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)

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(2) According to this model how many hours of daylight will there be on July 19 (day 200)?
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(2)
We can do this by plugging in the 200 into the equation above, using desmos I go to the next line and put x=200 representing the 200th day of the year as x and the y value will be the hous of daylight. When I do that the answer I get is 14.236 hours of daylight.

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(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) To solve for this we simply use the given data, the 14.236 hours of sunlight from the graph and the actual given information of 14 hours 16 minutes and 53 seconds of daylight which as a decimal is 14.28138888888889 but rounded 14.2814. To solve we just subtract our actual value from the graph value and we get:
$14.2814-14.236= 0.0454$ hours.
In minutes this is 2.724 minutes. So our final answer is that our model was 2.724 minutes off of the actual value of hours of daylight.

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(4) Compute $D'(x)$. Show all work.
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(4) So to compute $D'(x)$ we first start by simplifying the $\frac{2\pi}{365}$ and that comes out to be 0.01721 then next what we do is we put it into the formula:
$D'\left(x\right)=\left(\frac{d}{dx}\right)\left(12.1-2.4\cos\left(0.01721\left(x+10\right)\right)\right)$ then that comes out to be: $D'\left(x\right)=0\left(2.4\right)\left(\frac{d}{dx}\right)\cos\left(0.01721\left(x+10\right)\right)$ now we can apply the chain rule and that formula is: $\frac{df\left(u\right)}{dx}=\frac{df}{dx}+\frac{du}{dx}$ and then get D'(x)=-2.4(-sin u)(0.01721) and that being the 0.006121 to be the sin u and the u= 0.01721(x+10) so putting this all together we get this: $D'\left(x\right)=0.06121\sin\left(0.01721\left(x+10\right)\right)$ and that would be our final answer.

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(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) July 19th is the 200th day in the year and proving this we can plug it into the equation used in the previous question so it would be: $D'(x)=0.04131sin(0.1721(210))$ and that equals $0.041(-0.455)=-0.0188$. The next multiply by 60 to make the final answer into the minutes per day, and it shows that is -1.128 but we ignore that negative sign so it ends up being that July 19th is 1.128 minutes per day.

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(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) This would be like: $D'\left(x\right)=0$
$=\frac{4.8\pi}{365}\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)=0$ $=\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)=0$ $=\frac{2\pi\left(x+10\right)}{365}=n\pi$
$=\frac{365n\pi-20\pi}{2\pi}$ $=\frac{365n-20}{2}$ $
$n=1\ and\ x=172.5$
So this being said June 21st would be the 172.5 day

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(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) How you would be able to do this is by looking at the increasing slopes and then figure out which one would be the steapist or finding the biggest tangent of the increasing sections.

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