Maya FrostBrophy
Professor Yurk
Math 131
September 21, 2016

Dear Ms. Taiga,
I have come to a conclusion on the statement you sent Dr. Cusack. If we are talking about two diametrically opposite points on the equator, then at some given time it is possible for them to have equal temperatures. If we have points P and Q somewhere on the equator, then there are three possibilities in regards to the temperature. Either temperature P(Tp) is equal to temperature Q(Tq), Tp is greater than Tq, or Tq is greater than Tp. When we find the difference of Tp minus Tq, it is desired that they are equal to 0 because this would prove that they would have too equal each other. So in the case that Tp>Tq than we would be able to conclude that Tp-Tq>0. It is also possible, to have two diametrically opposed points on the great circle that have the same temperature. If we assume the temperature function is continuous over the interval [0, 2π], we can assume that t(0) < t(π). If we define T(t)= t(t) - t(t- π), then T(0) = t(0) - t(π) <0, while T(π) = t(π) - t(2π) = t(π) - t(0) > 0. Since the function is continuous, it has to cross the x-axis at some time, meaning that there is a point t0 such that T(t0) = 0. That is a point such that t(t0) - t(t0 + tπ) = 0, and as a result, there are two exactly opposite points at the same temperature. The same equation can be applied to barometric pressure, and you can have the same barometric pressure at two diametrically opposed points on a great circle since it is also a continuous phenomenon. These equations cannot be applied to altitude above sea level since it is not a continuous phenomenon.