is 0 that the first cross product on the RHS of the
above equation is 0. So we find
ds
ds 0
r0 (t) × r00 (t) =
T(t) ×
T (t)
dt
dt
2
ds
=
T(t) × T0 (t)
dt
But these last two vectors are perpendicular by our construction of the normal vector (remember T0 (t) = n(t)
has the same direction as the unit normal). Therefore,
2
2
ds ds |r0 (t) × r00 (t)| = |T(t)| |T0 (t)| = |T0 (t)|
dt
dt
1
CALCULUS III: LECTURE 8
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If we divide by |ds/dt|3 = |r0 (t)|3 , we find
|ds/dt|2 |T0 (t)|
|dT0 /dt|
|r0 (t) × r00 (t)|
=
=
=κ
|r0 (t)|3
|ds/dt|3
|ds/dt|
as desired.
Example 1.3. We still haven’t computed the curvature of any curve! Let’s start with the following easy
example:
r(t) = (a cos t, a sin t)
We compute
r0 (t) = a(− sin t, cos t)
and so, since |r0 (t)| = a, we find
T(t) = (− sin t, cos t)
so
T0 (t) = −(sin t, cos t)
From yesterday we know
s(t) = at
so
s0 (t) = a
and thus
1
a
Thus, the curvature of a circle is the reciprocal of its radius.
κ(t) =
Example 1.4. Find the curvature of r(t) = (t, t2 , t3 ) at any time t.
We compute, using the formula of the proposition:
r0 (t)
00
r (t)
=
(1, 2t, 3t2 )
=
(0, 2, 6t)
so
r0 (t) × r00 (t) = (6t2 , −6t, 2)
and so
√
|r0 (t) × r00 (t)|
36t4 + 36t2 + 4
κ(t) =
=
0
3
|r (t)|
(1 + 4t2 + 9t4 )3/2
Example 1.5. Find the curvature of a plane curve given by the graph of a function y = f (x) at the point
(x, f (x)).
We parameterize the curve by
r(t) = (t, f (t))
and we find, since
r0 (t)
= i + f 0 (t)j
r00 (t)
= f 00 (t)j
r0 (t) × r00 (t) = f 00 (t)k
p
so |r0 (t) × r00 (t)| = |f 000 (t)|,while |r0 (t)| = 1 + f 0 (t)2 . So
κ(t) =
|f 00 (t)|
(1 + f 0 (t)2 )3/2
and, since t = x in this parameterization, this gives us that the curvature at the point (x, f (x)) is
κ(x) =
|f 00 (x)|
(1 + f 0 (x)2 )3/2
What is a good geometric understanding of the curvature? We digress for a moment.
CALCULUS III: LECTURE 8
3
R
1.1. Normal and Osculating Planes. Given a C 1 curve C in 3 with C 1 parameterization r(t) and
non-vanishing T0 (t) = n(t) we can define certain special planes at every point r(t) of the curve. These are
given by the moving frame T(t), N(t), and B(t) for the curve. These are called the normal plane and
osculating plane. The normal plane is given as the plane going through the point r(t) on the curve which
is normal to the tangent vector T(t). The osculating plane is given as the plane going through the point r(t)
on the curve which is parallel to the tangent vector T(t) and N(t), hence normal to B(t). (Draw picture.)
Remark 1.6. Osculating comes from the Latin ocsulum, meaning kiss. The osculating plane is the plane
that the curve wants to lie in–loosely, it is the plane coming from the first and second derivatives of r(t).
The normal plane is the plane the curve punches through.
Within the osculating plane, one can put a circle which best approximates the behavior of the curve near
r(t). That is, we define the osculating circle to the curve at point r(t) to be the circle with curvature κ(t)
(hence radius 1/κ(t) by our previous example today) lying in the osculating plane tangent to the curve at
r(t) and lying on the concave side of the curve. So, geometrically, the curvature is the reciprocal of the best
circle approximation to C.
