# Book seems to be assuming too much

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• Apr 5th 2013, 02:12 AM
Paze
Book seems to be assuming too much
Hi. I have a problem:

The length of the days of Stockholm change according to the sine function:

$\displaystyle \frac{49}{4}+\frac{25}{4}\cdot sin\left(\frac{2\pi(x-82)}{365}\right)$

Where x = days, y is hours and x = 1 means January the first.

a) Determine the length of the longest and the shortest day.

b) When is day and night equally long?

I have learned that the highest value of these sine functions is 1 and the lowest is -1. I understand that, given all values for sine. However, here, the book seems to be assuming that x can take on any value and thus the highest and lowest value of the sine function is 1 and -1 in this example...But how can the book know whether the calculation inside the sine function will ever reach the values required to get the output 1 and -1 respectively?
• Apr 5th 2013, 02:55 AM
Gusbob
Re: Book seems to be assuming too much
$\displaystyle \sin(ax+b)$ has period $\displaystyle T=\frac{2\pi}{a}$. In your case, $\displaystyle a=\frac{2\pi}{365}$ so the period is exactly $\displaystyle 365$, how nice.
• Apr 5th 2013, 03:43 AM
Ruun
Re: Book seems to be assuming too much
OP: Does your $\displaystyle x$ actually means $\displaystyle t$? Look that you have a function $\displaystyle L(t)$ that tells you the length of the day for a given day in some particular units.

a) you need to know when a function does have a maximum or minimum, this is, when is its derivative with respect to $\displaystyle t$ equals zero.

b) as $\displaystyle L(t)$ is the length of the day, when that number is $\displaystyle 1/2$ of its maximum value then the day and the night will be equally long
• Apr 5th 2013, 11:12 AM
Paze
Re: Book seems to be assuming too much
Quote:

Originally Posted by Ruun
OP: Does your $\displaystyle x$ actually means $\displaystyle t$? Look that you have a function $\displaystyle L(t)$ that tells you the length of the day for a given day in some particular units.

a) you need to know when a function does have a maximum or minimum, this is, when is its derivative with respect to $\displaystyle t$ equals zero.

b) as $\displaystyle L(t)$ is the length of the day, when that number is $\displaystyle 1/2$ of its maximum value then the day and the night will be equally long

My x is indeed t. Sorry for the confusion.

Yes, but we haven't learned derivatives yet in school. I've learned differential calculus on my own and I can derive the function and find it's maxima and minima, but the method we learn to do it by is to simply say assume that the maximum and minum for the sine function is 1 and -1. Is the book making us assume numbers that have not been proven?
• Apr 5th 2013, 12:26 PM
Paze
Re: Book seems to be assuming too much
It seems that when I take the derivative of every sine function in my book and find the maxima and minima, I get 1 and -1 respectively.

How do I prove that for any x, the min and maxima of the sine function will be 1 and -1?
• Apr 5th 2013, 12:47 PM
emakarov
Re: Book seems to be assuming too much
Quote:

Originally Posted by Paze
I have learned that the highest value of these sine functions is 1 and the lowest is -1. I understand that, given all values for sine. However, here, the book seems to be assuming that x can take on any value and thus the highest and lowest value of the sine function is 1 and -1 in this example...But how can the book know whether the calculation inside the sine function will ever reach the values required to get the output 1 and -1 respectively?

There are values of x that cause sine to reach its maximum and minimum because the function inside sine, i.e., $\displaystyle \frac{2\pi(x-82)}{365}$, is a non-constant linear function, and such functions are bijections. In particular, they are surjections, which means that they assume all real values, in particular, $\displaystyle \pi/2+2\pi k$, where sine reaches its maximum, and $\displaystyle -\pi/2+2\pi k$, where sine reaches its minimum. Simply speaking, you can solve the equation $\displaystyle 2\pi(x-82)/365=\pi/2$ to find an x where sine equals 1.

