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title: 'Math 181 Miniproject 5: Hours of Daylight'

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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
===

**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.
:::

(1)
![](https://i.imgur.com/tOeuBv0.png)


:::info
(2) According to this model how many hours of daylight will there be on July 19 (day 200)?
:::

(2)
Using Desmos to calculate the output value of the provided function given x=200, there would be 14.236 hours of daylight on the 200th day of the year.

:::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?
:::

(3)
By converting the calculated amount of hours of daylight (14.236 hours from Question 2), one gets 854.16 minutes:

$14.235 hr (\frac{60min}{1hr})= 854.16$

By looking up the actual amount of hours (14:17:52) of daylight on July 19th and then converting to minutes, one gets 857.86 minutes: 

$14 hr  (\frac{60min}{1hr})= 840 min$

$17$ minutes was given in the value from the website.

$52 sec  (\frac{1 min}{60 sec})= 0.86 min$

$840+17+0.86=857.86$

Therefore, the difference between the two values is 1.7 minutes:

$857.86-854.16=3.7$

The model's prediction was off by an additional 3.7 minutes from that of the actual time of daylight recorded online. 

:::info
(4) Compute $D'(x)$. Show all work.
:::

(4)
$D(x)=12.1-2.4cos(\frac{2\pi(x+10)}{365})$

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

Distribute $2\pi$:
$D'(x)=2.4sin(\frac{2\pi(x+10)}{365})([\frac{d}{dx}]\frac{2\pi(x)+20\pi)}{365})$

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

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

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

:::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.
:::

(5)
The rate at which the amount of daylight is changing per day is -1.128 min/day (the following equation was put in a calculator in radian mode):

$D'(200)=2.4sin(\frac{2\pi((200)+10)}{365})(\frac{2\pi}{365})=-0.0188$

The above equation provides the rate of change in units of hours per day. Below shows the conversion to minutes per day:

$-0.0188 (\frac{60min}{1hr})=-1.128$

:::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.
:::

(6)
Setting $D'(x)=0$ and then solving for $x$:

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

Divide both sides by $\frac{2\pi}{365}$:
$2.4sin(\frac{2\pi(x+10)}{365})=0$

Divide both sides by 2.4:
$sin(\frac{2\pi(x+10)}{365})=0$

Take the arcsine of both sides:
$(\frac{2\pi(x+10)}{365})=arcsin(0)$

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

Multiply by 365:
$2\pi(x+10)=0$

Distribute $2\pi$:
$2\pi(x)+20(\pi)=0$

Subtract $20(\pi)$:
$2\pi(x)=-20(\pi)$

Divide by $2(\pi)$:
$x=-10$

The -10th day of the year would equate to a day at the end of December (the slope here is still 0 but this would actually be the day of the year with the least amount of daylight [the day with the lowest output value] and this can be confirmed by looking at the graph in Desmos). If you divide 365 days by 2 and add this to -10 you would get the day of the year that has the most hours of daylight. 

$\frac{365}{2}+(-10)=172.5$

This value is roughly day 172 ($x=172$) and can be confirmed by looking at the graph in Desmos (the slope at this point is roughly 0 hrs/day). 

:::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.
:::

(7)
The easiest way to determine what day of the year has the greatest slope (which day has the greatest rate of change of hours per day) would be to look at the graph in Desmos and find the inflection point of the graph where the function transitions from being concave up to concave down. The point immediately before this inflection point would have the greatest slope because D(x) would be the most steep in the positive direction here. 

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To submit this assignment click on the Publish button ![Publish button icon](https://i.imgur.com/Qk7vi9V.png). Then copy the url of the final document and submit it in Canvas.



