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

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amistre64Best ResponseYou've already chosen the best response.1
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(xp))
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
it looks like a normal period of 2pi is being squished into an interval of 1 1period = 2pi 1/2pi period = 1\[y=3sin(\frac{1}{2pi}x)\]
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
pfft, figures ... my period is incorrect
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
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)
 10 months ago

phiBest ResponseYou've already chosen the best response.0
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
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
the solution on your paper produces: http://www.wolframalpha.com/input/?i=y%3D3sin%283pi+x%2F2%29 a period of 4/3
 10 months ago

Summersnow8Best ResponseYou've already chosen the best response.0
i dont understand
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
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
 10 months ago

Summersnow8Best ResponseYou've already chosen the best response.0
okay..... so what does that mean
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
it means you did something wrong ...
 10 months ago

Summersnow8Best ResponseYou've already chosen the best response.0
could you walk me through what i did wrong then.... I don't understand how to solve for Bx
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
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
 10 months ago

Summersnow8Best ResponseYou've already chosen the best response.0
yeah i understand that 2pi is the normal period of the sine function
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
your solution attributed this to: 2pi/k = 3pi instead of 2pi/k = 1
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
there was also some other mismathing going on
 10 months ago

Summersnow8Best ResponseYou've already chosen the best response.0
well... that didn't help
 10 months ago

whpalmer4Best ResponseYou've already chosen the best response.0
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 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.
 10 months ago

whpalmer4Best ResponseYou've already chosen the best response.0
Some pictures will help, I think. First, \(y = \sin x\)
 10 months ago

Summersnow8Best ResponseYou've already chosen the best response.0
is there a certain equation that I could use to find this?
 10 months ago

whpalmer4Best ResponseYou've already chosen the best response.0
multiply argument by a factor of \(\pi\) closer, but still not enough
 10 months ago

whpalmer4Best ResponseYou've already chosen the best response.0
Multiplying argument by \(2\pi\) gets the right frequency: \(y = \sin 2\pi x\)
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
the equation/formula is the same one that weve been addressing: sin(kx), such that 2pi/k = period stated.
 10 months ago

whpalmer4Best ResponseYou've already chosen the best response.0
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:
 10 months ago

whpalmer4Best ResponseYou've already chosen the best response.0
The period business is just a simple proportion of the desired period and \(2\pi\)...
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.2
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.
 10 months ago

Summersnow8Best ResponseYou've already chosen the best response.0
how is the period 1?
 10 months ago

whpalmer4Best ResponseYou've already chosen the best response.0
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?
 10 months ago

whpalmer4Best 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"
 10 months ago

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

Summersnow8Best ResponseYou've already chosen the best response.0
no i don't understand.
 10 months ago

whpalmer4Best ResponseYou've already chosen the best response.0
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.
 10 months ago

Summersnow8Best ResponseYou've already chosen the best response.0
okay so there are 4 cycles
 10 months ago

Summersnow8Best ResponseYou've already chosen the best response.0
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
 10 months ago

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

Summersnow8Best ResponseYou've already chosen the best response.0
this isn't making any sense.....
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.2
dw:1370965653907:dwHere is a different sine curve. What is the length of one cycle?
 10 months ago

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

whpalmer4Best ResponseYou've already chosen the best response.0
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.
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.2
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.
 10 months ago

Summersnow8Best 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
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.2
What is the length of one cycle on the graph you were given? Look at the xaxis.
 10 months ago

Summersnow8Best ResponseYou've already chosen the best response.0
1, because you told me
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.2
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?
 10 months ago

phiBest ResponseYou've already chosen the best response.0
See http://www.khanacademy.org/math/trigonometry/basictrigonometry/trig_graphs_tutorial/v/weamplitudeandperiod it might help
 10 months ago

Summersnow8Best 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
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.2
If no numbers were given, you would be unable to determine the period.
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.2
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
 10 months ago

whpalmer4Best 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
 10 months ago

whpalmer4Best ResponseYou've already chosen the best response.0
I know that was for me, years ago!
 10 months ago

Summersnow8Best ResponseYou've already chosen the best response.0
dw:1370966450574:dw so what is the period of that?
 10 months ago

whpalmer4Best ResponseYou've already chosen the best response.0
don't you mean pi/2, pi, 3pi/2, 2pi? You've got numbers of decreasing size as you go to the right there...
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.2
Ah yes your labels are a little silly :) hehe
 10 months ago

Summersnow8Best ResponseYou've already chosen the best response.0
okay sure...... ugh!
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.2
dw:1370966654243:dwHow bout this one instead :D
 10 months ago

Summersnow8Best ResponseYou've already chosen the best response.0
but i have trouble with the fraction ones....
 10 months ago

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

zepdrixBest ResponseYou've already chosen the best response.2
Oh the fractions are the troublesome ones? ok my bad.
 10 months ago

whpalmer4Best 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!
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.2
dw:1370966770302:dw
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.2
here are some tricky fractions for ya :U
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.2
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?
 10 months ago

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

zepdrixBest ResponseYou've already chosen the best response.2
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. \[\large Period \qquad=\qquad \dfrac{2\pi}{B}\] \[\large \frac{\pi}{3} \qquad=\qquad \dfrac{2\pi}{B}\]
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.2
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,\[\large B \qquad=\qquad 2\cancel{\pi}\frac{3}{\cancel{\pi}}\]
 10 months ago

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

Summersnow8Best ResponseYou've already chosen the best response.0
dw:1370967157840:dw
 10 months ago

zepdrixBest 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?
 10 months ago

Summersnow8Best ResponseYou've already chosen the best response.0
dw:1370967349882:dw
 10 months ago

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

Summersnow8Best ResponseYou've already chosen the best response.0
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
 10 months ago
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