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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:::info
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)
There should be 14.23 hours of daylight on Julty 19th (Day 200) according to the .

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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 daylight hours on July 19 is 14 hours and 17 minutes.
according to the model there would be 14.23 hours of daylight.

Converting the actual number of hours to minutes there are 857 min-854.16 min = 2.84 minutes
so the model was off by only 2.84 minutes

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(4) Compute $D'(x)$. Show all work.
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(4)
$D'\left(x\right)=12.1-2.4\cos\left(\frac{2\pi\left(x+10\right)}{365}\right)$
$D'\left(x\right)=\frac{d}{dx}\left[12.1\right]-\frac{d}{dx}\left[\frac{2.4\cos\left(2\pi\left(x+10\right)\right)}{365}\right]$
$D'\left(x\right)=0-\frac{d}{dx}\left[\frac{2.4\cos\left(2\pi\left(x+10\right)\right)}{365}\right]$
$D'\left(x\right)=-\frac{d}{dx}\left[2.4\cos\left(2\pi\left(x+10\right)\right)\right]\cdot365-\left(2.4\cos\left(2\pi\left(x+10\right)\cdot\frac{d}{dx}\left[365\right]\right)\right)$
$D'\left(x\right)=2.4\sin\cdot\frac{2\pi\left(x+10\right)}{365}\cdot\frac{2\pi}{365}\left(\frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[10\right]\right)$
$D'\left(x\right)=\frac{4.8\pi\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)}{365}$

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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)
$D\left('x\right)=\frac{4.8\sin\left(\frac{\left(2\pi( x+10)\right)}{365}\right)}{365}=$
$\frac{4.8\pi\sin\left(\frac{2\pi\left(200+10\right)}{365}\right)}{365}= -0.01883537$
$(0.01883537)(60)=-1.130122$ $min/day$
The data shows that the number of hours of daylight are chainging by $-1.130122$ minutes per 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)
The slpoe will be zero when there is 14.5 hours of daylight and June 21st will be the longest day.

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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)
As the slope is increasing, we know that the derivative is going to be positive and increasing as well. Finding $D''(x)$ would let us know when the number of daylight hours will be increasing the fasest.

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