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)
$D(x)=12.1-2.4\cos \left(\frac{2\pi \left(x+10\right)}{365}\right)$
$D(200)=12.1-2.4\cos \left(\frac{2\pi \left(200+10\right)}{365}\right)$
$D(200)= 14.2361$

Using the model to compute, on July 19th there will be 14.2361 hours of daylight.

:::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) Model minutes: $(14.2361\cdot60=854.166)$
Minutes from the website: $(14\cdot60+16=856)$
According to timeanddate.com the model's prediction is off by about 2 minutes more of daylight.

:::info
(4) Compute $D'(x)$. Show all work.
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(4)
$D(x)=12.1-2.4\cos \left(\frac{2\pi \left(x+10\right)}{365}\right)$

$D'(x)=0+2.4\sin \left(\frac{2\pi \left(x+10\right)}{365}\right)\cdot\frac{d}{dx}[\frac{2\pi(x+10)}{365}]$

$D'(x)=2.4\sin \left(\frac{2\pi \left(x+10\right)}{365}\right)\cdot\frac{365\frac{d}{dx}[2\pi(x+10)]-2\pi(x+10)\frac{d}{dx}[365]}{365^2}$

$D'(x)=2.4\sin \left(\frac{2\pi \left(x+10\right)}{365}\right)\cdot\frac{365\cdot2\pi(1)-2\pi(x+10)(0)}{365^2}$

$D'(x)=2.4\sin \left(\frac{2\pi \left(x+10\right)}{365}\right)\cdot(\frac{365\cdot2\pi}{365^2})$

$D'(x)=2.4\sin \left(\frac{2\pi \left(x+10\right)}{365}\right)\cdot(\frac{2\pi}{365})$

:::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)
$D'(200)=2.4\sin \left(\frac{2\pi \left(200+10\right)}{365}\right)\cdot(\frac{2\pi}{365})$

$D'(200)= -0.018835$

The number of hours of daylight are decreasing by $0.018835$ $minutes/day^2$ on July 19th. This means that on July 19 of 2020 there was less daylight than on July 18th by about $0.0188$ $minutes/ day^2$.

:::info
(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)
$0=2.4\sin \left(\frac{2\pi \left(x+10\right)}{365}\right)\cdot(\frac{2\pi}{365})$

$0/(\frac{2\pi}{365})=2.4\sin \left(\frac{2\pi \left(x+10\right)}{365}\right)\cdot(\frac{2\pi}{365})/(\frac{2\pi}{365})$

$0=\frac{2.4\sin \left(\frac{2\pi \left(x+10\right)}{365}\right)}{2.4}$

$\arcsin(0)=\arcsin(\sin \left(\frac{2\pi \left(x+10\right)}{365}\right))$
I am actually stuck on solving the derivative for x. By using the graph of the derivative where y=0, I have found that the day of the year that will be the longest is the 172nd day of the year or June 21st.
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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) Besides using the graph of $D'(x)$ in desmos, and locating the highest y value, I am unsure how to find the day of the year when the number of hours of daylight is increasing most rapidly.

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