Summersnow8
  • Summersnow8
I was wondering if I did this problem right: find an equation for the graph shown in the form y= A sin (Bx)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Summersnow8
  • Summersnow8
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amistre64
  • amistre64
no
amistre64
  • amistre64
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))

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amistre64
  • amistre64
it looks like a normal period of 2pi is being squished into an interval of 1 1period = 2pi 1/2pi period = 1\[y=-3sin(\frac{1}{2pi}x)\]
amistre64
  • amistre64
pfft, figures ... my period is incorrect
Summersnow8
  • Summersnow8
ok, i will wait
amistre64
  • amistre64
-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)
phi
  • phi
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
amistre64
  • amistre64
the solution on your paper produces: http://www.wolframalpha.com/input/?i=y%3D3sin%283pi+x%2F2%29 a period of 4/3
Summersnow8
  • Summersnow8
i dont understand
amistre64
  • amistre64
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
Summersnow8
  • Summersnow8
okay..... so what does that mean
amistre64
  • amistre64
it means you did something wrong ...
Summersnow8
  • Summersnow8
could you walk me through what i did wrong then.... I don't understand how to solve for Bx
amistre64
  • amistre64
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|
Summersnow8
  • Summersnow8
yeah i understand that 2pi is the normal period of the sine function
amistre64
  • amistre64
your solution attributed this to: 2pi/k = 3pi instead of 2pi/k = 1
amistre64
  • amistre64
there was also some other mismathing going on
Summersnow8
  • Summersnow8
well... that didn't help
whpalmer4
  • whpalmer4
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.
whpalmer4
  • whpalmer4
Some pictures will help, I think. First, \(y = \sin x\)
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Summersnow8
  • Summersnow8
is there a certain equation that I could use to find this?
whpalmer4
  • whpalmer4
multiply argument by a factor of \(\pi\) closer, but still not enough
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whpalmer4
  • whpalmer4
Multiplying argument by \(2\pi\) gets the right frequency: \(y = \sin 2\pi x\)
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amistre64
  • amistre64
the equation/formula is the same one that weve been addressing: sin(kx), such that 2pi/k = period stated.
whpalmer4
  • whpalmer4
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:
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whpalmer4
  • whpalmer4
The period business is just a simple proportion of the desired period and \(2\pi\)...
zepdrix
  • zepdrix
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.
Summersnow8
  • Summersnow8
how is the period 1?
whpalmer4
  • whpalmer4
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?
whpalmer4
  • whpalmer4
"at the same point on the graph" means "the corresponding spot in the next cycle of the periodic function"
zepdrix
  • zepdrix
One full period of Sine cosists of 2 "loops". :) See how it loops down once then back up, and crosses x=1.
Summersnow8
  • Summersnow8
no i don't understand.
whpalmer4
  • whpalmer4
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.
Summersnow8
  • Summersnow8
okay so there are 4 cycles
Summersnow8
  • Summersnow8
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
zepdrix
  • zepdrix
Don't worry about how many cycles there are. The question is, how much distance along the x axis does ONE cycle cover?
Summersnow8
  • Summersnow8
this isn't making any sense.....
zepdrix
  • zepdrix
|dw:1370965653907:dw|Here is a different sine curve. What is the length of one cycle?
Summersnow8
  • Summersnow8
4?
zepdrix
  • zepdrix
Yes good :) Now look back at your graph, what is the length of one cycle? That is your period.
whpalmer4
  • whpalmer4
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.
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Summersnow8
  • Summersnow8
so [4=2\Pi/B\]
zepdrix
  • zepdrix
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.
Summersnow8
  • Summersnow8
@whpalmer4 you have confused me. I don't know how to these..... sorry for wasting your time
zepdrix
  • zepdrix
What is the length of one cycle on the graph you were given? Look at the x-axis.
Summersnow8
  • Summersnow8
1, because you told me
zepdrix
  • zepdrix
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?
phi
  • phi
See http://www.khanacademy.org/math/trigonometry/basic-trigonometry/trig_graphs_tutorial/v/we-amplitude-and-period it might help
Summersnow8
  • Summersnow8
@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
  • zepdrix
If no numbers were given, you would be unable to determine the period.
Summersnow8
  • Summersnow8
alright :/
zepdrix
  • zepdrix
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
whpalmer4
  • whpalmer4
@zepdrix probably that the don't involve pi, and the problems worked in class did is the confusing part, whether realized or not
zepdrix
  • zepdrix
ah, yes maybe :)
whpalmer4
  • whpalmer4
I know that was for me, years ago!
Summersnow8
  • Summersnow8
|dw:1370966450574:dw| so what is the period of that?
whpalmer4
  • whpalmer4
don't you mean pi/2, pi, 3pi/2, 2pi? You've got numbers of decreasing size as you go to the right there...
zepdrix
  • zepdrix
Ah yes your labels are a little silly :) hehe
Summersnow8
  • Summersnow8
okay sure...... ugh!
zepdrix
  • zepdrix
|dw:1370966654243:dw|How bout this one instead :D
Summersnow8
  • Summersnow8
but i have trouble with the fraction ones....
zepdrix
  • zepdrix
|dw:1370966692448:dw|This length is the period. See how the pattern starts to repeat itself after this length?
zepdrix
  • zepdrix
Oh the fractions are the troublesome ones? ok my bad.
whpalmer4
  • whpalmer4
@zepdrix be a mensch and do one with fractions for the pretty girl, eh? :-) I gotta run, good luck, all!
zepdrix
  • zepdrix
|dw:1370966770302:dw|
zepdrix
  • zepdrix
here are some tricky fractions for ya :U
zepdrix
  • zepdrix
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?
Summersnow8
  • Summersnow8
2pi, pi/3?
zepdrix
  • zepdrix
yah pi/3 sounds right. We started at x=0 ended at x=pi/3 after we did 2 loops.
zepdrix
  • zepdrix
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. \[\large Period \qquad=\qquad \dfrac{2\pi}{B}\] \[\large \frac{\pi}{3} \qquad=\qquad \dfrac{2\pi}{B}\]
Summersnow8
  • Summersnow8
6pi?
Summersnow8
  • Summersnow8
1/6 i mean
zepdrix
  • zepdrix
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,\[\large B \qquad=\qquad 2\cancel{\pi}\frac{3}{\cancel{\pi}}\]
zepdrix
  • zepdrix
I think it's going to be \(\large B=6\) Do those steps make sense?
Summersnow8
  • Summersnow8
|dw:1370967157840:dw|
zepdrix
  • zepdrix
\[\large \frac{\pi}{3}=\frac{2\pi}{B}\]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
  • Summersnow8
|dw:1370967349882:dw|
zepdrix
  • zepdrix
ya looks good :)
Summersnow8
  • Summersnow8
okay, thank you!
zepdrix
  • zepdrix
So in our example, we would be able to write a sine function like this: \(\large y=A\sin6x\)
Summersnow8
  • Summersnow8
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
zepdrix
  • zepdrix
Yes, good.

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