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)
<iframe src="https://www.desmos.com/calculator/cyplsnovdl?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>

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

:::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)
The website says the total hours of daylight for July 19 was 14:8 hrs. Our prediction was off by $0.563933816557$ minutes.

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

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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)
$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{730\pi}{365^{2}}\right)$
$D'\left(200\right)=2.4\sin\left(\frac{2\pi\left(200+10\right)}{365}\right)\cdot\left(\frac{730\pi}{365^{2}}\right)$
$D'(200)=-0.0188353725245\frac{hr}{day}$
$\frac{-0.0188353725245hr}{1\ day}\cdot\frac{60\min}{1\ hr}=-\frac{1.13012235147\min}{1\ day}$
The rate at which the numbers of hours of daylight are changing on July 19 is by -1.13012235147 daylight minutes per day. This just means that the number of daylight hours, or in this case minutes, on this day are decreasing.

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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)
$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{730\pi}{365^{2}}\right)$
$0=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{730\pi}{365^{2}}\right)$
$0=2.4\sin\left(\frac{2\pi x+20\pi}{365}\right)$
$0=\sin\left(\frac{2\pi x+20\pi}{365}\right)$
$\sin^{-1}\left(0\right)=\frac{2\pi x+20\pi}{365}$
$365\sin^{-1}\left(0\right)=2\pi x+20\pi$
$365\sin^{-1}\left(0\right)-20\pi=2\pi x$
$\frac{365\sin^{-1}\left(0\right)-20\pi}{2\pi}=x$
$-10=x$
But looking at the graph gives me $D'(x)=172.5$

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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)
I think that one way you can find the day of the year when the number of hours of daylight is increasing most rapidly is by reviewing the graph. More specifically, by looking for the part of the function where the graph seems to have the steepest positive slope.

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