Equation for moons illumination % per night

Equation for moons illumination % per night

Postby Guest » Tue May 19, 2020 7:22 pm

So..... favourite #2 daughter in grade 10 asked for math assignment help and i have no idea, but am now invested in finding out before she does... :D

She has to build a quadratic equation for the following data showing the moons illumination % per day in January.

My very limited skills fell apart as the data graphs to a sinusoidal wave and not a parabola, and i can not determine a second difference in the y data.

can it be done? show your working please.... apologies if this is a chump post..
Day of the month - % Illumination
1 30
2 39
3 48
4 58
5 67
6 76
7 84
8 90
9 96
10 99
11 100
12 98
13 94
14 87
15 78
16 68
17 56
18 45
19 34
20 24
21 16
22 9
23 4
24 1
25 0
26 1
27 4
28 9
29 15
30 22
31 31


salty
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Re: Equation for moons illumination % per night

Postby Guest » Mon May 25, 2020 2:00 pm

This specifically to be a quadratic equation?

Any quadratic equation can be written y= ax^2+ bx+ c. You need to determine the values or the there coefficients, a, b, and c. However, after looking in more detail at the data, I see that the vertex appears to be at x= 25, y= 0, I would be inclined to put it in "vertex" form, y= a(x- b)^2+ c. Immediately we have y= a(x- 25)^2 and only need to find a. I think I would do that by looking at the first value- when x= 0, y= 30 so 30= a(0- 25)^2= 625a so a= 30/625= 6/125. That would give y= 6(x- 25)^2/125.

Of course that probably won't give all the values in the list but should be close!

You could also use a "least squares approximation" that would give a different answer. (That would be a lot more work and there is no guarantee it would be any better.)
To do that, we, again, approximate the function by y= ax^2+ bx+ c. If we use "x" and "y" as the actual values given in the table, the "error", how much that calculation varies from the actual value, is y- ax^2- bx- c. The average of those errors will not necessarily tell us anything because some might be negative and cancel. Instead we average their squares. Calculate f(a, b, c)= [sum (y- ax^2+ bx+ c]/31] where the sum is over all those x and y values, and determine a, b, and c that minimize that sum.
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