Hint: In math, it is customary to write function names (as opposed to variable names) in upright font instead of italics. Thus, in sin(x), "sin" is upright because it is a name of a concrete function. In contrast, in f(x), "f" is italic because it is a variable that may take different values such as "sin". For this reason, LaTeX has commands like \sin and \cos for many standard functions, which typeset them in upright font and produce correct spacing between the function and its argument.
• Apr 5th 2013, 12:51 PM
Paze
Re: Book seems to be assuming too much
Quote:

Originally Posted by emakarov
There are values of x that cause sine to reach its maximum and minimum because the function inside sine, i.e., $\displaystyle \frac{2\pi(x-82)}{365}$, is a non-constant linear function, and such functions are bijections. In particular, they are surjections, which means that they assume all real values, in particular, $\displaystyle \pi/2+2\pi k$, where sine reaches its maximum, and $\displaystyle -\pi/2+2\pi k$, where sine reaches its minimum. Simply speaking, you can solve the equation $\displaystyle 2\pi(x-82)/365=\pi/2$ to find an x where sine equals 1.

Hint: In math, it is customary to write function names (as opposed to variable names) in upright font instead of italics. Thus, in sin(x), "sin" is upright because it is a name of a concrete function. In contrast, in f(x), "f" is italic because it is a variable that may take different values such as "sin". For this reason, LaTeX has commands like \sin and \cos for many standard functions, which typeset them in upright font and produce correct spacing between the function and its argument.

Thank you for a well descriptive answer. However, my x can only take on values from 1 - 365. How is that consistent with your explanation that the function can assume all real values?

For example the function $\displaystyle \sin\left(\frac{x}{2}\right), 0<x<1$ this function does not yield the min and max of 1 and -1 because I limited the values that x can take on.
• Apr 5th 2013, 01:07 PM
emakarov
Re: Book seems to be assuming too much
Quote:

Originally Posted by Paze
However, my x can only take on values from 1 - 365. How is that consistent with your explanation that the function can assume all real values?

Good point. But, as post #2 notes, the period of the whole function is 365, so on any segment [x, x + 365] the function assumes all its values. If we denote the argument of sine as g(x), you can also find the image of g(x) when 1 <= x <= 365 (or should it be 0 <= x <= 365?). Since g(x) is linear, the image is [g(1), g(365)] (it would be [g(365), g(1)] if the slope were negative). Check if the image contains any of the points $\displaystyle \pi/2+2\pi k$ and $\displaystyle -\pi/2+2\pi k$.
• Apr 5th 2013, 03:51 PM
Paze
Re: Book seems to be assuming too much
Quote:

Originally Posted by emakarov
Good point. But, as post #2 notes, the period of the whole function is 365, so on any segment [x, x + 365] the function assumes all its values. If we denote the argument of sine as g(x), you can also find the image of g(x) when 1 <= x <= 365 (or should it be 0 <= x <= 365?). Since g(x) is linear, the image is [g(1), g(365)] (it would be [g(365), g(1)] if the slope were negative). Check if the image contains any of the points $\displaystyle \pi/2+2\pi k$ and $\displaystyle -\pi/2+2\pi k$.

So that I understand this correctly. If the period had been 366 then, with 365 days, I would not reach a whole period, and thus my minima and maxima would not be consistent with 1 and -1, correct?
• Apr 5th 2013, 06:16 PM
semouey161
Re: Book seems to be assuming too much
The sine function has the amplitude that you describe. So $\displaystyle \sin \frac{2 \pi (x-82)}{365}$ by itself does have an amplitude of 1, but a period of 365.

Now, the important part is the $\displaystyle +\frac{49}{4}$, which is a vertical shift. This alters what the amplitude is for the entire funtion $\displaystyle \frac{25}{4} \sin \frac{2 \pi (x-82)}{365}+\frac{49}{4}$. Here's a graph of it:

Attachment 27823

The graph of the function makes it easier to see what's going on with the numbers. Also, I looked up the sunrise/sunset tables of Stockholm and this function produces accurate results (I have that kind of time tonight). So, no, the book isn't asking you to assume too much; just interpret the function correctly.
• Apr 5th 2013, 07:05 PM
Paze
Re: Book seems to be assuming too much
Quote:

Originally Posted by semouey161
The sine function has the amplitude that you describe. So $\displaystyle \sin \frac{2 \pi (x-82)}{365}$ by itself does have an amplitude of 1, but a period of 365.

Now, the important part is the $\displaystyle +\frac{49}{4}$, which is a vertical shift. This alters what the amplitude is for the entire funtion $\displaystyle \frac{25}{4} \sin \frac{2 \pi (x-82)}{365}+\frac{49}{4}$. Here's a graph of it:

Attachment 27823

The graph of the function makes it easier to see what's going on with the numbers. Also, I looked up the sunrise/sunset tables of Stockholm and this function produces accurate results (I have that kind of time tonight). So, no, the book isn't asking you to assume too much; just interpret the function correctly.

