Science:Math Exam Resources/Courses/MATH102/December 2016/Question A 04

MATH102 December 2016

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Question A 04

The body temperature $T(t)$ of an individual over a day is a rhythmic process. It reaches a maximum of 37.6 (degree Celsius) at 6:00 am and a minimum of 36.8 (degree Celsius) at 6:00 pm. Choose the trigonometric function that best describes $T(t)$, where $t$ is given in hours with $t=0$ taken at midnight (one minute after 11:59 pm). (i) $T(t)=37.2-0.4\cos \left({\frac {2\pi t}{12}}\right)$ (ii) $T(t)=37.2+0.4\cos \left({\frac {2\pi t}{12}}\right)$ (iii) $T(t)=37.2+0.4\sin \left({\frac {2\pi t}{12}}\right)$ (iv) $T(t)=37.2+0.4\sin \left({\frac {2\pi t}{24}}\right)$ (v) $T(t)=37.2-0.4\cos \left({\frac {2\pi t}{24}}\right)$ (vi) $T(t)=37.2+0.4\cos \left({\frac {2\pi t}{24}}\right)$

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
At 6:00 am, $t=6$ and at 6:00 pm $t=18$ . Translate the given information in terms of the given times. Also, note that at max/min times the derivative of $T$, $T'(t)$ , vanishes.

Hint 2
(Alternative solution) The trigonometric functions $\sin \left({\frac {2\pi t}{L}}\right)$ and $\cos \left({\frac {2\pi t}{L}}\right)$ are $L$-periodic. In other words, $\sin \left({\frac {2\pi t}{L}}\right)=\sin \left({\frac {2\pi (t+L)}{L}}\right)$ and $\cos \left({\frac {2\pi t}{L}}\right)=\cos \left({\frac {2\pi (t+L)}{L}}\right)$

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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. We need to check the following for each of the choices: $T(6)=37.6=37.2+0.4,\qquad T(18)=36.8=37.2-0.4$ , $T'(6)=0=T'(18)$ . Note that $\sin \left({\frac {2\pi \cdot 6}{12}}\right)=\sin \pi =0,\quad \sin \left({\frac {2\pi \cdot 18}{12}}\right)=\sin 3\pi =0$ , and $\cos \left({\frac {2\pi \cdot 6}{24}}\right)=\cos {\frac {\pi }{2}}=0,\quad \cos \left({\frac {2\pi \cdot 18}{24}}\right)=\cos {\frac {3\pi }{2}}=0$. This means that choices (iii), (v) and (vi) are out since for all of them the second term vanishes and $T(6)=37.2=T(18)$ (i) $T(6)=37.2-0.4\cos(\pi )=37.2+0.4=37.6,\qquad T(18)=37.2-0.4\cos(3\pi )=37.2+0.4=37.6\to \$ out (ii) $T(6)=37.2+0.4\cos(\pi )=37.2-0.4=36.8,\qquad T(18)=37.2+0.4\cos(3\pi )=37.2-0.4=36.8\to \$ out The only option left is (iv). We check the conditions: (iv) $T(6)=37.2+0.4\sin({\frac {\pi }{2}})=37.2+0.4=37.6,\qquad T(18)=37.2+0.4\sin({\frac {3\pi }{2}})=37.2-0.4=36.8$ , we can check the derivative too, but we don't really need. Answer: $\color {blue}(iv)$

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. (Alternative solution) By Hint 2, we know that $\sin \left({\frac {2\pi t}{L}}\right)$ and $\cos \left({\frac {2\pi t}{L}}\right)$ are $L$-periodic. This implies that for any constant $a$ and $b\neq 0$, both functions $a+b\sin \left({\frac {2\pi t}{L}}\right)$ and $a+b\cos \left({\frac {2\pi t}{L}}\right)$ are also $L$-periodic; Indeed, putting $T(t)=a+b\sin \left({\frac {2\pi t}{L}}\right)$, we have $T(t+L)=a+b\sin \left({\frac {2\pi (t+L)}{L}}\right)=a+b\sin \left({\frac {2\pi t}{L}}\right)=T(t)$. Similar argument holds for $a+b\cos \left({\frac {2\pi t}{L}}\right)$. Therefore, (i), (ii), (iii) are $12$-periodic, while (iv), (v), (vi) are $24$-periodic. However, since at $t=6$ (6 am) and at $t=18=6+12$ (6 pm), the function values of $T(t)$ are different, i.e., $T(6)=37.6\neq 36.8=T(18)$, $T(t)$ is not $12$-periodic---(i), (ii), (iii) are out. Then, using $\cos \left({\frac {2\pi \cdot 6}{24}}\right)=\cos \left({\frac {\pi }{2}}\right)=0$, $T$ given in (v) and (vi) satisfies $T(6)=37.2$, which is different from the given information $T(6)=37.6$---(v), (vi) are out. Answer $\color {blue}(iv)$