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

- Summersnow8

- Stacey Warren - Expert brainly.com

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- Summersnow8

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- amistre64

no

- 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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## More answers

- 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

pfft, figures ... my period is incorrect

- Summersnow8

ok, i will wait

- 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

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

the solution on your paper produces: http://www.wolframalpha.com/input/?i=y%3D3sin%283pi+x%2F2%29 a period of 4/3

- Summersnow8

i dont understand

- 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

okay..... so what does that mean

- amistre64

it means you did something wrong ...

- Summersnow8

could you walk me through what i did wrong then.... I don't understand how to solve for Bx

- 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

yeah i understand that 2pi is the normal period of the sine function

- amistre64

your solution attributed this to: 2pi/k = 3pi instead of 2pi/k = 1

- amistre64

there was also some other mismathing going on

- Summersnow8

well... that didn't help

- 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

Some pictures will help, I think. First, \(y = \sin x\)

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- Summersnow8

is there a certain equation that I could use to find this?

- whpalmer4

multiply argument by a factor of \(\pi\) closer, but still not enough

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- whpalmer4

Multiplying argument by \(2\pi\) gets the right frequency: \(y = \sin 2\pi x\)

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- amistre64

the equation/formula is the same one that weve been addressing: sin(kx), such that 2pi/k = period stated.

- 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

The period business is just a simple proportion of the desired period and \(2\pi\)...

- 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

how is the period 1?

- 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

"at the same point on the graph" means "the corresponding spot in the next cycle of the periodic function"

- zepdrix

One full period of Sine cosists of 2 "loops". :) See how it loops down once then back up, and crosses x=1.

- Summersnow8

no i don't understand.

- 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

okay so there are 4 cycles

- 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

Don't worry about how many cycles there are. The question is, how much distance along the x axis does ONE cycle cover?

- Summersnow8

this isn't making any sense.....

- zepdrix

|dw:1370965653907:dw|Here is a different sine curve. What is the length of one cycle?

- Summersnow8

4?

- zepdrix

Yes good :) Now look back at your graph, what is the length of one cycle? That is your period.

- 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

so [4=2\Pi/B\]

- 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

@whpalmer4 you have confused me. I don't know how to these..... sorry for wasting your time

- zepdrix

What is the length of one cycle on the graph you were given? Look at the x-axis.

- Summersnow8

1, because you told me

- 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

See http://www.khanacademy.org/math/trigonometry/basic-trigonometry/trig_graphs_tutorial/v/we-amplitude-and-period it might help

- 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

If no numbers were given, you would be unable to determine the period.

- Summersnow8

alright :/

- 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

@zepdrix probably that the don't involve pi, and the problems worked in class did is the confusing part, whether realized or not

- zepdrix

ah, yes maybe :)

- whpalmer4

I know that was for me, years ago!

- Summersnow8

|dw:1370966450574:dw| so what is the period of that?

- 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

Ah yes your labels are a little silly :) hehe

- Summersnow8

okay sure...... ugh!

- zepdrix

|dw:1370966654243:dw|How bout this one instead :D

- Summersnow8

but i have trouble with the fraction ones....

- zepdrix

|dw:1370966692448:dw|This length is the period. See how the pattern starts to repeat itself after this length?

- zepdrix

Oh the fractions are the troublesome ones? ok my bad.

- whpalmer4

@zepdrix be a mensch and do one with fractions for the pretty girl, eh? :-) I gotta run, good luck, all!

- zepdrix

|dw:1370966770302:dw|

- zepdrix

here are some tricky fractions for ya :U

- 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

2pi, pi/3?

- zepdrix

yah pi/3 sounds right. We started at x=0 ended at x=pi/3 after we did 2 loops.

- 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

6pi?

- Summersnow8

1/6 i mean

- 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

I think it's going to be \(\large B=6\) Do those steps make sense?

- Summersnow8

|dw:1370967157840:dw|

- 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

|dw:1370967349882:dw|

- zepdrix

ya looks good :)

- Summersnow8

okay, thank you!

- zepdrix

So in our example, we would be able to write a sine function like this: \(\large y=A\sin6x\)

- 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

Yes, good.

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