I realize the vertical shift. The thing is the amplitude (thank you for the English word). I have a limited number of values to assign to my x (namely 1-365) and that confuses me on how the sine function can take on 'any value' or how I can be sure that it reaches the minima and maxima of 1 and -1.

I realize how a function like $\displaystyle \sin\left(\frac{x}{2}\right)$ can take on any value and thus the maxima and minima become 1 and -1. However if I limit the values which can be assigned to x, like so: $\displaystyle \sin\left(\frac{x}{2}\right), 0<x<1$ then the maxima and minima are not 1 and -1. The thing I am trying to wrap my head around is why we are so sure that it takes on 1 and -1 in my example when we limit x to 1-365.
• Apr 5th 2013, 08:27 PM
semouey161
Re: Book seems to be assuming too much
Quote:

I have a limited number of values to assign to my x (namely 1-365) and that confuses me on how the sine function can take on 'any value' or how I can be sure that it reaches the minima and maxima of 1 and -1.
What are you talking about specifically? Are you talking about just $\displaystyle \sin{(x)}$ or are you talking about $\displaystyle \frac{25}{4}\sin\frac{2\pi(x-82)}{365}+\frac{49}{4}$ ? If you are referring to $\displaystyle \sin{(x)}$ , this function has a domain of all real numbers. This means I can take any real number to be the x value, plug it in, and I will get some number y that is between -1 and 1, inclusive. (That would be the range).

If we take a look at $\displaystyle \frac{2\pi(x-82)}{365}$ we can see that this expression results in a real number. This is so because the domain of this expression is all real numbers. Therefore, $\displaystyle \sin\frac{2\pi(x-82)}{365}$ is defined for all values of x that are real numbers. If that is true, then by the properties of sine functions and function transformations, we can determine that although the function is defined for all x such that x is a real number, it's amplitude is still 1 but it's period changed.

I can restrict your domain and tell you that $\displaystyle \frac{25}{4}\sin\frac{2\pi(x-82)}{365}+\frac{49}{4}$ is only defined on the interval $\displaystyle 1 \leq x \leq 365$ . Still, 1 is a real number, and so is 365 so I'm going to get real number results. The sine portion of the function can take on any number from its domain, but we aren't talking about any number; we're talking in the context of the problem, so we have to focus on a restricted domain, namely $\displaystyle 1 \leq x \leq 365$ . Does this restricted domain change what the sine function does? Of course not. Sine still does it's thing. I'm just limiting you on your choices of x values to plug in.

Quote:

The thing I am trying to wrap my head around is why we are so sure that it takes on 1 and -1 in my example when we limit x to 1-365.
To sum up, you have something*sin(stuff) + something, but the sin(stuff) still does its sine function thing, with a few alterations. Is the amplitude affected in just the sin(stuff) portion? No. Is its period affected? Yes. Does it still take on 1 and -1 if we limit the interval on which it is defined? No ... use a calculator and you'll see that we are at about -0.98 when x=1 and when x=365 . But careful! That's just the sin(stuff) portion!

Here's the graph of $\displaystyle \sin \frac{2\pi(x-82)}{365}$ on the interval $\displaystyle 1 \leq x \leq 365$ :
Attachment 27825

Quote:

However if I limit the values which can be assigned to x, like so: $\displaystyle \sin\left(\frac{x}{2}\right), 0<x<1$ then the maxima and minima are not 1 and -1.
Correct. To be clear why, let's look at $\displaystyle \sin x$ by itself. Again, the amplitude is 1. Oh, wait a second, I don't want you looking at the whole entire graph, so I'm going to restrict your domain to $\displaystyle \pi \leq x \leq 2 \pi$ . What happened? We're just looking at a portion of the graph of the function since we are only plotting points on the given interval. Does $\displaystyle \sin x$ on the interval $\displaystyle \pi \leq x \leq 2 \pi$ ever reach its normal maximum point of 1? No, of course not. That's because we're only looking at a specific set of x values defined by that interval.

