At 34 degrees north latitude, the longest day of the year has 14.273 hours of daylight, and the shortest day 9.726 hours. If the equinox falls on

Question

march 21(day 80), model the hours of the daylight by day. 

 

 

Sorry it all didn't fit and I really am stuck

Answer 1

Just like with your other question, you're going to need to use a sine wave to model this situation. Let's start with vertical shift and vertical stretch (amplitude).

 

Here, the sine wave's maximum is 14.273 and its minimum is 9.726. That means the sine wave's middle is at (14.273+9.726)/2 = 12. That's a vertical shift up of 12.

 

Since the sine wave reaches 14.273 - 12 = 2.273 units above its middle, its amplitude is 2.273. That's a vertical stretch of 2.273.

Now, let's work on the sine wave's horizontal shift and horizontal stretch (period).

Normally, the sine wave is at its middle point at x = 0, from which it increases to its maximum. Here, that happens at x = 80, the equinox, from which the daylight hours will increase to their maximum. That means the sine wave's middle point happened 80 late, which is a horizontal shift right of 80.

Since the sine wave models the daylight hours of a year, the sine wave's period is actually 365 days. That's how long it will take to get from one maximum to another. The stretch coefficient for the period, b, is:

(2π)/b = period
(2π)/b = 365
2π = 365b
(2π)/365 = b

That's a horizontal stretch of (2π)/365.


Let's take all that information and put it into the sine's equation. A normal sine wave has the equation:

y = sin(x)

A transformed sine wave has the equation:

y = asin(bx + c) + d

where a = vertical stretch (amplitude), b = horizontal stretch (period), c = horizontal shift (positive is left, negative is right), and d = vertical shift (positive is up, negative is down).

For us, a = 2.273, b = (2π)/365, c = -80 (shift right), and d = 12 (shift up). Plug those numbers in to get your equation:

y = 2.273sin(((2π)/365)x - 80) + 12