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Summersnow8
I was wondering if I did this problem right: find an equation for the graph shown in the form y= A sin (Bx)
3 is good, but we have an upside down sine wave to start with, so -3 sin(kx) unless you are adjusting it: 3 sin(k(x-p))
it looks like a normal period of 2pi is being squished into an interval of 1 1period = 2pi 1/2pi period = 1\[y=-3sin(\frac{1}{2pi}x)\]
pfft, figures ... my period is incorrect
-3 sin(kx) = 0, when x=1/2 might suit us better sin(0) or sin(pi) = 0 k/2 = 0, when k=0 ... not a good choice k/2 = pi when k=2pi, thats better y = -3 sin(2pi x)
when I do these problems, sin (k x) match 2pi/T = k where T is the length of 1 period from the graph, T=1, and we find k= 2 pi/1 = 2 pi
the solution on your paper produces: http://www.wolframalpha.com/input/?i=y%3D3sin%283pi+x%2F2%29 a period of 4/3
i think i see what you did: the graph you are given is marked of in intervals is: 1,2,3,... you mistakenly attributed this to 1pi, 2pi, 3pi
okay..... so what does that mean
it means you did something wrong ...
could you walk me through what i did wrong then.... I don't understand how to solve for Bx
a period is conventionally defined as the "shortest" interval that is repeated ... in this case; the normal period of sin(x) is 2pi |dw:1370963313143:dw|
yeah i understand that 2pi is the normal period of the sine function
your solution attributed this to: 2pi/k = 3pi instead of 2pi/k = 1
there was also some other mismathing going on
well... that didn't help
The normal period of a sine function is \(2\pi\) but your graph has a period going by in 1, rather than \(2\pi \approx 6.28\). Think of multiplying the argument to the sine function as speeding up or slowing the wiggling. If you want twice as many wiggles per unit length of the x-axis, you need to multiply the argument by 2, right? Well, here you are trying to get \(2\pi\) wiggles in the space of 1, so you need to multiply the argument by \(2\pi\) to speed up the wiggling.
Some pictures will help, I think. First, \(y = \sin x\)
is there a certain equation that I could use to find this?
multiply argument by a factor of \(\pi\) closer, but still not enough
Multiplying argument by \(2\pi\) gets the right frequency: \(y = \sin 2\pi x\)
the equation/formula is the same one that weve been addressing: sin(kx), such that 2pi/k = period stated.
But comparing with problem graph, we still don't have the right amplitude, and our graph is inverted. Fix that by multiplying by -3, which scales and inverts:
The period business is just a simple proportion of the desired period and \(2\pi\)...
An equation? Yes it's of this form: \(\large Period \qquad=\qquad \dfrac{2\pi}{B}\) We recognize that the graph shows a period of 1. \(\large 1\qquad=\qquad \dfrac{2\pi}{B}\) And from here you can solve for your missing B term. This is pretty much the same thing Ami said, but I wanted to include the B variable so you can match it up with your equation.
how is the period 1?
Look at the graph. the function crosses through x=0, y=0, and the next time it goes through y=0 at the same point on the graph is at x=1, right?
"at the same point on the graph" means "the corresponding spot in the next cycle of the periodic function"
One full period of Sine cosists of 2 "loops". :) See how it loops down once then back up, and crosses x=1.
no i don't understand.
Look at the graph in your problem. Start at x = -1. From there to x = 0 is one complete cycle of the sine function they want you to make, right? After that, from x = 0 to x = 1 is another identical cycle.
okay so there are 4 cycles
what now? I mean... you have told me the answer, but if I am given another problem like this, there is no way I would be able to solve it because all I know is what you told me..... its the first number where the second cycle starts
Don't worry about how many cycles there are. The question is, how much distance along the x axis does ONE cycle cover?
this isn't making any sense.....
|dw:1370965653907:dw|Here is a different sine curve. What is the length of one cycle?
Yes good :) Now look back at your graph, what is the length of one cycle? That is your period.
