January 26, 2013

http://coursework.tylerlogic.com/courses/math501/homework01

Assume for later contradiction that α′(t

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

Let’s rotate from 0 to 2π, fixing t

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)

Because

then α(0) = (0,0), and so α has a singular point at the origin (0,0).

Left Equation For α(t) = (x(t),y(t),z(t)) the Fundamental Theorem of Calculus gives us that there is a c ∈ I such that

but since v is a vector of magnitude 1, then

so

Right Inequality 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

Therefore |α′(t)| is

Hence α is parametrized by arc length.

Curvature. The curvature, κ, is |α′′(t)|. Using what we know from the previous part,

then

Torsion. 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

The following are for

(a) The parameter s is the arc length

Therefore s is the arc length.

The osculating plane of α at s = s

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 one.

Since n(s)(k(s))^{-1}α′′(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 n(s). Because

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.

Since

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