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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
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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.
:::

(1)


:::info
(2) According to this model how many hours of daylight will there be on July 19 (day 200)?
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(2)$$D(200)=12.1-2.4cos \frac{2pi(x+10)}{365}$$
$$D(200)=12.1-2.4cos\frac{2pi(200+10)}{365}$$
$$D(200)=14.24


:::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)For July the total number of dalight is 14.8 hrs. So based on my calculations I was off by 0.56 minutes which isn't that far.

:::info
(4) Compute $D'(x)$. Show all work.
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(4)$$D'(x)=2.4sin(\frac{2pi(x+10)}{365})$$
$$D'(x)=2.4sin(\frac{2pi(x+10)}{365})*\frac{d}{dx}[\frac{2pi(x+10)}{365}]$$

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

$$D'(x)=2.4sin(\frac{2pi(x+10)}{365})*(\frac{d}{dx}[\frac{2pi*365-(2pix+20pi)*0}{365^2}])$$

$$D'(x)=2.4sin(\frac{2pi(x+10)}{365})*(\frac{730pi}{365^2})$$

:::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.
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(5)$$D'(x)=2.4sin(\frac{2pi(x+10)}{365})*(\frac{730pi}{365^2})$$

$$D'(200)=2.4sin(\frac{2pi(200+10)}{365})*(\frac{730pi}{365^2})$$
$$D'(200)=-0.01884 or -0.02 \frac{hrs.}{day}$$
$$\frac{-0.01884 hrs.}{1 day}*\frac{60 mins.}{1 hr.}=\frac{-1.13012235147 mins}{1 day}$$

For July 19 the rate number of hours changed is about -1.13012235147 minutes per day, which is how many minutes are decreasing with 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.
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(6)$$D'(x)=2.4sin(\frac{2pi(x+10)}{365})*(\frac{730pi}{365^2})$$

$$0=2.4sin(\frac{2pi(x+10)}{365})*(\frac{730pi}{365^2})$$

$$0=2.4sin(\frac{2pix+20pi}{365})$$

$$0=sin(\frac{2pix+20pi}{365})$$

$$sin^-1(0)=(\frac{2pix+20pi}{365})$$

$$365sin^-1(0)=2pix+20pi$$

$$365sin^-1(0)-20pi=2pix$$

$$\frac{365sin^-1(0)-20pi}{2pi}=x$$
$$-10=x$$
 
The estimate from this, -10, does not match the number given by the graph for D'(x) since it equals 172.5.

:::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.
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(7)The first thing I would suggest is google. But going based off of the graph you will get a more accurate answer for the number of hours that daylight is increasing by. The steepest positiveslope of the function is where you would need to look.


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