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Summersnow8
 3 years ago
I was wondering if I did this problem right:
find an equation for the graph shown in the form y= A sin (Bx)
Summersnow8
 3 years ago
I was wondering if I did this problem right: find an equation for the graph shown in the form y= A sin (Bx)

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amistre64
 3 years ago
Best ResponseYou've already chosen the best response.13 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(xp))

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1it looks like a normal period of 2pi is being squished into an interval of 1 1period = 2pi 1/2pi period = 1\[y=3sin(\frac{1}{2pi}x)\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1pfft, figures ... my period is incorrect

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.13 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)

phi
 3 years ago
Best ResponseYou've already chosen the best response.0when 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1the solution on your paper produces: http://www.wolframalpha.com/input/?i=y%3D3sin%283pi+x%2F2%29 a period of 4/3

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1i 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

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0okay..... so what does that mean

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1it means you did something wrong ...

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0could you walk me through what i did wrong then.... I don't understand how to solve for Bx

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1a 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

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0yeah i understand that 2pi is the normal period of the sine function

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1your solution attributed this to: 2pi/k = 3pi instead of 2pi/k = 1

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1there was also some other mismathing going on

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0well... that didn't help

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.0The 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 xaxis, 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.

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.0Some pictures will help, I think. First, \(y = \sin x\)

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0is there a certain equation that I could use to find this?

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.0multiply argument by a factor of \(\pi\) closer, but still not enough

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.0Multiplying argument by \(2\pi\) gets the right frequency: \(y = \sin 2\pi x\)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1the equation/formula is the same one that weve been addressing: sin(kx), such that 2pi/k = period stated.

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.0But 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:

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.0The period business is just a simple proportion of the desired period and \(2\pi\)...

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2An 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.

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0how is the period 1?

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.0Look 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?

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.0"at the same point on the graph" means "the corresponding spot in the next cycle of the periodic function"

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2One full period of Sine cosists of 2 "loops". :) See how it loops down once then back up, and crosses x=1.

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0no i don't understand.

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.0Look 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.

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0okay so there are 4 cycles

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0what 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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Don't worry about how many cycles there are. The question is, how much distance along the x axis does ONE cycle cover?

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0this isn't making any sense.....

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1370965653907:dwHere is a different sine curve. What is the length of one cycle?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Yes good :) Now look back at your graph, what is the length of one cycle? That is your period.

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.0Well, 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.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2In 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.

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0@whpalmer4 you have confused me. I don't know how to these..... sorry for wasting your time

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2What is the length of one cycle on the graph you were given? Look at the xaxis.

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.01, because you told me

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Hmm 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?

phi
 3 years ago
Best ResponseYou've already chosen the best response.0See http://www.khanacademy.org/math/trigonometry/basictrigonometry/trig_graphs_tutorial/v/weamplitudeandperiod it might help

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0@zepdrix what if the tick marks were in the middle of the x axis, no number actually given though. then i would be screwed

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2If no numbers were given, you would be unable to determine the period.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2The 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

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.0@zepdrix probably that the don't involve pi, and the problems worked in class did is the confusing part, whether realized or not

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.0I know that was for me, years ago!

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1370966450574:dw so what is the period of that?

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.0don't you mean pi/2, pi, 3pi/2, 2pi? You've got numbers of decreasing size as you go to the right there...

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Ah yes your labels are a little silly :) hehe

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0okay sure...... ugh!

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1370966654243:dwHow bout this one instead :D

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0but i have trouble with the fraction ones....

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1370966692448:dwThis length is the period. See how the pattern starts to repeat itself after this length?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Oh the fractions are the troublesome ones? ok my bad.

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.0@zepdrix be a mensch and do one with fractions for the pretty girl, eh? :) I gotta run, good luck, all!

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2here are some tricky fractions for ya :U

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2So what is the period of this sine function? How far do we travel before we've completed one loop up and one loop down?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2yah pi/3 sounds right. We started at x=0 ended at x=pi/3 after we did 2 loops.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2So 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. \[\large Period \qquad=\qquad \dfrac{2\pi}{B}\] \[\large \frac{\pi}{3} \qquad=\qquad \dfrac{2\pi}{B}\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Solving 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,\[\large B \qquad=\qquad 2\cancel{\pi}\frac{3}{\cancel{\pi}}\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2I think it's going to be \(\large B=6\) Do those steps make sense?

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1370967157840:dw

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2\[\large \frac{\pi}{3}=\frac{2\pi}{B}\]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?

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1370967349882:dw

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2So in our example, we would be able to write a sine function like this: \(\large y=A\sin6x\)

Summersnow8
 3 years ago
Best ResponseYou've already chosen the best response.0yeah that makes sense. and if the sine function went up to 10 and down to 10, then it would be y= 10 sin 6x
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