00:01
Here is temperature as a function of the day x.
00:06
The derivative, d, y, d, x is the rate at which the temperature changes.
00:13
And we want to know when the temperature increases the fastest, meaning it has the maximum rate of change.
00:20
So we want to find x such that d, y, d x is maximized.
00:38
And d, y, dx is maximized when its derivative, which is d squared y, d x squared, so the second derivative of y is equal to zero.
00:54
Okay, so let's take the second derivative of y.
00:58
So we'll start by taking the first derivative.
01:00
We have that d, y, dx is equal to, here we need to use the chain rule.
01:11
So we have 37 times cosine of what's inside the brackets there.
01:19
So 2 pi over 365 times x minus 103.
01:32
And now we need to take the derivative of what's inside the brackets here.
01:36
So we have 2 pi over 365.
01:45
Okay.
01:47
And then 22 is a constant.
01:49
Okay.
01:49
So we, the derivative of that is just zero.
01:54
Okay, so this is our first derivative.
01:56
And now we'll take the second derivative.
02:02
So again, we have the chain rule.
02:04
The derivative of cosine is negative sign.
02:07
So we have negative 37, sine of 2 pi over 365 times x minus 103, multiplied by 2 pi over 36.
02:23
And we take the derivative of what's inside the brackets.
02:27
And it's another factor of 2 pi over 36, so i'll just square this.
02:33
Okay, so remember that dyx is maximized when the second derivative is zero.
02:40
So i'm going to set the second derivative equal to zero.
02:44
So zero equals negative 37 sign of 2 pi over 365 times x minus 103 times 2 pi over 36 squared.
02:59
Okay, i'm just going to divide out the negative 37 and the 2 pi over 36 squared.
03:07
So we just have 0 is equal to sign of 2 pi over 365 times x minus 103.
03:22
And the sign is zero when the angle is a is an integer, sorry, multiple of pi.
03:32
So we have like n pi where n is an integer.
03:38
So it's equal to dot dot dot negative 2, negative 1, 0, 1, 2, dot, dot, dot.
03:46
So a positive or negative integer or 0 or 0.
03:50
So i'll set what's inside the brackets equal to n pi.
03:53
So we have n pi is equal to 2 pi over 365 times x minus 103.
04:01
So i will divide by pi on both sides.
04:06
So that goes away.
04:08
And then i'm solving for x here.
04:11
So i'll multiply by 365.
04:13
So 365n and divide by 2.
04:18
Okay, so this is equal to x minus 103.
04:22
So i'll add 103 to both sides to get that x is equal to 365n over 2 plus 103.
04:32
And we're looking at a year, so that's 365 days.
04:37
So that means that x is greater than or equal to zero, sorry, one.
04:43
I meant to say one, because we start counting the first day as just one, and less than or equal to 365.
04:53
Okay, so it must fall within that range.
04:57
So i'll use this to solve for n.
04:59
So we have one is less than or equal to 365.
05:04
N over 2 plus 103 and i'm going to separate these two inequalities and we also have 365 n over 2 plus 103 is less than or equal to 365 okay so let's subtract i'll start with this inequality here i'm going to subtract 103 from both sides so i get negative 102 is less than or equal to 365 n divided by 2.
05:36
I'm going to multiply by 2 on both sides to get negative 204 is less than or equal to 365 n and divided by 365 so negative 204 divided by 365 is less than or equal to n.
05:56
Okay and this is approximately negative 0 .56 okay is less than or equal to n but remember remember the n is an integer.
06:07
So let's look at where negative 0 .56 is.
06:15
So negative 0 .56 is between negative 1 and 0.
06:22
Like those are the two nearest integers.
06:25
And n is greater than or equal to.
06:29
So it's negative 0 .56, so it's on this side.
06:32
Okay, so that means that since this is the nearest integer that n could possibly be is 0.
06:43
So we have that 0 is less than equal to n.
06:48
Okay, so n is greater than or equal to 0.
06:51
Okay, so n is greater than it equal to zero...