The sun always illuminates half of the moon’s surface, except during a lunar eclipse. The illuminated portion of the moon visible from Earth varies as it revolves around Earth resulting in the phases of the moon. The period from a full moon to a new moon and back to a full moon is called a synodic month and is 29 days, 12 hours, and 44.05 minutes long. Write a sine function that models the fraction of the moon’s surface which is seen to be illuminated during a synodic month as a function of the number of days, d, after a full moon. [Note: full moon equals 1/2 illuminated.]

I'm lost.

Question

The sun always illuminates half of the moon’s surface, except during a lunar eclipse. The illuminated portion of the moon visible from Earth varies as it revolves around Earth resulting in the phases of the moon. The period from a full moon to a new moon and back to a full moon is called a synodic month and is 29 days, 12 hours, and 44.05 minutes long. Write a sine function that models the fraction of the moon’s surface which is seen to be illuminated during a synodic month as a function of the number of days, d, after a full moon. [Note: full moon equals 1/2 illuminated.]

I'm lost.

whpalmer4 · Answer

Okay, you need to make a sine function with period 29 days, 12 hours, and 44.05 minutes (you'll need to express that as a decimal number of days, which will be slightly greater than 29.5).  Then you'll want to make sure the phase of your sine function is correct so that at t = 0 and at t = 29 days 12 hours 44.05 minutes the value of your sine function reaches its maximum, which should be 1/2 (at full moon, half the surface is illuminated).

So, you'll have something like
$$F = \frac{1}{2}\sin (kt + p)$$ where k is chosen to give you kt = the period of an ordinary sine function when t = 29 days 12 hours 44.05 minutes, and p is whatever fudge factor you need to add to make the sin function = 1 at t = 0.

I fear I didn't help with that explanation :-)

anonymous · Answer

Okai I got the first part, but the second and third part you just completely lost me

whpalmer4 · Answer

Well, let's break it down.  What is the period of the sine function $y=\sin t$?

anonymous · Answer

29=sin1/2?

whpalmer4 · Answer

no, ignore the moon part of the problem, we'll just do straight trig functions.  what is the period of sine function $y = \sin \theta$?

whpalmer4 · Answer

If we start at $\theta = 0$, what is the value of $\sin\theta$ there?

whpalmer4 · Answer

What is $\sin(0)$?

anonymous · Answer

sin(29) = .4848096202?

anonymous · Answer

and sin 0 is 0

whpalmer4 · Answer

Please, forget the moon problem, just answer the questions I ask, thanks.

Okay, $sin(0) = 0$.  What is $\sin(\pi/2)$?

anonymous · Answer

0.274121336

anonymous · Answer

And sorry (._. )

whpalmer4 · Answer

mmm...no.  Are you computing in degrees or radians?  Radians is what we want here.
$\sin(\pi/2) = 1$ — that's pointing straight up to the top of the unit circle.|dw:1360461986468:dw|

whpalmer4 · Answer

right?  I drew in the radius at $\theta = \pi/2$

anonymous · Answer

oh my bad I have it set on degrees

anonymous · Answer

okai I got 1 now

anonymous · Answer

Oh I'm suppose to be using the unit circle in this one?

whpalmer4 · Answer

We're just exploring the properties of the sine function, so we can figure out how to bend it to our will :-)

anonymous · Answer

So how would I set this up?

whpalmer4 · Answer

So, the period of the sine function (just like the cosine function) is $2\pi$ radians.  Every $2\pi$ radians the radius makes a complete trip around the unit circle:|dw:1360462848057:dw|