How to rotate a function about the origin?

Factual Questions
1

I am trying to model something as an exponential function. The exponential function I first came up with is

y = b^x

However, I needed to reflect that curve across the x axis. That was easy:

y = b^(-x)

Now I need to rotate it 90[sup]o[/sup]. I can’t remember how to rotate a function. I’m sure I learned it in high school analytic geometry. Searches turn up how to do much more complex rotations, like rotating a curve in 3-space to create a surface for computer graphics purposes. I’m looking for something that I think should be pretty basic.

2

I know how to rotate a linear function, and rotate a given point using a 2x2 rotation matrix. I just can’t figure out how to generalize it for an exponential function…

3

Doesn’t just swapping x and y do that?

4

Yes, although you will have to restrict to domain in order for it to remain a function.

5

It appears that switching to a logarithmic function works pretty well.

6

The graph of your function is the set of 2-vectors of the form (x, b^(-x)). You can rotate the graph by multiplying each of those vectors by the appropriate rotation matrix, but getting a new equation of the form y = f(x) will require some clever algebraic manipulation.

Swapping x and y is how you reflect the graph, not rotate it.

7

I ended up using

y = c*log(dx)

where c and d are arbitrary constants.

8

Terminology aside, though, it sounds like that’s what the OP is actually trying to do.

9

What you’re doing is switching from one function (the exponential) to its inverse function (the logarithm). If you want to do the same flip with y=x^2, you’ll end up using y=sqrt(x). This amounts to swapping x and y as beowulff said.

10

I’m a bit out of my depth here, but isn’t this an application of complex numbers? I think if you map onto the complex plane, and multiply by i, that will rotate everything by 90[sup]o[/sup]. If you’ve gone the wrong direction, try -i.

11

You can do a quick-and-dirty rotation by replacing y with x everywhere it appears, and replacing x with (-y). (The minus sign there is what makes it a rotation rather than a reflection.) In other words, if you plot all the points in the plane satisfying y = f(x), then the points satisfying x = f(-y) will look like the same graph rotated 90[sup]o[/sup] clockwise about the origin. In your example, if you take

y = b[sup]x[/sup]

and do this swap, you end up with

x = b[sup]-y[/sup]

which, rearranging things, gives

y = - log[sub]b[/sub] x.

Of course, you can’t always write the relation x = f(-y) in the form y = g(x) — it’s just luck that you can do so here, and even in this case the domain of the function is only the positive real numbers, not all real numbers.