A variable star is one whose brightness alternately increases and decr

A variable star is one whose brightness alternately increases and decreases. For the most visible variable star, Delta Cephei, the time between periods of maximum brightness is 5.4 days, the average brightness (or magnitude) of the star is 4.0, and its brightness varies by $±0.35$ magnitude. Find a function that models the brightness of Delta Cephei as a function of time.
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chumants6g

Step 1
Start with a sine function, that has average value 0, magnitude $±1$, period $2\pi$
$f\left(t\right)=\mathrm{sin}$
To make the period 5.4 days, multiply t by $\frac{2\pi }{5.4}$
$f\left(t\right)=\mathrm{sin}\left(\frac{2\pi }{5.4}t\right)$
Now whenever t is a multiple of 5.4, it will be the sine of an integer multiple of $2\pi$
Step 2
The make the average value (the center horizontal line that the sine function moves back and forth over) 4.0, add to the function 4.0, It shifts the graph up by 4 units
$f\left(t\right)=4+\mathrm{sin}\left(\frac{2\pi }{5.4}t\right)$
Step 3
To change the magnitude to $±0.35$ multiply the sine by 0.35. Since the sine bounces back and forth between -1 and 1, this makes it so it bounces back and forth between -0.35 and 0.35
$f\left(t\right)=4+0.35\mathrm{sin}\left(\frac{2\pi }{5.4}t\right)$
Since they don't give any information about what the brightness should be at a specific time, we don't need to add or subtract anything from t to horizontally shift the graph.

Lakisha Archer
Step 1
Rajdeep, First, lets

nick1337

Step 1
NSK
There are many functions that alternately increase and decrease (called periodic functions), but the simplest of these is the sine function,
$y=A\mathrm{sin}\left(\frac{2\pi }{T}\right)t-\varphi \right)+b$
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.
$\varphi$ 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
$\varphi =0$
Putting it all in, you get
$y=0.35\mathrm{sin}\left(\left(\frac{2\pi }{5.4}\right)t\right)+4.4$