---
# System prepended metadata

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

---

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/pxuzzxpeiw?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) The number x, *200*,  is plugged into the initial formual given. \\[
D(x)=12.1-2.4\cos \left(\frac{2\pi \left(x+10\right)}{365}\right).
\\]\\[
D(x)=12.1-2.4\cos \left(\frac{2\pi \left(200+10\right)}{365}\right).
\\]\\[
D(x)=12.1-2.4\cos \left(\frac{2\pi \left(210\right)}{365}\right).
\\]\\[
D(x)=12.1-2.4\cos \left(\frac{420\pi}{365}\right).
\\]\\[
D(x)=12.1-2.4\cos \left(\frac{84\pi}{73}\right).
\\]\\[
D(x)= 14.236 hours
\\] There will be 14.24 hours of daylight on July 19 (day 200).



:::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) The actual number of hours of daylight for July 19 of this year is 14.16 hours. That makes the prediction off by 0.08 hours. When 0.08 hours gets converted to minutes that results in 4.8 minutes.
$0.08hours$ x $60mins$ = $4.8mins$

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

(4) \\[
D(x)=12.1-2.4\cos \left(\frac{2\pi \left(x+10\right)}{365}\right).
\\]first, \\[\frac{2\pi}{365}= 0.01721\\]\\[
D(x)=12.1-2.4\cos(0.01721(x+10))
\\]\\[
D'(x)=\frac{d}{dx}(12.1-2.4\frac{d}{dx}\cos \left(\frac{2\pi \left(x+10\right)}{365}\right).
\\]\\[
D(x)=0-2.4\cos \left(\frac{2\pi \left(x+10\right)}{365}\right).
\\] next, \\[cos (0.01721(x+10)\\] apply the chain rule of \\[let, u=0.01721 (x+10)\\] \\[\frac{df(u)}{dx}=\frac{df}{du}x\frac{du}{dx}\\]\\[D'(x)=-2.4\frac{d}{du}(cos(u))x
\frac{d}{dx}(0.01721(x+10))\\]\\[D'(x)=-2.4(sinu)(0.01721)=0.04131sinu\\]\\[
D'(x)= 0.04131 sin(0.01721(x+10))
\\]
:::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 derivative is used to find this result. \\[D'(x)= 0.04131 sin(0.01721(x+10))
\\]\\[
D'(x)= 0.04131 sin(0.01721(200+10))
\\] \\[
D'(x)= 0.041(-0.455)\\]\\[
D'(x)= -0.0188\frac{hours}{day}
\\] We have to convert to make the result into minutes. -0.0188 hours x 60 minutes= -1.128min/day
The rate at which the number of hours of daylight is 1.128 minutes/day 

:::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)\\[
D'(x)= 0.04131 sin(0.01721(x+10))=0
\\]\\[
D'(x)= sin(0.01721(x+10))=0
\\]\\[
D'(x)=0.01721(x+10)=\pi
\\]\\[
D'(x)= \frac{2\pi}{365}(x+10)=\pi
\\]\\[
D'(x)= x+10=\frac{365}{2}\\]\\[
D'(x)= x=\frac{365}{2}-10\\]\\[
D'(x)=\frac{345}{2}=172.5\\] The longest day of the year was around June 21st or 22nd. 
:::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 number of hours of daylight are increasing most rapidly when D'(x) is maximum. When D'(x)=0, D(x) is increasing. But, when D'(x) is maximum, that means D(x) is increasing most rapidly. 

---

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.



<script src="https://gist.github.com/sergeballif/fd914e775b298a847925269f6752445d.js"></script>