Math 181 Miniproject 5: Hours of Daylight.md
---
---
tags: MATH 181
---

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).

---

:::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)
<iframe src="https://www.desmos.com/calculator/cbr3ooyxay?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>

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

(2)
According to this model, it shows that on day 200 there will be 14.236 hours of daylight, which would be 14 hours and 14 minutes and 10 seconds of daylight.(14:14:10)

:::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)
According to the webpage, the actual number of hours of daylight was 14 hours 17 minutes and 52 seconds. The models prediction was only off by 3 minutes and 42 seconds of daylight
:::info
(4) Compute $D'(x)$. Show all work.
:::

(4)
$D(x)=12.1-2.4cos(\frac{2π(x+10)}{365})$
$D'(x)=0-2.4cos(\frac{2π(x+10)}{365})$
$D'(x)=2.4sin(\frac{2π(x+10)}{365})*\frac{d}{dx}(\frac{2πx+20π)}{365})$
$D'(x)=2.4sin(\frac{2πx+20π}{365})*\frac{d}{dx}(\frac{2πx}{365}+\frac{20π}{365})$
$D'(x)=2.4sin(\frac{2πx+20π)}{365})*(\frac{2π(1)}{365}+0)$
$D'(x)=2.4sin(\frac{2πx+20π}{365})*(\frac{2π}{365})$
:::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)
$D'(200)=2.4sin(\frac{2π(200)+20π}{365})*(\frac{2π}{365})$
$D'(200)=-1.094176061*(\frac{2π}{365})$
$D'(200)=-1.094176061*(0.172142063)$
$D'(200)=-0.0188353724*60min/day$
$D'(200)=-1.13 min/day$
This means that on July 19 the day loses 1 minute and 13 seconds of daylight.

:::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)
$0=2.4sin(\frac{2π(200)+20π}{365})*(\frac{2π}{365})$
Looking at the graph below, we can see in the middle of the year when D'(x)=0, the slope is reaching it's maximum length at day 172. Day 172 end up being June 21, 2019. This means that this is the longest day of they year.

:::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)
To find the day of the year where the number of daylight hours is increasing most rapidly, you could do this by looking at the graph, and when the slopes is increasing at it's greatest amount, that shows that the daylight hours is increasing the most rapidly. In other words, when the graph reaches the maximum value, or it's peak.

---

To submit this assignment click on the Publish button . Then copy the url of the final document and submit it in Canvas.