Science:Math Exam Resources/Courses/MATH100/December 2010/Question 08

MATH100 December 2010

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Question 08

Full-Solution Problems. In questions 2-8, justify your answers and show all your work. If a box is provided, write your final answer there. Simplification of answers is not required unless explicitly requested. Two points on the surface of the Earth are called antipodal if they are at exactly opposite points (for example, the North Pole and South Pole are antipodal points). Prove that, at any given moment, there are two antipodal points on the equator with exactly the same temperature. Hint: Let $T(\theta )$ be the temperature, at any given moment, at the point on the equator with longitudinal angle $\theta$ measured in radians, $0\leq \theta \leq 2\pi$ (that is, in one complete trip around the equator, $\theta$ goes from 0 to $2\pi$) and consider $f(\theta )=T(\theta +\pi )-T(\theta )$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
There are two key facts to this problem. 1) The function T is periodic (like $\sin(x)$). 2) We are looking for a root of f. What theorem helps us guarantee the existence of roots? Use it!

Hint 2
The intermediate value theorem states that if a function is continuous on $\displaystyle [a,b]$ and $\displaystyle f(a)<0<f(b)$ or $\displaystyle f(b)<0<f(a)$, then there exists a point $\displaystyle c\in (a,b)$ such that $\displaystyle f(c)=0$.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Notice that the function $\displaystyle T(\theta )$ is $\displaystyle 2\pi$ periodic since the equator is shaped like a circle. Mathematically this in particular means $\displaystyle T(2\pi )=T(0).$ We exploit this fact by looking at $\displaystyle f(0)=T(\pi )-T(0)$ and ${\begin{aligned}\displaystyle f(\pi )&=T(\pi +\pi )-T(\pi )\\&=T(2\pi )-T(\pi )\\&=T(0)-T(\pi )\\&=-f(0).\end{aligned}}$ At this point we want to distinguish two cases: Case 1. f(0) = 0. Then $\displaystyle 0=f(0)=T(\pi )-T(0)$ and hence $\displaystyle T(\pi )=T(0)$, which is what we wanted to show. Case 2. f(0) ≠ 0. Then we have that one of $\displaystyle f(0)$ and $\displaystyle f(\pi )$ is positive and one is negative. As T is continuous, we have that f is continuous and hence we may invoke the intermediate value theorem to see that there exists a point $\displaystyle c\in (0,\pi )$ such that $\displaystyle 0=f(c)=T(c+\pi )-T(c)$. Thus, we have $\displaystyle T(c+\pi )=T(c)$ and this completes the proof.

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