period of trig function f(x) = a*sin(bx) is (2(pi))/b

[jwpaine]

period of function f(x) = a*sin(bx) is (2(pi))/b

so what would the period of y = -3sin(x/3) be?
would it be 2(pi)*3 ?

It must be...because if your dividing by b... than I should prob multiple 2(pi) by 3

Now what if I had y = -3sin(3/x)
would it be the same?

confused with the division.

 

[pka]

Did you graph the function?

 

[jwpaine]

Ok.

I graphed y = sin(x/3) and I see that the period is close to the x intercept 19
so the period is in fact 6(pi) = 18.8496...

I graphed y = sin(3/x) and I'm not sure how to conclude an expression for it's period....graphically, it looks like the period is increasing? (we have not had any function f(x) = sin(b/x) I was just curious.

 

[pka]

Re: period of trig function

period of function f(x) = a*sin(bx) is (2(pi))/b
so what would the period of y = -3sin(x/3) be?
would it be 2(pi)*3 ?
It must be...because if your dividing by b... than I should prob multiple 2(pi) by 3

All of that is correct.
The period of \(\displaystyle \sin (bx)\) is \(\displaystyle \frac{{2\pi }}{b}\).

 

[morson]

Ok.

I graphed y = sin(3/x) and I'm not sure how to conclude an expression for it's period....graphically, it looks like the period is increasing?

Interesting. The y-axis, just from looking at it, is a vertical asymptote. The graph is discontinuous at some parts (for example the origin). There are certainly an infinite number of intersections of the curve with the x-axis, but I don't know if they follow periodicity, nor do I know of a rule for it in this situation. The peak of the curve is obviously 1. Have you measured the distance between x-intercepts? Doesn't look periodic to me.

 

[jwpaine]

That's the plot.

The period seems to be increasing.....I'll ask my teacher today in class.