Well, to get one cycle of \sin x, your x value has to go from 0 to \(2\pi\), right? If you need that to happen in more or less space along the x axis, you just have to multiply or divide to make that happen. Say the next problem asked you to make a sine function with period \(\pi/2\). That means that your new multiplier and the old multiplier (which is 1 by default, as in \(y = \sin 1x\)) have to satisfy a proportion: \[\frac{1}{B} = \frac{T}{2\pi}\] where \(T\) is the new period, and \(B\) is the multiplier. If our period we want is \(2\pi\) then our value of \(B\) must be 1. That's just our starting point with \(y=\sin 1x\). If we want two cycles in the space normally occupied by 1, that means our period will be \(\pi\) instead of \(2\pi\) and \[\frac{1}{B} = \frac{T}{2\pi} = \frac{\pi}{2\pi} = \frac{1}{2}\] so \(B=2\). And if you graph \(y = \sin x\) and \(y = \sin 2x\) together you'll see that it does the right thing.
In the example I gave yes, that would be the correct way to find our missing \(\large B\) term summer. But as I said, that was just an example. That doesn't match the graph you were given.
@whpalmer4 you have confused me. I don't know how to these..... sorry for wasting your time
What is the length of one cycle on the graph you were given? Look at the x-axis.
1, because you told me
Hmm because I told you? :\ But here's the thing.. you recognized that there are 4 cycles on your own. So can't you measure the length of one of those cycles?
See http://www.khanacademy.org/math/trigonometry/basic-trigonometry/trig_graphs_tutorial/v/we-amplitude-and-period it might help
@zepdrix what if the tick marks were in the middle of the x axis, no number actually given though. then i would be screwed
If no numbers were given, you would be unable to determine the period.
The numbers along the axis are a requirement :) Or the problem isn't solvable. Is the fact that they're nice easy numbers 1,2,3 the part that's confusing you? :3
@zepdrix probably that the don't involve pi, and the problems worked in class did is the confusing part, whether realized or not
I know that was for me, years ago!
|dw:1370966450574:dw| so what is the period of that?
don't you mean pi/2, pi, 3pi/2, 2pi? You've got numbers of decreasing size as you go to the right there...
Ah yes your labels are a little silly :) hehe
okay sure...... ugh!
|dw:1370966654243:dw|How bout this one instead :D
but i have trouble with the fraction ones....
|dw:1370966692448:dw|This length is the period. See how the pattern starts to repeat itself after this length?
Oh the fractions are the troublesome ones? ok my bad.
@zepdrix be a mensch and do one with fractions for the pretty girl, eh? :-) I gotta run, good luck, all!
here are some tricky fractions for ya :U
So what is the period of this sine function? How far do we travel before we've completed one loop up and one loop down?
yah pi/3 sounds right. We started at x=0 ended at x=pi/3 after we did 2 loops.
So in this little example, if we wanted to find the \(\large B\) coefficient we would go back to our formula, now with our period value in hand. \[\large Period \qquad=\qquad \dfrac{2\pi}{B}\] \[\large \frac{\pi}{3} \qquad=\qquad \dfrac{2\pi}{B}\]
Solving for B? Hmmm let's see. Multiplying both sides by B,\[\large B\frac{\pi}{3} \qquad=\qquad 2\pi\]Divide each side by pi/3,\[\large B \qquad=\qquad 2\cancel{\pi}\frac{3}{\cancel{\pi}}\]
I think it's going to be \(\large B=6\) Do those steps make sense?
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\[\large \frac{\pi}{3}=\frac{2\pi}{B}\]Dividing both sides by 2pi gives us,\[\large \frac{\pi}{3}\cdot\frac{1}{2\pi}=\frac{1}{B}\] See how the B should still be in the denominator?
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So in our example, we would be able to write a sine function like this: \(\large y=A\sin6x\)
yeah that makes sense. and if the sine function went up to 10 and down to -10, then it would be y= 10 sin 6x