Math 501: Differential Geometry
January 26, 2013
1 do Carmo pg. 5 exercise 2
Assume for later contradiction that α′(t0) is not orthogonal to α(t0). Since α′(t0)
is nonzero, then we are able to find a ϵ > 0 such that either |α(t0 -ϵ)| < |α(t0)| or |α(t0 + ϵ)| < |α(t0)|. However, either case
contradicts the fact that α(t0) is the point of the trace of α which is closest to the origin. Thus α′(t0)⊥α(t0) must be
2 do Carmo pg. 7 exercise 2
Figure 1: Cycloid at a Glance.
We will parametrize the cycloid by the angle of rotation, denote it by t. From
the diagram in Figure 1 it can be seen that Thus the value of h is determined by the following sequence of equations. With these things, we can finally derive the point p. and
Let’s rotate from 0 to 2π, fixing t0 at 0. By making use of the double angle
formula for cos, we have the following.
3 do Carmo pg. 7 exercise 3
Figure 2: Cissoid of Diocles at a Glance.
For this problem we will use Figure 2 as a guide. Using some trigonometry,
from the figure we get the following (remember the diameter is length 2a) These allow us to have the following Hence letting t = tanθ we get
then α(0) = (0,0), and so α has a singular point at the origin (0,0).
4 do Carmo pg. 11 exercise 10
For α(t) = (x(t),y(t),z(t)) the Fundamental Theorem of Calculus gives us that there is a c ∈ I such
but since v is a vector of magnitude 1, then
By the definition of dot product
Now since cosθ ranges between -1 and 1 then
To show that α(t) is parametrized by arc lenght, we must show |α′(t)| = 1 for
all t. So proceeding that way, we have giving us
Therefore |α′(t)| is
Hence α is parametrized by arc length.
The curvature, κ, is |α′′(t)|. Using what we know from the previous part,
The torsion of α, τ, is the magnitude of the derivative of the binormal vector, b, as given by the Frenet equation b′ = τn.
The binomial vector is so and thus
6 do Carmo pg. 22 exercise 1
The following are for
(a) The parameter s is the arc length
Therefore s is the arc length.
(b) Curvature and torsion.
The osculating plane of α at s = s0 is defined by the following point, normal
the equations for which are given and determined in the previous problem.
The angle between the z-axis and a line passing through α(s)
and containing n(s) will be the same as the angle between n(s) and any vector contain by the z-axis, say
(0,0,1). And that angle will be the inverse cosine of their dot product, since both vectors have magnitude of
Since n(s)(k(s))-1α′′(s) = , then the dot product of it and (0,0,1) is zero, implying
that the cosine of the angle between them is zero, which in turn confirms that the angle between them is
This is similar to the previous problem except that we will use t(s) in place of
i.e. the cosine of the angle between them is a constant, then the angle bewteen t and (0,0,1) is constant, since they both
have a magnitude of one.
7 do Carmo pg. 25 exercise 12
then by the fundamental theorem of calculus we have ds∕dt = |α′(t)|, which implies dt∕ds = 1∕|α′(t)|.
Taking the derivative of both sides of the above equation leads to the following sequence of equations.
Assuming that the yet-to-be-proven part (b) of this problem is true, then the
signed curvature is