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)[
](https://)(2)[](https://)
14.236 hours

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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)According to the website the actual number of hours for July 19 of this year is
14:18 hrs.
$14.236-14.18=0.056hrsx60=3.36min off$

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(4) Compute $D'(x)$. Show all work.
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(4)
$D(x)=12.1-2.4\cos \left(\frac{2\pi \left(x+10\right)}{365}\right)$
$D(x)=0+2.4\sin \left(\frac{2\pi \left(x\right)}{365}\right)$+$\left(\frac{20\pi \left(\right)}{365}\right)$
$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\frac{2\pi }{365}$

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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)
$D'(200)=-0.019x60=-1.13$ min/day
The rate of hours of daylihgt is decreasing at a rate of 1.13 minutes per day on July 19.

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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)
$D'(x)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\frac{2\pi}{365}$
$0=\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\frac{4.8\pi}{365}$
$\frac{365}{4.8\pi}\cdot0=\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\frac{4.8\pi}{365}\cdot\frac{365}{4.8\pi}$
$\arcsin\cdot0=\arcsin\cdot\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)$
From the problem near the center of the year indicates the middle of year = pie on the unit circle , Thus,
$pi=\left(\frac{2\pi}{365}\right)\cdot\left(x+10\right)$
$\pi\cdot\frac{365}{2\pi}=\left(\frac{2\pi}{365}\right)\cdot\left(x+10\right)\cdot\frac{365}{2\pi}$
$\frac{365}{2}-10=x$
$172.5=x$
Approximatly June 21 or 22 will be the longest.

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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)
The number of hours of daylight are increasing most rapidly when D'(x) is maximum. Thus, when D'(x)>0, D(x) is increasing. When D''(x)=0 D'(x)is maximum or in other words as D'(x) is increasing before D''(x)=0 and concavity changes from up to down.

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