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/ci4dzsixq1?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 the graph above, there will be approximately $14.236$ $hours$ of daylight or 14 hours and 14 minutes 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 according to the website on July 19 of this year was about 14 hours and 18 minutes. Which is extremely close to the value we got from D(x), and was only off by about 4 minutes. :::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). \\] Use the Chain Rule $D'\left(x\right)=\frac{d}{dx}\left[12.1-2.4\cos\right]\cdot\frac{\left(2\pi\left(x+10\right)\right)}{365}\cdot\frac{d}{dx}\left[\frac{\left(2\pi\left(x+10\right)\right)}{365}\right]$ $D'(x)=0\cdot-2.4-\sin\left[\frac{\left(2\pi\left(x+10\right)\right)}{365}\right]\cdot\frac{2\pi\left(2.4\right)}{365}$ $D'\left(x\right)=0-\left[-\sin\left[0.0172142063\left(x+10\right)\right]0.0413140952\right]$ $D'\left(x\right)=0.0413140952\sin\left[0.0172142063\left(x+10\right)\right]$ This means that there is $D'\left(x\right)=0.0413140952\sin\left[0.0172142063\left(x+10\right)\right]$ hours of daylight/ X-th day of the year. :::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) $D'\left(x\right)=0.0413140952\sin\left[0.0172142063\left(x+10\right)\right]$ $D'\left(200\right)=0.0413140952\sin\left[0.0172142063\left(200+10\right)\right]$ $D'\left(200\right)=-0.01883537$ Convert to minutes by multiplying by 60. $D'\left(200\right)=-1.1301222$ $minutes/day$ This means that the slope/rate of change at which the hours of daylight on July 19th are changing at $-1.1301222$ minutes per 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) $D'\left(x\right)=0.0413140952\sin\left[0.0172142063\left(x+10\right)\right]$ $0=0.0413140952\sin\left[0.0172142063\left(x+10\right)\right]$ $0=\sin\left[0.0172142063\left(x+10\right)\right]$ The longest day would be day 172 and at the slope of 0 it would reach its maximum length. :::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) In order to find the day of the year in which the number of hours of daylight is increasing most rapidly is to calculate the second derivative, this is the amount of increase in sunlight in minutes/day/day. This lets you compare the days to find the exact day in the year where the hours of daylight 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.