Example 1.7. Find the osculating plane to the helix r(t) = (cos t, sin t, t) at time t =
We compute
r0 (t)
T(t)
π
2.
=
(− sin t, cos t, 1)
1
= √ (− sin t, cos t, 1)
2
andm
1
= √ (− sin t, − cos t, 0)
2
N(t) = (− cos t, − sin t, 0)
n(t)
so we find
1
B(t) = T(t) × N(t) = √ (sin t, − cos t, 1)
2
π
π
π
π
1
√
So at t = 2 , r( 2 ) = (0, 1, 2 ) and B( 2 ) = 2 (1, 0, 1), so the equation of the plane is (using the simpler
normal vector (1, 01) instead of B)
π
x + (z − ) = 0
2
2. Physics, Again
Let’s briefly go over the basics here: since this is a math class and not a physics class, we’ll be brief.
Suppose an object (particle) moves in space with position given by a vector function of time
r(t)
(Draw picture.) Then the velocity vector of the particle at time t is given by
v(t) = r0 (t)
the acceleration vector of the particle at time t is given by
a(t) = v0 (t) = r00 (t)
and the speed of the particle is given by
ds
dt
If the particle has mass m(t), then we define the momentum vector to be
v(t) = |v(t)| = |r0 (t)| =
p(t) = m(t)v(t)
The fundamental law of Newtonian physics is the following. If a particle of mass m(t) is being acted on
by a single force (or net force) of F(t), changing in time, then the acceleration of that particle is given by
F(t) =
d
d
p(t) = (m(t)v(t)) = m0 (t)v(t) + m(t)a(t)
dt
dt
CALCULUS III: LECTURE 8
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When the mass remains constant, this becomes the well known formula
F(t) = ma(t)
known as Newton’s second law of motion.
Remark 2.1. There are situations in which mass is not constant. Suppose, for instance, that you are trying
to understand the acceleration of a rocket launching from the ground into space. Much of a rocket’s mass
comes from the fuel it needs to carry to get it into space. But then once the rocket’s engine fires, the mass
of the rocket m(t) drops with time as fuel gets used up and ejected behind the rocket. However, for the
purposes of this section and for this class, we will always assume that the mass of the relevant particle doesn’t
change with time, i.e. that m(t) = m is a constant.
Example 2.2. (Centripetal force) A test tube in a centrifuge of radius R has mass m. The centrifuge has
angular speed ω radians per second, hence the tube moves quickly in a circular path described by
r(t) = (R cos ωt, R sin ωt)
Find the force acting on the object (i.e. the force the walls of the centrifuge are exerting to keep the test
tube moving in a circle instead of flying off in a tangential direction).
We compute
v(t)
= Rω(− sin ωt, cos ωt)
a(t)
= −Rω(cos ωt, sin ωt)
so by Newton’s second law,
F(t) = −mRω 2 (cos ωt, sin ωt)
This is pointing towards the origin (center of the centrifuge) and has length mRω 2
Example 2.3. (Projectiles) A projectile of mass m is fired with angle of elevation α and initial velocity v0 .
Assuming resistance is negligible and the only external force on the projectile is gravity, find the position
vector of the projectile. (Draw picture.)
After recalling that the force of gravity if given by F = (0, −mg), we find
a
=
(0, −g)
(0, −gt) + v0
1
r = (0, − gt2 ) + v0 t + r0
2
since we can assume r0 = 0 for this problem, we are done once we write
v
=
v0 = (v0 cos α, v0 sin α)
2.1. Tangential and Normal Components of Acceleration. We can break down acceleration a(t) into
its tangential and normal components. We compute
v = vT
so
a = v 0 T + vT0
0
0
0
but since T = |T |N and |T | = κv, we find
a = v 0 T + κv 2 N
so in fact the tangential component of acceleration is given by the derivative of the speed, while the normal
component is given by the square of the speed times the curvature, i.e.
aT
= v0
aN
= κv 2
Note that this whole discussion shows that the acceleration vector always lies in the osculating plane of curve
the particle traces out.