00:01
From this equation of a sine function, we can determine what the parameters of the sign function are, figure out what the period is, the maximum and minima of this function, and other special values.
00:15
So first, we can relate our equation to the general sign function, which is sum a times the sign of b times t minus c, all plus d.
00:32
Where a is the amplitude, b is related to the period, c is the phase shift, and d is the vertical shift.
00:40
So from our equation, we know that a is the number out front, which is 2 .8.
00:48
B is the number in front of the time minus the phase shift, which in this case is 2 pi over 365.
00:58
C is the number subtracted from time, which is 81, and d is, is the last number added on after the sign, which will be 12.
01:12
And from this, we can figure out the period if we know b, since the period t is just 2 pi over the parameter b.
01:23
And we know b to be 2 pi over 365.
01:26
So we can plug that in.
01:28
And note that in the numerator, 2 pi is the same thing as 2 pi divided by 1.
01:33
So it's be 2 pi over 1 divided by 2 pi over 365.
01:39
And if you have two fractions being divided by each other, this is equivalent to the first fraction multiplied by the reciprocal of the bottom fraction, which would be 365 over 2 pi.
01:52
The factors of 2 pi are going to cancel each other out, so we'll just be left with a period of 365.
02:00
Next, we can look at the maxima and minima of this function, since we know that for any sine function, the sine function, the sine of theta, will also, always be less than 1 and greater than negative 1, as in the smallest value that sign can have is negative 1, and the largest value that sign can have is 1.
02:22
So if we use that in the equation that we have, the 2 .8 sign of 2 pi over 365, t minus 81 plus 12.
02:40
If the sign, since the sign in this equation, the sign is the only thing that will be changing with time.
02:48
If the sign is a minimum of negative 1, then this will be negative 2 .8 plus 12, which is just equal to 9 .2.
03:05
And if the sign is 1, then this will become 2 .8 plus 12, which is 14 .8.
03:17
So these are the maxima and minima of the sign function.
03:21
We can ask ourselves at what time do these values occur? and we can figure this out because we know that the sign of pi over 2 is equal to 1.
03:39
Therefore, if the sign at pi over 2 is equal to 1, then in this equation, the maximum has to occur when the inside of the sign function is equal to pi over 2.
03:53
So therefore, 2 pi over 365 times t minus 81 is equal to pi over 2 when the entire sign equation is equal to 14 .8.
04:09
So here we can solve for t by multiplying by 365 over 2 pi on both sides to get that t minus 81 is equal to 365 over 2 pi.
04:27
Times pi over two.
04:30
And now we can add 81 on both sides to get t alone...