Question

I was wondering if I did this problem right:

find an equation for the graph shown in the form y= A sin (Bx)

Answer 1

no

Answer 2

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))

Answer 3

it looks like a normal period of 2pi is being squished into an interval of 1

1period = 2pi

1/2pi period = 1\[y=-3sin(\frac{1}{2pi}x)\]

Answer 4

pfft, figures ... my period is incorrect

Answer 5

ok, i will wait

Answer 6

-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)

Answer 7

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

Answer 8

the solution on your paper produces:

http://www.wolframalpha.com/input/?i=y%3D3sin%283pi+x%2F2%29

a period of 4/3

Answer 9

i dont understand

Answer 10

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

Answer 11

okay..... so what does that mean

Answer 12

it means you did something wrong ...

Answer 13

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

Answer 14

a period is conventionally defined as the "shortest" interval that is repeated ...

in this case; the normal period of sin(x) is 2pi

Answer 15

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

Answer 16

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

Answer 17

there was also some other mismathing going on

Answer 18

well... that didn't help

Answer 19

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.

Answer 20

Some pictures will help, I think.

First, \(y = \sin x\)

Answer 21

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

Answer 22

multiply argument by a factor of \(\pi\)

closer, but still not enough

Answer 23

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

Answer 24

the equation/formula is the same one that weve been addressing:

sin(kx), such that 2pi/k = period stated.

Answer 25

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:

Answer 26

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

Answer 27

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.

Answer 28

how is the period 1?

Answer 29

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?

Answer 30

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

Answer 31

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

Answer 32

no i don't understand.

Answer 33

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.

Answer 34

okay so there are 4 cycles

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

Here is a different sine curve.
What is the length of one cycle?