1 do Carmo pg 47 exercise 3
Vertices Using k, we can determine the vertices of α by finding the values of t which make k′(t) zero. Since
then k′(t) can only be zero when sin(2t) is zero, i.e. at t = for n ∈ ℤ. Since the domain is t ∈ [0,2π], then there are vertices at 0, , π, , and 2π, however, α(0) = α(2π) so there are only four unqiue vertices at α(0) = (a,0), α() = (0,b), α(π) = (-a,0), and α() = (0,-b).
2 do Carmo pg 47 exercise 4
With these changes, let’s evaluate the second order Taylor series exansion of α(s), given by
where R is the sum of the higher order terms. We know that α(0) is p = (0,0). Because s parametrized α by arc length, then α′(0) is the unit tangent vector at zero, which, since we aligned T with the x-axis, means simply that α′(0) = (1,0). Furthermore, we know α′′(0) = kn to be perpendicular to α′(0), meaning the normal vector n is (0,1), with the sign and magnitude of α′′(0) being determined by k. Substituting these realizations into Equation 2.1, we get
where d and h are as defined in the problem’s statement. Thus denoting R by (Rx,Ry) we have that d = s + Rx and h = ± k + Ry. The first equation can be rearranged as d-s = Rx, which tells us that s → d as s → 0 since Rx → 0 as s → 0. The second equation can be rearranged like
and it informs us that
since → 0 as s → 0, but because we saw above that s → d as s → 0, then
3 do Carmo pg 47 exercise 5
Now, making use of the Fundamental Theorem of The Local Theory of Curves, orient both α and C similarly to the previous problem, i.e. make p be the orgin and align the tangent vector of C at p with the x-axis. With this orientation, for a given d (defined as in the previous problem) hC ≤ hα. Because of this we have
by the previous problem, yielding |kα|≥ 1∕r as needed.
4 do Carmo pg 48 exercise 8 part a
and then making use of the Isoperimetric Inequality we get
or in other words, ℓ ≥, given the positivity of ℓ and c.