Applications of Trigonometric Functions (Graphing)
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skeeter
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The problem restricted the graph to [MATH]\left[0,\frac{3\pi}{2}\right][/MATH], which is 1.5 times the function's actual period.
If you were asked to graph [MATH]y = \sin{x}[/MATH] over the interval [MATH][0,\pi][/MATH], does that mean its period has changed?
Yes but if one period is pi, then the graph would look like this:
(refer to attachment)
Attachments
Are you imagining that a graph of a trigonometric function is restricted to just one period?
It is frequently the case that the graph of a strictly periodic function is displayed for just one period because everything before and after that is repetition. But if you have a practical problem that specifies a particular domain, THAT domain is what your graph is supposed to cover.
In any case, the graph of your function does not look like your attachment.
Are you mixing up symmetrical around the midpoint of the period with periodic?
It is frequently the case that the graph of a strictly periodic function is displayed for just one period because everything before and after that is repetition. But if you have a practical problem that specifies a particular domain, THAT domain is what your graph is supposed to cover.
In any case, the graph of your function does not look like your attachment.
Are you mixing up symmetrical around the midpoint of the period with periodic?
No, you still do not get it.
The following has nothing to do with trig functions specifically.
Given that f(x) is ANY differentiable function on an OPEN interval containing a.
[MATH]\text {If } f(a) \text { is a local extremum of } f(x), \text { then } f'(a) = 0.[/MATH]
[MATH]\text {If } f'(a) = 0, \text { then } f(a) \text { may be or not be a local extremum of } f(x).[/MATH]
We set the first derivative equal to zero and solve to find where we must TEST for local extrema on an open interval. A first derivative of zero does not guarantee a local extremum.
Do you get this? It really is basic stuff.
On a closed interval, it may be true that either or both of the endpoints of the interval are also local extrema.
So if we are talking about a closed interval
[MATH]\text {If } f(a) \text { is a local extremum of } f(x), \text { then } f'(a) = 0 \text { or } a \text { is an endpoint.}[/MATH]
[MATH]\text {If } a \text { is an endpoint or } f'(a) = 0, \text { then } f(a) \text { may be or not be a local extremum of } f(x).[/MATH]
We set the first derivative equal to zero and solve to find where we must TEST for local extrema on a closed interval. But we must also test the endpoints
Noooo I just meant that's what the FIRST period would look like...
Please refer to my attachment:
Attachments
If f′(a)=0, then f(a) may be or not be a local extremum of f(x).If f′(a)=0, then f(a) may be or not be a local extremum of f(x).
^ I think I understand this - that just because the derivative is 0, it does not necessarily mean that it will give a maximum or minimum value. It COULD though; it could also mean an inflection point. Is my understanding correct?
We set the first derivative equal to zero and solve to find where we must TEST for local extrema on an open (/closed) interval. A first derivative of zero does not guarantee a local extremum.
^ This part does not click with me - probably because we have not touched on first derivative content since last year...
YES
Ignoring endpoints for the moment, every local extremum will have a first derivative equal to zero. Therefore, if we find every point where the first derivative equals zero, we have found every point (other than an endpoint) that MAY BE A LOCAL EXTREMUM. Because some or all of those may NOT be extrema, we must next test each one to see whether or not that one is an extremum. In other words, finding extrema is at least a two stage process. Does the first derivative equal zero? if not, we are done unless it is an endpoint of a closed interval.If yes, we must inquire further. And if we are talking a closed interval, we must in addition consider endpoints.
If you were introduced to first derivatives last year, all this should have been explained back then. As it is, I fear you were given a misleadingly simplified introduction.
Hmmm, I see. We only spent one lesson on first derivatives, and it was only regarding how to use the formula to differentiate by first principles. However, we did cover what you said, with finding where the derivative is 0 and then checking to see if it is a max/min/inflection by drawing up a little table to see what direction the curve is going.
Also, could you please help me to understand about the period being pi issue (post #27)? I am still confused...
Does the book says the period is not [MATH]\pi[/MATH]?
No, it says that the period is pi and I also calculated the period to be pi, but if you look at the graph, the period if actually 3pi/4 (refer to attachment)
The period starts from a certain position, and it starts again from the same position following exactly the same path as before.
Your attachment, yeah it starts again from the same position, but not following the same path as before, so it is not the period.
Your attachment, yeah it starts again from the same position, but not following the same path as before, so it is not the period.
Sorry, I don't understand... Shouldn't the period start from 0 and end at pi for this sine graph? It does not do that though...
skeeter
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it's not a "sine" graph, it's a [MATH]\sin(2x)+\sin^2(2x)[/MATH] graph
look again at post #21 ... the graph shows two periods in the interval, [MATH][0,2\pi][/MATH], one red, the second blue.
What are you studying? None of it seems to be coordinated and thorough. Or perhaps you are skimming rather than studying your text. Skimming will not teach you math. Every word counts.
A qualitative understanding of a function comes from knowing a number of things such as in what intervals the signs of the first and second derivatives are positive and negative.
A periodic function f(x) has the following attribute:
[MATH]\exists \text { a positive real number } p \text { such that } f(x \pm p) = f(x) \\ \text { provided } x \pm p \text { is in the domain of } f(x).[/MATH]
The period of f(x) is p.
If you look at skeeter’s graph in 21, it is apparent that this function has a period of pi.
You seem, as I wondered in 25, to be mixing up the definition of period with an informal definition of symmetry.
The sine function has a simple symmetry within the period. That is not what makes it periodic. It may be (I have not proved or disproved the hypothesis) that any function that can be decomposed into only sine functions has at least one symmetrical aspect, but it certainly is not true that the symmetry of any composition of sine functions is simple or replicates the symmetry of the basic sine function.
Symmetry of the sine function
[MATH]sin(x \pm \pi) = - sin(x).[/MATH]
Period of the sine function
[MATH]sin(x \pm 2 \pi) = sin(x).[/MATH]
Uh oh. Skeeter and I are not saying the same thing. That probably means that skeeter is correct and I am wrong. I would not use the term “period” the way he seems to. But he and I mean much the same thing. In this function, there are “layers“ of symmetries. I am not sure what the vocabulary for that is.
In this function, we have a symmetrical pattern followed by a different symmetrical pattern and then the succession of symmetries repeats.
In this function, we have a symmetrical pattern followed by a different symmetrical pattern and then the succession of symmetries repeats.
Ah I see! NOT the same as a sine graph, silly me. I did not know what a sin^2(x) graph looked like...
Thank you!
I understand what you mean about the symmetrical pattern repeating - I think it was just that I did not realise the graph is not the same as a sine graph...
Thank you!