# Quick Sinusoidal Problem Help

• May 22nd 2010, 01:19 PM
falconskid007
Quick Sinusoidal Problem Help
Hey guys can you help me with this one question please and maybe explain the process.

The daylight time (time between sunrise and sunset) varies sinusoidally throughout a year. The longest day of the year is the 170th day with a daylight time of 15 hours. The Shortest daylight time occurs on the 353rd day of the year with a daylight time of 6 hours.

It's a two part question:

What is the period and amplitude: I figured A=4.5 and P=365

And then it asks to create a graph of this model and generate a specific equation with time t=0 being the first day of the year.

So could you guys explain how to graph this equation? She also said there is a vertical and horizontal shift and that C/B=170.

I was sick the day my teacher was explaining problems like these and am totally lost.
• May 22nd 2010, 02:04 PM
pickslides
Where T is total length of the day and D is day of the year you have

$\displaystyle T= A\sin b\left(D-c \right) +d$

You need to find $\displaystyle a,b,c,d$

$\displaystyle T= 4.5\sin \frac{2\pi}{365}\left(D-c \right) +10.5$

Now to find $\displaystyle c$ use the fact that

Quote:

Originally Posted by falconskid007
The longest day of the year is the 170th day with a daylight time of 15 hours. The Shortest daylight time occurs on the 353rd day of the year with a daylight time of 6 hours.

What do you get?
• May 22nd 2010, 02:33 PM
skeeter
$\displaystyle y = 4.5 \sin\left[\frac{2\pi}{365}(t - c)\right] + 10.5$

$\displaystyle y_{max} = 15$ occurs at $\displaystyle t = 170$, and when $\displaystyle \sin\left[\frac{2\pi}{365}(170 - c)\right] = 1$

$\displaystyle \frac{2\pi}{365}(170 - c) = \frac{\pi}{2}$

$\displaystyle 170 - c = \frac{365}{4}$

$\displaystyle c = 170 - \frac{365}{4} = \frac{315}{4}$

$\displaystyle y = 4.5 \sin\left[\frac{2\pi}{365}\left(t - \frac{315}{4}\right)\right] + 10.5$

graph attached