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)**
According to the model there is about 14.24 hours of daylight on July 19th (day 200).

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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 amount of daylight hours for July 19, 2020 equated to 14 hours and 17 minutes. This is slightly higher than the calculated estimate of 14 hours and 14.2 minutes, resulting in a small difference of 2.8 minutes.

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(4) Compute $D'(x)$. Show all work.
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**(4)**
d(x) = 12.1 - 2.4cos($\frac{2\pi(x+10)}{365}$)
d'(x) = 0 - 2.4(-sin)($\frac{2\pi(x+10)}{365}$)
d'(x) = 2.4sin($\frac{2\pi(x+10)}{365}$)
d'(x) = ($\frac{2\pi}{365}$) $\times$ 2.4sin($\frac{2\pi(x+10)}{365}$)
d'(x) = (($\frac{2\pi(x+10)}{365}$) $\times$ 2.4)sin ($\frac{2\pi(x+10)}{365}$)
d'(x) = 0.041314095170 sin($\frac{2\pi(x+10)}{365}$)

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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)**
The result for d'(200) represents the rate of change in the hours of daylight on July 19th. For July 19th there is a loss of about 1.13 minutes.
d'(200) = 0.041314095170 sin($\frac{2\pi(x+10)}{365}$)
d'(200) = -0.01883537251 $\times$ 60 min
d'(200) = -1.1301223506 mins of daylight / 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)**
When setting d'(x) = 0 and then solving, we determined which day is the longest as having the most hours of daylight hours within a day to be July 21st, the 172.5th day of the year.

d'(x) = 0 = 0.04134095170 sin($\frac{2\pi(x+10)}{365}$)
d'(x) = sin ($\frac{2\pi(x+10)}{365}$)
d'(x) = $\frac{2\pi(x+10)}{365}$
x = $\frac{365h\pi - 20h}{2h}$ = $\frac{(365\pi - 20)h}{2h}$ = $\frac{365h - 20}{2}$

h = 1 --> $\frac{365(1) - 20}{2}$
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 we determind the longest day of the year, the slope of that point when graphed was at, or very close to, zero. To find the day of the year in which the hours of daylight are rapidly increasing, we could simply look at that same graph to determine where the line is rising most quickly.
*(see area circled in green in below image)*

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