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)

:::info
(2) According to this model how many hours of daylight will there be on July 19 (day 200)?
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(2)$D\left(x\right)=12.1-2.4\cos\left(\frac{2\pi\left(x+10\right)}{365}\right)$
$D\left(200\right)=12.1-2.4\cos\left(\frac{2\pi\left(200+10\right)}{365}\right)$
$D(200)=14.2360661834hrs$

:::info
(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)According to the website, the actual hours of daylight for July 19th was 14:17:52. Our prediction was about 6 minutes off.

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(4) Compute $D'(x)$. Show all work.
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(4)$D\left(x\right)=12.1-2.4\cos\left(\frac{2\pi\left(x+10\right)}{365}\right)$
$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\frac{d}{dx}\left[\frac{2\pi\left(x\cdot10\right)}{365}\right]$
$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\frac{\left(\frac{d}{dx}\left[2\pi x+20\pi\right]\cdot365-\left(2\pi x+20\pi\right)\cdot\frac{d}{dx}\left[365\right]\right)}{\left(365\right)^{2}}$
$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\frac{\left(2\pi\cdot365-\left(2\pi x+20\pi\right)\cdot0\right)}{\left(365\right)^{2}}$
$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{730\pi}{365^{2}}\right)$

:::info
(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)$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{730\pi}{365^{2}}\right)$
$D'\left(200\right)=2.4\sin\left(\frac{2\pi\left(200+10\right)}{365}\right)\cdot\left(\frac{730\pi}{365^{2}}\right)$
$D'\left(200\right)=-0.01883537\frac{hr}{day}$
$=-\frac{0.01883537hr}{1\ day}\cdot\ \frac{60\min}{1\ hr}=-\frac{1.1301222\min}{1\ day}$
This number means that daylght is decreasing by -1.1301222 minutes per day, as of Juy 19th.

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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)$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{730\pi}{365^{2}}\right)$
$0=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{730\pi}{365^{2}}\right)$
$0=2.4\sin\left(\frac{2\pi x+20\pi}{365}\right)$
$0=\sin\left(\frac{2\pi x+20\pi}{365}\right)$
$\sin^{-1}\left(0\right)=\frac{2\pi x+20\pi}{365}$
$365\sin^{-1}\left(0\right)=2\pi x+20\pi$
$365\sin^{-1}\left(0\right)-20=2\pi x$
$\frac{\left(365\sin^{-1}\left(0\right)-20\right)}{2\pi}=x$
$x=-10$
$365/2 = 182.10$
$182.10-10=172.10$
The longest day of the year would by June 20th, it has 14:37:11 hours of daylight. June 20th is also day 172.5 out 365.

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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)We could find the day of the year where daylight is increasing the most by taking a look at the graph from above. We would want to find the place where the slope is increasing the fastest. When we locate that spot, we can find the corresponsing day of the year on the graph. Once you know that day, you can count what day that is out of the year & that is when you could find the day when daylight is increasing most rapidly.

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