Science:Math Exam Resources/Courses/MATH102/December 2013/Question A 07

MATH102 December 2013

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Question A 07

The number of sunspots (solar storms on the sun) varies with a period of roughly 11 years reaching a high of 120 and a low of 0 sunspots detected. A peak of 120 sunspots was detected in the year 2000. Which of the following trigonometric functions could be used to approximate this cycle? (a) $N=60+120\sin \left({\frac {2\pi }{11}}(t-2000)+{\frac {\pi }{2}}\right)$ (b) $N=60+60\sin \left({\frac {11}{2\pi }}(t+2000)\right)$ (c) $N=60+60\cos \left({\frac {11}{2\pi }}(t+2000)\right)$ (d) $N=60+60\sin \left({\frac {2\pi }{11}}(t-2000)\right)$ (e) $N=60+60\cos \left({\frac {2\pi }{11}}(t-2000)\right)$

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
One way of solving this problem is to simply plug $t=2000$ into the given possibilities and check if you get the number of sunspots that were detected in the year 2000.

Hint 2

A more methodical method of approaching the problem is as follows: We know that the minimum of the function is 0 and the maximum is 120, which means that we are looking for a trigonometric function of the form $\displaystyle 60+60\sin(f(t))$ or $\displaystyle 60+60\cos(f(t))$ for some function $\displaystyle f(t)$. Then we use the fact that the period is 11 and that the maximum occurs for $t=2000$: The period of $\sin(t)$ and $\cos(t)$ is $2\pi$. What factor is the argument multiplied by to get a period of 1? A period of 11? $\sin(t)$ has a maximum at $t=\pi /2$ while $\cos(t)$ has a maximum at $t=0$. How can you shift the function left or right to shift the maximum to $t=2000$?

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 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Substituting $t=2000$ into the given possibilities gives (a) $N=60+120\sin \left({\frac {2\pi }{11}}(t-2000)+{\frac {\pi }{2}}\right)=60+120\sin \left({\frac {\pi }{2}}\right)={\color {red}180}$ (b) $N=60+60\sin \left({\frac {11}{2\pi }}(t+2000)\right)={\color {red}60+60\sin \left({\frac {22000}{\pi }}\right)}$ (c) $N=60+60\cos \left({\frac {11}{2\pi }}(t+2000)\right)={\color {red}60+60\cos \left({\frac {22000}{\pi }}\right)}$ (d) $N=60+60\sin \left({\frac {2\pi }{11}}(t-2000)\right)=60+60\sin(0)={\color {red}60}$ (e) $N=60+60\cos \left({\frac {2\pi }{11}}(t-2000)\right)=60+60\cos(0)={\color {OliveGreen}120}$ Therefore, the answer is ${\color {blue}{\text{(E)}}}$.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. A more methodical method of approaching the problem is as follows: We know that the minimum of the function is 0 and the maximum is 120, which means that we are looking for a trigonometric function of the form $\displaystyle 60+60\sin(f(t))$ or $\displaystyle 60+60\cos(f(t))$ for some function $\displaystyle f(t)$. Since the period is 11 years, we see that $\displaystyle f(t)={\frac {2\pi }{11}}\cdot g(t)$ for some other function $\displaystyle g(t)$. Lastly, we notice that a peak was detected in 2000. If we used the cosine function, we would have $\displaystyle g(t)=t-2000$ since the peak of the cosine function is at 0 (up to integer multiples of $2\pi$). If we used the sine function, we would have $\displaystyle f(t)={\frac {2\pi }{11}}\cdot g(t)+{\frac {\pi }{2}}$ and $\displaystyle g(t)=t-2000$ since the peak of the sine function occurs at ${\tfrac {\pi }{2}}$ up to integer multiples of $2\pi$. Thus, the possible answers are: $\displaystyle N=60+60\sin \left({\frac {2\pi }{11}}(t-2000)+{\frac {\pi }{2}}\right)$ or $\displaystyle N=60+60\cos \left({\frac {2\pi }{11}}(t-2000)\right)$ The latter is a possible choice in the question, and so the answer is ${\color {blue}{\text{(E)}}}$.