Math 181 Miniproject 5: Hours of Daylight.md
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tags: MATH 181
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Math 181 Miniproject 5: Hours of Daylight
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**Overview:** This miniproject will apply what you've learned about derivatives so far, especially the Chain Rule, to analyze the change the hours of daylight.

**Prerequisites:** The computational methods of Sections 2.1--2.5 of *Active Calculus*, especially Section 2.5 (The Chain Rule).

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The number of hours of daylight in Las Vegas on the $x$-th day of the year ($x=1$ for Jan 1) is given by the function together with a best fit curve from Desmos.}[^first]
[^first]: The model comes from some data at http://www.timeanddate.com/sun/usa/las-vegas?

\\[
D(x)=12.1-2.4\cos \left(\frac{2\pi \left(x+10\right)}{365}\right).
\\]
(1) Plot a graph of the function $D(x)$. Be sure to follow the guidelines for formatting graphs from the specifications page for miniprojects.
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(1)
https://www.desmos.com/calculator/zmyfy9yht1

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(2) According to this model how many hours of daylight will there be on July 19 (day 200)?
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(2) There will be 14.236 hours of sunlight on day 200.

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(3) Go to http://www.timeanddate.com/sun/usa/las-vegas? and look up the actual number of hours of daylight for July 19 of this year. By how many minutes is the model's prediction off of the actual number of minutes of daylight?
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(3) The actual number of hours are 14 hours 10 min and 12 seconds. The model's prediction says it will be 14 hours 14 minutes and 10 seconds. The model is off by 4 minutes.

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(4) Compute $D'(x)$. Show all work.
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(4) $D(x)=12.1-2.4\cos\left(\frac{2\pi\left(x+10\right)}{365}\right)$
$D'(x)=\frac{d}{dx}(12.1-2.4\cos\left(\frac{2\pi\left(x+10\right)}{365}\right))$
$=\frac{d}{dx}\left(\frac{121}{10}-\frac{12}{5}\cdot\cos\left(\frac{\left(2\pi x+20\pi\right)}{365}\right)\right)$
$=\frac{d}{dx}\left(\frac{121}{10}\right)-\frac{d}{dx}\left(\frac{12\cos\left(\frac{\left(2\pi x+20\pi\right)}{365}\right)}{5}\right)$
$=0-\frac{12}{5}\cdot\left(-\sin\left(\frac{\left(2\pi x+20\pi\right)}{365}\right)\cdot\frac{1}{365}\cdot2\pi\right)$
$D'\left(x\right)=\left(\frac{24\pi\cdot\sin\left(\frac{\left(2\pi x+20\pi\right)}{365}\right)}{1825}\right)$

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(5) Find the rate at which the number of hours of daylight are changing on July 19. Give your answer in minutes/day and interpret the results.
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(5) The answer I got is -0.0194655288 minutes/day is the rate of daylight hours that's changing on July 19 in minutes/day. I did this by plugging 200 into x for the D'(x) formula by representing day 200. D'(x) represents the rate of change of hours of daylight in minutes/ day so that is why I plugged 200 into x in this equation and not D(x).

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(6) Note that near the center of the year the day will reach its maximum length when the slope of $D(x)$ is zero. Find the day of the year that will be longest by setting $D'(x)=0$ and solving.
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(6) The day of the year it is the longest day is June 20th, day 172, actually point 172.5 on graph (x). The hour is 14.5 (y) where the slope is zero on the graph of D(x).

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(7) Write an explanation of how you could find the day of the year when the number of hours of daylight is increasing most rapidly.
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(7) An easy method to finding the day of the year when the numbers of daylight is increasing rapidly is looking at the longest days of the year (summer months) and look up the summer solstice, which is usually June 20-22nd and the longest day of the year. Go to the graph of the function and look at this day (use your mouse to find the x value of this day number (in this case day 172.5) and find where it peaks on the hump. The exact point where it peaks and the tangent line's slope is zero, is the longest day of the year and the number of hours of daylight on this day. You can double check it that way. Look up the summer solstice, then check your graph for that point and see if the tangent line's slope is zero. If it's not clear, you could use the central difference to see the slope (technically finding the secant line) but use two values very close to the number you are checking, for example .001 difference both positive and negative away from your x value point.

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