I hope that helps. (Nod)
• Apr 5th 2013, 09:25 PM
Paze
Re: Book seems to be assuming too much
Quote:

Originally Posted by semouey161
What are you talking about specifically? Are you talking about just $\displaystyle \sin{(x)}$ or are you talking about $\displaystyle \frac{25}{4}\sin\frac{2\pi(x-82)}{365}+\frac{49}{4}$ ? If you are referring to $\displaystyle \sin{(x)}$ , this function has a domain of all real numbers. This means I can take any real number to be the x value, plug it in, and I will get some number y that is between -1 and 1, inclusive. (That would be the range).

If we take a look at $\displaystyle \frac{2\pi(x-82)}{365}$ we can see that this expression results in a real number. This is so because the domain of this expression is all real numbers. Therefore, $\displaystyle \sin\frac{2\pi(x-82)}{365}$ is defined for all values of x that are real numbers. If that is true, then by the properties of sine functions and function transformations, we can determine that although the function is defined for all x such that x is a real number, it's amplitude is still 1 but it's period changed.

I can restrict your domain and tell you that $\displaystyle \frac{25}{4}\sin\frac{2\pi(x-82)}{365}+\frac{49}{4}$ is only defined on the interval $\displaystyle 1 \leq x \leq 365$ . Still, 1 is a real number, and so is 365 so I'm going to get real number results. The sine portion of the function can take on any number from its domain, but we aren't talking about any number; we're talking in the context of the problem, so we have to focus on a restricted domain, namely $\displaystyle 1 \leq x \leq 365$ . Does this restricted domain change what the sine function does? Of course not. Sine still does it's thing. I'm just limiting you on your choices of x values to plug in.

To sum up, you have something*sin(stuff) + something, but the sin(stuff) still does its sine function thing, with a few alterations. Is the amplitude affected in just the sin(stuff) portion? No. Is its period affected? Yes. Does it still take on 1 and -1 if we limit the interval on which it is defined? No ... use a calculator and you'll see that we are at about -0.98 when x=1 and when x=365 . But careful! That's just the sin(stuff) portion!

Here's the graph of $\displaystyle \sin \frac{2\pi(x-82)}{365}$ on the interval $\displaystyle 1 \leq x \leq 365$ :
Attachment 27825

Correct. To be clear why, let's look at $\displaystyle \sin x$ by itself. Again, the amplitude is 1. Oh, wait a second, I don't want you looking at the whole entire graph, so I'm going to restrict your domain to $\displaystyle \pi \leq x \leq 2 \pi$ . What happened? We're just looking at a portion of the graph of the function since we are only plotting points on the given interval. Does $\displaystyle \sin x$ on the interval $\displaystyle \pi \leq x \leq 2 \pi$ ever reach its normal maximum point of 1? No, of course not. That's because we're only looking at a specific set of x values defined by that interval.

I hope that helps. (Nod)

Thank you, your answer was very helpful and it confirmed many of my suspicions. However I would like to point to:

Quote:

Does it still take on 1 and -1 if we limit the interval on which it is defined? No ... use a calculator and you'll see that we are at about -0.98 when x=1 and when x=365 . But careful! That's just the sin(stuff) portion!
Are you saying that if we limit the interval to 1-365, then the function's minima is at about -0.98?
• Apr 5th 2013, 09:28 PM
Paze
Re: Book seems to be assuming too much
And just on a side-note. Is it because my function is designed in such a way that by limiting it to this extent, it just happens to converge on the maxima 1 and minima -1?

If so...If I am to identify another likewise function, do I need to derive it to get a correct minima/maxima result?
• Apr 5th 2013, 11:27 PM
semouey161
Re: Book seems to be assuming too much
Quote:

Originally Posted by Paze
However I would like to point to:

Are you saying that if we limit the interval to 1-365, then the function's minima is at about -0.98?

Here are the results I got:
Attachment 27826

Looks like the result on the left would be the minimum. Again, these results are only for the expression

$\displaystyle \sin\frac{2\pi(x-82)}{365}$ evaluated at x=1 and at x=365.

This means that your other expression $\displaystyle \frac{25}{4}\sin\frac{2\pi(x-82)}{365}+\frac{49}{4}$ has different values at x =1 and x =365 . Go ahead, plug in x = 1 and x = 365 into this one, and post the results. Also, I think I showed you the graph earlier for this expression on the interval in question. If I didn't, let me know and I'll get a picture of it here.
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