Chapter 5.6, Problem 40E

Variable Stars Astronomers believe that the radius of a variable star increases and decreases with the brightness of the star. The variable star Delta Cephei (Example 4) has an average radius of 20 million miles and changes by a maximum of 1.5 million miles from this average during a single pulsation. Find an equation that describes the radius of this star as a function of time.

To determine

To find: The equation that describe the radius of star as a function of time.

Answer to Problem 40E

The equation is $y=1.5\cos \left(\frac{2\pi t}{T}\right)+20$.

Explanation of Solution

Given:

The variable star has period $T$, average radius of star is $20\text{million}\text{miles}$ magnitudes and the maximum variation from the average is $1.5\text{million}\text{miles}$ magnitude.

Calculations:

Let’s find a function in the form,

$y=a\cos \left(\omega \left(t-c\right)\right)+b$

The amplitude is the maximum variation from the average radius, so the amplitude is $a=1.5\text{magnitutde}$ and the period is T days.

So the angular velocity,

$\omega =\frac{2\pi }{T}$.

Since the radius varies from an average value of $3.8$ magnitude, the graph is shifted upward by $b=20$.

If we take $t=0$ to be a time when the star is at maximum radius, there is no horizontal shift, so $c=0$ because a cosine curve achieves its maximum at $T=0$.

Thus, the function we want is

$y=1.5\cos \left(\frac{2\pi t}{T}\right)+20$.

So, the equation is $y=1.5\cos \left(\frac{2\pi t}{T}\right)+20$.