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)

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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)
According to the graph there will be 14.236 hours on July 19.

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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) On July 19 the daylight hours was 14 hours 17 minutes 34 seconds, which equals 14.292778 hrs.
To find how many minutes the model is off we have to subtract and convert hrs to min.
So, $14,292778 - 14.236+ 0.0567hrs$
$0.0567 hrs =3.4min$
The model's prediction is 3.4 min off the actual number of min of daylight.

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(4) Compute $D'(x)$. Show all work.
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(4)
$D'\left(x\right)=-2.4\cos\left(\frac{2\pi\left(x+10\right)}{365}\right)$
$=-2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\frac{2\pi}{365}$
$=\frac{4.8\pi}{365}\cdot\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)$
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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)
Rate of which the number of hours of daylight changing on July 19 is $D'(200)$ which is the $200th$ day.
$D'\left(200\right)=\frac{4.8\pi}{365}\sin\ \frac{2\pi\left(210\right)}{365}=0$
$=\frac{4.8\pi}{365}\sin\ \frac{420\pi}{\left(365\right)}$
$=0.0188353 hours/day$
$=1.13 min/day$
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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)$D'\left(x\right)=0$
$=\frac{4.8\pi}{365}\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)$
$=\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)$
$\frac{2\pi\left(x+10\right)}{395}=n\pi$
$x=\frac{365n\pi-20\pi}{2\pi}$
$x=172.5$
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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) When the hours of the daylight are most increasing we can find the day of the year by graphing the equation from above and entering the hours as $y$ and the $x$ will be the day.

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