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

$D(200)=12.1-2.4\cos\left(\frac{2\pi\left(200+10\right)}{365}\right)$
$D(x)=12.1-2.4(-.89003)$
$D(x)=12-(-2.36)$
$D(x)=14.24$

On July 19 in Las Vegas, there will be 14.236 hours of day light.

:::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 actual number of hours of daylight for july 19 is 14.8 hr.Based on my calcuations I was off by .56 minutes.

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

$D'(x)=\frac{d}{dx}\left[12.1\right]\cdot\left(-2.4\cos\left(\frac{2\pi\left(x+10\right)}{365}\right)\right)+12.1\ \cdot\frac{d}{dx}\left[2.4\cos\left(\frac{2\pi\left(x+10\right)}{365}\right)\right]$

$D'(x)=0\left(-2.4\cos\left(\frac{2\pi\left(x+10\right)}{365}\right)\right)+12.1\ \cdot\ \left(\frac{d}{dx}\left[-2.4\right]\cdot\cos\left(\frac{2\pi\left(x+10\right)}{365}\right)-2.4\ \cdot\ \frac{d}{dx}\left[\cos\left(\frac{2\pi\left(x+10\right)}{365}\right)\right]\right)$

$D'(x)=12.1\left(0\cdot\cos\left(\frac{2\pi\left(x+10\right)}{365}\right)+\left(-2.4*-\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\frac{d}{dx}\left[\frac{2\pi\left(x+10\right)}{365}\right]\right)\right)$

$D'(x)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{\frac{d}{dx}\left[2\pi x+20\right]\cdot365-\left(2\pi+20\right)\cdot\frac{d}{dx}\left[365\right]}{\left(365\right)^{2}}\right)$

$D'(x)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{730\pi}{133225}\right)$

:::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'(x)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{730\pi}{133225}\right)$

$D'(200)=2.4\sin\left(\frac{2\pi\left(200+10\right)}{365}\right)\cdot\left(\frac{730\pi}{133225}\right)$
$D'(200)=−0.0188353725245\frac{hour}{day}$

Converting:

$\frac{-0.0188353725245\ hr}{1\ day}\cdot\frac{60\ \min}{1\ hr}$ $=-1.13012235147\ \frac{\min}{day}$
The rate in which the number of hours of daylight within a day after July 19 is -1.13012235147 minutes per day. This means the number of hours of daylight is decreasing.

:::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)
$$D'(x)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{730\pi}{133225}\right)$

$0=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{730\pi}{133225}\right)$

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

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

$\sin^{-1}\left(0\right)=\frac{2\pi x+20\pi}{365}$

$365\sin^{-1}\left(0\right)=2\pi x+20\pi$

$365\sin^{-1}\left(0\right)-20\pi=2\pi x$

$x=\frac{365\sin^{-1}\left(0\right)-20\pi}{2\pi}$

$x=-10$

The eztimate from this -10, doesn't match the number given by the graph for D'(x) since is equals 172.5.

:::info
(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)
One way to find the day of year when the number of hours of daylight is increasing is by looking at the graph. By looking at the part of the graph where the function is increasing or when it's the steepest positive slope.
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