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A variable star is one whose brightness alternately increases and decreases.

 A variable star is one whose brightness alternately increases and decreases. For one such star, the time between periods of maximum brightness is 5.4 days, the average brightness (or magnitude) of the star is 4.4, and its brightness varies by ±0.35 magnitude. Find a function that models the brightness of the star as a function of time (in days), t. (Assume that at t = 0 the brightness of the star is 4.4 and that it is increasing.)
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2 Answers

Rajdeep, First, let's create a table to find when the star will be at its maximum, minimum and its average, and what direction the magnitude will be moving at that time.
 
t = 0, M = 4.4 +
t = 1.35, M = 4.75 -
t = 2.7, M = 4.4 -
t = 4.05, M = 4.05 +
t = 5.4, M = 4.4 +
t = 6.75, M = 4.75 -
 
The t=0, entry is given by the parenthetical.  The period is given as 5.4 days (the time between the start of each cycle), so the star will reach its maximum at 1.35 days, return to average at 2.7 days, reach its minimum at 4.05 days, return to average at the period interval, 5.4 days, and then reach maximum again in 6.75 days.  The difference between 6.75 and 1.35 days (the time between maximums) is 5.4 days, as required.
 
There are many functions that might be chosen, I would choose a sine or cosine function.  The choice between sine and cosine is usually best done on whether you want the initial value of the function to be 0 (sine) or 1 (cosine).  Here, we start at the average value, so we are better off starting with sine.  The amplitude of the sine function needs to be the change from the average value (positive or negative) or .35, since the sine of 0, pi, 2pi, 3pi ... is 0, we need to move the center line of the sine function up by 4.4, the average value of the magnitude.  Finally, we need multiply the time value in order to adjust the time into the appropriate radian measure for the 5.4 day period.  So we want
 
0 days -> 0 radians
1.35 days -> pi/2 radians
2.7 days -> pi radians
4.05 days -> 3pi/2 radians
5.4 days -> 2pi radians
 
So we need to divide each time value by half the period (2.7) and multiply it by pi or pi/2.7.
 
Thus, the best sinusoidal equation is
M = .35 sin (t • pi/2.7) + 4.4
 
This assumes that a magnitude of 4.75 is brighter than 4.4, which would be brighter than 4.05.  If magnitude works the other way, then all you need to do is change the sign or the equation, i.e.
M = -.35 sin (t • pi/2.7) + 4.4
 
There are many functions that alternately increase and decrease (called periodic functions), but the simplest of these is the sine function,
y = A sin( (2pi/T) t - φ ) + b.
 
Here y and t are the variables: magnitude and time in your case.
A is called the amplitude; it tells us the maximum variation of y. In your case, A = 0.35.
T is called the period; it tells us the duration of one cycle. In your case, T = 5.4 days.
b is the off-set in the y-direction, or average value of y. In your case, b = 4.4.
φ is called the phase shift; it tells us by how much the initial value of y is off-set from the average value, y=b. In your case, φ=0.
Putting it all in, you get
 
y = 0.35 sin( (2pi/5.4) t ) + 4.4.
 
 

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